Name:______________________________________________________________________
Hour__________
2.3 PS1-7 IC2 Quiz Review
“I can calculate percent composition and use it to determine empirical formulas and molecular formulas.”
Determine the percent composition of the given compounds.
Note: to check your work on these, %s should add up to about 100%
1. Al(NO3)3
Al: 27/213 = 12.67%
N: 42/213 = 19.72% O: 144/213 = 67.61%
2. Ca(H2PO2)2
Ca: 40/170 = 23.53%
P: 62/170 = 36.47%
H: 4/170 = 2.35%
O: 64/170 = 37.65%
3. F3NO3S
F: 57/151 = 37.75%
O: 48/151 = 31.79%
N: 14/151 = 9.27%
S: 32/151 = 21.19%
Molar mass:
Al: 1 x 27 = 27
N: 3 x 14 = 42
O: 9 x 16 = 144
Total:
213
Molar mass:
Ca: 1 x 40 = 40
H: 4 x 1 = 4
P: 2 x 31 = 62
O: 4 x 16 = 64
Total:
170
Molar mass:
F: 3 x 19 = 57
N: 1 x 14 = 14
O: 3 x 16 = 48
S: 1 x 32 = 32
Total:
151
4. Describe how to determine an empirical formula if you know the amount of each element used in a
chemical reaction.
If the amount is in a per-cent, drop the % sign and change to grams.
Once in grams, convert to moles (use the amount x 1mol/atomic mass)
Divide each result by the smallest result
Use that number (round if necessary) as the subscript in the formula
5. Describe how to use an empirical formula to determine a molecular formula.
You need the molar mass of the compound.
Determine the molar mass of the empirical formula
Divide the molar mass of the compound by the molar mass of the empirical formula
Round that ratio to the nearest whole number, then multiply it by each subscript in the empirical
formula.
6. You have a compound that is composed of 58.8% carbon, 9.9% hydrogen, and 31.3% oxygen. Its molecular
mass is 306 g/mol.
a. What is the empirical formula?
C: 58.8 g x 1 mol/12 g = 4.9 mol/1.96 mol = 2.5 ≈ 3
Empirical formula = C3H5O
H: 9.9 g x 1 mol/1 g = 9.9 mol/1.96 mol = 5.06 ≈ 5
O: 31.3 g x 1 mol/16 g = 1.96/1.96 = 1
b. What is the molecular formula?
Molar mass of C3H5O
C: 3 x 12 = 36
H: 5 x 1 = 5
O: 1 x 16 = 16
Total
= 57 → 306/57 = 5.37 ≈ 5 Molecular formula = C15H25O5
7. A compound has a molecular mass of 612g and is composed of 22.5% sodium, 30.4% phosphorous, and
47.1% oxygen.
a. What is the empirical formula?
Na: 22.5 g x 1 mol/23 g = 0.978 mol/0.978 mol = 1
Empirical formula = NaPO3
P: 30.4 g x 1 mol/31 g = 0.981 mol/0.978 mol = 1.002 ≈ 1
O: 47.1 g x 1 mol/16 g = 2.944 mol/0.978 mol = 3.009 ≈ 3
b.
What is the molecular formula?
Molar mass of NaPO3
Na: 1 x 23 = 23
P: 1 x 31 = 31
O: 3 x 16 = 48
Total
= 102 → 612/102 = 6 Molecular formula = Na6P6O18
8. A compound is made of 24.2% carbon, 4.0% hydrogen, and 71.7% chlorine. It has a molecular mass of 99
g/mol.
a. What is the empirical formula?
C: 24.2 g x 1 mol/12 g = 2.017 mol/2.017 mol = 1
Empirical formula = CH2Cl
H: 4.0 g x 1 mol/1 g = 4.0 mol/2.017 mol = 1.98 ≈ 2
Cl: 71.7 g x 1 mol/35 g = 2.05 mol/2.017 mol = 1.02 ≈ 1
b.
What is the molecular formula?
Molar mass of CH2Cl
C: 1 x 12 = 12
H: 2 x 1 = 2
Cl: 1 x 35 = 35
Total:
49 → 99/49 = 2.02 ≈ 2 Molecular formula = C2H4Cl2