>> View the other Key Equations and Concepts in this chapter

Chemical Formulas and Percent Composition

>> Parts of this equation/concept include:

A -	Finding Percent Composition from Chemical Formula
B -	Finding Empirical Formulas from Percent Composition
C -	Finding Molecular Formulas from Percent Composition

Percent composition implies mass percent. So percent composition is

%

=

grams element grams compound

100

Because percent is a ratio, any quantity of compound will give the same ratio and the same percent composition. Therefore you can choose any convenient amount of compound to work with. In determining percent composition from chemical formula, a convenient amount to work with is 1 mole. For instance, the mass of 1 mole of a compound is its molar mass. The mass of any element in a compound is its molar mass times the number of atoms per molecule in that compound.

The sum of the percent composition of all elements in a compound should add up to 100.

>> Example 1

What is the percent composition of each element in MnS?

Solution:

The molar mass of MnS is 54.94 + 32.07 = 87.01 g/mol.

% Mn

=

54.94 87.01

100

=

63.14%

% S

=

32.07 87.01

100

=

36.85%

or

% S = 100% – 63.14% = 36.85%

>> Example 2

What is the percent composition of each element in Al(OH)₃?

Solution:

The molar mass of Al(OH)₃ is 26.98 + 3(16.00) + 3(1.008) = 78.00 g/mol.

% Al

=

26.98 78.00

100

=

34.59%

% O

=

3(16.00) 78.00

100

=

61.54%

% H

=

3(1.008) 78.00

100

=

3.877%

  >> back to the Top of the Page

An empirical formula is the simplest mole ratio of elements. As with percent composition, since it is a ratio any amount of overall compound can be used. When starting with percent composition, it is convenient to start with 100 g of compound.

Since one meaning of percent is number of grams in 100 grams, the percent value is also the number of grams in 100 grams. The molar mass of the element is used to convert grams to moles. The last step is to get the mole ratios into simple whole numbers. The easiest way to do this is to divide by the smallest value. Normally, you just round this value to the nearest whole number. Since the answer must be a whole number (by definition), you don't have to be as careful with significant figures.

Before rounding to the nearest whole number, consider the possibility that the value is a simple fraction, such as 1/2 (0.5) or 1/3 (0.333), and round to that value instead. If you round to a fraction, multiply by the denominator to get the whole-number ratio.

Empirical also means "experimental," so don't expect the numbers to work perfectly.

The best order of the elements can be determined from the periodic table. The leftmost element goes first. If two elements are in the same column, use the lower element first. The position of hydrogen is more difficult to predict. (It even turns up on various places, depending on the periodic table.)

>> Example 3

What is the empirical formula of a compound whose percent composition is 40.0% sulfur and 60.0% oxygen?

Solution:

If 100 g of compound is assumed, there are 40.0 g sulfur and 60.0 g oxygen.

To convert to moles

40.0 g S

1 mol 32.07 g

=

1.247 mol

40.0 g O

1 mol 16.00 g

=

3.75 mol

To get the smallest ratio, divide by the smallest mole value. In this case, that is 1.247.

1.247 mol S 1.247

=

1

3.75 mol O 1.247

=

3.007

Therefore the ratio is one sulfur to three oxygens. Sulfur and oxygen are in the same group, and sulfur is lower. Therefore sulfur will be first in the formula—SO₃.

>> Example 4

What is the empirical formula of a molecule with percent composition of 15.8% Al, 28.1% S, and 56.1% O?

Solution:

In 100 g of compound, there are 15.8 g Al, 28.1 g S, and 56.1 g oxygen. To convert to moles

15.8 g Al

1 mol 26.98 g

=

0.5856 mol Al

28.1 g S

1 mol 32.07 g

=

0.8762 mol S

56.1 g O

1 mol 16.00 g

=

3.506 mol O

To get the simplest ratio, divide by the smallest number of moles.

0.5856 mol Al 0.5856

=

1 Al

0.8762 mol S 0.5856

=

1.496 S

3.506 mol O 0.5856

=

5.987 O

Rounding whole numbers, you should notice that sulfur rounds better to 1.5 than to either 1 or 2. Since 1.5 is a simple fraction (1 1/2), it is better to round to that than to either 1 or 2. So that gives the ratio of 1 Al:1.5 S:6 O. Since fractions are not acceptable in a formula, to get rid of 1/2, multiply each 1 by 2. Thus the formula is Al₂S₃O₁₂.

Note the order of the elements. On the periodic table, Al is to the left of sulfur and oxygen. Sulfur is lower on the periodic table than oxygen, so it is next.

  >> back to the Top of the Page

An empirical formula is the simplest ratio of elements. For some molecules the simplest ratio is not the actual ratio. However, to determine the actual ratio, more information is needed. The extra information is the molar mass. Like percent composition, molar mass can be determined experimentally, although it is a different experiment.

Once the empirical formula is determined, the molecular formula must be some multiple of that formula. To determine that multiplication factor, compare the actual molar mass (given in the problem) to the apparent molar mass (based on the empirical formula).

factor

=

actual molar mass apparent molar mass

To determine the molecular formula, multiply each subscript by the factor. (Remember that if no subscript appears, it is understood to be a 1.) Of course, if the factor is 1, the molecular formula is the same as the empirical formula.

>> Example 5

What is the molecular formula of a compound that is 86% C and 14% H and that has a molar mass of 56 g/mol?

Solution:

First determine the empirical formula in the usual way.

86 g C

1 mol 12.01 g

=

7.2 mol 7.2

=

1

14 g H

1 mol 1.008 g

=

13.88 mol 7.2

=

1.92

The empirical formula is CH₂. The apparent molar mass is 12.01 + 2(1.008) = 14.02 g. The multiplication factor is 56/14 = 4. So the actual molecular formula is C₄H₈.

>> View the other Key Equations and Concepts in this chapter