AP Chemistry Review
Name:______________________________
Topic 2: Stoichiometry and Chemical Equations
__E___1. Which expression gives percent by mass of carbon in oxalic acid, H2C2O4 • 2H2O?
(A)
2
 100
14
(B)
12
 100
90
(C)
24
 100
66
(D)
24
 100
90
(E)
24
 100
126
__C___2. A compound contains 1.10 mol of K, 0.55 mol of Te, and 1.65 mol of O. What is the simplest
formula of this compound?
(A) KTeO
(B) KTe2O
(C) K2TeO3
(D) K2TeO6
(E) K4TeO6
__B___3. When a hydrate of Na2CO3 is heated until all the water is removed, it loses 54.3 percent of its
mass. The formula of the hydrate is…
(A) Na2CO3 •10 H2O (B) Na2CO3 • 7 H2O (C) Na2CO3 • 5 H2O
(D) Na2CO3 •3 H2O
(E) Na2CO3 • H2O
__B___4. Propane gas, C3H8, burns in excess oxygen gas. When the equation for this reaction is
correctly balanced and all coefficients are reduced to their lowest whole-number terms, the
coefficient for O2 is…
(A) 4
(B) 5
(C) 7
(D) 10
(E) 22
_2_ CH3CH2COOH (l) + _7_ O2(g)  _6_ CO2(g) + _6_ H2O(l)
__D___5. How many moles of O2 are required to oxidize 1 mole of CH3CH2COOH according to the
unbalanced reaction represented above?
(A) 2 moles
(B) 5/2 moles
(C) 3 moles
(D) 7/2 moles
(E) 9/2 moles
__D___6. The limiting reagent in a chemical reaction is one that:
(A) has the largest molar mass (formula weight).
(B) has the smallest molar mass (formula weight).
(C) has the smallest coefficient.
(D) is consumed completely.
(E) is in excess.
Compound
Molar Mass (g/mol)
NaCl
58.5
KCl
74.6
AgCl
143.3
CuCl
99.0
LiCl
42.44
__E___7. A 1.0 gram sample of which of the compounds listed above would contain the highest number
Cl atoms?
(A) NaCl
(B) KCl
(C) AgCl
(D) CuCl
(E) LiCL
AP Chemistry Review
Name:______________________________
1. When chlorine gas is combined with fluorine gas, a compound is formed that is 38% chlorine and 62%
fluorine. The compound has a molar mass of 277.5g What is the empirical formula of the compound?
What is the molecular formula of the compound?
38% Cl and 62% F
Assume a 100 gram sample 38 g Cl and 62 g F
Convert to mols:
1 𝑚𝑜𝑙 𝐶𝑙
35.5 𝑔 𝐶𝑙
38 g Cl •
= 1.07 mol Cl
Divide by the smallest number of mols:
3.26 𝑚𝑜𝑙 𝐹
1.07 𝑚𝑜𝑙
62 g F •
1 𝑚𝑜𝑙 𝐹
19.0 𝑔 𝐹
= 3.26 mol F
≈3
1.07 𝑚𝑜𝑙 𝐶𝑙
1.07 𝑚𝑜𝑙
≈1
Empirical formula: ClF3
Molecular Formula:
Divide the molar mass by the empirical formula mass.
277.5 𝑔/𝑚𝑜𝑙
92.5 𝑔/𝑚𝑜𝑙
=3
ClF3 → Cl3F9
2. A 27.0-gram sample of an unknown hydrocarbon was burned in excess oxygen to form 88.0 grams of
carbon dioxide and 27.0 grams of water. What is a possible molecular formula of the hydrocarbon?
Assume all of the carbon became carbon dioxide
Assume all of the hydrogen became water
CxHy + O2 → CO2 + H2O
88.0 g CO2 •
27.0 g H2O •
1 𝑚𝑜𝑙 𝐶𝑂2
44 𝑔 𝐶𝑂2
1 𝑚𝑜𝑙 𝐻2 𝑂
18 𝑔 𝐻2 𝑂
•
1 𝑚𝑜𝑙 𝐶
1 𝑚𝑜𝑙 𝐶𝑂2
= 2 mol C
2 𝑚𝑜𝑙 𝐻
2𝑂
= 3 mol H
• 1 𝑚𝑜𝑙 𝐻
Empirical formula: C2H3 Possible Molecular formula: C4H6
3. Pure silicon is required for computer chips and solar cells. It can be produced when sand (SiO2) reacts
with magnesium metal. The other product is magnesium oxide.
a.
Write the balanced chemical equation.
SiO2 (s) + 2 Mg (s) → Si (s) + 2 MgO (s)
AP Chemistry Review
b.
7.00 g SiO2 •
7.00 g Mg •
Name:______________________________
What mass of silicon (Si) can be produced when 7.00g of sand (SiO2) reacts with 7.00 g
of magnesium?
1 𝑚𝑜𝑙 𝑆𝑖𝑂2
60 𝑔 𝑆𝑖𝑂2
1 𝑚𝑜𝑙 𝑀𝑔
24.3 𝑔 𝑀𝑔
1 𝑚𝑜𝑙 𝑆𝑖
28 𝑔 𝑆𝑖
• 1 𝑚𝑜𝑙 𝑆𝑖𝑂 • 1 𝑚𝑜𝑙 𝑆𝑖 = 3.27g Si
2
1 𝑚𝑜𝑙 𝑆𝑖
28 𝑔 𝑆𝑖
• 2 𝑚𝑜𝑙 𝑀𝑔• 1 𝑚𝑜𝑙 𝑆𝑖 = 4.03 g Si
SiO2 is the limiting reactant. 3.27 g Si are formed
4. An ionic compound MX3 is prepared according to the following unbalanced chemical equation.
M + X2  MX3
A 0.105-g sample of X2 contains 8.92 × 1020 molecules. The compound MX3 consists of 54.47% X by
mass. What is the identity of X? What is the identity of M?
First find the identity of X. This is best accomplished by finding the molar mass of X2.
8.92 x 1020 molecules X •
0.105 𝑔 𝑋2
0.001481 𝑚𝑜𝑙𝑠 𝑋2
1 𝑚𝑜𝑙 𝑋
6.023 𝑥 1023 𝑚𝑜𝑙𝑒𝑐𝑢𝑙𝑒𝑠
= 70.9 g/mol X2
= 0.001481 mol X2
X = 35.45 g/mol
X represent chlorine.
Second Find M using the mass percent of MX3
%X in MX3 is 54.47%
%X =
𝑚𝑎𝑠𝑠 𝑜𝑓 𝑋
𝑀𝑎𝑠𝑠 𝑡𝑜𝑡𝑎𝑙
54.47 =
54.47 =
x 100 =
𝑚𝑎𝑠𝑠 𝑜𝑓 𝑋
𝑀𝑎𝑠𝑠 𝑜𝑓 𝑀+ 𝑚𝑎𝑠𝑠 𝑜𝑓 𝑋
3• 35.45
(𝑚𝑎𝑠𝑠 𝑜𝑓 𝑀)+3• 35.45
106.35
(𝑚𝑎𝑠𝑠 𝑜𝑓 𝑀)+106.35
x 100
x 100
Mass of M = 88.9 g/mol M is Yttrium
x 100
=
3• 35.45
(𝑚𝑎𝑠𝑠 𝑜𝑓 𝑀)+3• 35.45
x 100