Honors Chemistry
Chapter 10 Worksheet 2
Name___________________
1. Convert each of the following using the factor-label method: watch units/sig figs!
43.15 g of Iron (III) Sulfite to atoms of Sulfur answer = 2.215 x 1023 atoms S
43.15 g Fe2(SO3)3
1
1 mole Fe2(SO3)3 X 6.02 x 1023 formula units Fe2(SO3)3
351.9 g Fe2(SO3)3
1 mole Fe2(SO3)3
X
X
3 atoms S =
1fu Fe2(SO3)3
1.7 X 1024 atoms of Oxygen to L of Oxygen gas answer = 32 L O2
1.7 x 1024 atoms O
1
X
1 molecule O2
2 atoms O
X
1 mole O2
=
X 22.4 L O2
23
6.02 x 10 molecules O2
1 mole O2
2.40 X 1023 molecules of Dinitrogen Trioxide to the total number of atoms
2.40 x 1023 molecules N2O3
1
X
5 atoms
= 1.20 x 1024 atoms
1 molecules N2O3
2. Calculate the empirical formula for a compound that contains 14.0 grams of Nitrogen and 16.0
grams of Oxygen.
N
14.0 = 1
14.0
O
1/1 = 1
16.0 = 1
16.0
so….
1/1 = 1
NO
3. Determine the molecular formula of a compound that is made up of an unknown percentage of
C, 53.3% O, and 6.7% Hydrogen, if it has a molecular mass of 180 g/mole
C
40.0
12.0
= 3.33
3.33/3.33 = 1
H
6.7
1.0
= 6.7
6.7/3.33 = 2
O
53.3
16.0
= 3.33
3.33/3.33 = 1
Empirical = CH2O (mm = 30.0 g/mole) so 30.0 X ____?_____ = 180.0 ( ? = 6)
So…. CH2O X 6 = molecular formula = C6H12O6
4. How many moles of Hydrogen molecules in 7.0 g? How many moles of Hydrogen atoms in
the same amount of substance?
7.0 g H2
1 1
X
1 mole H2
2.0 g H2
7.0 g H2
1
X
1 mole H2
2.0 g H2
= 3.5 moles H2
X 2 mole H atoms
= 7.0 moles H
1 mole H2 molecules
5. What is the percentage composition of H2O?
H
2.0 x 100 = 11%
18.0
O
16.0
18.0
x 100
= 88.9%
6. If a substance decomposes into 44.8 L of Oxygen gas and 4.0 g of Hydrogen gas, what is the
substance’s empirical formula?
44.8 L O2
1
4.0 g H2
1 1
O
1 mole O2
22.4 L O2
X
X
2 mole O = 4.00 moles O
1 mole O2
1 mole H2 x 2 mole H
2.0 g H2
1 mole H2
X
4.00/4.0 = 1
= 4.0 moles H
so…. Empirical = HO
H 4.0/4.0 = 1
7. A new compound is created in the lab from the synthesis of 11.68 grams of C, 2.928 X 1023
molecules of Oxygen, and 21.8 L of Hydrogen.
what is the percent composition?
2.928 x 1023 molecules O2
1
21.8 L H2
1
X
1 mole H2
22.4 L H2
X
X
C = 40.01 %
H = 6.68 % O = 53.31 %
1 mole O2
X 32.0 g O2
6.02 x 1023 molecules O2 1 mole O2
= 15.56 g O2
2.0 g H2
= 1.95 g H2
1 mole H2
C 11.68/29.19 x 100 = 40.01 % H 1.95/29.19 x 100 = 6.68% O15.56/29.19 x 100 = 53.31%
what is it’s molecular formula if it’s gram formula mass is 360 g?
C 40.01/12.0 = 3.33 H 6.68/1.0 = 6.7 O 53.31/16.0 = 3.33
C 3.33/3.33 = 1
H 6.7/3.33 = 2
C12H24O12
O 3.33/3.33 = 1 so… CH2O (mm = 30.0)
Empirical = CH2O (mm = 30.0 g/mole) so 30.0 X ____?_____ = 360( ? = 12)
So…. CH2O X 12 = molecular formula = C12H24O12
8. Explain when you have to use the gram atomic mass of Hydrogen of 1.0 g/mol in problems vs
a molecular mass of 2.0 g/mol.
When dealing with diatomic molecules, the weight of both atoms together must be used when the
substance is by itself (outside of a compound). When calculations are performed where the
substance is currently in the compound, the weight of the single atom is used.
Another important point: when calculating empirical formula, it really is best to find moles of a
single atom – you would do it for carbon, sulfur, etc. Why not do the same when dealing with
hydrogen, oxygen, nitrogen, etc.? We should also be cognizant of this when working with
diatomic substances. For example: if data indicates that there are 22.4 L of oxygen released
when a compound decomposes, then that means 1 mole of O2 is released. That means there were
actually 2 moles of O originally in the compound.