Percent Composition &
Formulas Notes
5. Percent Composition of
Compounds
Mass Percent for = mass of the element present in 1 mole of the compound x100%
a given element
mass of 1 mol of the compound
Steps for Calculating Percent Composition
1. Calculate the molar mass of the compound.
2. Divide the mass of each element in the compound
by the mass of the compound.
3. Multiply each by 100%.
4. Double check. The sum of the mass percents
should be 100.
Part x 100%
______
Whole
1
Examples:
CCl4
12.0 g x 100%
% C = ______
= 7.79 %
154.0 g
C = 1 x 12.0 = 12.0
g x 100% = 92.21 %
Cl = 4 x 35.5 = 142.0
% Cl = 142.0
______
154.0 g
154.0 g/mol
NaOH
23.0 g x 100%
% Na = ______
= 57.5 %
40.0 g
Na = 1 x 23.0 = 23.0
O = 1 x 16.0 = 16.0
16.0 g x 100% = 40.0 %
H = 1 x 1.0 = 1.0
% O = ______
40.0 g
40.0 g/mol
1.0 g x 100%
% H = ______
= 2.5 %
40.0 g
6. Formulas of Compounds
Empirical formula – the formula of a compound that
expresses the smallest whole-number ratio of the
atoms present.
Molecular formula – the actual formula of a
compound, the formula that tells the actual
composition of the molecules that are present.
Examples: For each pair of formulas, circle the formula that
best represents the empirical formula.
CH2
a. C2H4
b. CH2O
C4H8O4
c. C4H9
C8H18
2
7. Calculation of Empirical
Formulas
Steps for Calculating the Empirical Formula
1. Obtain the mass of each element, generally given, but
may involve a subtraction step.
– For percentages assume a 100 gram sample, so the % = g
2. Convert grams to moles.
3. Find the ratio of elements: Divide the number of moles
of each element by the smallest number of moles.
– If all calculated values are whole numbers, these are the
subscripts in the empirical formula.
– If NOT whole numbers go to step four.
4. Multiply all the numbers from step three by the smallest
whole number that will convert all of them to whole
numbers.
Calculate the empirical formula of a compound for a sample that
contains 4.151 g of Al and 3.692 g O.
4.151 g Al x _________
1 mol Al
= 0.154 mol Al / 0.154 = 1 x 2 = 2
27.0 g Al
3.692 g O x _________
1 mol O
= 0.231 mol O / 0.154 = 1.5 x 2 = 3
16.0 g O
Al2O3
0.3545 g V reacts with oxygen to achieve a final mass of 0.6330
g. Calculate the empirical formula of the compound.
1 mol V
0.6330 g 0.3545 g V x _________
= 0.00696 mol V = 1 x 2 = 2
50.9 g V
/ 0.00696
- 0.3545 g
0.2785 g O x _________
1 mol O
0.2785 g
= 0.0174 mol O
= 2.5 x 2 = 5
16.0 g O
/ 0.00696
V2O5
Calculate the empirical formula of a compound that contains
65.02% Pt, 9.34% N, 2.02% H, and 23.63% Cl.
65.02 g Pt x ________
1 mol Pt
= 0.333 mol Pt / 0.333 = 1
195.1 g Pt
23.63 g Cl x ________
1 mol Cl
= 0.666 mol Cl
9.34 g N x _________
1 mol N
= 0.667 mol N/ 0.333 = 2 35.5 g Cl
/ 0.333 = 2
14.0 g N
2.02 g H x _______
1 mol H
= 2.0 mol H / 0.333 = 6
PtN2H6Cl2
1.0 g H
3
8. Calculation of Molecular
Formulas
To calculate the molecular formula, the empirical formula and molecular
molar mass are needed.
Molecular Formula = (empirical formula)n
n = Molecular molar mass
Empirical molar mass
Steps for Calculating the Molecular Formula
1. Calculate the empirical formula, if necessary.
2. Find the molar mass of the empirical formula.
3. Divide the molecular molar mass by the empirical molar mass.
4. multiply the subscripts in the empirical formula by the result of #3.
Example:
Calculate the molecular formula of a compound that has a molar mass
of 283.88 g and an empirical formula of P2O5.
P 2 x 31.0 = 62.0
O 5 x 16.0 = 80.0
142.0 g/mol
n =_________
283.88 g = 2
P4O10
142.0 g
4