Question

What is the empirical formula of a compound containing $60\%$ Sulphur and $40\%$ Oxygen?

Answer 1

Step 1: Given,

• Amount of Sulphur $\text{(S)}=60\%$
• Amount of Oxygen $\text{(O)}=40\%$
• Molar mass of Sulphur $=32{\text{g mol}}^{-1}$
• Molar mass of Oxygen $=16{\text{g mol}}^{-1}$

Step 2: Mass of Sulphur and oxygen

• If percentages are given, then we are taking the total mass as $100\text{g}$.
• So, the mass of each element is as follows:
• Mass of Sulphur $=60\text{g}$
• Mass of Oxygen $=40\text{g}$

Step 3: Convert given masses of Sulphur and Oxygen into moles.

Formula used:

$\text{Moles =}\frac{\text{Given mass}}{\text{Molar mass}}$

$$\begin{gathered} \text{Moles of Sulphur =}\frac{\text{Given mass of Sulphur}}{\text{Molar mass of Sulphur}}=\frac{60\text{g}}{{\text{32 g mol}}^{-1}}=\text{1.875 mol} \\ \text{Moles of Oxygen =}\frac{\text{Given mass of Oxygen}}{\text{Molar mass of Oxygen}}=\frac{\text{40 g}}{{\text{16 g mol}}^{-1}}=\text{2.5 mol} \end{gathered}$$

Step 4: For mole ratio, divide each value of moles by the lowest number of moles calculated.

$$\begin{gathered} \text{For Sulphur =}\frac{1.875}{1.875}=1 \\ \text{For Oxygen =}\frac{2.5}{1.875}=1.33 \end{gathered}$$

Step 5: Converting the ratios into a simple whole number by multiplying by $\mathbf{3}$.

The ratio of Sulphur and Oxygen will be $3:4$.

${\text{Empirical formula = S}}_3{\text{O}}_4$

Therefore, the empirical formula of the compound is ${\mathbf{\text{S}}}_{\mathbf{3}}{\mathbf{\text{O}}}_{\mathbf{4}}$.