Answer to Mole Problems
Mole Worksheet
KEY
Name____________________
Molar Relationships
Period____
Directions: Answer the following questions on a separate sheet of paper. Set-up all problems
using the factor-label method of dimensional analysis and show all your work and units.
1. What is the mass of 7.50 moles of sulfur dioxide (SO2)?480g
7.5mol 64g
x
 480g
1
1mol
2. How many moles are there in 250.0 grams of sodium phosphate (Na3PO4)? 1.52mol
250g 1mol
x
 1.52mol
1
164g
3. How many grams of potassium sulfate (K2SO4) are there in 25.3 moles? 4402g
25.3mol 176g
x
 4402g
1
1mol
4. What is the volume of 0.38 moles of any gas at STP? 8.5L
0.38mol 22.4L
x
 8.5L
1
1mol

7. What is the mass of 51 liters of oxygen gas? 73g
51L 1mol 32g
x
x
 73g
1 22.4L 1mol

1. Calculate the number of moles in 32.2-L of NH3


32.3L 1mol
x
1.44mol
1
22.4L
2. Calculate the number of grams in 3.25-mol of AgNO3

MM  107  14  3x16 
169g
1mol
3.25mol 169g
x
 549g
1
1mol
8. Calculate the number of liters in 3.25-g of NH3
17g
1mol
3.26g 1mol 22.4L
x
x
 4.30L
1
17g 1mol
MM  14  3 

9. Calculate

the number of grams in 3.54-L of CO2
44g
MM  12  16x2 
1mol
3.54L 1mol 44g
x
x
 6.95g
1
22.4L 1mol
Part II
1. Calculate the % composition of Li2O.

Li2O :: MM  7x2  16  30g /mol
14
Li 
x100  46.7%
30
16
O
x100  53.3%
30
2. What is the percentage composition of a carbon-oxygen compound, given that a 95.2 g
sample
of the compound contains 40.8 g of carbon and 54.4 g of oxygen?

40.8
x100  42.9%
95.2
54.4
O
x100  57.1%
95.2
C
Part E: Empirical and Molecular Formulas
1. Determine the empirical formula of a compound with 72.4% Fe and 27.6% Oxygen.
72.4gFe 1molFe
x
 1.29molFe
1
56gFe
27.6gO 1molO
O:
x
 1.725molO
1
16gO
1.29molFe
 1x3  3
1.29mol
1.725molO
 1.33x3  4
1.29mol
Fe3O4
Fe :
+6
2. Determine the empirical formula of a compound with 52.8% Sn, 12.4% Fe, 16% C and
18.8% N.


52.8gSc 1molSc 0.444molSn
x

2
1
119gSc
0.221
12.4gFe 1molFe 0.221molFe
Fe :
x

1
1
56gFe
0.221
16gC 1molC 1.33molC
C:
x

6
1
12gC
0.221
18.8gC 1molN 1.34molN
N:
x

6
1
14gN
0.221
Sn :
+8
Sn 2 FeC6 N 6
4. Determine the molecular formula for a compound that contains 12.2-g Nitrogen, 27.8-g
Oxygen, and a molecular mass of 92.0 g/mol.


12.2gN 1molN
0.87molN
x
 0.87molN ::
1
1
14gN
0.87mol
27.8gO 1molO
1.74molO
O:
x
 1.74molO ::
2
1
16gO
0.87mol
EF  NO2 :: EM  46 :: MM  92
Factor  2
N:
+8
NO2 x2  N 2O4
5. Determine the molecular formula for a compound that contains 94.1% oxygen and 5.9%
hydrogen and a molecular mass of 34 g/mol.
5.9gH 1molH
5.9molN
H:
x
 5.9molH ::
1
1
1gH
5.9mol
94.1O 1molO
5.9molO
O:
x
 5.9molO ::
1
1
16gO
5.9mol
EF  HO :: EM  17 :: MM  34
Factor  2
+8
HOx 2  H 2O2
6. A sample of TNT, a common explosive is analyzed and found to contain 1.03-g of
nitrogen, 0.220-g hydrogen, and 1.76-g of carbon. The molar mass is 123 g/mol. What is the
molecular formula?


1.03 g 1mol
N 

x
 0.0736mol / 0.0736mol  1
1
14 g




0.220 g 1mol
H 
x
 0.220mol / 0.0736mol  2.99 
1
1g




1.76 g 1mol
 C 

x
 0.147mol / 0.0736mol  2
1
12
g


NH 3C 2
MM  123 : EM  41(multiple3)
MF  N 3 H 9C6
8. Azobenzene is an important intermediate in the manufacture of dyes. It contains
79.1% carbon, 5.95% hydrogen, and 15.4% nitrogen. It has a molar mass of 182g/mol. What is the molecular formula?


79.1g 1mol
C 
x
 6.59mol / 1.1mol  6 
1
12 g




5.95 g 1mol
H 
x
 5.95mol / 1.1mol  5 
1
1g




15.4 g 1mol
 N 
x
 1.1mol / 1.1mol  1 
1
14
g


C6 H 5 N
MM  182 : EM  91(multiple 2)
MF  C12 H 10 N 2