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Chemical calculations used in medicine part 1

Pavla BaliacutenovaacutePavla Baliacutenovaacute

Prefixes for Prefixes for uunitsnits

giga-giga- GG 101099

mega- mega- MM 101066

kilo-kilo- kk 101033

deci-deci- dd 1010-1-1

centi-centi- cc 1010-2-2

milli-milli- mm 1010-3-3

micro-micro- μμ 1010-6-6

nano-nano- nn 1010-9-9

pico-pico- pp 1010-12-12

femto-femto- ff 1010-15-15

atto-atto- aa 1010-18-18

Basic terms

MW = molecular weight (gmol)= mass of 1 mole of substance in gramsor relative molecular weight Mr

Avogadroacutes number N = 6022 x 1023 particles in 1 mol

n = substance amount in moles (mol)n = mMW (m = mass (g))Also used mmol micromol nmol pmol hellip

Concentration ndash amount of a substance in specified final volumeMolar concentration or molarity (c) ndash number of

moles of a substance per liter of solution

unit molL = moldm3 = M

c = n (mol) V (L)

Molality (molkg) ndash concentration of moles of substance per 1 kg of solvent

Molar concentration - examples

1) 174 g NaCl in 300 mL of solution MW (NaCl) = 58 c =

2) 45 g glucose in 25 L of solution MW (glucose) = 180 c =

3) Solution of glycine c = 3 mM V = 100 mL MW (glycine ) = 75 m = mg of glycine in the solution

Number of ions in a certain volume

Problem 1 2 litres of solution contain 142 g of Na2HPO4 How many mmol Na+ ions are found in 20 mL of this solution Mr (Na2HPO4) = 142

Substance amount of Na2HPO4 in 2 L of solution n = 142142 = 1 mol

1 mol of Na2HPO4 in 2 L05 mol of Na2HPO4 in 1 L rarr05 mol Na2HPO4 gives 1 mol of Na+ and

05 mol of HPO42-

1 mol of Na+ in 1 LX mol of Na+ in 002 L X = 0021 x 1 = 002 mol = 20 mmol

Problem 2 Molarity of CaCl2 solution is 01 M Calculate the volume of solution containing 4 mmol of Cl-

01 M CaCl2 = 01 mol in 1 L 01 mol of CaCl2 gives 01 mol of Ca2+ and 02 mol of Cl-

02 mol of Cl- in 1 L0004 mol of Cl- in X LX = 000402 x 1 = 002 L = 20 mL

Osmotic pressureOsmotic pressure π is a hydrostatic pressure produced by solution in a space divided by a

semipermeable membrane due to a differential in the concentrations of solute

unit pascal PaΠ = i x c x R x T

Osmosis

= the movement of solvent

from an area of low concentration of

solute to an area of high concentration

Free diffusion

= the movement of solute from the

site of higher concentration to

the site of lower concentration

Oncotic pressure

= is a form of osmotic pressure exerted by proteins

in blood plasma

Osmolarity Osmolarity is a number of moles of a substance that contribute to

osmotic pressure of solution (osmolL) The concentration of body fluids is typically reported in mosmolLOsmolarity of blood is 290 ndash 300 mosmolL

π = i c R TThe figure is found at httpenwikipediaorgwikiOsmotic_pressureThe figure is found at httpenwikipediaorgwikiOsmotic_pressure

Osmolarity - examplesExample 1 A 1 M NaCl solution contains 2 osmol of solute per

liter of solution NaCl rarr Na+ + Cl-

1 M does dissociate 1 osmolL 1 osmolL 2 osmolL in total

Example 2 A 1 M CaCl2 solution contains 3 osmol of solute per liter of solution CaCl2 rarr Ca 2+ + 2 Cl-

1 M does dissociate 1 osmolL 2 osmolL 3 osmolL in totalExample 3 The concentration of a 1 M glucose solution is 1

osmolL

C6H12O6 rarr C6H12O6

1 M does not dissociate rarr 1 osmolL

Osmolarity - examplesOsmolarity - examples1 What is an osmolarity of 015 molL solution of1 What is an osmolarity of 015 molL solution of

a) NaCla) NaCl

b) MgClb) MgCl22

c) Nac) Na22HPOHPO44

d) glucosed) glucose

2 Saline is 150 mM solution of NaCl Which solutions are isotonic with 2 Saline is 150 mM solution of NaCl Which solutions are isotonic with saline saline [= 150 mM = 300 mosmol[= 150 mM = 300 mosmolLL]] a) 300 mM glucosea) 300 mM glucose

b) 50 mM CaClb) 50 mM CaCl22

c) 300 mM KClc) 300 mM KCl

d) 015 M NaHd) 015 M NaH22POPO44

3 What is molarity of 900 mosmoll solution of MgCl3 What is molarity of 900 mosmoll solution of MgCl22 in molL in molL

Percent concentration eexpressed as xpressed as part of solute per 100 part of solute per 100

parts of total solutionparts of total solution () () = = mass of solute mass of solute xx 100 100 mass of solutionmass of solution

it hit has 3 formsas 3 forms1 weight per weight (ww)1 weight per weight (ww)

10 of KCl = 1010 of KCl = 10 g of KCl + 90 g of g of KCl + 90 g of HH22O O ==100 g of solution100 g of solution

2 volume per volume2 volume per volume (vv)(vv)5 HCl = 5 5 HCl = 5 mLmL HCl in 100 m HCl in 100 mLL of of solutionsolution

3 weight per volume (wv)3 weight per volume (wv)the the most common expressionmost common expression09 NaCl = 09 g of NaCl in 100 m09 NaCl = 09 g of NaCl in 100 mLL of of solutionsolution

Percent concentrations - examples

1) 600 g 5 NaCl mass of NaCl mass of H2O

2) 250 g 8 Na2CO3 mass of Na2CO3 (purity 96)

3) 250 mL 39 ethanol solution mL of ethanol mL of H2O

4) Saline is 150 mM solution of NaCl Calculate the percent concentration by mass of this solution Mr(NaCl) = 585

Density ρ

- is defined as the amount of mass per unit of volume

ρ = mV rarr m = ρ x V and V = m ρ - these equations are useful for calculations

units gcm3 or gmL

- density of water = 1 g cm3

- density of lead (Pb) = 1134 gcm3

Conversions of concentrations ( and c) with density

1) What is a percent concentration of 2 M HNO3 solution Density (HNO3) = 1076 gml

Mr (HNO3) = 6301 Conversion of molar concentration to concentration 2 M HNO3 solution means that 2 mol of HNO3 are dissolved in 1 L of solution Mass of HNO3 = n x Mr = 2 x 6301 = 12602 g of HNO3

Mass of solution = ρ x V = 1076 x 1000 = 1076 gW = 12602 x 100 = 1171 1076

2) What is the molarity of 38 HCl solution Density (38 HCl) = 11885 gml and Mr(HCl) = 3645 Conversion of percent by mass concentration to molar concentration 38 HCl solution means that 38 g of HCl in 100 g of solution One liter of solution has a mass m = V x ρ = 1000 x 11885 = 11885 g

38 g HCl -------gt 100 g of solutionx g HCl -------gt 11885 g of solutionx = 45163 g HCl in 1 L of solution rarr n = m M = 45163 3645 = 124 mol of HCl

c(HCl) = n V = 124 1 = 124 molL

Conversions of concentrations ( and c) with density

Conversion of molarity to percent concentration

= c (molL) x Mr 10 x ρ (gcm3)

Conversion of percent concentration to molarity

c = x 10 x ρ (gcm3) Mr

Conversions of concentrations ( and c) with density - examples

1) (ww) of HNO3 ρ = 136 gcm3 if 1dm3 of solution contains 08 kg of HNO3

2) c (HNO3) = 562 M ρ = 118 gcm3 MW = 63 gmol 3) 10 HCl ρ = 1047 gcm3 MW = 365 c (HCl)

Dilution= concentration of a substance lowers number of moles of the substance remains the same

1) mix equation m1 x p1 + m2 x p2 = p x ( m1 + m2 ) m = mass of mixed solution p = concentration

2) expression of dilutionIn case of a liquid solute the ratio is presented as a dilution factor For example 1 5 is presented as 15 (1 mL of solute in 5 mL of solution)Example c1 = 025 M (original concentration) x 15 = 005 M (final concentration c2)

3) useful equation n1 = n2

V1 x c1 = V2 x c2

Dilution - examples

1) Mix 50 g 3 solution with 10 g 5 solution final concentration = ()

2) Final solution 190 g 10 sol

m (g) of 38 HCl + m (g) H2O 3) Dilute 300 g of 40 solution to 20 solution g

of solvent do you need4) preparation of 250 mL of 01 M HCl from stock 1

M HCl5) 10 M NaOH is diluted 1 20 final concentration6) 1000 mgL glucose is diluted 1 10 and then 1 2

final concentration

Conversions of unitsbull concentrationie gdL rarr mgLie molL harr osmolLbull units of pressure used in medicine1 mmHg (millimeter of mercury) = 1 Torr1 mmHg = 13322 Pabull units of energy1 cal = 41868 joule (J) 1 J = 0238 calcalorie (cal) is used in nutrition

Conversions of units ndash examples

1) Concentration of cholesterol in patient blood sample is 180 mgdL Convert this value into mmolL if MW of cholesterol is 387 gmol

2) Partial pressure of CO2 is 533 kPa Convert this value into mmHg

3) Red Bull energy drink (1 can) contains 160 cal Calculate the amount of energy in kJ

Calculations in spectrophotometry

spectrophotometry is an analytical method based on the interaction between an electromagnetic radiation of a known intenzity (Io) and the analyzed sample

a part of the radiation is absorbed by the analyzed substance found in the of sample

the intensity of the radiation which passed (I) through the solution

is detected (I lt I0) rarrthe transmittance (T) of a solution is defined as the proportion T = I I0

transmittance can be expressed in percentage T() = (II0) x 100

values of the measured transmittance are found from 0 - 1 = 0 - 100

Calculations in spectrophotometry

How to calculate a concentration of substance in analyzed sample

in the laboratory it is more convenient to calculate concentration from values of the absorbance (A) which is directly proportional to the concentration than from the transmittance Calculation from the Beeracutes law A = c x l x ε

c = molar concentration (molL)

l = inner width of the cuvette in centimeters

ε = molar absorption coefficient (tabelated value)

Relationship between A and T A = log (1T)= -log T

Calculations in spectrophotometry - examples1) Ast = 040 cst = 4 mgL

Asam = 025 csam = mgL

2) Standard solution of glucose (conc = 1000 mgL) reads a transmittance 49 Unknown sample of glucose reads T = 55 Calculate the concentration of glucose in the sample in mgL and mmolL

Mr (glucose) = 180

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Prefixes for Prefixes for uunitsnits

giga-giga- GG 101099

mega- mega- MM 101066

kilo-kilo- kk 101033

deci-deci- dd 1010-1-1

centi-centi- cc 1010-2-2

milli-milli- mm 1010-3-3

micro-micro- μμ 1010-6-6

nano-nano- nn 1010-9-9

pico-pico- pp 1010-12-12

femto-femto- ff 1010-15-15

atto-atto- aa 1010-18-18

Basic terms

MW = molecular weight (gmol)= mass of 1 mole of substance in gramsor relative molecular weight Mr

Avogadroacutes number N = 6022 x 1023 particles in 1 mol

n = substance amount in moles (mol)n = mMW (m = mass (g))Also used mmol micromol nmol pmol hellip

Concentration ndash amount of a substance in specified final volumeMolar concentration or molarity (c) ndash number of

moles of a substance per liter of solution

unit molL = moldm3 = M

c = n (mol) V (L)

Molality (molkg) ndash concentration of moles of substance per 1 kg of solvent

Molar concentration - examples

1) 174 g NaCl in 300 mL of solution MW (NaCl) = 58 c =

2) 45 g glucose in 25 L of solution MW (glucose) = 180 c =

3) Solution of glycine c = 3 mM V = 100 mL MW (glycine ) = 75 m = mg of glycine in the solution

Number of ions in a certain volume

Problem 1 2 litres of solution contain 142 g of Na2HPO4 How many mmol Na+ ions are found in 20 mL of this solution Mr (Na2HPO4) = 142

Substance amount of Na2HPO4 in 2 L of solution n = 142142 = 1 mol

1 mol of Na2HPO4 in 2 L05 mol of Na2HPO4 in 1 L rarr05 mol Na2HPO4 gives 1 mol of Na+ and

05 mol of HPO42-

1 mol of Na+ in 1 LX mol of Na+ in 002 L X = 0021 x 1 = 002 mol = 20 mmol

Problem 2 Molarity of CaCl2 solution is 01 M Calculate the volume of solution containing 4 mmol of Cl-

01 M CaCl2 = 01 mol in 1 L 01 mol of CaCl2 gives 01 mol of Ca2+ and 02 mol of Cl-

02 mol of Cl- in 1 L0004 mol of Cl- in X LX = 000402 x 1 = 002 L = 20 mL

Osmotic pressureOsmotic pressure π is a hydrostatic pressure produced by solution in a space divided by a

semipermeable membrane due to a differential in the concentrations of solute

unit pascal PaΠ = i x c x R x T

Osmosis

= the movement of solvent

from an area of low concentration of

solute to an area of high concentration

Free diffusion

= the movement of solute from the

site of higher concentration to

the site of lower concentration

Oncotic pressure

= is a form of osmotic pressure exerted by proteins

in blood plasma

Osmolarity Osmolarity is a number of moles of a substance that contribute to

osmotic pressure of solution (osmolL) The concentration of body fluids is typically reported in mosmolLOsmolarity of blood is 290 ndash 300 mosmolL

π = i c R TThe figure is found at httpenwikipediaorgwikiOsmotic_pressureThe figure is found at httpenwikipediaorgwikiOsmotic_pressure

Osmolarity - examplesExample 1 A 1 M NaCl solution contains 2 osmol of solute per

liter of solution NaCl rarr Na+ + Cl-

1 M does dissociate 1 osmolL 1 osmolL 2 osmolL in total

Example 2 A 1 M CaCl2 solution contains 3 osmol of solute per liter of solution CaCl2 rarr Ca 2+ + 2 Cl-

1 M does dissociate 1 osmolL 2 osmolL 3 osmolL in totalExample 3 The concentration of a 1 M glucose solution is 1

osmolL

C6H12O6 rarr C6H12O6

1 M does not dissociate rarr 1 osmolL

Osmolarity - examplesOsmolarity - examples1 What is an osmolarity of 015 molL solution of1 What is an osmolarity of 015 molL solution of

a) NaCla) NaCl

b) MgClb) MgCl22

c) Nac) Na22HPOHPO44

d) glucosed) glucose

2 Saline is 150 mM solution of NaCl Which solutions are isotonic with 2 Saline is 150 mM solution of NaCl Which solutions are isotonic with saline saline [= 150 mM = 300 mosmol[= 150 mM = 300 mosmolLL]] a) 300 mM glucosea) 300 mM glucose

b) 50 mM CaClb) 50 mM CaCl22

c) 300 mM KClc) 300 mM KCl

d) 015 M NaHd) 015 M NaH22POPO44

3 What is molarity of 900 mosmoll solution of MgCl3 What is molarity of 900 mosmoll solution of MgCl22 in molL in molL

Percent concentration eexpressed as xpressed as part of solute per 100 part of solute per 100

parts of total solutionparts of total solution () () = = mass of solute mass of solute xx 100 100 mass of solutionmass of solution

it hit has 3 formsas 3 forms1 weight per weight (ww)1 weight per weight (ww)

10 of KCl = 1010 of KCl = 10 g of KCl + 90 g of g of KCl + 90 g of HH22O O ==100 g of solution100 g of solution

2 volume per volume2 volume per volume (vv)(vv)5 HCl = 5 5 HCl = 5 mLmL HCl in 100 m HCl in 100 mLL of of solutionsolution

3 weight per volume (wv)3 weight per volume (wv)the the most common expressionmost common expression09 NaCl = 09 g of NaCl in 100 m09 NaCl = 09 g of NaCl in 100 mLL of of solutionsolution

Percent concentrations - examples

1) 600 g 5 NaCl mass of NaCl mass of H2O

2) 250 g 8 Na2CO3 mass of Na2CO3 (purity 96)

3) 250 mL 39 ethanol solution mL of ethanol mL of H2O

4) Saline is 150 mM solution of NaCl Calculate the percent concentration by mass of this solution Mr(NaCl) = 585

Density ρ

- is defined as the amount of mass per unit of volume

ρ = mV rarr m = ρ x V and V = m ρ - these equations are useful for calculations

units gcm3 or gmL

- density of water = 1 g cm3

- density of lead (Pb) = 1134 gcm3

Conversions of concentrations ( and c) with density

1) What is a percent concentration of 2 M HNO3 solution Density (HNO3) = 1076 gml

Mr (HNO3) = 6301 Conversion of molar concentration to concentration 2 M HNO3 solution means that 2 mol of HNO3 are dissolved in 1 L of solution Mass of HNO3 = n x Mr = 2 x 6301 = 12602 g of HNO3

Mass of solution = ρ x V = 1076 x 1000 = 1076 gW = 12602 x 100 = 1171 1076

2) What is the molarity of 38 HCl solution Density (38 HCl) = 11885 gml and Mr(HCl) = 3645 Conversion of percent by mass concentration to molar concentration 38 HCl solution means that 38 g of HCl in 100 g of solution One liter of solution has a mass m = V x ρ = 1000 x 11885 = 11885 g

38 g HCl -------gt 100 g of solutionx g HCl -------gt 11885 g of solutionx = 45163 g HCl in 1 L of solution rarr n = m M = 45163 3645 = 124 mol of HCl

c(HCl) = n V = 124 1 = 124 molL

Conversions of concentrations ( and c) with density

Conversion of molarity to percent concentration

= c (molL) x Mr 10 x ρ (gcm3)

Conversion of percent concentration to molarity

c = x 10 x ρ (gcm3) Mr

Conversions of concentrations ( and c) with density - examples

1) (ww) of HNO3 ρ = 136 gcm3 if 1dm3 of solution contains 08 kg of HNO3

2) c (HNO3) = 562 M ρ = 118 gcm3 MW = 63 gmol 3) 10 HCl ρ = 1047 gcm3 MW = 365 c (HCl)

Dilution= concentration of a substance lowers number of moles of the substance remains the same

1) mix equation m1 x p1 + m2 x p2 = p x ( m1 + m2 ) m = mass of mixed solution p = concentration

2) expression of dilutionIn case of a liquid solute the ratio is presented as a dilution factor For example 1 5 is presented as 15 (1 mL of solute in 5 mL of solution)Example c1 = 025 M (original concentration) x 15 = 005 M (final concentration c2)

3) useful equation n1 = n2

V1 x c1 = V2 x c2

Dilution - examples

1) Mix 50 g 3 solution with 10 g 5 solution final concentration = ()

2) Final solution 190 g 10 sol

m (g) of 38 HCl + m (g) H2O 3) Dilute 300 g of 40 solution to 20 solution g

of solvent do you need4) preparation of 250 mL of 01 M HCl from stock 1

M HCl5) 10 M NaOH is diluted 1 20 final concentration6) 1000 mgL glucose is diluted 1 10 and then 1 2

final concentration

Conversions of unitsbull concentrationie gdL rarr mgLie molL harr osmolLbull units of pressure used in medicine1 mmHg (millimeter of mercury) = 1 Torr1 mmHg = 13322 Pabull units of energy1 cal = 41868 joule (J) 1 J = 0238 calcalorie (cal) is used in nutrition

Conversions of units ndash examples

1) Concentration of cholesterol in patient blood sample is 180 mgdL Convert this value into mmolL if MW of cholesterol is 387 gmol

2) Partial pressure of CO2 is 533 kPa Convert this value into mmHg

3) Red Bull energy drink (1 can) contains 160 cal Calculate the amount of energy in kJ

Calculations in spectrophotometry

spectrophotometry is an analytical method based on the interaction between an electromagnetic radiation of a known intenzity (Io) and the analyzed sample

a part of the radiation is absorbed by the analyzed substance found in the of sample

the intensity of the radiation which passed (I) through the solution

is detected (I lt I0) rarrthe transmittance (T) of a solution is defined as the proportion T = I I0

transmittance can be expressed in percentage T() = (II0) x 100

values of the measured transmittance are found from 0 - 1 = 0 - 100

Calculations in spectrophotometry

How to calculate a concentration of substance in analyzed sample

in the laboratory it is more convenient to calculate concentration from values of the absorbance (A) which is directly proportional to the concentration than from the transmittance Calculation from the Beeracutes law A = c x l x ε

c = molar concentration (molL)

l = inner width of the cuvette in centimeters

ε = molar absorption coefficient (tabelated value)

Relationship between A and T A = log (1T)= -log T

Calculations in spectrophotometry - examples1) Ast = 040 cst = 4 mgL

Asam = 025 csam = mgL

2) Standard solution of glucose (conc = 1000 mgL) reads a transmittance 49 Unknown sample of glucose reads T = 55 Calculate the concentration of glucose in the sample in mgL and mmolL

Mr (glucose) = 180

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Basic terms

MW = molecular weight (gmol)= mass of 1 mole of substance in gramsor relative molecular weight Mr

Avogadroacutes number N = 6022 x 1023 particles in 1 mol

n = substance amount in moles (mol)n = mMW (m = mass (g))Also used mmol micromol nmol pmol hellip

Concentration ndash amount of a substance in specified final volumeMolar concentration or molarity (c) ndash number of

moles of a substance per liter of solution

unit molL = moldm3 = M

c = n (mol) V (L)

Molality (molkg) ndash concentration of moles of substance per 1 kg of solvent

Molar concentration - examples

1) 174 g NaCl in 300 mL of solution MW (NaCl) = 58 c =

2) 45 g glucose in 25 L of solution MW (glucose) = 180 c =

3) Solution of glycine c = 3 mM V = 100 mL MW (glycine ) = 75 m = mg of glycine in the solution

Number of ions in a certain volume

Problem 1 2 litres of solution contain 142 g of Na2HPO4 How many mmol Na+ ions are found in 20 mL of this solution Mr (Na2HPO4) = 142

Substance amount of Na2HPO4 in 2 L of solution n = 142142 = 1 mol

1 mol of Na2HPO4 in 2 L05 mol of Na2HPO4 in 1 L rarr05 mol Na2HPO4 gives 1 mol of Na+ and

05 mol of HPO42-

1 mol of Na+ in 1 LX mol of Na+ in 002 L X = 0021 x 1 = 002 mol = 20 mmol

Problem 2 Molarity of CaCl2 solution is 01 M Calculate the volume of solution containing 4 mmol of Cl-

01 M CaCl2 = 01 mol in 1 L 01 mol of CaCl2 gives 01 mol of Ca2+ and 02 mol of Cl-

02 mol of Cl- in 1 L0004 mol of Cl- in X LX = 000402 x 1 = 002 L = 20 mL

Osmotic pressureOsmotic pressure π is a hydrostatic pressure produced by solution in a space divided by a

semipermeable membrane due to a differential in the concentrations of solute

unit pascal PaΠ = i x c x R x T

Osmosis

= the movement of solvent

from an area of low concentration of

solute to an area of high concentration

Free diffusion

= the movement of solute from the

site of higher concentration to

the site of lower concentration

Oncotic pressure

= is a form of osmotic pressure exerted by proteins

in blood plasma

Osmolarity Osmolarity is a number of moles of a substance that contribute to

osmotic pressure of solution (osmolL) The concentration of body fluids is typically reported in mosmolLOsmolarity of blood is 290 ndash 300 mosmolL

π = i c R TThe figure is found at httpenwikipediaorgwikiOsmotic_pressureThe figure is found at httpenwikipediaorgwikiOsmotic_pressure

Osmolarity - examplesExample 1 A 1 M NaCl solution contains 2 osmol of solute per

liter of solution NaCl rarr Na+ + Cl-

1 M does dissociate 1 osmolL 1 osmolL 2 osmolL in total

Example 2 A 1 M CaCl2 solution contains 3 osmol of solute per liter of solution CaCl2 rarr Ca 2+ + 2 Cl-

1 M does dissociate 1 osmolL 2 osmolL 3 osmolL in totalExample 3 The concentration of a 1 M glucose solution is 1

osmolL

C6H12O6 rarr C6H12O6

1 M does not dissociate rarr 1 osmolL

Osmolarity - examplesOsmolarity - examples1 What is an osmolarity of 015 molL solution of1 What is an osmolarity of 015 molL solution of

a) NaCla) NaCl

b) MgClb) MgCl22

c) Nac) Na22HPOHPO44

d) glucosed) glucose

2 Saline is 150 mM solution of NaCl Which solutions are isotonic with 2 Saline is 150 mM solution of NaCl Which solutions are isotonic with saline saline [= 150 mM = 300 mosmol[= 150 mM = 300 mosmolLL]] a) 300 mM glucosea) 300 mM glucose

b) 50 mM CaClb) 50 mM CaCl22

c) 300 mM KClc) 300 mM KCl

d) 015 M NaHd) 015 M NaH22POPO44

3 What is molarity of 900 mosmoll solution of MgCl3 What is molarity of 900 mosmoll solution of MgCl22 in molL in molL

Percent concentration eexpressed as xpressed as part of solute per 100 part of solute per 100

parts of total solutionparts of total solution () () = = mass of solute mass of solute xx 100 100 mass of solutionmass of solution

it hit has 3 formsas 3 forms1 weight per weight (ww)1 weight per weight (ww)

10 of KCl = 1010 of KCl = 10 g of KCl + 90 g of g of KCl + 90 g of HH22O O ==100 g of solution100 g of solution

2 volume per volume2 volume per volume (vv)(vv)5 HCl = 5 5 HCl = 5 mLmL HCl in 100 m HCl in 100 mLL of of solutionsolution

3 weight per volume (wv)3 weight per volume (wv)the the most common expressionmost common expression09 NaCl = 09 g of NaCl in 100 m09 NaCl = 09 g of NaCl in 100 mLL of of solutionsolution

Percent concentrations - examples

1) 600 g 5 NaCl mass of NaCl mass of H2O

2) 250 g 8 Na2CO3 mass of Na2CO3 (purity 96)

3) 250 mL 39 ethanol solution mL of ethanol mL of H2O

4) Saline is 150 mM solution of NaCl Calculate the percent concentration by mass of this solution Mr(NaCl) = 585

Density ρ

- is defined as the amount of mass per unit of volume

ρ = mV rarr m = ρ x V and V = m ρ - these equations are useful for calculations

units gcm3 or gmL

- density of water = 1 g cm3

- density of lead (Pb) = 1134 gcm3

Conversions of concentrations ( and c) with density

1) What is a percent concentration of 2 M HNO3 solution Density (HNO3) = 1076 gml

Mr (HNO3) = 6301 Conversion of molar concentration to concentration 2 M HNO3 solution means that 2 mol of HNO3 are dissolved in 1 L of solution Mass of HNO3 = n x Mr = 2 x 6301 = 12602 g of HNO3

Mass of solution = ρ x V = 1076 x 1000 = 1076 gW = 12602 x 100 = 1171 1076

2) What is the molarity of 38 HCl solution Density (38 HCl) = 11885 gml and Mr(HCl) = 3645 Conversion of percent by mass concentration to molar concentration 38 HCl solution means that 38 g of HCl in 100 g of solution One liter of solution has a mass m = V x ρ = 1000 x 11885 = 11885 g

38 g HCl -------gt 100 g of solutionx g HCl -------gt 11885 g of solutionx = 45163 g HCl in 1 L of solution rarr n = m M = 45163 3645 = 124 mol of HCl

c(HCl) = n V = 124 1 = 124 molL

Conversions of concentrations ( and c) with density

Conversion of molarity to percent concentration

= c (molL) x Mr 10 x ρ (gcm3)

Conversion of percent concentration to molarity

c = x 10 x ρ (gcm3) Mr

Conversions of concentrations ( and c) with density - examples

1) (ww) of HNO3 ρ = 136 gcm3 if 1dm3 of solution contains 08 kg of HNO3

2) c (HNO3) = 562 M ρ = 118 gcm3 MW = 63 gmol 3) 10 HCl ρ = 1047 gcm3 MW = 365 c (HCl)

Dilution= concentration of a substance lowers number of moles of the substance remains the same

1) mix equation m1 x p1 + m2 x p2 = p x ( m1 + m2 ) m = mass of mixed solution p = concentration

2) expression of dilutionIn case of a liquid solute the ratio is presented as a dilution factor For example 1 5 is presented as 15 (1 mL of solute in 5 mL of solution)Example c1 = 025 M (original concentration) x 15 = 005 M (final concentration c2)

3) useful equation n1 = n2

V1 x c1 = V2 x c2

Dilution - examples

1) Mix 50 g 3 solution with 10 g 5 solution final concentration = ()

2) Final solution 190 g 10 sol

m (g) of 38 HCl + m (g) H2O 3) Dilute 300 g of 40 solution to 20 solution g

of solvent do you need4) preparation of 250 mL of 01 M HCl from stock 1

M HCl5) 10 M NaOH is diluted 1 20 final concentration6) 1000 mgL glucose is diluted 1 10 and then 1 2

final concentration

Conversions of unitsbull concentrationie gdL rarr mgLie molL harr osmolLbull units of pressure used in medicine1 mmHg (millimeter of mercury) = 1 Torr1 mmHg = 13322 Pabull units of energy1 cal = 41868 joule (J) 1 J = 0238 calcalorie (cal) is used in nutrition

Conversions of units ndash examples

1) Concentration of cholesterol in patient blood sample is 180 mgdL Convert this value into mmolL if MW of cholesterol is 387 gmol

2) Partial pressure of CO2 is 533 kPa Convert this value into mmHg

3) Red Bull energy drink (1 can) contains 160 cal Calculate the amount of energy in kJ

Calculations in spectrophotometry

spectrophotometry is an analytical method based on the interaction between an electromagnetic radiation of a known intenzity (Io) and the analyzed sample

a part of the radiation is absorbed by the analyzed substance found in the of sample

the intensity of the radiation which passed (I) through the solution

is detected (I lt I0) rarrthe transmittance (T) of a solution is defined as the proportion T = I I0

transmittance can be expressed in percentage T() = (II0) x 100

values of the measured transmittance are found from 0 - 1 = 0 - 100

Calculations in spectrophotometry

How to calculate a concentration of substance in analyzed sample

in the laboratory it is more convenient to calculate concentration from values of the absorbance (A) which is directly proportional to the concentration than from the transmittance Calculation from the Beeracutes law A = c x l x ε

c = molar concentration (molL)

l = inner width of the cuvette in centimeters

ε = molar absorption coefficient (tabelated value)

Relationship between A and T A = log (1T)= -log T

Calculations in spectrophotometry - examples1) Ast = 040 cst = 4 mgL

Asam = 025 csam = mgL

2) Standard solution of glucose (conc = 1000 mgL) reads a transmittance 49 Unknown sample of glucose reads T = 55 Calculate the concentration of glucose in the sample in mgL and mmolL

Mr (glucose) = 180

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Concentration ndash amount of a substance in specified final volumeMolar concentration or molarity (c) ndash number of

moles of a substance per liter of solution

unit molL = moldm3 = M

c = n (mol) V (L)

Molality (molkg) ndash concentration of moles of substance per 1 kg of solvent

Molar concentration - examples

1) 174 g NaCl in 300 mL of solution MW (NaCl) = 58 c =

2) 45 g glucose in 25 L of solution MW (glucose) = 180 c =

3) Solution of glycine c = 3 mM V = 100 mL MW (glycine ) = 75 m = mg of glycine in the solution

Number of ions in a certain volume

Problem 1 2 litres of solution contain 142 g of Na2HPO4 How many mmol Na+ ions are found in 20 mL of this solution Mr (Na2HPO4) = 142

Substance amount of Na2HPO4 in 2 L of solution n = 142142 = 1 mol

1 mol of Na2HPO4 in 2 L05 mol of Na2HPO4 in 1 L rarr05 mol Na2HPO4 gives 1 mol of Na+ and

05 mol of HPO42-

1 mol of Na+ in 1 LX mol of Na+ in 002 L X = 0021 x 1 = 002 mol = 20 mmol

Problem 2 Molarity of CaCl2 solution is 01 M Calculate the volume of solution containing 4 mmol of Cl-

01 M CaCl2 = 01 mol in 1 L 01 mol of CaCl2 gives 01 mol of Ca2+ and 02 mol of Cl-

02 mol of Cl- in 1 L0004 mol of Cl- in X LX = 000402 x 1 = 002 L = 20 mL

Osmotic pressureOsmotic pressure π is a hydrostatic pressure produced by solution in a space divided by a

semipermeable membrane due to a differential in the concentrations of solute

unit pascal PaΠ = i x c x R x T

Osmosis

= the movement of solvent

from an area of low concentration of

solute to an area of high concentration

Free diffusion

= the movement of solute from the

site of higher concentration to

the site of lower concentration

Oncotic pressure

= is a form of osmotic pressure exerted by proteins

in blood plasma

Osmolarity Osmolarity is a number of moles of a substance that contribute to

osmotic pressure of solution (osmolL) The concentration of body fluids is typically reported in mosmolLOsmolarity of blood is 290 ndash 300 mosmolL

π = i c R TThe figure is found at httpenwikipediaorgwikiOsmotic_pressureThe figure is found at httpenwikipediaorgwikiOsmotic_pressure

Osmolarity - examplesExample 1 A 1 M NaCl solution contains 2 osmol of solute per

liter of solution NaCl rarr Na+ + Cl-

1 M does dissociate 1 osmolL 1 osmolL 2 osmolL in total

Example 2 A 1 M CaCl2 solution contains 3 osmol of solute per liter of solution CaCl2 rarr Ca 2+ + 2 Cl-

1 M does dissociate 1 osmolL 2 osmolL 3 osmolL in totalExample 3 The concentration of a 1 M glucose solution is 1

osmolL

C6H12O6 rarr C6H12O6

1 M does not dissociate rarr 1 osmolL

Osmolarity - examplesOsmolarity - examples1 What is an osmolarity of 015 molL solution of1 What is an osmolarity of 015 molL solution of

a) NaCla) NaCl

b) MgClb) MgCl22

c) Nac) Na22HPOHPO44

d) glucosed) glucose

2 Saline is 150 mM solution of NaCl Which solutions are isotonic with 2 Saline is 150 mM solution of NaCl Which solutions are isotonic with saline saline [= 150 mM = 300 mosmol[= 150 mM = 300 mosmolLL]] a) 300 mM glucosea) 300 mM glucose

b) 50 mM CaClb) 50 mM CaCl22

c) 300 mM KClc) 300 mM KCl

d) 015 M NaHd) 015 M NaH22POPO44

3 What is molarity of 900 mosmoll solution of MgCl3 What is molarity of 900 mosmoll solution of MgCl22 in molL in molL

Percent concentration eexpressed as xpressed as part of solute per 100 part of solute per 100

parts of total solutionparts of total solution () () = = mass of solute mass of solute xx 100 100 mass of solutionmass of solution

it hit has 3 formsas 3 forms1 weight per weight (ww)1 weight per weight (ww)

10 of KCl = 1010 of KCl = 10 g of KCl + 90 g of g of KCl + 90 g of HH22O O ==100 g of solution100 g of solution

2 volume per volume2 volume per volume (vv)(vv)5 HCl = 5 5 HCl = 5 mLmL HCl in 100 m HCl in 100 mLL of of solutionsolution

3 weight per volume (wv)3 weight per volume (wv)the the most common expressionmost common expression09 NaCl = 09 g of NaCl in 100 m09 NaCl = 09 g of NaCl in 100 mLL of of solutionsolution

Percent concentrations - examples

1) 600 g 5 NaCl mass of NaCl mass of H2O

2) 250 g 8 Na2CO3 mass of Na2CO3 (purity 96)

3) 250 mL 39 ethanol solution mL of ethanol mL of H2O

4) Saline is 150 mM solution of NaCl Calculate the percent concentration by mass of this solution Mr(NaCl) = 585

Density ρ

- is defined as the amount of mass per unit of volume

ρ = mV rarr m = ρ x V and V = m ρ - these equations are useful for calculations

units gcm3 or gmL

- density of water = 1 g cm3

- density of lead (Pb) = 1134 gcm3

Conversions of concentrations ( and c) with density

1) What is a percent concentration of 2 M HNO3 solution Density (HNO3) = 1076 gml

Mr (HNO3) = 6301 Conversion of molar concentration to concentration 2 M HNO3 solution means that 2 mol of HNO3 are dissolved in 1 L of solution Mass of HNO3 = n x Mr = 2 x 6301 = 12602 g of HNO3

Mass of solution = ρ x V = 1076 x 1000 = 1076 gW = 12602 x 100 = 1171 1076

2) What is the molarity of 38 HCl solution Density (38 HCl) = 11885 gml and Mr(HCl) = 3645 Conversion of percent by mass concentration to molar concentration 38 HCl solution means that 38 g of HCl in 100 g of solution One liter of solution has a mass m = V x ρ = 1000 x 11885 = 11885 g

38 g HCl -------gt 100 g of solutionx g HCl -------gt 11885 g of solutionx = 45163 g HCl in 1 L of solution rarr n = m M = 45163 3645 = 124 mol of HCl

c(HCl) = n V = 124 1 = 124 molL

Conversions of concentrations ( and c) with density

Conversion of molarity to percent concentration

= c (molL) x Mr 10 x ρ (gcm3)

Conversion of percent concentration to molarity

c = x 10 x ρ (gcm3) Mr

Conversions of concentrations ( and c) with density - examples

1) (ww) of HNO3 ρ = 136 gcm3 if 1dm3 of solution contains 08 kg of HNO3

2) c (HNO3) = 562 M ρ = 118 gcm3 MW = 63 gmol 3) 10 HCl ρ = 1047 gcm3 MW = 365 c (HCl)

Dilution= concentration of a substance lowers number of moles of the substance remains the same

1) mix equation m1 x p1 + m2 x p2 = p x ( m1 + m2 ) m = mass of mixed solution p = concentration

2) expression of dilutionIn case of a liquid solute the ratio is presented as a dilution factor For example 1 5 is presented as 15 (1 mL of solute in 5 mL of solution)Example c1 = 025 M (original concentration) x 15 = 005 M (final concentration c2)

3) useful equation n1 = n2

V1 x c1 = V2 x c2

Dilution - examples

1) Mix 50 g 3 solution with 10 g 5 solution final concentration = ()

2) Final solution 190 g 10 sol

m (g) of 38 HCl + m (g) H2O 3) Dilute 300 g of 40 solution to 20 solution g

of solvent do you need4) preparation of 250 mL of 01 M HCl from stock 1

M HCl5) 10 M NaOH is diluted 1 20 final concentration6) 1000 mgL glucose is diluted 1 10 and then 1 2

final concentration

Conversions of unitsbull concentrationie gdL rarr mgLie molL harr osmolLbull units of pressure used in medicine1 mmHg (millimeter of mercury) = 1 Torr1 mmHg = 13322 Pabull units of energy1 cal = 41868 joule (J) 1 J = 0238 calcalorie (cal) is used in nutrition

Conversions of units ndash examples

1) Concentration of cholesterol in patient blood sample is 180 mgdL Convert this value into mmolL if MW of cholesterol is 387 gmol

2) Partial pressure of CO2 is 533 kPa Convert this value into mmHg

3) Red Bull energy drink (1 can) contains 160 cal Calculate the amount of energy in kJ

Calculations in spectrophotometry

spectrophotometry is an analytical method based on the interaction between an electromagnetic radiation of a known intenzity (Io) and the analyzed sample

a part of the radiation is absorbed by the analyzed substance found in the of sample

the intensity of the radiation which passed (I) through the solution

is detected (I lt I0) rarrthe transmittance (T) of a solution is defined as the proportion T = I I0

transmittance can be expressed in percentage T() = (II0) x 100

values of the measured transmittance are found from 0 - 1 = 0 - 100

Calculations in spectrophotometry

How to calculate a concentration of substance in analyzed sample

in the laboratory it is more convenient to calculate concentration from values of the absorbance (A) which is directly proportional to the concentration than from the transmittance Calculation from the Beeracutes law A = c x l x ε

c = molar concentration (molL)

l = inner width of the cuvette in centimeters

ε = molar absorption coefficient (tabelated value)

Relationship between A and T A = log (1T)= -log T

Calculations in spectrophotometry - examples1) Ast = 040 cst = 4 mgL

Asam = 025 csam = mgL

2) Standard solution of glucose (conc = 1000 mgL) reads a transmittance 49 Unknown sample of glucose reads T = 55 Calculate the concentration of glucose in the sample in mgL and mmolL

Mr (glucose) = 180

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Molar concentration - examples

1) 174 g NaCl in 300 mL of solution MW (NaCl) = 58 c =

2) 45 g glucose in 25 L of solution MW (glucose) = 180 c =

3) Solution of glycine c = 3 mM V = 100 mL MW (glycine ) = 75 m = mg of glycine in the solution

Number of ions in a certain volume

Problem 1 2 litres of solution contain 142 g of Na2HPO4 How many mmol Na+ ions are found in 20 mL of this solution Mr (Na2HPO4) = 142

Substance amount of Na2HPO4 in 2 L of solution n = 142142 = 1 mol

1 mol of Na2HPO4 in 2 L05 mol of Na2HPO4 in 1 L rarr05 mol Na2HPO4 gives 1 mol of Na+ and

05 mol of HPO42-

1 mol of Na+ in 1 LX mol of Na+ in 002 L X = 0021 x 1 = 002 mol = 20 mmol

Problem 2 Molarity of CaCl2 solution is 01 M Calculate the volume of solution containing 4 mmol of Cl-

01 M CaCl2 = 01 mol in 1 L 01 mol of CaCl2 gives 01 mol of Ca2+ and 02 mol of Cl-

02 mol of Cl- in 1 L0004 mol of Cl- in X LX = 000402 x 1 = 002 L = 20 mL

Osmotic pressureOsmotic pressure π is a hydrostatic pressure produced by solution in a space divided by a

semipermeable membrane due to a differential in the concentrations of solute

unit pascal PaΠ = i x c x R x T

Osmosis

= the movement of solvent

from an area of low concentration of

solute to an area of high concentration

Free diffusion

= the movement of solute from the

site of higher concentration to

the site of lower concentration

Oncotic pressure

= is a form of osmotic pressure exerted by proteins

in blood plasma

Osmolarity Osmolarity is a number of moles of a substance that contribute to

osmotic pressure of solution (osmolL) The concentration of body fluids is typically reported in mosmolLOsmolarity of blood is 290 ndash 300 mosmolL

π = i c R TThe figure is found at httpenwikipediaorgwikiOsmotic_pressureThe figure is found at httpenwikipediaorgwikiOsmotic_pressure

Osmolarity - examplesExample 1 A 1 M NaCl solution contains 2 osmol of solute per

liter of solution NaCl rarr Na+ + Cl-

1 M does dissociate 1 osmolL 1 osmolL 2 osmolL in total

Example 2 A 1 M CaCl2 solution contains 3 osmol of solute per liter of solution CaCl2 rarr Ca 2+ + 2 Cl-

1 M does dissociate 1 osmolL 2 osmolL 3 osmolL in totalExample 3 The concentration of a 1 M glucose solution is 1

osmolL

C6H12O6 rarr C6H12O6

1 M does not dissociate rarr 1 osmolL

Osmolarity - examplesOsmolarity - examples1 What is an osmolarity of 015 molL solution of1 What is an osmolarity of 015 molL solution of

a) NaCla) NaCl

b) MgClb) MgCl22

c) Nac) Na22HPOHPO44

d) glucosed) glucose

2 Saline is 150 mM solution of NaCl Which solutions are isotonic with 2 Saline is 150 mM solution of NaCl Which solutions are isotonic with saline saline [= 150 mM = 300 mosmol[= 150 mM = 300 mosmolLL]] a) 300 mM glucosea) 300 mM glucose

b) 50 mM CaClb) 50 mM CaCl22

c) 300 mM KClc) 300 mM KCl

d) 015 M NaHd) 015 M NaH22POPO44

3 What is molarity of 900 mosmoll solution of MgCl3 What is molarity of 900 mosmoll solution of MgCl22 in molL in molL

Percent concentration eexpressed as xpressed as part of solute per 100 part of solute per 100

parts of total solutionparts of total solution () () = = mass of solute mass of solute xx 100 100 mass of solutionmass of solution

it hit has 3 formsas 3 forms1 weight per weight (ww)1 weight per weight (ww)

10 of KCl = 1010 of KCl = 10 g of KCl + 90 g of g of KCl + 90 g of HH22O O ==100 g of solution100 g of solution

2 volume per volume2 volume per volume (vv)(vv)5 HCl = 5 5 HCl = 5 mLmL HCl in 100 m HCl in 100 mLL of of solutionsolution

3 weight per volume (wv)3 weight per volume (wv)the the most common expressionmost common expression09 NaCl = 09 g of NaCl in 100 m09 NaCl = 09 g of NaCl in 100 mLL of of solutionsolution

Percent concentrations - examples

1) 600 g 5 NaCl mass of NaCl mass of H2O

2) 250 g 8 Na2CO3 mass of Na2CO3 (purity 96)

3) 250 mL 39 ethanol solution mL of ethanol mL of H2O

4) Saline is 150 mM solution of NaCl Calculate the percent concentration by mass of this solution Mr(NaCl) = 585

Density ρ

- is defined as the amount of mass per unit of volume

ρ = mV rarr m = ρ x V and V = m ρ - these equations are useful for calculations

units gcm3 or gmL

- density of water = 1 g cm3

- density of lead (Pb) = 1134 gcm3

Conversions of concentrations ( and c) with density

1) What is a percent concentration of 2 M HNO3 solution Density (HNO3) = 1076 gml

Mr (HNO3) = 6301 Conversion of molar concentration to concentration 2 M HNO3 solution means that 2 mol of HNO3 are dissolved in 1 L of solution Mass of HNO3 = n x Mr = 2 x 6301 = 12602 g of HNO3

Mass of solution = ρ x V = 1076 x 1000 = 1076 gW = 12602 x 100 = 1171 1076

2) What is the molarity of 38 HCl solution Density (38 HCl) = 11885 gml and Mr(HCl) = 3645 Conversion of percent by mass concentration to molar concentration 38 HCl solution means that 38 g of HCl in 100 g of solution One liter of solution has a mass m = V x ρ = 1000 x 11885 = 11885 g

38 g HCl -------gt 100 g of solutionx g HCl -------gt 11885 g of solutionx = 45163 g HCl in 1 L of solution rarr n = m M = 45163 3645 = 124 mol of HCl

c(HCl) = n V = 124 1 = 124 molL

Conversions of concentrations ( and c) with density

Conversion of molarity to percent concentration

= c (molL) x Mr 10 x ρ (gcm3)

Conversion of percent concentration to molarity

c = x 10 x ρ (gcm3) Mr

Conversions of concentrations ( and c) with density - examples

1) (ww) of HNO3 ρ = 136 gcm3 if 1dm3 of solution contains 08 kg of HNO3

2) c (HNO3) = 562 M ρ = 118 gcm3 MW = 63 gmol 3) 10 HCl ρ = 1047 gcm3 MW = 365 c (HCl)

Dilution= concentration of a substance lowers number of moles of the substance remains the same

1) mix equation m1 x p1 + m2 x p2 = p x ( m1 + m2 ) m = mass of mixed solution p = concentration

2) expression of dilutionIn case of a liquid solute the ratio is presented as a dilution factor For example 1 5 is presented as 15 (1 mL of solute in 5 mL of solution)Example c1 = 025 M (original concentration) x 15 = 005 M (final concentration c2)

3) useful equation n1 = n2

V1 x c1 = V2 x c2

Dilution - examples

1) Mix 50 g 3 solution with 10 g 5 solution final concentration = ()

2) Final solution 190 g 10 sol

m (g) of 38 HCl + m (g) H2O 3) Dilute 300 g of 40 solution to 20 solution g

of solvent do you need4) preparation of 250 mL of 01 M HCl from stock 1

M HCl5) 10 M NaOH is diluted 1 20 final concentration6) 1000 mgL glucose is diluted 1 10 and then 1 2

final concentration

Conversions of unitsbull concentrationie gdL rarr mgLie molL harr osmolLbull units of pressure used in medicine1 mmHg (millimeter of mercury) = 1 Torr1 mmHg = 13322 Pabull units of energy1 cal = 41868 joule (J) 1 J = 0238 calcalorie (cal) is used in nutrition

Conversions of units ndash examples

1) Concentration of cholesterol in patient blood sample is 180 mgdL Convert this value into mmolL if MW of cholesterol is 387 gmol

2) Partial pressure of CO2 is 533 kPa Convert this value into mmHg

3) Red Bull energy drink (1 can) contains 160 cal Calculate the amount of energy in kJ

Calculations in spectrophotometry

spectrophotometry is an analytical method based on the interaction between an electromagnetic radiation of a known intenzity (Io) and the analyzed sample

a part of the radiation is absorbed by the analyzed substance found in the of sample

the intensity of the radiation which passed (I) through the solution

is detected (I lt I0) rarrthe transmittance (T) of a solution is defined as the proportion T = I I0

transmittance can be expressed in percentage T() = (II0) x 100

values of the measured transmittance are found from 0 - 1 = 0 - 100

Calculations in spectrophotometry

How to calculate a concentration of substance in analyzed sample

in the laboratory it is more convenient to calculate concentration from values of the absorbance (A) which is directly proportional to the concentration than from the transmittance Calculation from the Beeracutes law A = c x l x ε

c = molar concentration (molL)

l = inner width of the cuvette in centimeters

ε = molar absorption coefficient (tabelated value)

Relationship between A and T A = log (1T)= -log T

Calculations in spectrophotometry - examples1) Ast = 040 cst = 4 mgL

Asam = 025 csam = mgL

2) Standard solution of glucose (conc = 1000 mgL) reads a transmittance 49 Unknown sample of glucose reads T = 55 Calculate the concentration of glucose in the sample in mgL and mmolL

Mr (glucose) = 180

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Number of ions in a certain volume

Problem 1 2 litres of solution contain 142 g of Na2HPO4 How many mmol Na+ ions are found in 20 mL of this solution Mr (Na2HPO4) = 142

Substance amount of Na2HPO4 in 2 L of solution n = 142142 = 1 mol

1 mol of Na2HPO4 in 2 L05 mol of Na2HPO4 in 1 L rarr05 mol Na2HPO4 gives 1 mol of Na+ and

05 mol of HPO42-

1 mol of Na+ in 1 LX mol of Na+ in 002 L X = 0021 x 1 = 002 mol = 20 mmol

Problem 2 Molarity of CaCl2 solution is 01 M Calculate the volume of solution containing 4 mmol of Cl-

01 M CaCl2 = 01 mol in 1 L 01 mol of CaCl2 gives 01 mol of Ca2+ and 02 mol of Cl-

02 mol of Cl- in 1 L0004 mol of Cl- in X LX = 000402 x 1 = 002 L = 20 mL

Osmotic pressureOsmotic pressure π is a hydrostatic pressure produced by solution in a space divided by a

semipermeable membrane due to a differential in the concentrations of solute

unit pascal PaΠ = i x c x R x T

Osmosis

= the movement of solvent

from an area of low concentration of

solute to an area of high concentration

Free diffusion

= the movement of solute from the

site of higher concentration to

the site of lower concentration

Oncotic pressure

= is a form of osmotic pressure exerted by proteins

in blood plasma

Osmolarity Osmolarity is a number of moles of a substance that contribute to

osmotic pressure of solution (osmolL) The concentration of body fluids is typically reported in mosmolLOsmolarity of blood is 290 ndash 300 mosmolL

π = i c R TThe figure is found at httpenwikipediaorgwikiOsmotic_pressureThe figure is found at httpenwikipediaorgwikiOsmotic_pressure

Osmolarity - examplesExample 1 A 1 M NaCl solution contains 2 osmol of solute per

liter of solution NaCl rarr Na+ + Cl-

1 M does dissociate 1 osmolL 1 osmolL 2 osmolL in total

Example 2 A 1 M CaCl2 solution contains 3 osmol of solute per liter of solution CaCl2 rarr Ca 2+ + 2 Cl-

1 M does dissociate 1 osmolL 2 osmolL 3 osmolL in totalExample 3 The concentration of a 1 M glucose solution is 1

osmolL

C6H12O6 rarr C6H12O6

1 M does not dissociate rarr 1 osmolL

Osmolarity - examplesOsmolarity - examples1 What is an osmolarity of 015 molL solution of1 What is an osmolarity of 015 molL solution of

a) NaCla) NaCl

b) MgClb) MgCl22

c) Nac) Na22HPOHPO44

d) glucosed) glucose

2 Saline is 150 mM solution of NaCl Which solutions are isotonic with 2 Saline is 150 mM solution of NaCl Which solutions are isotonic with saline saline [= 150 mM = 300 mosmol[= 150 mM = 300 mosmolLL]] a) 300 mM glucosea) 300 mM glucose

b) 50 mM CaClb) 50 mM CaCl22

c) 300 mM KClc) 300 mM KCl

d) 015 M NaHd) 015 M NaH22POPO44

3 What is molarity of 900 mosmoll solution of MgCl3 What is molarity of 900 mosmoll solution of MgCl22 in molL in molL

Percent concentration eexpressed as xpressed as part of solute per 100 part of solute per 100

parts of total solutionparts of total solution () () = = mass of solute mass of solute xx 100 100 mass of solutionmass of solution

it hit has 3 formsas 3 forms1 weight per weight (ww)1 weight per weight (ww)

10 of KCl = 1010 of KCl = 10 g of KCl + 90 g of g of KCl + 90 g of HH22O O ==100 g of solution100 g of solution

2 volume per volume2 volume per volume (vv)(vv)5 HCl = 5 5 HCl = 5 mLmL HCl in 100 m HCl in 100 mLL of of solutionsolution

3 weight per volume (wv)3 weight per volume (wv)the the most common expressionmost common expression09 NaCl = 09 g of NaCl in 100 m09 NaCl = 09 g of NaCl in 100 mLL of of solutionsolution

Percent concentrations - examples

1) 600 g 5 NaCl mass of NaCl mass of H2O

2) 250 g 8 Na2CO3 mass of Na2CO3 (purity 96)

3) 250 mL 39 ethanol solution mL of ethanol mL of H2O

4) Saline is 150 mM solution of NaCl Calculate the percent concentration by mass of this solution Mr(NaCl) = 585

Density ρ

- is defined as the amount of mass per unit of volume

ρ = mV rarr m = ρ x V and V = m ρ - these equations are useful for calculations

units gcm3 or gmL

- density of water = 1 g cm3

- density of lead (Pb) = 1134 gcm3

Conversions of concentrations ( and c) with density

1) What is a percent concentration of 2 M HNO3 solution Density (HNO3) = 1076 gml

Mr (HNO3) = 6301 Conversion of molar concentration to concentration 2 M HNO3 solution means that 2 mol of HNO3 are dissolved in 1 L of solution Mass of HNO3 = n x Mr = 2 x 6301 = 12602 g of HNO3

Mass of solution = ρ x V = 1076 x 1000 = 1076 gW = 12602 x 100 = 1171 1076

2) What is the molarity of 38 HCl solution Density (38 HCl) = 11885 gml and Mr(HCl) = 3645 Conversion of percent by mass concentration to molar concentration 38 HCl solution means that 38 g of HCl in 100 g of solution One liter of solution has a mass m = V x ρ = 1000 x 11885 = 11885 g

38 g HCl -------gt 100 g of solutionx g HCl -------gt 11885 g of solutionx = 45163 g HCl in 1 L of solution rarr n = m M = 45163 3645 = 124 mol of HCl

c(HCl) = n V = 124 1 = 124 molL

Conversions of concentrations ( and c) with density

Conversion of molarity to percent concentration

= c (molL) x Mr 10 x ρ (gcm3)

Conversion of percent concentration to molarity

c = x 10 x ρ (gcm3) Mr

Conversions of concentrations ( and c) with density - examples

1) (ww) of HNO3 ρ = 136 gcm3 if 1dm3 of solution contains 08 kg of HNO3

2) c (HNO3) = 562 M ρ = 118 gcm3 MW = 63 gmol 3) 10 HCl ρ = 1047 gcm3 MW = 365 c (HCl)

Dilution= concentration of a substance lowers number of moles of the substance remains the same

1) mix equation m1 x p1 + m2 x p2 = p x ( m1 + m2 ) m = mass of mixed solution p = concentration

2) expression of dilutionIn case of a liquid solute the ratio is presented as a dilution factor For example 1 5 is presented as 15 (1 mL of solute in 5 mL of solution)Example c1 = 025 M (original concentration) x 15 = 005 M (final concentration c2)

3) useful equation n1 = n2

V1 x c1 = V2 x c2

Dilution - examples

1) Mix 50 g 3 solution with 10 g 5 solution final concentration = ()

2) Final solution 190 g 10 sol

m (g) of 38 HCl + m (g) H2O 3) Dilute 300 g of 40 solution to 20 solution g

of solvent do you need4) preparation of 250 mL of 01 M HCl from stock 1

M HCl5) 10 M NaOH is diluted 1 20 final concentration6) 1000 mgL glucose is diluted 1 10 and then 1 2

final concentration

Conversions of unitsbull concentrationie gdL rarr mgLie molL harr osmolLbull units of pressure used in medicine1 mmHg (millimeter of mercury) = 1 Torr1 mmHg = 13322 Pabull units of energy1 cal = 41868 joule (J) 1 J = 0238 calcalorie (cal) is used in nutrition

Conversions of units ndash examples

1) Concentration of cholesterol in patient blood sample is 180 mgdL Convert this value into mmolL if MW of cholesterol is 387 gmol

2) Partial pressure of CO2 is 533 kPa Convert this value into mmHg

3) Red Bull energy drink (1 can) contains 160 cal Calculate the amount of energy in kJ

Calculations in spectrophotometry

spectrophotometry is an analytical method based on the interaction between an electromagnetic radiation of a known intenzity (Io) and the analyzed sample

a part of the radiation is absorbed by the analyzed substance found in the of sample

the intensity of the radiation which passed (I) through the solution

is detected (I lt I0) rarrthe transmittance (T) of a solution is defined as the proportion T = I I0

transmittance can be expressed in percentage T() = (II0) x 100

values of the measured transmittance are found from 0 - 1 = 0 - 100

Calculations in spectrophotometry

How to calculate a concentration of substance in analyzed sample

in the laboratory it is more convenient to calculate concentration from values of the absorbance (A) which is directly proportional to the concentration than from the transmittance Calculation from the Beeracutes law A = c x l x ε

c = molar concentration (molL)

l = inner width of the cuvette in centimeters

ε = molar absorption coefficient (tabelated value)

Relationship between A and T A = log (1T)= -log T

Calculations in spectrophotometry - examples1) Ast = 040 cst = 4 mgL

Asam = 025 csam = mgL

2) Standard solution of glucose (conc = 1000 mgL) reads a transmittance 49 Unknown sample of glucose reads T = 55 Calculate the concentration of glucose in the sample in mgL and mmolL

Mr (glucose) = 180

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Osmotic pressureOsmotic pressure π is a hydrostatic pressure produced by solution in a space divided by a

semipermeable membrane due to a differential in the concentrations of solute

unit pascal PaΠ = i x c x R x T

Osmosis

= the movement of solvent

from an area of low concentration of

solute to an area of high concentration

Free diffusion

= the movement of solute from the

site of higher concentration to

the site of lower concentration

Oncotic pressure

= is a form of osmotic pressure exerted by proteins

in blood plasma

Osmolarity Osmolarity is a number of moles of a substance that contribute to

osmotic pressure of solution (osmolL) The concentration of body fluids is typically reported in mosmolLOsmolarity of blood is 290 ndash 300 mosmolL

π = i c R TThe figure is found at httpenwikipediaorgwikiOsmotic_pressureThe figure is found at httpenwikipediaorgwikiOsmotic_pressure

Osmolarity - examplesExample 1 A 1 M NaCl solution contains 2 osmol of solute per

liter of solution NaCl rarr Na+ + Cl-

1 M does dissociate 1 osmolL 1 osmolL 2 osmolL in total

Example 2 A 1 M CaCl2 solution contains 3 osmol of solute per liter of solution CaCl2 rarr Ca 2+ + 2 Cl-

1 M does dissociate 1 osmolL 2 osmolL 3 osmolL in totalExample 3 The concentration of a 1 M glucose solution is 1

osmolL

C6H12O6 rarr C6H12O6

1 M does not dissociate rarr 1 osmolL

Osmolarity - examplesOsmolarity - examples1 What is an osmolarity of 015 molL solution of1 What is an osmolarity of 015 molL solution of

a) NaCla) NaCl

b) MgClb) MgCl22

c) Nac) Na22HPOHPO44

d) glucosed) glucose

2 Saline is 150 mM solution of NaCl Which solutions are isotonic with 2 Saline is 150 mM solution of NaCl Which solutions are isotonic with saline saline [= 150 mM = 300 mosmol[= 150 mM = 300 mosmolLL]] a) 300 mM glucosea) 300 mM glucose

b) 50 mM CaClb) 50 mM CaCl22

c) 300 mM KClc) 300 mM KCl

d) 015 M NaHd) 015 M NaH22POPO44

3 What is molarity of 900 mosmoll solution of MgCl3 What is molarity of 900 mosmoll solution of MgCl22 in molL in molL

Percent concentration eexpressed as xpressed as part of solute per 100 part of solute per 100

parts of total solutionparts of total solution () () = = mass of solute mass of solute xx 100 100 mass of solutionmass of solution

it hit has 3 formsas 3 forms1 weight per weight (ww)1 weight per weight (ww)

10 of KCl = 1010 of KCl = 10 g of KCl + 90 g of g of KCl + 90 g of HH22O O ==100 g of solution100 g of solution

2 volume per volume2 volume per volume (vv)(vv)5 HCl = 5 5 HCl = 5 mLmL HCl in 100 m HCl in 100 mLL of of solutionsolution

3 weight per volume (wv)3 weight per volume (wv)the the most common expressionmost common expression09 NaCl = 09 g of NaCl in 100 m09 NaCl = 09 g of NaCl in 100 mLL of of solutionsolution

Percent concentrations - examples

1) 600 g 5 NaCl mass of NaCl mass of H2O

2) 250 g 8 Na2CO3 mass of Na2CO3 (purity 96)

3) 250 mL 39 ethanol solution mL of ethanol mL of H2O

4) Saline is 150 mM solution of NaCl Calculate the percent concentration by mass of this solution Mr(NaCl) = 585

Density ρ

- is defined as the amount of mass per unit of volume

ρ = mV rarr m = ρ x V and V = m ρ - these equations are useful for calculations

units gcm3 or gmL

- density of water = 1 g cm3

- density of lead (Pb) = 1134 gcm3

Conversions of concentrations ( and c) with density

1) What is a percent concentration of 2 M HNO3 solution Density (HNO3) = 1076 gml

Mr (HNO3) = 6301 Conversion of molar concentration to concentration 2 M HNO3 solution means that 2 mol of HNO3 are dissolved in 1 L of solution Mass of HNO3 = n x Mr = 2 x 6301 = 12602 g of HNO3

Mass of solution = ρ x V = 1076 x 1000 = 1076 gW = 12602 x 100 = 1171 1076

2) What is the molarity of 38 HCl solution Density (38 HCl) = 11885 gml and Mr(HCl) = 3645 Conversion of percent by mass concentration to molar concentration 38 HCl solution means that 38 g of HCl in 100 g of solution One liter of solution has a mass m = V x ρ = 1000 x 11885 = 11885 g

38 g HCl -------gt 100 g of solutionx g HCl -------gt 11885 g of solutionx = 45163 g HCl in 1 L of solution rarr n = m M = 45163 3645 = 124 mol of HCl

c(HCl) = n V = 124 1 = 124 molL

Conversions of concentrations ( and c) with density

Conversion of molarity to percent concentration

= c (molL) x Mr 10 x ρ (gcm3)

Conversion of percent concentration to molarity

c = x 10 x ρ (gcm3) Mr

Conversions of concentrations ( and c) with density - examples

1) (ww) of HNO3 ρ = 136 gcm3 if 1dm3 of solution contains 08 kg of HNO3

2) c (HNO3) = 562 M ρ = 118 gcm3 MW = 63 gmol 3) 10 HCl ρ = 1047 gcm3 MW = 365 c (HCl)

Dilution= concentration of a substance lowers number of moles of the substance remains the same

1) mix equation m1 x p1 + m2 x p2 = p x ( m1 + m2 ) m = mass of mixed solution p = concentration

2) expression of dilutionIn case of a liquid solute the ratio is presented as a dilution factor For example 1 5 is presented as 15 (1 mL of solute in 5 mL of solution)Example c1 = 025 M (original concentration) x 15 = 005 M (final concentration c2)

3) useful equation n1 = n2

V1 x c1 = V2 x c2

Dilution - examples

1) Mix 50 g 3 solution with 10 g 5 solution final concentration = ()

2) Final solution 190 g 10 sol

m (g) of 38 HCl + m (g) H2O 3) Dilute 300 g of 40 solution to 20 solution g

of solvent do you need4) preparation of 250 mL of 01 M HCl from stock 1

M HCl5) 10 M NaOH is diluted 1 20 final concentration6) 1000 mgL glucose is diluted 1 10 and then 1 2

final concentration

Conversions of unitsbull concentrationie gdL rarr mgLie molL harr osmolLbull units of pressure used in medicine1 mmHg (millimeter of mercury) = 1 Torr1 mmHg = 13322 Pabull units of energy1 cal = 41868 joule (J) 1 J = 0238 calcalorie (cal) is used in nutrition

Conversions of units ndash examples

1) Concentration of cholesterol in patient blood sample is 180 mgdL Convert this value into mmolL if MW of cholesterol is 387 gmol

2) Partial pressure of CO2 is 533 kPa Convert this value into mmHg

3) Red Bull energy drink (1 can) contains 160 cal Calculate the amount of energy in kJ

Calculations in spectrophotometry

spectrophotometry is an analytical method based on the interaction between an electromagnetic radiation of a known intenzity (Io) and the analyzed sample

a part of the radiation is absorbed by the analyzed substance found in the of sample

the intensity of the radiation which passed (I) through the solution

is detected (I lt I0) rarrthe transmittance (T) of a solution is defined as the proportion T = I I0

transmittance can be expressed in percentage T() = (II0) x 100

values of the measured transmittance are found from 0 - 1 = 0 - 100

Calculations in spectrophotometry

How to calculate a concentration of substance in analyzed sample

in the laboratory it is more convenient to calculate concentration from values of the absorbance (A) which is directly proportional to the concentration than from the transmittance Calculation from the Beeracutes law A = c x l x ε

c = molar concentration (molL)

l = inner width of the cuvette in centimeters

ε = molar absorption coefficient (tabelated value)

Relationship between A and T A = log (1T)= -log T

Calculations in spectrophotometry - examples1) Ast = 040 cst = 4 mgL

Asam = 025 csam = mgL

2) Standard solution of glucose (conc = 1000 mgL) reads a transmittance 49 Unknown sample of glucose reads T = 55 Calculate the concentration of glucose in the sample in mgL and mmolL

Mr (glucose) = 180

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Osmolarity Osmolarity is a number of moles of a substance that contribute to

osmotic pressure of solution (osmolL) The concentration of body fluids is typically reported in mosmolLOsmolarity of blood is 290 ndash 300 mosmolL

π = i c R TThe figure is found at httpenwikipediaorgwikiOsmotic_pressureThe figure is found at httpenwikipediaorgwikiOsmotic_pressure

Osmolarity - examplesExample 1 A 1 M NaCl solution contains 2 osmol of solute per

liter of solution NaCl rarr Na+ + Cl-

1 M does dissociate 1 osmolL 1 osmolL 2 osmolL in total

Example 2 A 1 M CaCl2 solution contains 3 osmol of solute per liter of solution CaCl2 rarr Ca 2+ + 2 Cl-

1 M does dissociate 1 osmolL 2 osmolL 3 osmolL in totalExample 3 The concentration of a 1 M glucose solution is 1

osmolL

C6H12O6 rarr C6H12O6

1 M does not dissociate rarr 1 osmolL

Osmolarity - examplesOsmolarity - examples1 What is an osmolarity of 015 molL solution of1 What is an osmolarity of 015 molL solution of

a) NaCla) NaCl

b) MgClb) MgCl22

c) Nac) Na22HPOHPO44

d) glucosed) glucose

2 Saline is 150 mM solution of NaCl Which solutions are isotonic with 2 Saline is 150 mM solution of NaCl Which solutions are isotonic with saline saline [= 150 mM = 300 mosmol[= 150 mM = 300 mosmolLL]] a) 300 mM glucosea) 300 mM glucose

b) 50 mM CaClb) 50 mM CaCl22

c) 300 mM KClc) 300 mM KCl

d) 015 M NaHd) 015 M NaH22POPO44

3 What is molarity of 900 mosmoll solution of MgCl3 What is molarity of 900 mosmoll solution of MgCl22 in molL in molL

Percent concentration eexpressed as xpressed as part of solute per 100 part of solute per 100

parts of total solutionparts of total solution () () = = mass of solute mass of solute xx 100 100 mass of solutionmass of solution

it hit has 3 formsas 3 forms1 weight per weight (ww)1 weight per weight (ww)

10 of KCl = 1010 of KCl = 10 g of KCl + 90 g of g of KCl + 90 g of HH22O O ==100 g of solution100 g of solution

2 volume per volume2 volume per volume (vv)(vv)5 HCl = 5 5 HCl = 5 mLmL HCl in 100 m HCl in 100 mLL of of solutionsolution

3 weight per volume (wv)3 weight per volume (wv)the the most common expressionmost common expression09 NaCl = 09 g of NaCl in 100 m09 NaCl = 09 g of NaCl in 100 mLL of of solutionsolution

Percent concentrations - examples

1) 600 g 5 NaCl mass of NaCl mass of H2O

2) 250 g 8 Na2CO3 mass of Na2CO3 (purity 96)

3) 250 mL 39 ethanol solution mL of ethanol mL of H2O

4) Saline is 150 mM solution of NaCl Calculate the percent concentration by mass of this solution Mr(NaCl) = 585

Density ρ

- is defined as the amount of mass per unit of volume

ρ = mV rarr m = ρ x V and V = m ρ - these equations are useful for calculations

units gcm3 or gmL

- density of water = 1 g cm3

- density of lead (Pb) = 1134 gcm3

Conversions of concentrations ( and c) with density

1) What is a percent concentration of 2 M HNO3 solution Density (HNO3) = 1076 gml

Mr (HNO3) = 6301 Conversion of molar concentration to concentration 2 M HNO3 solution means that 2 mol of HNO3 are dissolved in 1 L of solution Mass of HNO3 = n x Mr = 2 x 6301 = 12602 g of HNO3

Mass of solution = ρ x V = 1076 x 1000 = 1076 gW = 12602 x 100 = 1171 1076

2) What is the molarity of 38 HCl solution Density (38 HCl) = 11885 gml and Mr(HCl) = 3645 Conversion of percent by mass concentration to molar concentration 38 HCl solution means that 38 g of HCl in 100 g of solution One liter of solution has a mass m = V x ρ = 1000 x 11885 = 11885 g

38 g HCl -------gt 100 g of solutionx g HCl -------gt 11885 g of solutionx = 45163 g HCl in 1 L of solution rarr n = m M = 45163 3645 = 124 mol of HCl

c(HCl) = n V = 124 1 = 124 molL

Conversions of concentrations ( and c) with density

Conversion of molarity to percent concentration

= c (molL) x Mr 10 x ρ (gcm3)

Conversion of percent concentration to molarity

c = x 10 x ρ (gcm3) Mr

Conversions of concentrations ( and c) with density - examples

1) (ww) of HNO3 ρ = 136 gcm3 if 1dm3 of solution contains 08 kg of HNO3

2) c (HNO3) = 562 M ρ = 118 gcm3 MW = 63 gmol 3) 10 HCl ρ = 1047 gcm3 MW = 365 c (HCl)

Dilution= concentration of a substance lowers number of moles of the substance remains the same

1) mix equation m1 x p1 + m2 x p2 = p x ( m1 + m2 ) m = mass of mixed solution p = concentration

2) expression of dilutionIn case of a liquid solute the ratio is presented as a dilution factor For example 1 5 is presented as 15 (1 mL of solute in 5 mL of solution)Example c1 = 025 M (original concentration) x 15 = 005 M (final concentration c2)

3) useful equation n1 = n2

V1 x c1 = V2 x c2

Dilution - examples

1) Mix 50 g 3 solution with 10 g 5 solution final concentration = ()

2) Final solution 190 g 10 sol

m (g) of 38 HCl + m (g) H2O 3) Dilute 300 g of 40 solution to 20 solution g

of solvent do you need4) preparation of 250 mL of 01 M HCl from stock 1

M HCl5) 10 M NaOH is diluted 1 20 final concentration6) 1000 mgL glucose is diluted 1 10 and then 1 2

final concentration

Conversions of unitsbull concentrationie gdL rarr mgLie molL harr osmolLbull units of pressure used in medicine1 mmHg (millimeter of mercury) = 1 Torr1 mmHg = 13322 Pabull units of energy1 cal = 41868 joule (J) 1 J = 0238 calcalorie (cal) is used in nutrition

Conversions of units ndash examples

1) Concentration of cholesterol in patient blood sample is 180 mgdL Convert this value into mmolL if MW of cholesterol is 387 gmol

2) Partial pressure of CO2 is 533 kPa Convert this value into mmHg

3) Red Bull energy drink (1 can) contains 160 cal Calculate the amount of energy in kJ

Calculations in spectrophotometry

spectrophotometry is an analytical method based on the interaction between an electromagnetic radiation of a known intenzity (Io) and the analyzed sample

a part of the radiation is absorbed by the analyzed substance found in the of sample

the intensity of the radiation which passed (I) through the solution

is detected (I lt I0) rarrthe transmittance (T) of a solution is defined as the proportion T = I I0

transmittance can be expressed in percentage T() = (II0) x 100

values of the measured transmittance are found from 0 - 1 = 0 - 100

Calculations in spectrophotometry

How to calculate a concentration of substance in analyzed sample

in the laboratory it is more convenient to calculate concentration from values of the absorbance (A) which is directly proportional to the concentration than from the transmittance Calculation from the Beeracutes law A = c x l x ε

c = molar concentration (molL)

l = inner width of the cuvette in centimeters

ε = molar absorption coefficient (tabelated value)

Relationship between A and T A = log (1T)= -log T

Calculations in spectrophotometry - examples1) Ast = 040 cst = 4 mgL

Asam = 025 csam = mgL

2) Standard solution of glucose (conc = 1000 mgL) reads a transmittance 49 Unknown sample of glucose reads T = 55 Calculate the concentration of glucose in the sample in mgL and mmolL

Mr (glucose) = 180

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Osmolarity - examplesExample 1 A 1 M NaCl solution contains 2 osmol of solute per

liter of solution NaCl rarr Na+ + Cl-

1 M does dissociate 1 osmolL 1 osmolL 2 osmolL in total

Example 2 A 1 M CaCl2 solution contains 3 osmol of solute per liter of solution CaCl2 rarr Ca 2+ + 2 Cl-

1 M does dissociate 1 osmolL 2 osmolL 3 osmolL in totalExample 3 The concentration of a 1 M glucose solution is 1

osmolL

C6H12O6 rarr C6H12O6

1 M does not dissociate rarr 1 osmolL

Osmolarity - examplesOsmolarity - examples1 What is an osmolarity of 015 molL solution of1 What is an osmolarity of 015 molL solution of

a) NaCla) NaCl

b) MgClb) MgCl22

c) Nac) Na22HPOHPO44

d) glucosed) glucose

2 Saline is 150 mM solution of NaCl Which solutions are isotonic with 2 Saline is 150 mM solution of NaCl Which solutions are isotonic with saline saline [= 150 mM = 300 mosmol[= 150 mM = 300 mosmolLL]] a) 300 mM glucosea) 300 mM glucose

b) 50 mM CaClb) 50 mM CaCl22

c) 300 mM KClc) 300 mM KCl

d) 015 M NaHd) 015 M NaH22POPO44

3 What is molarity of 900 mosmoll solution of MgCl3 What is molarity of 900 mosmoll solution of MgCl22 in molL in molL

Percent concentration eexpressed as xpressed as part of solute per 100 part of solute per 100

parts of total solutionparts of total solution () () = = mass of solute mass of solute xx 100 100 mass of solutionmass of solution

it hit has 3 formsas 3 forms1 weight per weight (ww)1 weight per weight (ww)

10 of KCl = 1010 of KCl = 10 g of KCl + 90 g of g of KCl + 90 g of HH22O O ==100 g of solution100 g of solution

2 volume per volume2 volume per volume (vv)(vv)5 HCl = 5 5 HCl = 5 mLmL HCl in 100 m HCl in 100 mLL of of solutionsolution

3 weight per volume (wv)3 weight per volume (wv)the the most common expressionmost common expression09 NaCl = 09 g of NaCl in 100 m09 NaCl = 09 g of NaCl in 100 mLL of of solutionsolution

Percent concentrations - examples

1) 600 g 5 NaCl mass of NaCl mass of H2O

2) 250 g 8 Na2CO3 mass of Na2CO3 (purity 96)

3) 250 mL 39 ethanol solution mL of ethanol mL of H2O

4) Saline is 150 mM solution of NaCl Calculate the percent concentration by mass of this solution Mr(NaCl) = 585

Density ρ

- is defined as the amount of mass per unit of volume

ρ = mV rarr m = ρ x V and V = m ρ - these equations are useful for calculations

units gcm3 or gmL

- density of water = 1 g cm3

- density of lead (Pb) = 1134 gcm3

Conversions of concentrations ( and c) with density

1) What is a percent concentration of 2 M HNO3 solution Density (HNO3) = 1076 gml

Mr (HNO3) = 6301 Conversion of molar concentration to concentration 2 M HNO3 solution means that 2 mol of HNO3 are dissolved in 1 L of solution Mass of HNO3 = n x Mr = 2 x 6301 = 12602 g of HNO3

Mass of solution = ρ x V = 1076 x 1000 = 1076 gW = 12602 x 100 = 1171 1076

2) What is the molarity of 38 HCl solution Density (38 HCl) = 11885 gml and Mr(HCl) = 3645 Conversion of percent by mass concentration to molar concentration 38 HCl solution means that 38 g of HCl in 100 g of solution One liter of solution has a mass m = V x ρ = 1000 x 11885 = 11885 g

38 g HCl -------gt 100 g of solutionx g HCl -------gt 11885 g of solutionx = 45163 g HCl in 1 L of solution rarr n = m M = 45163 3645 = 124 mol of HCl

c(HCl) = n V = 124 1 = 124 molL

Conversions of concentrations ( and c) with density

Conversion of molarity to percent concentration

= c (molL) x Mr 10 x ρ (gcm3)

Conversion of percent concentration to molarity

c = x 10 x ρ (gcm3) Mr

Conversions of concentrations ( and c) with density - examples

1) (ww) of HNO3 ρ = 136 gcm3 if 1dm3 of solution contains 08 kg of HNO3

2) c (HNO3) = 562 M ρ = 118 gcm3 MW = 63 gmol 3) 10 HCl ρ = 1047 gcm3 MW = 365 c (HCl)

Dilution= concentration of a substance lowers number of moles of the substance remains the same

1) mix equation m1 x p1 + m2 x p2 = p x ( m1 + m2 ) m = mass of mixed solution p = concentration

2) expression of dilutionIn case of a liquid solute the ratio is presented as a dilution factor For example 1 5 is presented as 15 (1 mL of solute in 5 mL of solution)Example c1 = 025 M (original concentration) x 15 = 005 M (final concentration c2)

3) useful equation n1 = n2

V1 x c1 = V2 x c2

Dilution - examples

1) Mix 50 g 3 solution with 10 g 5 solution final concentration = ()

2) Final solution 190 g 10 sol

m (g) of 38 HCl + m (g) H2O 3) Dilute 300 g of 40 solution to 20 solution g

of solvent do you need4) preparation of 250 mL of 01 M HCl from stock 1

M HCl5) 10 M NaOH is diluted 1 20 final concentration6) 1000 mgL glucose is diluted 1 10 and then 1 2

final concentration

Conversions of unitsbull concentrationie gdL rarr mgLie molL harr osmolLbull units of pressure used in medicine1 mmHg (millimeter of mercury) = 1 Torr1 mmHg = 13322 Pabull units of energy1 cal = 41868 joule (J) 1 J = 0238 calcalorie (cal) is used in nutrition

Conversions of units ndash examples

1) Concentration of cholesterol in patient blood sample is 180 mgdL Convert this value into mmolL if MW of cholesterol is 387 gmol

2) Partial pressure of CO2 is 533 kPa Convert this value into mmHg

3) Red Bull energy drink (1 can) contains 160 cal Calculate the amount of energy in kJ

Calculations in spectrophotometry

spectrophotometry is an analytical method based on the interaction between an electromagnetic radiation of a known intenzity (Io) and the analyzed sample

a part of the radiation is absorbed by the analyzed substance found in the of sample

the intensity of the radiation which passed (I) through the solution

is detected (I lt I0) rarrthe transmittance (T) of a solution is defined as the proportion T = I I0

transmittance can be expressed in percentage T() = (II0) x 100

values of the measured transmittance are found from 0 - 1 = 0 - 100

Calculations in spectrophotometry

How to calculate a concentration of substance in analyzed sample

in the laboratory it is more convenient to calculate concentration from values of the absorbance (A) which is directly proportional to the concentration than from the transmittance Calculation from the Beeracutes law A = c x l x ε

c = molar concentration (molL)

l = inner width of the cuvette in centimeters

ε = molar absorption coefficient (tabelated value)

Relationship between A and T A = log (1T)= -log T

Calculations in spectrophotometry - examples1) Ast = 040 cst = 4 mgL

Asam = 025 csam = mgL

2) Standard solution of glucose (conc = 1000 mgL) reads a transmittance 49 Unknown sample of glucose reads T = 55 Calculate the concentration of glucose in the sample in mgL and mmolL

Mr (glucose) = 180

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Osmolarity - examplesOsmolarity - examples1 What is an osmolarity of 015 molL solution of1 What is an osmolarity of 015 molL solution of

a) NaCla) NaCl

b) MgClb) MgCl22

c) Nac) Na22HPOHPO44

d) glucosed) glucose

2 Saline is 150 mM solution of NaCl Which solutions are isotonic with 2 Saline is 150 mM solution of NaCl Which solutions are isotonic with saline saline [= 150 mM = 300 mosmol[= 150 mM = 300 mosmolLL]] a) 300 mM glucosea) 300 mM glucose

b) 50 mM CaClb) 50 mM CaCl22

c) 300 mM KClc) 300 mM KCl

d) 015 M NaHd) 015 M NaH22POPO44

3 What is molarity of 900 mosmoll solution of MgCl3 What is molarity of 900 mosmoll solution of MgCl22 in molL in molL

Percent concentration eexpressed as xpressed as part of solute per 100 part of solute per 100

parts of total solutionparts of total solution () () = = mass of solute mass of solute xx 100 100 mass of solutionmass of solution

it hit has 3 formsas 3 forms1 weight per weight (ww)1 weight per weight (ww)

10 of KCl = 1010 of KCl = 10 g of KCl + 90 g of g of KCl + 90 g of HH22O O ==100 g of solution100 g of solution

2 volume per volume2 volume per volume (vv)(vv)5 HCl = 5 5 HCl = 5 mLmL HCl in 100 m HCl in 100 mLL of of solutionsolution

3 weight per volume (wv)3 weight per volume (wv)the the most common expressionmost common expression09 NaCl = 09 g of NaCl in 100 m09 NaCl = 09 g of NaCl in 100 mLL of of solutionsolution

Percent concentrations - examples

1) 600 g 5 NaCl mass of NaCl mass of H2O

2) 250 g 8 Na2CO3 mass of Na2CO3 (purity 96)

3) 250 mL 39 ethanol solution mL of ethanol mL of H2O

4) Saline is 150 mM solution of NaCl Calculate the percent concentration by mass of this solution Mr(NaCl) = 585

Density ρ

- is defined as the amount of mass per unit of volume

ρ = mV rarr m = ρ x V and V = m ρ - these equations are useful for calculations

units gcm3 or gmL

- density of water = 1 g cm3

- density of lead (Pb) = 1134 gcm3

Conversions of concentrations ( and c) with density

1) What is a percent concentration of 2 M HNO3 solution Density (HNO3) = 1076 gml

Mr (HNO3) = 6301 Conversion of molar concentration to concentration 2 M HNO3 solution means that 2 mol of HNO3 are dissolved in 1 L of solution Mass of HNO3 = n x Mr = 2 x 6301 = 12602 g of HNO3

Mass of solution = ρ x V = 1076 x 1000 = 1076 gW = 12602 x 100 = 1171 1076

2) What is the molarity of 38 HCl solution Density (38 HCl) = 11885 gml and Mr(HCl) = 3645 Conversion of percent by mass concentration to molar concentration 38 HCl solution means that 38 g of HCl in 100 g of solution One liter of solution has a mass m = V x ρ = 1000 x 11885 = 11885 g

38 g HCl -------gt 100 g of solutionx g HCl -------gt 11885 g of solutionx = 45163 g HCl in 1 L of solution rarr n = m M = 45163 3645 = 124 mol of HCl

c(HCl) = n V = 124 1 = 124 molL

Conversions of concentrations ( and c) with density

Conversion of molarity to percent concentration

= c (molL) x Mr 10 x ρ (gcm3)

Conversion of percent concentration to molarity

c = x 10 x ρ (gcm3) Mr

Conversions of concentrations ( and c) with density - examples

1) (ww) of HNO3 ρ = 136 gcm3 if 1dm3 of solution contains 08 kg of HNO3

2) c (HNO3) = 562 M ρ = 118 gcm3 MW = 63 gmol 3) 10 HCl ρ = 1047 gcm3 MW = 365 c (HCl)

Dilution= concentration of a substance lowers number of moles of the substance remains the same

1) mix equation m1 x p1 + m2 x p2 = p x ( m1 + m2 ) m = mass of mixed solution p = concentration

2) expression of dilutionIn case of a liquid solute the ratio is presented as a dilution factor For example 1 5 is presented as 15 (1 mL of solute in 5 mL of solution)Example c1 = 025 M (original concentration) x 15 = 005 M (final concentration c2)

3) useful equation n1 = n2

V1 x c1 = V2 x c2

Dilution - examples

1) Mix 50 g 3 solution with 10 g 5 solution final concentration = ()

2) Final solution 190 g 10 sol

m (g) of 38 HCl + m (g) H2O 3) Dilute 300 g of 40 solution to 20 solution g

of solvent do you need4) preparation of 250 mL of 01 M HCl from stock 1

M HCl5) 10 M NaOH is diluted 1 20 final concentration6) 1000 mgL glucose is diluted 1 10 and then 1 2

final concentration

Conversions of unitsbull concentrationie gdL rarr mgLie molL harr osmolLbull units of pressure used in medicine1 mmHg (millimeter of mercury) = 1 Torr1 mmHg = 13322 Pabull units of energy1 cal = 41868 joule (J) 1 J = 0238 calcalorie (cal) is used in nutrition

Conversions of units ndash examples

1) Concentration of cholesterol in patient blood sample is 180 mgdL Convert this value into mmolL if MW of cholesterol is 387 gmol

2) Partial pressure of CO2 is 533 kPa Convert this value into mmHg

3) Red Bull energy drink (1 can) contains 160 cal Calculate the amount of energy in kJ

Calculations in spectrophotometry

spectrophotometry is an analytical method based on the interaction between an electromagnetic radiation of a known intenzity (Io) and the analyzed sample

a part of the radiation is absorbed by the analyzed substance found in the of sample

the intensity of the radiation which passed (I) through the solution

is detected (I lt I0) rarrthe transmittance (T) of a solution is defined as the proportion T = I I0

transmittance can be expressed in percentage T() = (II0) x 100

values of the measured transmittance are found from 0 - 1 = 0 - 100

Calculations in spectrophotometry

How to calculate a concentration of substance in analyzed sample

in the laboratory it is more convenient to calculate concentration from values of the absorbance (A) which is directly proportional to the concentration than from the transmittance Calculation from the Beeracutes law A = c x l x ε

c = molar concentration (molL)

l = inner width of the cuvette in centimeters

ε = molar absorption coefficient (tabelated value)

Relationship between A and T A = log (1T)= -log T

Calculations in spectrophotometry - examples1) Ast = 040 cst = 4 mgL

Asam = 025 csam = mgL

2) Standard solution of glucose (conc = 1000 mgL) reads a transmittance 49 Unknown sample of glucose reads T = 55 Calculate the concentration of glucose in the sample in mgL and mmolL

Mr (glucose) = 180

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Percent concentration eexpressed as xpressed as part of solute per 100 part of solute per 100

parts of total solutionparts of total solution () () = = mass of solute mass of solute xx 100 100 mass of solutionmass of solution

it hit has 3 formsas 3 forms1 weight per weight (ww)1 weight per weight (ww)

10 of KCl = 1010 of KCl = 10 g of KCl + 90 g of g of KCl + 90 g of HH22O O ==100 g of solution100 g of solution

2 volume per volume2 volume per volume (vv)(vv)5 HCl = 5 5 HCl = 5 mLmL HCl in 100 m HCl in 100 mLL of of solutionsolution

3 weight per volume (wv)3 weight per volume (wv)the the most common expressionmost common expression09 NaCl = 09 g of NaCl in 100 m09 NaCl = 09 g of NaCl in 100 mLL of of solutionsolution

Percent concentrations - examples

1) 600 g 5 NaCl mass of NaCl mass of H2O

2) 250 g 8 Na2CO3 mass of Na2CO3 (purity 96)

3) 250 mL 39 ethanol solution mL of ethanol mL of H2O

4) Saline is 150 mM solution of NaCl Calculate the percent concentration by mass of this solution Mr(NaCl) = 585

Density ρ

- is defined as the amount of mass per unit of volume

ρ = mV rarr m = ρ x V and V = m ρ - these equations are useful for calculations

units gcm3 or gmL

- density of water = 1 g cm3

- density of lead (Pb) = 1134 gcm3

Conversions of concentrations ( and c) with density

1) What is a percent concentration of 2 M HNO3 solution Density (HNO3) = 1076 gml

Mr (HNO3) = 6301 Conversion of molar concentration to concentration 2 M HNO3 solution means that 2 mol of HNO3 are dissolved in 1 L of solution Mass of HNO3 = n x Mr = 2 x 6301 = 12602 g of HNO3

Mass of solution = ρ x V = 1076 x 1000 = 1076 gW = 12602 x 100 = 1171 1076

2) What is the molarity of 38 HCl solution Density (38 HCl) = 11885 gml and Mr(HCl) = 3645 Conversion of percent by mass concentration to molar concentration 38 HCl solution means that 38 g of HCl in 100 g of solution One liter of solution has a mass m = V x ρ = 1000 x 11885 = 11885 g

38 g HCl -------gt 100 g of solutionx g HCl -------gt 11885 g of solutionx = 45163 g HCl in 1 L of solution rarr n = m M = 45163 3645 = 124 mol of HCl

c(HCl) = n V = 124 1 = 124 molL

Conversions of concentrations ( and c) with density

Conversion of molarity to percent concentration

= c (molL) x Mr 10 x ρ (gcm3)

Conversion of percent concentration to molarity

c = x 10 x ρ (gcm3) Mr

Conversions of concentrations ( and c) with density - examples

1) (ww) of HNO3 ρ = 136 gcm3 if 1dm3 of solution contains 08 kg of HNO3

2) c (HNO3) = 562 M ρ = 118 gcm3 MW = 63 gmol 3) 10 HCl ρ = 1047 gcm3 MW = 365 c (HCl)

Dilution= concentration of a substance lowers number of moles of the substance remains the same

1) mix equation m1 x p1 + m2 x p2 = p x ( m1 + m2 ) m = mass of mixed solution p = concentration

2) expression of dilutionIn case of a liquid solute the ratio is presented as a dilution factor For example 1 5 is presented as 15 (1 mL of solute in 5 mL of solution)Example c1 = 025 M (original concentration) x 15 = 005 M (final concentration c2)

3) useful equation n1 = n2

V1 x c1 = V2 x c2

Dilution - examples

1) Mix 50 g 3 solution with 10 g 5 solution final concentration = ()

2) Final solution 190 g 10 sol

m (g) of 38 HCl + m (g) H2O 3) Dilute 300 g of 40 solution to 20 solution g

of solvent do you need4) preparation of 250 mL of 01 M HCl from stock 1

M HCl5) 10 M NaOH is diluted 1 20 final concentration6) 1000 mgL glucose is diluted 1 10 and then 1 2

final concentration

Conversions of unitsbull concentrationie gdL rarr mgLie molL harr osmolLbull units of pressure used in medicine1 mmHg (millimeter of mercury) = 1 Torr1 mmHg = 13322 Pabull units of energy1 cal = 41868 joule (J) 1 J = 0238 calcalorie (cal) is used in nutrition

Conversions of units ndash examples

1) Concentration of cholesterol in patient blood sample is 180 mgdL Convert this value into mmolL if MW of cholesterol is 387 gmol

2) Partial pressure of CO2 is 533 kPa Convert this value into mmHg

3) Red Bull energy drink (1 can) contains 160 cal Calculate the amount of energy in kJ

Calculations in spectrophotometry

spectrophotometry is an analytical method based on the interaction between an electromagnetic radiation of a known intenzity (Io) and the analyzed sample

a part of the radiation is absorbed by the analyzed substance found in the of sample

the intensity of the radiation which passed (I) through the solution

is detected (I lt I0) rarrthe transmittance (T) of a solution is defined as the proportion T = I I0

transmittance can be expressed in percentage T() = (II0) x 100

values of the measured transmittance are found from 0 - 1 = 0 - 100

Calculations in spectrophotometry

How to calculate a concentration of substance in analyzed sample

in the laboratory it is more convenient to calculate concentration from values of the absorbance (A) which is directly proportional to the concentration than from the transmittance Calculation from the Beeracutes law A = c x l x ε

c = molar concentration (molL)

l = inner width of the cuvette in centimeters

ε = molar absorption coefficient (tabelated value)

Relationship between A and T A = log (1T)= -log T

Calculations in spectrophotometry - examples1) Ast = 040 cst = 4 mgL

Asam = 025 csam = mgL

2) Standard solution of glucose (conc = 1000 mgL) reads a transmittance 49 Unknown sample of glucose reads T = 55 Calculate the concentration of glucose in the sample in mgL and mmolL

Mr (glucose) = 180

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Percent concentrations - examples

1) 600 g 5 NaCl mass of NaCl mass of H2O

2) 250 g 8 Na2CO3 mass of Na2CO3 (purity 96)

3) 250 mL 39 ethanol solution mL of ethanol mL of H2O

4) Saline is 150 mM solution of NaCl Calculate the percent concentration by mass of this solution Mr(NaCl) = 585

Density ρ

- is defined as the amount of mass per unit of volume

ρ = mV rarr m = ρ x V and V = m ρ - these equations are useful for calculations

units gcm3 or gmL

- density of water = 1 g cm3

- density of lead (Pb) = 1134 gcm3

Conversions of concentrations ( and c) with density

1) What is a percent concentration of 2 M HNO3 solution Density (HNO3) = 1076 gml

Mr (HNO3) = 6301 Conversion of molar concentration to concentration 2 M HNO3 solution means that 2 mol of HNO3 are dissolved in 1 L of solution Mass of HNO3 = n x Mr = 2 x 6301 = 12602 g of HNO3

Mass of solution = ρ x V = 1076 x 1000 = 1076 gW = 12602 x 100 = 1171 1076

2) What is the molarity of 38 HCl solution Density (38 HCl) = 11885 gml and Mr(HCl) = 3645 Conversion of percent by mass concentration to molar concentration 38 HCl solution means that 38 g of HCl in 100 g of solution One liter of solution has a mass m = V x ρ = 1000 x 11885 = 11885 g

38 g HCl -------gt 100 g of solutionx g HCl -------gt 11885 g of solutionx = 45163 g HCl in 1 L of solution rarr n = m M = 45163 3645 = 124 mol of HCl

c(HCl) = n V = 124 1 = 124 molL

Conversions of concentrations ( and c) with density

Conversion of molarity to percent concentration

= c (molL) x Mr 10 x ρ (gcm3)

Conversion of percent concentration to molarity

c = x 10 x ρ (gcm3) Mr

Conversions of concentrations ( and c) with density - examples

1) (ww) of HNO3 ρ = 136 gcm3 if 1dm3 of solution contains 08 kg of HNO3

2) c (HNO3) = 562 M ρ = 118 gcm3 MW = 63 gmol 3) 10 HCl ρ = 1047 gcm3 MW = 365 c (HCl)

Dilution= concentration of a substance lowers number of moles of the substance remains the same

1) mix equation m1 x p1 + m2 x p2 = p x ( m1 + m2 ) m = mass of mixed solution p = concentration

2) expression of dilutionIn case of a liquid solute the ratio is presented as a dilution factor For example 1 5 is presented as 15 (1 mL of solute in 5 mL of solution)Example c1 = 025 M (original concentration) x 15 = 005 M (final concentration c2)

3) useful equation n1 = n2

V1 x c1 = V2 x c2

Dilution - examples

1) Mix 50 g 3 solution with 10 g 5 solution final concentration = ()

2) Final solution 190 g 10 sol

m (g) of 38 HCl + m (g) H2O 3) Dilute 300 g of 40 solution to 20 solution g

of solvent do you need4) preparation of 250 mL of 01 M HCl from stock 1

M HCl5) 10 M NaOH is diluted 1 20 final concentration6) 1000 mgL glucose is diluted 1 10 and then 1 2

final concentration

Conversions of unitsbull concentrationie gdL rarr mgLie molL harr osmolLbull units of pressure used in medicine1 mmHg (millimeter of mercury) = 1 Torr1 mmHg = 13322 Pabull units of energy1 cal = 41868 joule (J) 1 J = 0238 calcalorie (cal) is used in nutrition

Conversions of units ndash examples

1) Concentration of cholesterol in patient blood sample is 180 mgdL Convert this value into mmolL if MW of cholesterol is 387 gmol

2) Partial pressure of CO2 is 533 kPa Convert this value into mmHg

3) Red Bull energy drink (1 can) contains 160 cal Calculate the amount of energy in kJ

Calculations in spectrophotometry

spectrophotometry is an analytical method based on the interaction between an electromagnetic radiation of a known intenzity (Io) and the analyzed sample

a part of the radiation is absorbed by the analyzed substance found in the of sample

the intensity of the radiation which passed (I) through the solution

is detected (I lt I0) rarrthe transmittance (T) of a solution is defined as the proportion T = I I0

transmittance can be expressed in percentage T() = (II0) x 100

values of the measured transmittance are found from 0 - 1 = 0 - 100

Calculations in spectrophotometry

How to calculate a concentration of substance in analyzed sample

in the laboratory it is more convenient to calculate concentration from values of the absorbance (A) which is directly proportional to the concentration than from the transmittance Calculation from the Beeracutes law A = c x l x ε

c = molar concentration (molL)

l = inner width of the cuvette in centimeters

ε = molar absorption coefficient (tabelated value)

Relationship between A and T A = log (1T)= -log T

Calculations in spectrophotometry - examples1) Ast = 040 cst = 4 mgL

Asam = 025 csam = mgL

2) Standard solution of glucose (conc = 1000 mgL) reads a transmittance 49 Unknown sample of glucose reads T = 55 Calculate the concentration of glucose in the sample in mgL and mmolL

Mr (glucose) = 180

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Density ρ

- is defined as the amount of mass per unit of volume

ρ = mV rarr m = ρ x V and V = m ρ - these equations are useful for calculations

units gcm3 or gmL

- density of water = 1 g cm3

- density of lead (Pb) = 1134 gcm3

Conversions of concentrations ( and c) with density

1) What is a percent concentration of 2 M HNO3 solution Density (HNO3) = 1076 gml

Mr (HNO3) = 6301 Conversion of molar concentration to concentration 2 M HNO3 solution means that 2 mol of HNO3 are dissolved in 1 L of solution Mass of HNO3 = n x Mr = 2 x 6301 = 12602 g of HNO3

Mass of solution = ρ x V = 1076 x 1000 = 1076 gW = 12602 x 100 = 1171 1076

2) What is the molarity of 38 HCl solution Density (38 HCl) = 11885 gml and Mr(HCl) = 3645 Conversion of percent by mass concentration to molar concentration 38 HCl solution means that 38 g of HCl in 100 g of solution One liter of solution has a mass m = V x ρ = 1000 x 11885 = 11885 g

38 g HCl -------gt 100 g of solutionx g HCl -------gt 11885 g of solutionx = 45163 g HCl in 1 L of solution rarr n = m M = 45163 3645 = 124 mol of HCl

c(HCl) = n V = 124 1 = 124 molL

Conversions of concentrations ( and c) with density

Conversion of molarity to percent concentration

= c (molL) x Mr 10 x ρ (gcm3)

Conversion of percent concentration to molarity

c = x 10 x ρ (gcm3) Mr

Conversions of concentrations ( and c) with density - examples

1) (ww) of HNO3 ρ = 136 gcm3 if 1dm3 of solution contains 08 kg of HNO3

2) c (HNO3) = 562 M ρ = 118 gcm3 MW = 63 gmol 3) 10 HCl ρ = 1047 gcm3 MW = 365 c (HCl)

Dilution= concentration of a substance lowers number of moles of the substance remains the same

1) mix equation m1 x p1 + m2 x p2 = p x ( m1 + m2 ) m = mass of mixed solution p = concentration

2) expression of dilutionIn case of a liquid solute the ratio is presented as a dilution factor For example 1 5 is presented as 15 (1 mL of solute in 5 mL of solution)Example c1 = 025 M (original concentration) x 15 = 005 M (final concentration c2)

3) useful equation n1 = n2

V1 x c1 = V2 x c2

Dilution - examples

1) Mix 50 g 3 solution with 10 g 5 solution final concentration = ()

2) Final solution 190 g 10 sol

m (g) of 38 HCl + m (g) H2O 3) Dilute 300 g of 40 solution to 20 solution g

of solvent do you need4) preparation of 250 mL of 01 M HCl from stock 1

M HCl5) 10 M NaOH is diluted 1 20 final concentration6) 1000 mgL glucose is diluted 1 10 and then 1 2

final concentration

Conversions of unitsbull concentrationie gdL rarr mgLie molL harr osmolLbull units of pressure used in medicine1 mmHg (millimeter of mercury) = 1 Torr1 mmHg = 13322 Pabull units of energy1 cal = 41868 joule (J) 1 J = 0238 calcalorie (cal) is used in nutrition

Conversions of units ndash examples

1) Concentration of cholesterol in patient blood sample is 180 mgdL Convert this value into mmolL if MW of cholesterol is 387 gmol

2) Partial pressure of CO2 is 533 kPa Convert this value into mmHg

3) Red Bull energy drink (1 can) contains 160 cal Calculate the amount of energy in kJ

Calculations in spectrophotometry

spectrophotometry is an analytical method based on the interaction between an electromagnetic radiation of a known intenzity (Io) and the analyzed sample

a part of the radiation is absorbed by the analyzed substance found in the of sample

the intensity of the radiation which passed (I) through the solution

is detected (I lt I0) rarrthe transmittance (T) of a solution is defined as the proportion T = I I0

transmittance can be expressed in percentage T() = (II0) x 100

values of the measured transmittance are found from 0 - 1 = 0 - 100

Calculations in spectrophotometry

How to calculate a concentration of substance in analyzed sample

in the laboratory it is more convenient to calculate concentration from values of the absorbance (A) which is directly proportional to the concentration than from the transmittance Calculation from the Beeracutes law A = c x l x ε

c = molar concentration (molL)

l = inner width of the cuvette in centimeters

ε = molar absorption coefficient (tabelated value)

Relationship between A and T A = log (1T)= -log T

Calculations in spectrophotometry - examples1) Ast = 040 cst = 4 mgL

Asam = 025 csam = mgL

2) Standard solution of glucose (conc = 1000 mgL) reads a transmittance 49 Unknown sample of glucose reads T = 55 Calculate the concentration of glucose in the sample in mgL and mmolL

Mr (glucose) = 180

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Conversions of concentrations ( and c) with density

1) What is a percent concentration of 2 M HNO3 solution Density (HNO3) = 1076 gml

Mr (HNO3) = 6301 Conversion of molar concentration to concentration 2 M HNO3 solution means that 2 mol of HNO3 are dissolved in 1 L of solution Mass of HNO3 = n x Mr = 2 x 6301 = 12602 g of HNO3

Mass of solution = ρ x V = 1076 x 1000 = 1076 gW = 12602 x 100 = 1171 1076

2) What is the molarity of 38 HCl solution Density (38 HCl) = 11885 gml and Mr(HCl) = 3645 Conversion of percent by mass concentration to molar concentration 38 HCl solution means that 38 g of HCl in 100 g of solution One liter of solution has a mass m = V x ρ = 1000 x 11885 = 11885 g

38 g HCl -------gt 100 g of solutionx g HCl -------gt 11885 g of solutionx = 45163 g HCl in 1 L of solution rarr n = m M = 45163 3645 = 124 mol of HCl

c(HCl) = n V = 124 1 = 124 molL

Conversions of concentrations ( and c) with density

Conversion of molarity to percent concentration

= c (molL) x Mr 10 x ρ (gcm3)

Conversion of percent concentration to molarity

c = x 10 x ρ (gcm3) Mr

Conversions of concentrations ( and c) with density - examples

1) (ww) of HNO3 ρ = 136 gcm3 if 1dm3 of solution contains 08 kg of HNO3

2) c (HNO3) = 562 M ρ = 118 gcm3 MW = 63 gmol 3) 10 HCl ρ = 1047 gcm3 MW = 365 c (HCl)

Dilution= concentration of a substance lowers number of moles of the substance remains the same

1) mix equation m1 x p1 + m2 x p2 = p x ( m1 + m2 ) m = mass of mixed solution p = concentration

2) expression of dilutionIn case of a liquid solute the ratio is presented as a dilution factor For example 1 5 is presented as 15 (1 mL of solute in 5 mL of solution)Example c1 = 025 M (original concentration) x 15 = 005 M (final concentration c2)

3) useful equation n1 = n2

V1 x c1 = V2 x c2

Dilution - examples

1) Mix 50 g 3 solution with 10 g 5 solution final concentration = ()

2) Final solution 190 g 10 sol

m (g) of 38 HCl + m (g) H2O 3) Dilute 300 g of 40 solution to 20 solution g

of solvent do you need4) preparation of 250 mL of 01 M HCl from stock 1

M HCl5) 10 M NaOH is diluted 1 20 final concentration6) 1000 mgL glucose is diluted 1 10 and then 1 2

final concentration

Conversions of unitsbull concentrationie gdL rarr mgLie molL harr osmolLbull units of pressure used in medicine1 mmHg (millimeter of mercury) = 1 Torr1 mmHg = 13322 Pabull units of energy1 cal = 41868 joule (J) 1 J = 0238 calcalorie (cal) is used in nutrition

Conversions of units ndash examples

1) Concentration of cholesterol in patient blood sample is 180 mgdL Convert this value into mmolL if MW of cholesterol is 387 gmol

2) Partial pressure of CO2 is 533 kPa Convert this value into mmHg

3) Red Bull energy drink (1 can) contains 160 cal Calculate the amount of energy in kJ

Calculations in spectrophotometry

spectrophotometry is an analytical method based on the interaction between an electromagnetic radiation of a known intenzity (Io) and the analyzed sample

a part of the radiation is absorbed by the analyzed substance found in the of sample

the intensity of the radiation which passed (I) through the solution

is detected (I lt I0) rarrthe transmittance (T) of a solution is defined as the proportion T = I I0

transmittance can be expressed in percentage T() = (II0) x 100

values of the measured transmittance are found from 0 - 1 = 0 - 100

Calculations in spectrophotometry

How to calculate a concentration of substance in analyzed sample

in the laboratory it is more convenient to calculate concentration from values of the absorbance (A) which is directly proportional to the concentration than from the transmittance Calculation from the Beeracutes law A = c x l x ε

c = molar concentration (molL)

l = inner width of the cuvette in centimeters

ε = molar absorption coefficient (tabelated value)

Relationship between A and T A = log (1T)= -log T

Calculations in spectrophotometry - examples1) Ast = 040 cst = 4 mgL

Asam = 025 csam = mgL

2) Standard solution of glucose (conc = 1000 mgL) reads a transmittance 49 Unknown sample of glucose reads T = 55 Calculate the concentration of glucose in the sample in mgL and mmolL

Mr (glucose) = 180

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Conversions of concentrations ( and c) with density

Conversion of molarity to percent concentration

= c (molL) x Mr 10 x ρ (gcm3)

Conversion of percent concentration to molarity

c = x 10 x ρ (gcm3) Mr

Conversions of concentrations ( and c) with density - examples

1) (ww) of HNO3 ρ = 136 gcm3 if 1dm3 of solution contains 08 kg of HNO3

2) c (HNO3) = 562 M ρ = 118 gcm3 MW = 63 gmol 3) 10 HCl ρ = 1047 gcm3 MW = 365 c (HCl)

Dilution= concentration of a substance lowers number of moles of the substance remains the same

1) mix equation m1 x p1 + m2 x p2 = p x ( m1 + m2 ) m = mass of mixed solution p = concentration

2) expression of dilutionIn case of a liquid solute the ratio is presented as a dilution factor For example 1 5 is presented as 15 (1 mL of solute in 5 mL of solution)Example c1 = 025 M (original concentration) x 15 = 005 M (final concentration c2)

3) useful equation n1 = n2

V1 x c1 = V2 x c2

Dilution - examples

1) Mix 50 g 3 solution with 10 g 5 solution final concentration = ()

2) Final solution 190 g 10 sol

m (g) of 38 HCl + m (g) H2O 3) Dilute 300 g of 40 solution to 20 solution g

of solvent do you need4) preparation of 250 mL of 01 M HCl from stock 1

M HCl5) 10 M NaOH is diluted 1 20 final concentration6) 1000 mgL glucose is diluted 1 10 and then 1 2

final concentration

Conversions of unitsbull concentrationie gdL rarr mgLie molL harr osmolLbull units of pressure used in medicine1 mmHg (millimeter of mercury) = 1 Torr1 mmHg = 13322 Pabull units of energy1 cal = 41868 joule (J) 1 J = 0238 calcalorie (cal) is used in nutrition

Conversions of units ndash examples

1) Concentration of cholesterol in patient blood sample is 180 mgdL Convert this value into mmolL if MW of cholesterol is 387 gmol

2) Partial pressure of CO2 is 533 kPa Convert this value into mmHg

3) Red Bull energy drink (1 can) contains 160 cal Calculate the amount of energy in kJ

Calculations in spectrophotometry

spectrophotometry is an analytical method based on the interaction between an electromagnetic radiation of a known intenzity (Io) and the analyzed sample

a part of the radiation is absorbed by the analyzed substance found in the of sample

the intensity of the radiation which passed (I) through the solution

is detected (I lt I0) rarrthe transmittance (T) of a solution is defined as the proportion T = I I0

transmittance can be expressed in percentage T() = (II0) x 100

values of the measured transmittance are found from 0 - 1 = 0 - 100

Calculations in spectrophotometry

How to calculate a concentration of substance in analyzed sample

in the laboratory it is more convenient to calculate concentration from values of the absorbance (A) which is directly proportional to the concentration than from the transmittance Calculation from the Beeracutes law A = c x l x ε

c = molar concentration (molL)

l = inner width of the cuvette in centimeters

ε = molar absorption coefficient (tabelated value)

Relationship between A and T A = log (1T)= -log T

Calculations in spectrophotometry - examples1) Ast = 040 cst = 4 mgL

Asam = 025 csam = mgL

2) Standard solution of glucose (conc = 1000 mgL) reads a transmittance 49 Unknown sample of glucose reads T = 55 Calculate the concentration of glucose in the sample in mgL and mmolL

Mr (glucose) = 180

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Conversions of concentrations ( and c) with density - examples

1) (ww) of HNO3 ρ = 136 gcm3 if 1dm3 of solution contains 08 kg of HNO3

2) c (HNO3) = 562 M ρ = 118 gcm3 MW = 63 gmol 3) 10 HCl ρ = 1047 gcm3 MW = 365 c (HCl)

Dilution= concentration of a substance lowers number of moles of the substance remains the same

1) mix equation m1 x p1 + m2 x p2 = p x ( m1 + m2 ) m = mass of mixed solution p = concentration

2) expression of dilutionIn case of a liquid solute the ratio is presented as a dilution factor For example 1 5 is presented as 15 (1 mL of solute in 5 mL of solution)Example c1 = 025 M (original concentration) x 15 = 005 M (final concentration c2)

3) useful equation n1 = n2

V1 x c1 = V2 x c2

Dilution - examples

1) Mix 50 g 3 solution with 10 g 5 solution final concentration = ()

2) Final solution 190 g 10 sol

m (g) of 38 HCl + m (g) H2O 3) Dilute 300 g of 40 solution to 20 solution g

of solvent do you need4) preparation of 250 mL of 01 M HCl from stock 1

M HCl5) 10 M NaOH is diluted 1 20 final concentration6) 1000 mgL glucose is diluted 1 10 and then 1 2

final concentration

Conversions of unitsbull concentrationie gdL rarr mgLie molL harr osmolLbull units of pressure used in medicine1 mmHg (millimeter of mercury) = 1 Torr1 mmHg = 13322 Pabull units of energy1 cal = 41868 joule (J) 1 J = 0238 calcalorie (cal) is used in nutrition

Conversions of units ndash examples

1) Concentration of cholesterol in patient blood sample is 180 mgdL Convert this value into mmolL if MW of cholesterol is 387 gmol

2) Partial pressure of CO2 is 533 kPa Convert this value into mmHg

3) Red Bull energy drink (1 can) contains 160 cal Calculate the amount of energy in kJ

Calculations in spectrophotometry

spectrophotometry is an analytical method based on the interaction between an electromagnetic radiation of a known intenzity (Io) and the analyzed sample

a part of the radiation is absorbed by the analyzed substance found in the of sample

the intensity of the radiation which passed (I) through the solution

is detected (I lt I0) rarrthe transmittance (T) of a solution is defined as the proportion T = I I0

transmittance can be expressed in percentage T() = (II0) x 100

values of the measured transmittance are found from 0 - 1 = 0 - 100

Calculations in spectrophotometry

How to calculate a concentration of substance in analyzed sample

in the laboratory it is more convenient to calculate concentration from values of the absorbance (A) which is directly proportional to the concentration than from the transmittance Calculation from the Beeracutes law A = c x l x ε

c = molar concentration (molL)

l = inner width of the cuvette in centimeters

ε = molar absorption coefficient (tabelated value)

Relationship between A and T A = log (1T)= -log T

Calculations in spectrophotometry - examples1) Ast = 040 cst = 4 mgL

Asam = 025 csam = mgL

2) Standard solution of glucose (conc = 1000 mgL) reads a transmittance 49 Unknown sample of glucose reads T = 55 Calculate the concentration of glucose in the sample in mgL and mmolL

Mr (glucose) = 180

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Dilution= concentration of a substance lowers number of moles of the substance remains the same

1) mix equation m1 x p1 + m2 x p2 = p x ( m1 + m2 ) m = mass of mixed solution p = concentration

2) expression of dilutionIn case of a liquid solute the ratio is presented as a dilution factor For example 1 5 is presented as 15 (1 mL of solute in 5 mL of solution)Example c1 = 025 M (original concentration) x 15 = 005 M (final concentration c2)

3) useful equation n1 = n2

V1 x c1 = V2 x c2

Dilution - examples

1) Mix 50 g 3 solution with 10 g 5 solution final concentration = ()

2) Final solution 190 g 10 sol

m (g) of 38 HCl + m (g) H2O 3) Dilute 300 g of 40 solution to 20 solution g

of solvent do you need4) preparation of 250 mL of 01 M HCl from stock 1

M HCl5) 10 M NaOH is diluted 1 20 final concentration6) 1000 mgL glucose is diluted 1 10 and then 1 2

final concentration

Conversions of unitsbull concentrationie gdL rarr mgLie molL harr osmolLbull units of pressure used in medicine1 mmHg (millimeter of mercury) = 1 Torr1 mmHg = 13322 Pabull units of energy1 cal = 41868 joule (J) 1 J = 0238 calcalorie (cal) is used in nutrition

Conversions of units ndash examples

1) Concentration of cholesterol in patient blood sample is 180 mgdL Convert this value into mmolL if MW of cholesterol is 387 gmol

2) Partial pressure of CO2 is 533 kPa Convert this value into mmHg

3) Red Bull energy drink (1 can) contains 160 cal Calculate the amount of energy in kJ

Calculations in spectrophotometry

spectrophotometry is an analytical method based on the interaction between an electromagnetic radiation of a known intenzity (Io) and the analyzed sample

a part of the radiation is absorbed by the analyzed substance found in the of sample

the intensity of the radiation which passed (I) through the solution

is detected (I lt I0) rarrthe transmittance (T) of a solution is defined as the proportion T = I I0

transmittance can be expressed in percentage T() = (II0) x 100

values of the measured transmittance are found from 0 - 1 = 0 - 100

Calculations in spectrophotometry

How to calculate a concentration of substance in analyzed sample

in the laboratory it is more convenient to calculate concentration from values of the absorbance (A) which is directly proportional to the concentration than from the transmittance Calculation from the Beeracutes law A = c x l x ε

c = molar concentration (molL)

l = inner width of the cuvette in centimeters

ε = molar absorption coefficient (tabelated value)

Relationship between A and T A = log (1T)= -log T

Calculations in spectrophotometry - examples1) Ast = 040 cst = 4 mgL

Asam = 025 csam = mgL

2) Standard solution of glucose (conc = 1000 mgL) reads a transmittance 49 Unknown sample of glucose reads T = 55 Calculate the concentration of glucose in the sample in mgL and mmolL

Mr (glucose) = 180

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Dilution - examples

1) Mix 50 g 3 solution with 10 g 5 solution final concentration = ()

2) Final solution 190 g 10 sol

m (g) of 38 HCl + m (g) H2O 3) Dilute 300 g of 40 solution to 20 solution g

of solvent do you need4) preparation of 250 mL of 01 M HCl from stock 1

M HCl5) 10 M NaOH is diluted 1 20 final concentration6) 1000 mgL glucose is diluted 1 10 and then 1 2

final concentration

Conversions of unitsbull concentrationie gdL rarr mgLie molL harr osmolLbull units of pressure used in medicine1 mmHg (millimeter of mercury) = 1 Torr1 mmHg = 13322 Pabull units of energy1 cal = 41868 joule (J) 1 J = 0238 calcalorie (cal) is used in nutrition

Conversions of units ndash examples

1) Concentration of cholesterol in patient blood sample is 180 mgdL Convert this value into mmolL if MW of cholesterol is 387 gmol

2) Partial pressure of CO2 is 533 kPa Convert this value into mmHg

3) Red Bull energy drink (1 can) contains 160 cal Calculate the amount of energy in kJ

Calculations in spectrophotometry

spectrophotometry is an analytical method based on the interaction between an electromagnetic radiation of a known intenzity (Io) and the analyzed sample

a part of the radiation is absorbed by the analyzed substance found in the of sample

the intensity of the radiation which passed (I) through the solution

is detected (I lt I0) rarrthe transmittance (T) of a solution is defined as the proportion T = I I0

transmittance can be expressed in percentage T() = (II0) x 100

values of the measured transmittance are found from 0 - 1 = 0 - 100

Calculations in spectrophotometry

How to calculate a concentration of substance in analyzed sample

in the laboratory it is more convenient to calculate concentration from values of the absorbance (A) which is directly proportional to the concentration than from the transmittance Calculation from the Beeracutes law A = c x l x ε

c = molar concentration (molL)

l = inner width of the cuvette in centimeters

ε = molar absorption coefficient (tabelated value)

Relationship between A and T A = log (1T)= -log T

Calculations in spectrophotometry - examples1) Ast = 040 cst = 4 mgL

Asam = 025 csam = mgL

2) Standard solution of glucose (conc = 1000 mgL) reads a transmittance 49 Unknown sample of glucose reads T = 55 Calculate the concentration of glucose in the sample in mgL and mmolL

Mr (glucose) = 180

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Conversions of unitsbull concentrationie gdL rarr mgLie molL harr osmolLbull units of pressure used in medicine1 mmHg (millimeter of mercury) = 1 Torr1 mmHg = 13322 Pabull units of energy1 cal = 41868 joule (J) 1 J = 0238 calcalorie (cal) is used in nutrition

Conversions of units ndash examples

1) Concentration of cholesterol in patient blood sample is 180 mgdL Convert this value into mmolL if MW of cholesterol is 387 gmol

2) Partial pressure of CO2 is 533 kPa Convert this value into mmHg

3) Red Bull energy drink (1 can) contains 160 cal Calculate the amount of energy in kJ

Calculations in spectrophotometry

spectrophotometry is an analytical method based on the interaction between an electromagnetic radiation of a known intenzity (Io) and the analyzed sample

a part of the radiation is absorbed by the analyzed substance found in the of sample

the intensity of the radiation which passed (I) through the solution

is detected (I lt I0) rarrthe transmittance (T) of a solution is defined as the proportion T = I I0

transmittance can be expressed in percentage T() = (II0) x 100

values of the measured transmittance are found from 0 - 1 = 0 - 100

Calculations in spectrophotometry

How to calculate a concentration of substance in analyzed sample

in the laboratory it is more convenient to calculate concentration from values of the absorbance (A) which is directly proportional to the concentration than from the transmittance Calculation from the Beeracutes law A = c x l x ε

c = molar concentration (molL)

l = inner width of the cuvette in centimeters

ε = molar absorption coefficient (tabelated value)

Relationship between A and T A = log (1T)= -log T

Calculations in spectrophotometry - examples1) Ast = 040 cst = 4 mgL

Asam = 025 csam = mgL

2) Standard solution of glucose (conc = 1000 mgL) reads a transmittance 49 Unknown sample of glucose reads T = 55 Calculate the concentration of glucose in the sample in mgL and mmolL

Mr (glucose) = 180

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Conversions of units ndash examples

1) Concentration of cholesterol in patient blood sample is 180 mgdL Convert this value into mmolL if MW of cholesterol is 387 gmol

2) Partial pressure of CO2 is 533 kPa Convert this value into mmHg

3) Red Bull energy drink (1 can) contains 160 cal Calculate the amount of energy in kJ

Calculations in spectrophotometry

spectrophotometry is an analytical method based on the interaction between an electromagnetic radiation of a known intenzity (Io) and the analyzed sample

a part of the radiation is absorbed by the analyzed substance found in the of sample

the intensity of the radiation which passed (I) through the solution

is detected (I lt I0) rarrthe transmittance (T) of a solution is defined as the proportion T = I I0

transmittance can be expressed in percentage T() = (II0) x 100

values of the measured transmittance are found from 0 - 1 = 0 - 100

Calculations in spectrophotometry

How to calculate a concentration of substance in analyzed sample

in the laboratory it is more convenient to calculate concentration from values of the absorbance (A) which is directly proportional to the concentration than from the transmittance Calculation from the Beeracutes law A = c x l x ε

c = molar concentration (molL)

l = inner width of the cuvette in centimeters

ε = molar absorption coefficient (tabelated value)

Relationship between A and T A = log (1T)= -log T

Calculations in spectrophotometry - examples1) Ast = 040 cst = 4 mgL

Asam = 025 csam = mgL

2) Standard solution of glucose (conc = 1000 mgL) reads a transmittance 49 Unknown sample of glucose reads T = 55 Calculate the concentration of glucose in the sample in mgL and mmolL

Mr (glucose) = 180

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Calculations in spectrophotometry

spectrophotometry is an analytical method based on the interaction between an electromagnetic radiation of a known intenzity (Io) and the analyzed sample

a part of the radiation is absorbed by the analyzed substance found in the of sample

the intensity of the radiation which passed (I) through the solution

is detected (I lt I0) rarrthe transmittance (T) of a solution is defined as the proportion T = I I0

transmittance can be expressed in percentage T() = (II0) x 100

values of the measured transmittance are found from 0 - 1 = 0 - 100

Calculations in spectrophotometry

How to calculate a concentration of substance in analyzed sample

in the laboratory it is more convenient to calculate concentration from values of the absorbance (A) which is directly proportional to the concentration than from the transmittance Calculation from the Beeracutes law A = c x l x ε

c = molar concentration (molL)

l = inner width of the cuvette in centimeters

ε = molar absorption coefficient (tabelated value)

Relationship between A and T A = log (1T)= -log T

Calculations in spectrophotometry - examples1) Ast = 040 cst = 4 mgL

Asam = 025 csam = mgL

2) Standard solution of glucose (conc = 1000 mgL) reads a transmittance 49 Unknown sample of glucose reads T = 55 Calculate the concentration of glucose in the sample in mgL and mmolL

Mr (glucose) = 180

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Calculations in spectrophotometry

How to calculate a concentration of substance in analyzed sample

in the laboratory it is more convenient to calculate concentration from values of the absorbance (A) which is directly proportional to the concentration than from the transmittance Calculation from the Beeracutes law A = c x l x ε

c = molar concentration (molL)

l = inner width of the cuvette in centimeters

ε = molar absorption coefficient (tabelated value)

Relationship between A and T A = log (1T)= -log T

Calculations in spectrophotometry - examples1) Ast = 040 cst = 4 mgL

Asam = 025 csam = mgL

2) Standard solution of glucose (conc = 1000 mgL) reads a transmittance 49 Unknown sample of glucose reads T = 55 Calculate the concentration of glucose in the sample in mgL and mmolL

Mr (glucose) = 180

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Calculations in spectrophotometry - examples1) Ast = 040 cst = 4 mgL

Asam = 025 csam = mgL

2) Standard solution of glucose (conc = 1000 mgL) reads a transmittance 49 Unknown sample of glucose reads T = 55 Calculate the concentration of glucose in the sample in mgL and mmolL

Mr (glucose) = 180

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