Question

A sample of natural gas consisting of methane and ethylene was burnt in excess of oxygen yielding 14.5 g of carbon dioxide .Calculate the weight %of ethylene

Answer 1

Let x gm ethylene (C2H4) and y gm methane (CH4) is present in mixture.

combustion of ethylene....

$C_2H_4+3O_2\rightarrow 2CO_2+2H_2O$

here it is clear that , 1 mole of ethylene forms 2 mole of carbon dioxide.

or, 28g of ethylene forms 88g of carbon dioxide.

[ as we know, molecular weight of C2H4 = 28g/mol and molecular weight of CO2 is 44g/mol]

or, x gm of ethylene forms 88x/28 = 22x/7 gm of carbon dioxide.

similarly, combustion of methane...

$CH_4+2O_2\rightarrow CO_2+2H_2O$

1 mole of methane forms 1 mole of carbon dioxide.

so, 16g of methane forms 44g of carbon dioxide. [ as molecular weight of methane is 16g/mol]

or, y gm of methane forms 44y/16 gm = 22y/8 of carbon dioxide.

now, it is given that mixture yields 14.5g of carbon dioxide.

22x/7 + 22y/8 = 14.5

or, 22(8x + 7y)/56 = 14.5

or, (8x + 7y) = 14.5 × 28/11

or, (8x + 7y) = 29 × 28/11 = 36.9 ≈ 37

or, 8x + 7y = 16 + 21 = 8 × 2 + 7 × 3

comparing both sides,

x = 2 and y = 3

so, weight % of ethylene = x/(x + y) × 100

= 2/(2 + 3) × 100

= 40%