Percent Composition Notes
Procedure:
How to calculate the percent composition of a compound using the periodic table: percentage of that element compared to the total.
% composition of element =(total molar mass of element ÷ molar mass of the compound) x 100%
NaCl Example:
Calculate the percent by mass of sodium (Na) and chlorine (Cl) in sodium chloride
(NaCl). a. Calculate the molar mass of NaCl: 22.99 g/mol (Na)+ 35.45 g/mol (Cl) =
58.44 g/mol b. Calculate the molar mass of Na: molar mass of Na = 22.99 g/mol c. Calculate the percent by mass of Na in NaCl: %Na = (molar mass Na ÷ molar mass NaCl) x 100 % = (22.99 g/mol ÷ 58.44 g/mol) x 100 %= 39.34% d. Calculate the molar mass of Cl present: molar mass of Cl = 35.45 g/mol e. Calculate the percent by mass of Cl in NaCl: %Cl = (mass Cl ÷ molar mass
NaCl) x 100% = (35.45 g/mol ÷ 58.44 g/mol) x 100 %= 60.66%
Check your answer: 39.34% Na + 60.66 % Cl = 100.00%
(Note that in some calculations the total may not be exactly 100.00 % due to rounding.)

Empirical and Molecular Formula Notes
Empirical formula" is a REAL IMPORTANT concept. Here's the definition: the formula of a compound expressed as the smallest possible whole-number ratio of subscripts of the elements in the formula
For example, CH
3
COOH has two carbons, four hydrogens and two oxygens. So we could write the formula like this: C
2
H
4
O
2
and so it reduces to CH
2
O.
Contrast the above definition to this one for "molecular formula:" the formula of a compound in which the subscripts give the actual number of each element in the formula
Here are the four formulas being used as examples:
Molecular Formula Empirical Formula
H
2
O H
2
O
CH
3
COOH
CH
2
C
6
H
12
O
O
6
CH
2
O
CH
2
O
CH
2
O
Notice two things:
1. The molecular formula and the empirical formula can be identical.
2. You scale up from the empirical formula to the molecular formula by a whole number factor.
When converting percentage composition to an empirical formula, remember the following rhyme:
Percent to mass
Mass to mole
Divide by small
Multiply 'til whole
Here's an example of how it works. A compound consists of 72.2% magnesium and 27.8% nitrogen by mass. What is the empirical formula?

(1) Percent to mass:
Assume 100 g of the substance, then 72.2 g magnesium and 27.8 g nitrogen.
(2) Mass to moles: for Mg: 72.2 g Mg x (1 mol Mg/24.3 g Mg) = 2.97 mol Mg for N: 27.8 g N x (1 mol N/14.0 g N) = 1.99 mol N
(3) Divide by small: for Mg: 2.97 mol / l.99 mol = 1.49 for N: 1.99 mol / l.99 mol = 1.00
(4) Multiply 'til whole: …multiple both by the same number by trail and error! for Mg: 2 x 1.49 = 2.98 (i.e., 3) for N: 2 x 1.00 = 2.00 and the formula of the compound is Mg
3
N
2
.
Example Problem: A compound is analyzed and found to contain 68.54% carbon, 8.63% hydrogen, and 22.83% oxygen. The molecular weight of this compound is known to be approximately 140 g/mol. What is the empirical formula? What is the molecular formula? 1) Percent to mass. Assume 100 grams of the substance is present, therefore its composition is: carbon: 68.54 grams hydrogen: 8.63 grams oxygen: 22.83 grams
(2) Mass to moles. Divide each mass by the proper atomic weight. carbon: 68.54 / 12.011 = 5.71 mol hydrogen: 8.63 / 1.008 = 8.56 mol oxygen: 22.83 / 16.00 = 1.43 mol
(3) Divide by small:

carbon: 5.71 ÷ 1.43 = 3.99 hydrogen: 8.56 ÷ 1.43 = 5.99 oxygen: 1.43 ÷ 1.43 = 1.00
(4) Multiply 'til whole. Not needed since all values came out whole.
The empirical formula of the compound is C
4
H
6
O.
Next we need to determine the molecular formula, knowing the empirical formula and the molecular weight.
Here's how:
1) Calculate the "empirical formula weight." This is not a standard chemical term, but the ChemTeam believes it is understandable.
C
4
H
6
O gives an "EFW" of 70.092.
2) Divide the molecular weight by the "EFW."
140 ÷ 70 = 2
3) Multiply the subscripts of the empirical formula by the factor just computed.
C
4
H
6
O times 2 gives a formula of C
8
H
12
O
2
.<-- molecular formula.