Review Sheet – Chemistry, CCP –Moles
Name: ____________________
Date:___________
Period:___
This test covers materials in the textbook in Chapters 10 (pp. 311-339). You are always responsible for the
topics on the previous tests.
. Test Expectations
The student should be able:
 Correctly use the terminology of mass and moles.
 Convert between moles, particles, mass, and volume of a gas (at STP).
 Calculate the percent composition of a compound.
 Determine the empirical and/or molecular formula of a compound from given data.
1. Define each of the following terms:
(a) Formula Unit - One formula
weight of an ionic compound.
(b) Molar Mass – Mass in grams of 1
mole of substance.
(c) Avogadro’s number - The
number of particles in one mole,
equal to 6.02214199 × 1023 mol.
(d) Molar Volume – The volume
occupied by one mole of a gas
under standard conditions (22.4
L at STP)
(e) Mole - SI unit for amount of
substance, defined as the
number of atoms in exactly 12 g
of carbon-12.
(f) Empirical Formula – Simplest
whole number ratio of atoms in
a compound.
(g) Molecular Formula – Actual
whole number ratio of atoms in
a compound.
(h) Atomic Mass – Weighted
average of the masses of the
isotopes of an element.
(i) STP – Standard condition for
gases, 0°C and 1 atm.
(j) Formula Mass – Sum of the all
the atomic masses of the atoms
in a compound (in amu).
(k) Carbon 12 - Standard for atomic
mass and is used to define the
mole.
(l) Percent Composition - The
percent of each element present
in a compound by mass.
(m) Elemental Analysis – Analytical
method that determines the
percent composition (by mass)
of a compound.
(n) Molecule - The smallest particle
of a covalent compound that
retains the chemical properties
of the element or compound.
2. Find the molar mass of the following compounds:
A. Ba(OH)2
1 x Ba 137.33 =
2 x O 16.00 =
2 x H 1.01 =
137.33
32.00
2.02
171.35 g/mole
B. C9H20
9 x C 12.01 =
20 x H 1.01 =
108.09
20.20
128.29 g/mole
3. An accident in a chemistry laboratory releases 7.4 moles of the war gas phosgene (COCl 2). How many grams of
phosgene were released?
1-step problem
 98.91 g 
7.4 moles 
  730 g
 1 mole 
4. A similar accident the previous year released 9.07 x 1024 molecules of phosgene. How much volume did this first
accident release? (Assuming this happens at STP…)
2-step problem
1 mole

 22.4 L 
9.07 x 10 24 molecules 

  337 L
23
 6.02 x 10 molecules  1 mole 
5. What is the percent composition by mass of phosgene (COCl2)?
1 x C 12.01
1 x O 16.00
2 x Cl 35.45
=
=
=
=
12.01
16.00
70.90
98.91
12.01
x 100  12.14 %
98.91
16.00
%O 
x 100  16.18 %
98.91
70.90
%Cl 
x 100  71.68 %
98.91
%C 
6. Mustard gas (C4H8Cl2S) is a poisonous gas that was used in World War I and banned afterward. It causes general
destruction of tissues, resulting in the formation of large water blisters. There is no effective antidote. Calculate the
percent composition by mass of the elements in mustard gas.
4 x C 12.01
8 x H 1.01
2 x Cl 35.45
1 x S 32.07
=
=
=
=
48.04
8.08
70.90
32.07
48.04
%C 
x 100  30.20 %
159.09
8.08
%H 
x 100  5.08 %
159.09
70.90
%Cl 
x 100  44.57 %
159.09
32.07
%S 
x 100  20.16 %
159.09
159.09 g/mole
7. Ascorbic acid (vitamin C) cures scurvy and may help to prevent the common cold. It is composed of 40.92 % carbon,
4.58 percent hydrogen, and 54.50 percent oxygen by mass. Determine its empirical formula.
STEP 1: Assuming 100 g of ascorbic acid:
 1 mole 
  3.41 moles
C : 40.92 g
12
.
01
g


 1 mole 
  4.53 mole
H : 4.58 g
 1.01 g 
 1 mole 
  3.41 moles
O : 54.50 g
 16.00 g 
STEP 2 & 3:
C:H:O
3.41 : 4.53 : 3.41
3.41
1 : 1.33 : 1
STEP 4: Multiply by 3 in order to get 1.33 to be a whole number.
3:4:3
C3H4O3
8. The molar mass of caffeine is 194.19 g/mole. Is the molecular formula of caffeine C4H5N2O or C8H10N4O2?
C4H5N2O
Molar Mass = 4 x 12 + 5 x 1 + 2 x 14 + 16 = 97 g/mole
MM MF 194.19

 2.002  2
MM EF
97
The answer is C8H10N4O2, then.
9. The empirical formula of a compound is NaO, and the approximate molar mass is 76 g. What is the molecular formula?
NaO
Molar Mass = 23.0 + 16.0 = 39.0 g/mole
MM MF 76

 1.95  2
MM EF 39
The answer then is Na2O2.
10. A compound containing boron and hydrogen has the composition 6.444 g of B and 1.803 g of H. The molar mass of
this compound is 30 g. What is its molecular formula?
STEP 1:
 1 mole 
  0.596 moles
B : 6.444 g
 10.81g 
 1 mole 
  1.79 mole
H :1.803 g
 1.01 g 
STEP 2 & 3:
B:H
0.596 : 1.79
0.596
1 : 3.003
1: 3
STEP 4: BH3 is the empirical formula.
STEP 5:
MM MF
30

 2.17  2
MM EF 13.8
B2H6 is the molecular formula.
11. Allicin is the compound responsible for the characteristic smell of garlic. It contains carbon, hydrogen, sulfur, and
oxygen. An analysis of the compound gives the following percent composition by mass: carbon, 44.4%, hydrogen, 6.21%;
sulfur 39.5%. What is its molecular formula given that its molar mass is about 162 g/mole.
Let’s try a slightly different method for this problem. Let’s assume a mass equal to the molar mass 162 g/mole.
This will make more work for us up front, but simplify the later steps dramatically.
 1mole 
44.4 %
  5.99moles
C:
* 162  71.93 g 
100
 12.01g 
H:
 1mole 
6.21 %
  9.96moles
* 162  10.06 g 
100
 1.01g 
S:
 1mole 
39.5 %
  2.00moles
* 162  63.99 g 
100
 32.07 g 
O:
 1mole 
100 - (44.4  6.21  39.5) %
  1.00moles
* 162  16.02 g 
100
 16.00 g 
Looking at these values, the actual ratio is clear. The molecular formula is C6H10S2O. To find the empirical
formula, the subscripts need to be reduced to its simplest form. In this case, the empirical formula will be the
same. This method works for any molecular formula problem (you need the elements molar mass).