Question

The principal component of mothballs is naphthalene, a compound with a molecular mass of about 130 amu, containing only carbon and hydrogen. A 3.000-mg sample of naphthalene burns to give 10.3 mg of CO2. Determine its empirical and molecular formulas.

Short answer

Empirical formula is C₅H₄.

Molecular Formula is C₁₀H₈.

Step-by-step solution

Data given 

Assume the combustion reaction as follows

C _xH _y + O₂ \( \to \) CO₂ + H₂O

3.00 mg 10.3 mg

3.00 mg = 0.003 g of C _xH _y

10.3 mg = 0.0103 g of CO2

Determine the required formula

0.0103 g of CO₂ x 1 mol CO₂ x 1 mol C x 12 g of C

44 g of CO₂ 1 mol CO₂ 1 mol C

= 0.00281 g C

C _xH _y = 0.003 g

Then as we know carbon amount, we can calculate H by subtraction.

H = 0.003 - 0.00281 = 1.9 x 10^-4 g of H

We will use the formula

Grams \( \to \) mols \( \to \) mol ratio \( \to \) empirical formula \( \to \) molecular formula

0.00281 g of C x 1 mol C = 2.3397 x 10 ^-4 mol C

12 g C

1.9 x 10^-4 g of H x 1 mol H = 1.8849 x 10 ^-4 mol H

1 g H

Mole ratio

1.25 mol C has to be whole number . So multiply it to get whole number.

2.3397 x 10 ^-4 mol C = 1.25 mol C x 4 = 5 mol C

1.8849 x 10 ^-4 mol

1.8849 x 10 ^-4 mol H = 1 mol H x 4 = 4 mol H

1..8849 x 10 ^-4 mol

So empirical formula is C₅H₄.

Molecular mass/ Empirical mass = 130/ 64.082 = 2

Molecular formula = C₁₀ H₈