7.3 Percent composition and
chemical formulas
Percent composition
• The relative amount of mass of each element
in a compound, expressed in %
Calculating percent composition
% composition of an element =
mass of element/mass of compound X 100
• The sum of all the % compositions of all the elements
must equal 100%
Percent composition
• We don’t know the mass of the element , but
we can find the percent composition using the
formula and the molar mass of a substance
percent composition =
grams of element in 1 mole
molar mass of compound
100
8.20 g Mg combines completely with 5.40 g
O. What is the percent composition of this
compound?
8.20 g Mg combines completely with
5.40 g O. What is the percent
composition of this compound?
• Add the masses of the elements
• 8.20 g + 5.40 g = 13.60 g
• Divide the mass of each element by the total
mass and multiply by 100 (make it a percent)
• 8.20 g Mg / 13.60 g = 60.3 % Mg
• That makes the % composition of O 39.7 %
222.6 g N combines completely with 77.4 g O. What
is the percent composition of each element?
222.6 g N combines completely with 77.4 g O. What
is the percent composition of each element?
• Total weight of compound is 300 g
• 222.6 g/300 g X 100 = 74.2 % N
• 77.4 g/300 g X 100 = 25.8% O
• 74.2 %+ 25.8 % = 100%
Calculate the percent composition of HCN.
Calculate the percent composition of HCN.
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Molar mass H = 1 g
Molar mass C = 12 g
Molar mass N = 14 g
Total molar mass = 27 g
1 g H/27 g X 100 = 3.7% H
12 g C/27 g X 100 = 44.4% C
14 g N/27 g X 100 = 51.9% N
Using the % composition of H in HCN (3.7%),
calculate the amount of H in 378 g HCN.
Using the % composition of H in HCN (3.7%),
calculate the amount of H in 378 g HCN.
• 378 g HCN X 3.7 g H/100 g HCN
378 g HCN
3.7 g H
100 g HCN
14 g H
Conversion factor…
There are 3.7 g of H in
each 100 g HCN
Empirical formulas
• Percent composition can be used to calculate
the empirical formula for a compound
– Empirical formulas are the lowest whole
number ratios of the atoms in a compound
– Empirical formulas may or may not be the
actual formula when we are dealing with
molecules
• For example, hydrogen peroxide, H2O2 has an
empirical formula of HO, but doesn’t occur in
nature that way
Other cases where empirical formulas
don’t tell us the composition of a molecule
• Empirical formula CH
– Molecular formula C2H2 (ethyne or acetylene)
– Molecular formula C6H6 (benzene)
• Empirical formula CH2O
– Molecular formula C2H4O2 (ethanoic acid)
– Molecular formula C6H12O6 (glucose)
What is the empirical formula of a
compound which is 25.9% N and 74.1% O?
25.9 g N
1 mol N
14 g N
1.85 mol N
74.1 g O
1 mol O
16 g O
4.63 mol O
The ratio of moles of N to moles of O is
1.85/4.63, or 1/2.5. In whole numbers, this
is a ratio of 2/5. Therefore, the empirical
formula for this compound is N2O5.
Calculate the empirical formula of a compound
which is 43.64% P and 56.36% O.
Calculate the empirical formula of a compound
which is 43.64% P and 56.36% O.
43.64 g P
1 mol P
30.97 g P
1.409 mol P
56.36 g O
1 mol O
16 g O
3.523 mol O
The ratio of moles of P to moles of O is
1.409/3.523, or 1/2.5. In whole numbers,
this is a ratio of 2/5. Therefore, the empirical
formula for this compound is P2O5.
A compound is 2.477 g Mn and 1.323 g O. What
is the empirical formula?
A compound is 2.477 g Mn and 1.323 g O. What
is the empirical formula?
2.477 g Mn
1 mol Mn
59.94 g Mn
1.323 g O
1 mol O
16 g O
.0413 mol Mn
.0827 mol O
The ratio of moles of Mn to moles of O is
.0413/.0827, or 1/2. Therefore, the empirical
formula for this compound is MnO2.
Methyl butanoate has a percent composition of:
58.8 % C; 9.8 % H; and 31.4 % O. Its molar mass is 102
grams. Want is its molecular formula?
Methyl butanoate has a percent composition of:
58.8 % C; 9.8 % H; and 31.4 % O. Its molar mass is 102
grams. Want is its molecular formula?
• .588 X 102 g = 59.98 g C
• .098 X 102 g = 9.99 g H
• .314 X 102 g = 32.03 g O
These numbers look awfully suspicious!
The first is 5 moles C, the second is 10 moles H,
and the third is 2 moles O. So the molecular
formula is C5H10O2