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 (12)/(44)* 2.492=0.680~g`

mass of H in H₂O =

`\tt (2.1)/(18)* 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 (0.68)/(12)=0.056`

mol of H :

`\tt (0.072)/(1)=0.072`

mol of O :

`\tt (0.248)/(16)=0.0155`

divide by 0.0155(the lowest ratio)

C : H : O ⇒

`\tt (0.056)/(0.0155)/ (0.072)/(0.0155)/ (0.0155)/(0.0155)=3.6/ 4.6/ 1\\\\(11)/(3)/ (14)/(3)/ (3)/(3)=11:14:3`