Question

17. 1.42 g of a compound containing carbon, hydrogen and oxygen (CXHYOZ) is subjected to combustion analysis. The results show that 1.95 g of CO2 and 1.68 g of H2O were produced. What is the empirical formula for the compound? If the molecular weight of the compound is 160 g/mol, what is the molecular formula of the compound?

Answer 1

Given:

weight of compound = 1.42gm

CO2 produced after combusition = 1.95gm

H2O produced after combusition = 1.68gm

Molecular weight of compound = 160g/mol

Calculate the mass of atoms involved in chemical reaction

C= 12 O=16, H= 1

molecular weight weight of CO2 = 44.0 and for H2O= 18

From above data we can calculate the moles of product generated

1.95 / 44 = 0.044 mole of CO2

1.68/ 18 = 0.0933 moles of H2O

1.42/ 160 = 0.008875 mole of compound

If we calculate the molar ratio to organic compound

we found

0.044 / 0.008 = 5.

It means that five molar equivalent of CO2 generated during combusion that is five carbons are in organic compound.

Similarlly

If we calculate the molar ratio of H2O to organic compound we found

0.0933 / 0.0088 = 10.60

It means that 10.60 molar equivalent of H2O generated during combusion. that is 10 equiv H present.

The partial formuila is,

C₅H₁₀ = (12 x 5) + (1 x 10) = 70

Now the Remaining mass contributed by an Oygen

Therefore 160-70 = 90 = 6 Oxygen atoms

Hence The Empirical formula of compound would be C₅H₁₀O₆