Chapter 3 – Mass Relationships in Chemical Reactions
1. Balance the following equations:
a) C5H10O2 + O2  CO2 + H2O
b) PCl5 + H2O  H3PO4 + HCl
c) Al(OH)3 + H2SO4  ____________+
d) Na + ________  NaOH + H2
Dr. Pahlavan
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2. Write balanced chemical equations to correspond to each of the following descriptions:
a)
b)
c)
d)
Heptane, C7H16, is combusted in air
When sulfur trioxide gas reacts with water, a solution of sulfuric acid forms
Lithium metal is added to water
ZnCO3(s) is decomposed upon heating
3. Determine the formula weights of each of the following compounds:
a) Ca(C2H3O2)2
b) CH3CH2COOH
c) (NH4)3PO4
4. Calculate the percentage compositions in the following compounds;
a) Benzoic acid, C7H6O2
b) Urea, (NH2)2CO
c) Laughing gas, N2O
5. Gallium (Ga) consists of two naturally occurring isotopes with masses of 68.926 amu and
70.926 amu. Calculate the abundance of each isotope. The average atomic mass of Ga is
69.726 amu.
6. Complete the following conversions;
a) moles of 0.005 g water
c) grams mass of 2.50 mole O2
b) atoms in 3.50 mole NaOH
d) carbon atoms in 10.5 g C6H12O6
7. Determine the empirical formulas of each of the compounds if a sample contains the
following compositions:
a) 62.1% C, 5.21% H, 12.1% N, and 20.7% O
b) 0.104 mol K, 0.052 mol C, and 0.152 mol O
c) 11.66g iron and 5.01g oxygen
8. Determine the molecular formula of each of the following compounds?
a) empirical formula HCO2, molar mass = 90.0 g/mol
b) 75.69% C, 8.80% H, and 15.51% O by mass; molar mass 206g
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9. Combustion analysis of toluene, a common organic solvent, produces 5.86 mg of CO2 and
1.37 mg of H2O. If the compound contains only carbon and hydrogen, what is its
molecular formula? Its molecular mass is 92 g/ mol.
10. A 0.1005 g sample of organic solvent is combusted, producing 0.2829 g of CO2 and
0.1159 g of water. If the compound has a molar mass of 156 g/mol, what is its
molecular formula?
11. The complete combustion of octane, C8H18, a component of gasoline, proceeds as
follows:
2 C8H18 + 25 O2  16 CO2 + 18 H2O
a) how many moles of CO2 are produced when 1.50 mol octane reacted?
b) How many grams water produced in this reaction?
c) How many moles oxygen required to form 90.0 g water?
12. In a certain experiment 2.50 g of NH3 reacts with 2.85 g of O2. Which reactant is the
limiting reactant? How many grams of NO form? How much excess reactant remains
after the liming reactant is completely consumed?
13. Given 30.0 g benzene and 65.0 g bromine in the following reaction:
C6H6 + Br2  C6H5Br + HBr
If the actual yield of C6H5Br, bromobenzene, is 56.7 g, what is the percentage yield?
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Chapter 3–Mass Relationships in Chemical Reactions(answers)
Dr. Pahlavan
1. a) C5H10O2 + O2 → CO2 + H2O
(2, 13, 10, 10)
b) PCl5 + H2O → H3PO4 + HCl
(1, 4, 1, 5)
c) Al(OH)3 + H2SO4 → Al2(SO4)3 + H2O
(2, 3, 1, 6)
d) Na + H2O → NaOH + H2
(2, 2, 2, 1)
2. C7H16 + O2 → CO2 + H2O (1, 11, 7, 8)
Li + H2O → LiOH + H2 (2, 2, 2, 1)
SO3 + H2O → H2SO4 (balanced)
ZnCO3 → ZnO + CO2 (balanced)
3. Determine formula weight: Ca(C2H3O2)2 : 1(40) + 4(12) + 6(1) + 4(16) = 158g
4. Percentage compositions:
a) benzoic acid, C7H6O2 mol wt. is:
7(12) + 6(1) + 2(16) = 84 + 6 + 32 = 122g
%comp. C  (84/122) x 100 = 68.85%, H => (6/122) x 100 = 4.92%
O  (32/122) x 100 = 26.23%
5. Let x be the fractional abundance of isotope 70.926-Ga. Then:
69.726 = 70.926x + 68.926(1-x)  x = 0.8000/2 = 0.4000
% abundance of isotope 70.926 is 0.4000(100) = 40%
% abundance of isotope 68.926 is 60%.
6. (a) 0.005g H2O │ 1mole H2O
│ 18gH2O
= 2.77 x 10-4 mole = 3 x 10-4 mole
(b) 3.5moles │6.022 x 1023 molecules│3atoms
= 6.32 x 1024 atoms
│ 1 mole
│1molecule
(c) 2.5 moles │ 32g O2 = 80g O2
│ 1 mole
(d) F.W. C6H12O6 = 180g = glucose = glu
(10.5g-glu)│1mole-glu│6.02 x 1023molecules-glu│6C = 2.1 x 1023 C atoms
│180g-glu) │1 mole-glu
│1 molecule-glu)
7. (a) CWHXNYOZ Epirical Formula? 62.19% C, 5.21% H, 12.1% N, 20.7% O
Assume 100g sample. Then:
62.1g/12g = 5.175moles C; 5.21g/1g = 5.21moles H; 12.1g/14g = 0.86 mole N
20.7g/16g = 1.29 moles O
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Divide smallest mole number into each calculated mole:
5.175/0.86 = 6;
5.21/0.86 = 6; 0.86/0.86 = 1; 1.29/0.86 = 1.5
multiply all subscripts by 2 to get all of them as whole numbers.
Therefore: Empirical Formula of CWHXNYOZ is C12H12N2O3
(b) KXCYOZ Empirical Formula?
Divide smallest mole number into each mole.
i.e. (0.104/.052) = 2moles K; (0.052/0.052) = 1mole C; (0.152/0.052) = 3moles O ; Therefore
K2CO3
(c) Get equivalent mole value for each mass.
Fe 11.66g/56g = 0.21 mole; O 5.01g/16 = 0.313 mole
Divide smallest mole into each calculated mole.
Fe  1,
O  1.5 , Multiply subscripts by 2 : Empirical formula For FexOy is Fe2O3
8. (a) First find empirical formula mass. HCO2  1 + 12 + 2(16) = 45g
If molecular formula mass is 90g  90g/45g = 2  (HCO2)2 = H2C2O4
(b) % composition given, Molar mass given  206g for CxHyOz
Moles of C is 75.69g/12g = 6.31moles
Moles of H is 8.8g/1g = 8.8moles
Moles of O is 15.51g/16g = 0.97mole
C  6.31/0.97 = 6.5; H  8.8/0.97 = 9.07; O  0.97/0.97 = 1.00
Multiply by 2. Then C  13; H  18; O  2
Then CxHyOz is C13H18O2 = empirical and molecular formula. OR
mole of C = (206g/12g)( 0.7569) = 12.99 mole  13 moles = x
mole of H = (206g/1g)( 0.088) = 18.12 mole  18 moles = y
mole of O = (206g/16g)( 0.1551) = 1.99 moles  2 moles = z
 C13H18O2 = molecular formula.
9. (3/11)5.86g = 1.60g C; (1/9)1.37g = 0.15g H  toluene sample mass is 1.75g.
% mass C is 1.60g/1.75g x 100 =91.43%
% mass H is 0.15/1.75 x 100 = 8.6%
So moles C in 92g/mol is (0.9143 x 92g)/12g = 7.01 moles  7.00 moles = x
So moles H in 92g/mol is (0.086 x 92g)/1g = 7.9 moles  8.00 moles = y
So for CxHy molecular formula is C7H8
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10. (3/11)(0.2829g) = 0.07715g C; (1/9)(0.1159g) = 0.01288g H.
Mass of C & H in sample is (0.07715 + 0.01288) = 0.09003g
Therefore sample contains O of mass (0.1005- 0.09003) = 0.01047g
So the compound is CxHyOz
Moles of C in molecular mass is (0.07715g x 156g)/(0.1005 x 12g) = 9.97  10
Moles of H in molar mass is (0.01288g x 156g)/(0.1005g x 1g) = 19.99  20
Moles of O in molar mass is (0.01047g x 156g)/(0.1005g x 16g) = 1.01  1.00
Then x = 10; y = 20; z = 1
Therefore molecular formula of CxHyOz is C10H20O
11. (a) 16 CO2/2 C8H18 = X/1.5; X = 12.0 moles CO2
(b) 18H2O/2 C8H18 = X/1.5; X = 13.5 moles H2O  13.5x18g/mol = 243g H2O
(c) (90g water)/18g/mol = 5 moles; (25mol)/(18mol) = (Xmoles O2)/5moles H2O; X = 6.94
moles oxygen gas.
12. 4NH3 + 5O2  4NO + 6H2O
(a) 2.50g NH3 is 0.147mole  0.147 mole NO produced
2.85g O2 is 0.089mole  (4/5)(0.089mole) = 0.071mole NO produced.
Therefore O2 is limiting reagent.
(b) Grams NO formed is (0.071mole x 30g) = 2.13g NO
(c) NH3 is excess reactant. How much? 0.147mole – (4/5)(0.089mole)= 0.147 – 0.071= 0.076
mole = 17g(0.076mol) = 1.29 grams excess NH3.
13. C6H6 + Br2  C6H5Br + HBr (30g benzene, 65g bromine gas)
C6H6 mol wt. is 78g; Br2 mol wt. is 160g; C6H5Br mol wt. is 157g
All stoichiometric relationships are 1 to 1. Limiting reagent?
For Benzene (157g/78g) = (X/30g); X = 60.38g bromobemzene produced.
For Bromine gas (157g/160g) = (X/65g); X = 63.78g bromobenzene produced.
Therefore benzene is limiting reagent. Then theoretically, 30g benzene should
produce 60.38g of bromobenzene for a 100% yield. But since only 56.7g was
produced, percent yield is (56.7g/60.38g) x 100 = 94% yield.
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