Question

0.29g of hydrocarbon with vapour density 29 when burnt completely in oxygen produces 448 ml of CO2 at stp. find empirical formula of hydrocarbon and molecular formula of hydrocarbon.

Answer 1

vapour density of hydrocarbon is 29
we know,
vapor density = molecular mass/2
molecular mass = 2 × vapor density = 2 × 29 = 58 g/mol

Given,
0.29 g of hydrocarbon produces 448ml of CO2 at STP.

mole of hydrocarbon = given weight/molecular weight
= 0.29/58 = 0.05

mole of CO2 = given volume/22400
= 448/22400 = 0.02

means 0.05 mole of hydrocarbon produces 0.02 mole of CO2 .---(1)

Now see the combustion reaction of hydrocarbon,

CxHy + (x + y/4)O2 ---->xCO2 + (y/2)H2O

here, 1 mole of hydrocarbon produces x mole of CO2,
so, 0.05 mole of hydrocarbon produces 0.05x mole of CO2.---(2)

Compare equation (1) and (2)

0.05x = 0.02
x = 2/5

Let y = 1, then, x/y = 2/5
then, C2H5 is the emperical formula of hydrocarbon .
weight of emperical formula = 2 × 12 + 5 = 29

molecular formula = n × emperical formula

Because, molecular weight = 58g/mol
so, 58 = n × 29
n = 2 , hence, molecular formula will be C4H10.