Empirical and Molecular Formula

As another example, let's look at phosphorus oxide \(P_4O_{10}\)

Find the empirical formula of phosphorus oxide.

Empirical formula of phosphorus oxide = \(P_2O_5\)

For every two phosphorus atoms, there are five oxygen atoms.

Here's a Tip:

You can discover the empirical formula by counting the number of each atom in a compound and dividing it by the lowest number.

In the phosphorus oxide example ( \(P_4O_{10}\) ) the lowest number is 4.

4 ÷ 4 = 1

10 ÷ 4 = 2.5

Since the empirical formula must be a whole number, you must choose a factor to multiply them by that will give a whole number.

1 x 2 = 2

2.5 x 2 = 5

\(P_4O_{10}\) → \(P_2O_5\)

Determine the empirical formula of a compound that contains 10 g of hydrogen and 80 g of oxygen.

Find the atomic mass of oxygen and hydrogen

O = 16

H = 1

Divide the masses of each element by their atomic mass to find the number of moles.

80g ÷ 16g = 5 mol. of oxygen

10g ÷ 1g = 10 mol. of hydrogen

Divide the number of moles by the lowest figure to get the ratio.

5 ÷ 5 = 1

10 ÷ 5 = 2

Empirical formula = \(H_2O\)

0.273g of Mg is heated in a Nitrogen (\(N_2\)) environment. The product of the reaction has a mass of 0.378g . Calculate the empirical formula.

Find the mass percentage of the elements in the compound.

N = 0.3789 - 0.273g = 0.105g

N = (0.105 ÷ 0.378) x 100 = 27.77%

Mg = (0.273 ÷ 0.378) x 100 = 77.23%

Change the percent composition to grams.

27.77% → 27.77g

77.23% → 77.23g

Divide the percent compositions by their atomic mass.

N = 14g

27.77g ÷ 14g = 1.98 mol

Mg = 24.31g

77.23g ÷ 24.31g = 2.97 mol

Divide the number of moles by the smallest number.

1.98 ÷ 1.98 = 1

2.97 ÷ 1.98 = 1.5

Remember we need whole number ratios, choose a factor to multiply that will give a whole number.

1 x 2 = 2

1.5 x 2 = 3

Empirical formula = \(Mg_3N_2\) [Magnesium Nitride]

Determine the empirical formula of a compound that contains 85.7% carbon and 14.3% hydrogen.

% mass C = 85.7

% mass H = 14.3

Divide the percentages by the atomic mass.

C = 12

H = 1

85.7 ÷ 12 = 7.142 mol

14.3 ÷ 1 = 14.3 mol

Divide by the lowest number.

7.142 ÷ 7.142 = 1

14.3 ÷ 7.142 = 2

Empirical formula = \(CH_2\)

A substance has the empirical formula \(C_4H_{10}S\) and relative formula mass (Mr) of 180. What is its molecular formula?

Find the relative formula mass (Mr) of \(C_4H_{10}S\) (the empirical formula).

Ar of C = 12

Ar of H = 1

Ar of S = 32

Mr = (12 x 4) + (10 x 1) + 32 = 90

Divide the Mr of the molecular formula by the Mr of the empirical formula.

180 ÷ 90 = 2

The ratio between the Mr of the substance and the empirical formula is 2.

Multiply each number of elements by two.

(C4 x 2 H10 x 2 S1 x2)

Molecular formula = \(C_8H_{10}S_2\)

A substance has the empirical formula \(C_2H_6O\) and a molar mass of 46g.

Find the mass of one mole of the empirical formula.

(Carbon 12 x 2) + (Hydrogen 1 x 2) + (Oxygen 16) = 46g

The molar mass of the empirical formula and the molecular formula are the same. The molecular formula must be the same as the empirical formula.

Molecular formula = \(C_2H_6O\)