c101 1 lecturenotes chapter4 Foster · 10/20/09 1 © 2008 Brooks/Cole 1 Chapter 4: Quantities of...

10/20/09

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© 2008 Brooks/Cole 1

Chapter 4: Quantities of Reactants and Products


C6H12O6(aq) 2 C2H5OH(l) + 2 CO2(g) glucose ethanol carbon dioxide

yeast

Conditions may be shown over the arrow. e.g.

heat (Δ) reflux catalyst present (yeast)

Physical states are often listed: (g) gas (s) solid (l) liquid (aq) aqueous (dissolved in water)

Chemical Equations


O2 molar mass

2 C2H6(g) + 7 O2(g) 4 CO2(g) + 6 H2O(l)

2(30.0)= 60.0 g 7(32.0)= 224.0 g 4(44.0)= 176.0 g 6(18.0)= 108.0 g

C2H6 molar mass

284.0 g 284.0 g

“Mass is neither created nor destroyed in a chemical reaction.”

Chemical Equations


CaCO3(s) + 2 HNO3(aq)

Ca(NO3)2(aq) + CO2(g) + H2O(l) A stoichiometric

coefficient

Chemical Equations


Combination Reactions

+ X Z XZ

Element plus halogen or O2: 2 Mg(s) + O2(g) → 2 MgO(s) I2(s) + Zn(s) → ZnI2(s)

There are other types: 2 SO2(g) + O2(g) → 2 SO3(g)


+ XZ X Z

Occasionally by shock: 4 C3H5(NO3)3(l) 12 CO2(g) + 10 H2O(l) + 6 N2(g) + O2(g)

Often initiated by heat: CaCO3(s) CaO(s) + CO2(g)

2 KNO3(s) 2 KNO2(s) + O2(g) heat

800 - 1000°C

Decomposition Reactions

nitroglycerin

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Decomposition Reactions


Displacement Reactions

Some other examples: F2(g) + 2 LiCl(s) 2 LiF(s) + Cl2(g)

Some metals displace another metal from its salt Fe(s) + CuSO4(aq) FeSO4(aq) + Cu(s)

+ + A XZ AZ X

Zn(s) + 2 AgNO3(aq) Zn(NO3)2(aq) + 2 Ag(s)

2 Na(s) + 2 H2O(l) 2 NaOH(aq) + H2(g)


Displacement Reactions


Exchange Reactions + + AD XZ AZ XD

AgNO3(aq) + HCl(aq) → AgCl(s) + HNO3(aq)

Pb(NO3)2(aq) + K2CrO4(aq) → PbCrO4(s) + 2 KNO3(aq)


1.  Write an unbalanced equation with correct formulas for all substances.

2.  Balance the atoms of one element. a. Start with the most complex molecule b. Change the coefficients in front of the molecules c.  Do NOT alter the chemical formulas

3.  Balance the remaining elements.

4.  Check the atoms are all balanced.

Balancing Chemical Equations


step 4 – balanced

Balance : Al + Fe2O3 Al2O3 + Fe Balancing Chemical Equations

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Balancing Chemical Equations Combustion of rocket fuel:

C2H8N2 + N2O4 N2 + H2O + CO2


NaNO3(s) + H2SO4(aq) Na2SO4(aq) + HNO3(aq)

Balancing Chemical Equations

Balance Na in Na2SO4

NaNO3(s) + H2SO4(aq) Na2SO4(aq) + HNO3(aq)


2 C2H6 (g) + 7 O2 (g) 4 CO2(g) + 6 H2O (l)

The Mole and Chemical Reactions

2 mol C2H6 7 mol O2

=1 7 mol O2

2 mol C2H6 =1

2 moles of C2H6 react with 7 moles of O2 2 moles of C2H6 produce 4 moles of CO2

2 mol C2H6 ≡ 7 mol O2 2 mol C2H6 ≡ 4 mol CO2 etc.



What mass of O2 and Br2 is produced by the reaction of 25.0 g of TiO2 with excess BrF3?

Notes: •  Check the equation is balanced!

•  Stoichiometric ratios: 3TiO2 ≡ 3O2 ; 3TiO2 ≡ 2Br2 ; and many others

3 TiO2(s) + 4 BrF3(l) 3 TiF4(s) + 2 Br2(l) + 3 O2(g)


= 25.0 g x = 0.3130 mol TiO2 1 mol

79.88 g

nTiO2 = mass TiO2 / FM TiO2

What mass of O2 and Br2 is produced by the reaction of 25.0g of TiO2 with excess BrF3? 3 TiO2(s) + 4 BrF3(l) 3 TiF4(s) + 2 Br2(l) + 3 O2 (g)

3 mol TiO2 ≡ 3 mol O2

0.3130 mol TiO2 = 0.3130 mol O2 3 mol O2 3 mol TiO2



Mass of O2 produced = nO2 (mol. wt. O2)

= 0.3130 mol x 32.00 g/mol = 10.0 g

nBr2 = 0.3130 mol TiO2 = 0.2087 mol Br2 2Br2

3 TiO2

Mass of Br2 = 0.2087 mol = 33.4 g Br2 159.81 g mol Br2

What mass of O2 and Br2 is produced by the reaction of 25.0g of TiO2 with excess BrF3? 3 TiO2(s) + 4 BrF3(l) 3 TiF4(s) + 2 Br2(l) + 3 O2(g)


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Practice Problem 4.8 The purity of Mg can be found by reaction with excess HCl (aq), evaporating the water from the resulting solution and weighing the solid MgCl2 formed.

Mg(s) + 2 HCl(aq) MgCl2(aq) + H2(g) Calculate the % Mg in a 1.72-g sample that produced 6.46 g of MgCl2 when reacted with excess HCl.

More difficult – What should you calculate?

– Express as a % of the original mass.


Practice Problem 4.8

FW of MgCl2 = 24.31 + 2(35.45) = 95.21 g/mol

nMgCl2 = 6.46 g MgCl2 1 mol

95.21 g = 0.06785 mol MgCl2

Mg required: 0.06785 mol MgCl2 1 Mg

1 MgCl2


0.06785 mol Mg x = 1.649 g Mg 24.31 g 1 mol

Practice Problem 4.8

1.649 g 1.72 g

Given 1.72 g of impure Mg.

Purity (as mass %) = x 100% = 95.9 %


Reactions with Reactant in Limited Supply

Given 10 slices of cheese and 14 slices of bread. How many sandwiches can you make?

Balanced equation 1 cheese + 2 bread 1 sandwich

1 cheese ≡ 2 bread 1 cheese ≡ 1 sandwich 2 bread ≡ 1 sandwich


Two methods can be used:

10 cheese x = 10 sandwiches 1 sandwich 1 cheese

14 bread x = 7 sandwiches 1 sandwich 2 bread

Correct answer Bread is limiting. It will

be used up first

Reactions with Limited Reactants


•  Yes = Your choice is the limiting reactant.

•  No = Another reactant is limiting.


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…base all other calculations on the limiting reactant.

Started with 10 cheese. Cheese remaining 10 – 7 = 3 slices



How much water will be produced by the combustion of 25.0 g of H2 in the presence of 100. g of O2?

Write a balanced equation:

nH2 = 25.0 g = 12.40 mol H2 1 mol H2 2.016 g

2 H2(g) + O2(g) 2 H2O(l)

nO2 = 100. g = 3.125 mol O2 1 mol O2 32.00 g



Using H2 12.40 mol H2 (2H2O /2H2 ) = 12.40 mol H2O Using O2 3.125 mol O2 (2 H2O /1 O2 ) = 6.250 mol H2O

How much water will be produced?



H2O formed: 3.125 mol O2 (2H2O/1O2) = 6.250 mol.

= 113. g



Consider : 4 NH3(g) + 5 O2(g) 4 NO(g) + 6 H2O(g) If 374 g of NH3 and 768 g of O2 are mixed, what mass of NO will form?

1 mol 17.03 g

nNH3 = 374 g = 21.96 mol

1 mol 32.00 g nO2 = 768g = 24.00 mol



4 NH3(g) + 5 O2(g) 4 NO(g) + 6 H2O(g) Mol available: 21.96 24.00

From NH3 NO formed: 21.96 mol NH3 = 21.96 mol NO 4 NO

4 NH3

4 NO 5 O2


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Mass of NO = 19.20 mol NO x = 576 g NO

4 NH3(g) + 5 O2(g) 4 NO(g) + 6 H2O(g)

Mass of NO formed?

30.01g 1 mol

NO formed = 19.20 mol NO

21.96 mol 24.00 mol 19.20 mol



What mass of MgI2 is made by the reaction of 75.0 g of Mg with 75.0 g of I2?

Mg + I2 → MgI2

•  Calculate moles  75.0 g of Mg = 75.0g/(24.31 g mol-1) = 3.085 mol Mg

 75.0 g of I2 = 75.0g/(253.9 g mol-1) = 0.2955 mol I2



% yield = x 100% Actual yield Theoretical yield

Percent Yield


Percent Yield Few reactions have 100% yield.


Percent Yield You heat 2.50 g of copper with an excess of sulfur and synthesize 2.53 g of copper(I) sulfide

16 Cu(s) + S8(s) 8 Cu2S(s) What was the percent yield for your reaction?

nCu used: = 0.03934 mol Cu 1 mol 63.55g 2.50 g

16 mol Cu used ≡ 8 mol Cu2S made

8 Cu2S 16 Cu


Percent Yield Heat 2.50 g of Cu with excess S8 and make 2.53 g of copper(I) sulfide: 16 Cu(s) + S8(s) → 8 Cu2S(s). What was the %-yield for your reaction?

159.2 g 1 mol = 0.01967 mol Cu2S = 3.131 g Cu2S

2.53 g 3.131 g

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Atom Economy Examines the fate of all starting-material atoms.

atomic mass of atoms in useful product(s) atomic mass of all reactants used x 100%

% atom economy =


C and H are converted to CO2 & H2O. •  Both are trapped and the weight gain measured. •  Other elements (N, O …) with other traps or by mass

difference.

H2O absorber

Mg(ClO4)2

CO2 absorber

NaOH

O2 Sample in a furnace

Percent Composition & Empirical Formulas


Vitamin C contains C, H & O only. 1.502 g of CO2 and 0.409 g of H2O are produced when 1.000 g is burned in O2. Given the molar mass of vitamin C is 176.12 g/mol, find its empirical & molecular formula.

1.502 g CO2 = 0.4099 g C 1 mol CO2 44.009 g CO2

12.011 g C 1 mol C

1 mol C 1 mol CO2

1.502 g CO2 = 0.4099 g C 12.011 g C 44.009 g CO2



0.409 g H2O 2.0158 g H

18.015 g H2O = 0.04577 g H

Vitamin C contains C, H & O only. Combustion of 1.000 g of vitamin C produced 1.502 g of CO2 and 0.409 g of H2O.

mass of O = sample mass – (mass of C + mass of H)



= 0.04541 mol H 0.04577 g H 1.0079 g/mol

Convert to moles:

= 0.03413 mol C 0.4099 g C 12.011 g/mol

= 0.0340 mol O 0. 544 g O 15.999 g/mol



Find the mole ratio (divide by smallest…):

C 0.03413 / 0.0340 = 1.00

H 0.04541 / 0.0340 = 1.34

O 0.0340 / 0.0340 = 1.00

Close to 1 : 1⅓ : 1 (C : H : O)

Multiply by 3 to get an integer ratio (3 : 4 : 3)


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Empirical mass = 3(12) + 4(1) + 3(16) g = 88 g

Molar mass = 176.12 g

Molar mass ≈ 2 x (empirical mass)

…Vitamin C contains C, H & O only… … molar mass of vitamin C is 176.12 g/mol, find its empirical and molecular formula.

Empirical formula = C3H4O3


c101 1 lecturenotes chapter4 Foster · 10/20/09 1 © 2008 Brooks/Cole 1 Chapter 4: Quantities of...

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