Question

A 100g of sample of a compound is combusted in excess oxygen and the products are 2.492g of CO2 and 0.6495 of H2O. Determine the empirical formula of the compound?

Answer 1

The empirical formula : C₁₁O₁₄O₃

Further explanation

The assumption of the compound consists of C, H, and O

mass of C in CO₂ =

$\tt \dfrac{12}{44}\times 2.492=0.680~g$

mass of H in H₂O =

$\tt \dfrac{2.1}{18}\times 0.6495=0.072~g$

mass of O :

mass sample-(mass C + mass H)

$\tt 1-(0.68+0.072)=0.248`g$

mol of  C :

$\tt \dfrac{0.68}{12}=0.056$

mol of H :

$\tt \dfrac{0.072}{1}=0.072$

mol of O :

$\tt \dfrac{0.248}{16}=0.0155$

divide by 0.0155(the lowest ratio)

C : H : O ⇒

$\tt \dfrac{0.056}{0.0155}\div \dfrac{0.072}{0.0155}\div \dfrac{0.0155}{0.0155}=3.6\div 4.6\div 1\\\\\dfrac{11}{3}\div \dfrac{14}{3}\div \dfrac{3}{3}=11:14:3$