A 3.87mg sample of an organic compound gave 5.80mg CO² and 1.58mg h²O on combustion. Deduce the empirical formula of the compound.

C₃H₄.

Explanation:

The organic compound gives CO₂ and H₂O.

The no. of moles of CO₂ produced = mass/molar mass = 5.80 mg/44.0 g/mol = 0.1318 mmol.

Which is corresponding to 0.1318 mmol of C.

The no. of moles of H₂O produced = mass/molar mass = 1.58 mg/18.0 g/mol = 0.08778 mmol.

which corresponds to (0.08778 x 2) = 0.1756 mmoles of H.

The ratio of the number of moles of hydrogen to carbon in the composition of the compound will be 0.1756/0.1318 = 1.332 = 1 + 1/3 = 4/3.

Therefore, the empirical formula of the compound under consideration is C₃H₄.