# Percent Composition (Section 11.4)

```Percent Composition
(Section 11.4)
• Helps determine identity of unknown compound
– Think CSI—they use a mass spectrometer
• Percent by mass of each element in a compound
mass of element
percent by mass 
x 100
mass of compound
To calculate percent composition
• H2O
percent by mass 
mass of element
x 100
mass of compound
Percent Composition of Water
• Even though there are 2 hydrogen
atoms in water, most of the mass of
water comes from the 1 oxygen atom
hydrogen
oxygen
Percent Composition Practice
• KNO2
Percent composition of N
mass of element
percent by mass 
x 100
mass of compound
2. % P in Al2(PO4)3
3. % Cl in LiCl
4. % Ca in CaSO4
a. 40.1 %
b. 29.44 %
c. 136.2 %
d. 70.5 %
5. % H in NaOH
a. 2.5 %
b. 5.0 %
c. 97.5 %
d. 95 %
6. % H in (NH4)3PO4
a. 0.67 %
b. 2.68 %
c. 4.70 %
d. 8.05 %
7. Cl in CuCl2
a. 26.39 %
b. 52.79 %
c. 1.89 %
d. 3.79 %
8. % Sn in SnSO3
a.
b.
c.
d.
0.60 %
59.71 %
1.67 %
167 %
9. % O in Mg(OH)2
a. 27.44 %
b. 18.22 %
c. 37.82 %
d. 54.89 %
• Calculate the percent composition
– Phosphorus in Al2(PO4)3
• P = 3 x 31.0 = 93 g/mol
• Al2(PO4)3 = 2 x 27 + 3 x 31 + 12 x 16 = 339 g/mol
• % P = (93 / 339) x 100 = 27.43%
– Chlorine in LiCl
• Cl = 1 x 35.5 = 35.5 g/mol
• LiCl= 1 x 6.9 + 1 x 35.5 = 42.4 g/mol
• % Cl = (35.5 / 42.4) x 100 = 83.73%
– Calcium in CaSO4
• Ca = 1 x 40.1 = 40.1 g/mol
• CaSO4 = 1 x 40.1 + 1 x 32.1 + 4 x 16 = 136.2 g/mol
• % Ca = (40.1 / 136.2) x 100 = 29.44%
• Calculate the percent composition
– Sodium in NaCl
• Na = 1 x 23 g/mol = 23 g/mol
• NaCl = 1 x 23 + 1 x 35.5 = 58.5 g/mol
• % Na = (23 / 58.5) x 100 = 39.32%
– Sulfur in H2SO4
• S = 1 x 32.1 = 32.1 g/mol
• H2SO4 = 2 x 1 + 1 x 32.1 + 4 x 16 = 98.1 g/mol
• % S = (32.1 / 98.1) x 100 = 32.72%
– Potassium in KNO3
• K = 1 x 39.1 = 39.1 g/mol
• KNO3 = 1 x 39.1 + 1 x 14 + 3 x 16 = 101.1 g/mol
• % K = (39.1 / 101.1) x 100 = 38.67%
Empirical vs Molecular Formulas
• Empirical formula = formula with smallest whole
number ratio of elements
– Recognize because it is the simplified formula
• Molecular formula = formula with actual number of
atoms of each element
Empirical
Molecular
C4H9
C8H18
H2O
H6O3
We use the percent composition to calculate the
empirical and molecular formulas.
Which is the empirical formula?
a. C3H6O2
b. C6H12O4
a. C22H24O12
b. C11H12O6
Which is the empirical formula?
a. C2H6
b. CH3
c. C4H12
a. N4O12
b. N2O6
c. NO3
How to calculate empirical formula
1. Use mass of elements to calculate moles
2. Pick the smallest number of moles
3. Divide each by the smallest number of
moles
•
•
•
should get whole numbers. If not, multiply to
get whole numbers
.25
.50
.75
.33
.66
4. Write empirical formula
•
A blue solid is found to contain 36.43% nitrogen and
63.16% oxygen. What is the empirical formula?
•
A blue solid is found to contain 36.43% nitrogen and
63.16% oxygen. What is the empirical formula?
If we assume sample is 100 grams, we have 36.43 g N
and 63.16 g O.
1. Calculate moles of each element
N = 36.43 / 14 = 2.60 mol
O = 63.16 / 16 = 3.95 mol
2. Smallest number of moles = nitrogen with 2.60 mol
3. Divide by smallest number of moles
N = 2.60 / 2.60 = 1 x 2 = 2
O = 3.95 / 2.60 = 1.5 x 2 = 3
we must get whole numbers as our answer—that’s why we
multiplied by two
4. Empirical formula = N2O3
•
Determine the empirical formula for a compound that
contains 35.98% aluminum and 64.02% sulfur
•
Determine the empirical formula for a compound that
contains 35.98% aluminum and 64.02% sulfur
If we assume sample is 100 grams, we have 35.98 g Al
and 64.02 g S.
1. Calculate moles of each element
Al = 35.98 / 27 = 1.33 mol
S = 64.02 / 32.1 = 2.00 mol
2. Smallest number of moles = aluminum with 1.33 mol
3. Divide by smallest number of moles
Al = 1.33 / 1.33 = 1 x 2 = 2
S = 2.00 / 1.33 = 1.5 x 2 = 3
we must get whole numbers as our answer—that’s why we
multiplied by two
4. Empirical formula = Al2S3
•
Find the empirical formula for a substance that
contains 48.64% carbon, 8.16% hydrogen, and
43.20% oxygen
•
Find the empirical formula for a substance that
contains 48.64% carbon, 8.16% hydrogen, and
43.20% oxygen
If we assume sample is 100 grams, we have 48.64 g C,
8.16 g H, and 43.20 g O.
1. Calculate moles of each element
C = 48.64 / 12 = 4.05 mol
H = 8.16 / 1 = 8.16 mol
O = 43.20 / 16 = 2.70 mol
2. Smallest number of moles = oxygen with 2.70 moles
3. Divide by smallest number of moles
C = 4.05 / 2.70 = 1.5 x 2 = 3
H = 8.16 / 2.70 = 3.0 x 2 = 6
O = 2.70 / 2.70 = 1.0 x 2 = 2
we must get whole numbers as our answer—that’s why we
multiplied by two
4. Empirical formula = C3H6O2
•
The chemical analysis of aspirin indicates that the
molecule is 60%carbon, 4.44% hydrogen, and
35.56% oxygen. Determine the empirical formula for
aspirin.
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