The Mole and Stoichiometry: Question 10

Worked solution

Part (a): Empirical formula from percentage composition

Convert each percentage into moles of atoms using $\text{moles}=\dfrac{\text{percentage mass}}{A_r}$:

$\text{moles of C} = \frac{85.7}{12} = 7.14$

$\text{moles of H} = \frac{14.3}{1} = 14.3$

Divide both values by the smaller one ($7.14$) to find the simplest whole-number ratio:

$\text{C}:\text{H} = \frac{7.14}{7.14} : \frac{14.3}{7.14} = 1:2.00 = 1:2$

So the empirical formula is $\boxed{\text{CH}_2}$.

Part (b): Molecular formula from the relative molecular mass

First find the mass of the empirical formula unit:

$\text{empirical formula mass of CH}_2 = 12+(2\times1)=14$

Divide the relative molecular mass by the empirical formula mass to find how many empirical units make up one molecule:

$n = \frac{M_r(\text{molecule})}{\text{empirical formula mass}} = \frac{42}{14} = 3$

Multiply every subscript in the empirical formula by $3$:

$\text{molecular formula} = (\text{CH}_2)_3 = \text{C}_3\text{H}_6$

Check: $M_r(\text{C}_3\text{H}_6) = (3\times12)+(6\times1) = 36+6=42$, which matches the given relative molecular mass, and $\dfrac{36}{42}\times100=85.7\%$ carbon, $\dfrac{6}{42}\times100=14.3\%$ hydrogen, matching the given percentages.

Final answers

• (a) Empirical formula $=\boxed{\text{CH}_2}$
• (b) Molecular formula $=\boxed{\text{C}_3\text{H}_6}$