Question

A compound contains 14.31 % sodium, 9.97% sulphur, 6.22% hydrogen and 69.5% oxygen. Calculate its empirical formula.

Answer 1

Step 1: Defining the empirical formula

The simplest ratio of atoms of different elements present in a molecule of the compound.

Step 2: Recalling the Atomic mass

• The atomic mass of sodium = 23u
• The atomic mass of sulphur = 32u
• The atomic mass of hydrogen = 1u
• The atomic mass of oxygen = 16u

Step 3: Identifying the relative number of moles

The relative number of moles = $\frac{\%\mathrm{of}\mathrm{composition}}{\mathrm{Atomic}\mathrm{mass}}$

The relative number of moles for sodium $=\frac{14.31}{23}=0.62$ mol

The relative number of moles for sulphur $=\frac{9.97}{32}=0.31$ mol

The relative number of moles for hydrogen $=\frac{6.22}{1}=6.22$ mol

The relative number of moles for oxygen $=\frac{69.5}{16}=4.34$ mol

Step 4: Identifying the simple ratio

To identify the simple ratio, divide each of the relative numbers of moles by the least value amongst them.

Simple ratio of sodium $=\frac{0.62}{0.31}=2$

Simple ratio of sulphur $=\frac{0.31}{0.31}=1$

Simple ratio of hydrogen $=\frac{6.22}{0.31}=20$

Simple ratio of oxygen $=\frac{4.34}{0.31}=14$

Step 5: Finding the empirical formula

From the simple ratio identified above, the empirical formula is ${\mathrm{Na}}_2{\mathrm{SH}}_{20}{\mathrm{O}}_{14}$