Unit 3 Review – Chem Honors
Moles & Stoichiometry
NAME Answer Key
Period
Date
Material covered on test:
• Labs: Percent Composition/Empirical Formula Lab, Cu3(PO4)2 Lab, Micro-Mole Rockets
• Coverage in Book: Empirical and Molecular Formulas: p 229-233;
Mole Conversions: p. 78-85, p221-233
Balancing Equations, p.241-253; Stoichiometry p.275-294
TEST Format: 70-80 pts total: multiple choice, short explanations, calculations, lab type questions
Topics covered on test:
• History of moles: Law of conservation of mass, Law of Definite proportions and Law of multiple
proportions; Theories of Democritus, Dalton, Gay-Lussac, and Avogadro
• Avogadro’s Hypothesis: How it’s used to balance equations (& determine formulas) and relative masses
of gases
• Mole Conversions between grams or liters ↔ moles ↔ particles (molecules or separate atoms)
• Percent composition, empirical formulas and molecular formulas
• Balancing equations
• Stoichiometry problems: Conversions between amounts of one substance to amounts of another (in a
reaction)
• Limiting reactants, theoretical yield and percent yield
• Lab type questions-- general lab separation techniques, concepts and calculations
I. Practice Questions
1) Compare and contrast the Law of Definite Proportions to the Law of Multiple Proportions.
The Law of Definite Proportions describes the constant composition of any two samples of the same
compound (e.g. H2O is always 2 parts H and 1 part O, or by mass 1 g H to 8 g O). The Law of Multiple
Proportions also describes the compositions of compounds, but compares the different ratios of the same
elements in two different compounds (e.g. H2O vs. H2O2 or CO2 vs. CO).
2) What did Gay Lussac contribute to our understanding of reactions of gases?
He showed that gases react in simple whole number ratios of volumes. (He did lots of reactions with
gases and measured the volumes involved. He collected the data, but he did not explain why these
results are obtained. )
3) 1 mole of O2 gas has the same volume of 1 mole of CH4 gas (at same T + P). However, 1 mole of liquid
O2 does not have the same volume as 1 mole of liquid CH4. Explain why. (Hint: think about the sizes of
the individual molecules.)
In gases, the particles are very far apart, and the volume occupied by the sample is large compared to the
size of each particle, so their size is inconsequential. In a liquid, however, the particles are in contact
with each other, so the volume occupied by the sample depends on the size of the particles.
4) When 1 liter of hydrogen gas and 1 liter of chlorine gas react, 2 liters of hydrogen chloride gas is
produced. Why can’t the balanced equation be H + Cl → HCl ? (Use Avogadro’s Hypothesis)
Since 2 L of HCl are produced, its coefficient must be 2 and there must be two atoms of H and two
atoms of Cl. However, the coefficients must match the volumes, so 1 volume of H must contain 2 H
atoms and 1 volume of Cl must contain 2 Cl atoms, so their formulas must be H2 and Cl2. Thus the
equation must be
H2 + Cl2 → 2 HCl
5) What is the difference between an empirical formula and a molecular formula?
An empirical formula is the simplest, most reduced whole-number ratio of the elements in a compound;
a molecular formula is the exact, unreduced formula showing the actual numbers of atoms of each
element in the compound.
6) What is the empirical formula for the substance with the molecular formula of C2H4Br2? CH2Br
7) Balance the following equations using the lowest whole number coefficients:
a)
Fe2(SO4)3 + 3
b)
2
C3H8O + 9
c)
3
Ca(OH)2 + 2
Ba(NO3)2 →
O2 →
8
2
Fe(NO3)3 + 3
H2 O + 6
H3PO4 →
BaSO4
CO2
Ca3(PO4)2 + 6
H2 O
II. Moles Calculations
8) How many molecules of Ca(ClO3)2 are there in 8.41 g of Ca(ClO3)2?
1 mol Ca(ClO3 )2
6.022 × 1023 mlcl Ca(ClO3 )2
? mlcl Ca(ClO3 )2 = 8.41 g Ca(ClO3 )2 ×
×
207.0 g Ca(ClO3 )2
1 mol Ca(ClO3 )2
22
= 2.45 × 10 mlcl Ca(ClO3 )2
9) How many grams of N2O gas are there in 59.2 L of N2O gas at STP?
1 mol N 2O 44.01 g N 2O
? g N 2O = 59.2 L N 2O ×
×
= 116 g N 2O
22.4 L N 2O 1 mol N 2O
10) What is the density (g/L) of CO2 gas at STP. (Hint: Assume you have 1 mole of CO2 gas)
Molar Mass
44.01 g/mol
DCO2 =
=
= 1.96 g/L
Molar Volume 22.4 g/mol
11) Aluminum has a density of 2.71 g/cm3. What is the volume of one Al atom? (What is its molar mass?)
1 mol Al
26.98 g Al 1 cm 3
? cm 3 = 1 atom Al ×
×
×
= 1.65 × 10−23 cm 3
23
1
mol
Al
2.71
g
6.022 × 10 atoms Al
12) An unknown sample contains only C & H. If 26.8 g of the sample contains 4.90 g H, what is its %
composition?
MassC = 26.8 g − 4.90 g = 21.9 g C
%C =
21.9 g
4.90 g
× 100% = 81.7% C ;%H =
× 100% = 18.3% H
26.8 g
26.8 g
13) What is the percent composition (by mass) of Sr3(PO4)2?
MM = 3(87.62) + 2(30.97) + 8(16.00) = 452.8 g/mol
%Sr =
2
262.86 g
61.94 g
128.0 g
× 100% = 58.05% Sr ; %P =
× 100% = 13.68% P ; %O =
× 100% = 28.27% O
452.8 g
452.8 g
452.8 g
Unit 3 Review: Moles & Stoichiometry
14) Xylose is an important sugar that is the building block for wood fibers.
a) Xylose contains 40.0 % carbon, 6.67 % hydrogen and 53.3% oxygen (all by mass). What is its
empirical formula?
⎫
1 mol C
40.0 g C ×
= 3.33 mol C ÷ 3.33 = 1⎪
12.01 mol C
⎪
1 mol H
⎪
6.67 g H ×
= 6.62 mol H ÷ 3.33 = 2 ⎬ EF = CH 2O
1.008 g H
⎪
⎪
1 mol O
53.3 g O ×
= 3.33 mol O ÷ 3.33 = 1 ⎪
16.00 g O
⎭
b) The molar mass of xylose is 150 g/mol. What is its molecular formula?
EF Mass = 12.01+ 2(1.008) + 16.00 = 30.03 g/mol
150 g/mol
= 5; Molecular Formula = (CH 2O) × 5 = C5H10O5
30.03 g/mol
III. Stoichiometry Calculations
15) Liquid octane (C8H18) burns in oxygen according to this unbalanced equation (balance it):
Multiplier n =
2 C8H18 (l) + 25 O2 (g) → 16 CO2 (g) + 18 H2O (g)
How many liters of oxygen (at STP) are required to burn 39.8 liters of octane? Density of octane =
0.702 g/mL
1 mL
0.702 g 1 mol C8 H18
25 mol O 2
22.4 L
? L O 2 = 39.8 L C8 H18 ×
×
×
×
×
= 68,500 L O 2
−3
1× 10 L 1 mL 114.2 g C8 H18 2 mol C8 H18 1 mol
16) This reaction is carried out:
P4 (s) + 6 Cl2 → 4 PCl3 (balance!)
0.130 moles of P4 is reacted with 1.392 moles of Cl2. What is the limiting reactant? Show a calc and a
sentence to explain.
0.130 mol P4
1.392 mol Cl2
= 0.130 equiv. P4 ;
= 0.232 equiv. Cl2
1 mol P4
6 mol Cl2
P4 is limiting since it has a smaller equiv. (less than needed for the equiv. Cl2 ).
17) Hydrofluoric acid solutions cannot be stored in glass containers because HF reacts readily with silica
(SiO2) in glass to produce hexafluorosilicic acid (H2SiF6): SiO2(s) + 6 HF(aq) → H2SiF6(aq) + 2 H2O(l)
Suppose 45.2 g SiO2 and 88.2 g HF are mixed together and 91.5 g H2SiF6 are produced.
a) What is the theoretical yield of H2SiF6 (in grams)?
1 mol SiO 2
0.752 mol SiO 2
45.2 g SiO 2 ×
= 0.752 mol SiO 2 ;
= 0.752 equiv. SiO 2
60.08 g SiO 2
1 mol SiO 2
1 mol HF
4.41 mol HF
88.2 g HF ×
= 4.41 mol HF;
= 0.735 equiv. HF *Limiting Reactant*
20.01 g HF
6 mol HF
1 mol H 2SiF6 144.1 g H 2SiF6
? g H 2SiF6 = 4.41 mol HF ×
×
= 105.9 g H 2SiF6 = 106 g H 2SiF6
6 mol HF
1 mol H 2SiF6
b) What is a limiting reactant? Why was it important to determine the limiting reactant in this
question?
HF is the limiting react. It must be determined because when it runs out no more product can be
made, so it controls the theoretical yield.
c) How many grams of the excess reactant are left unreacted?
1 mol SiO 2 60.08 g SiO 2
? g SiO 2 used = 4.41 mol HF ×
×
= 44.14 g SiO 2 = 44.1 g SiO 2 used
6 mol HF
1 mol SiO 2
? g SiO 2 remaining = 45.2 g SiO 2 − 44.1 g SiO 2 = 1.1 g SiO 2 remaining
Unit 3 Review: Moles & Stoichiometry
3
d) What is the percent yield of H2SiF6?
91.5 g H 2SiF6
% Yield =
× 100% = 86.3 %
106 g H 2SiF6
IV. Extra Review—More practice of the more difficult types of calculations
18) Suppose 6.00 L of hydrogen gas is reacted with excess nitrogen gas at standard temperature and
pressure to form ammonia by the following balanced chemical equation: N2 (g) + 3 H2 (g) → 2 NH3 (g)
If the ammonia gas produced is cooled until it liquefies, what volume of liquid ammonia should be
collected? Density of liquid ammonia = 0.674 g/mL
1 mol H 2 2 mol NH 3 17.03 g NH 3
1 mL NH 3
? mL NH 3 = 6.00 L H 2 ×
×
×
×
= 4.512 mL NH 3 = 4.51 mL NH 3
22.4 L H 2 3 mol H 2
1 mol NH 3 0.674 g NH 3
19) Hydrogen is generated by passing hot steam of iron, which oxidizes to form Fe3O4, in the following
unbalanced equation:
3 Fe (s) + 4 H2O (g) → 4 H2 (g) +
Fe3O4 (s)
Suppose 9.78 L of hydrogen gas is produced at STP when 21.5 g Fe reacts with 15.6 g H2O.
a) What is the theoretical yield of hydrogen gas (in L, at STP)?
1 mol Fe
0.385 mol Fe
21.5 g Fe ×
= 0.385 mol Fe;
= 0.128 eq. mol Fe *Limiting Reactant*
55.85 g Fe
3 mol Fe
1 mol H 2O
0.866 mol H 2O
15.6 g H 2O ×
= 0.866 mol H 2O;
= 0.216 eq. mol H 2O
18.02 g H 2O
4 mol H 2O
4 mol H 2 22.4 L H 2
? L H 2 = 0.385 mol Fe ×
×
= 11.497 L H 2 = 11.5 L H 2
3 mol Fe 1 mol H 2
b) What is the percent yield of hydrogen gas?
9.78 L H 2
× 100% = 85.0 % H 2
11.5 L H 2
c) How many grams of the excess reactant are left over?
4 mol H 2O 18.02 g H 2O
? g H 2O used = 0.385 mol Fe ×
×
= 9.249 g H 2O = 9.25 g H 2O used
3 mol Fe
1 mol H 2O
% Yield =
? g H 2O remaining = 15.6 g H 2O − 9.25 g H 2O = 6.35 g H 2O = 6.4 g H 2O remaining
20) What is the molecular formula for the organic compound with a molar mass of 208.2 g/mole and the
following percent composition: 46.15 % carbon, 7.75% hydrogen and 46.11% oxygen?
⎫
Find empirical formula first:
⎪
1 mol C
46.15 g C ×
= 3.843 mol C ÷ 2.882 = 1.333× 3 = 4 ⎪
12.01 g C
⎪
⎪
1 mol H
⎬ EF = C4 H 8O3
7.75 g H ×
= 7.69 mol H ÷ 2.882 = 2.668 × 3 = 8 ⎪
1.008 g H
⎪
1 mol O
⎪
46.11 g O ×
= 2.882 mol O ÷ 2.882 = 1× 3 = 3 ⎪
16.00 g O
⎭
EF mass = 4(12.01) + 8(1.008) + 3(16.00) = 104.1
multiplier n =
4
208.2 g/mol
= 2 so Molecular Formula = (C4 H 8O3 ) × 2 = C8 H16O6
104.1 g/mol
Unit 3 Review: Moles & Stoichiometry