Question

0.22g of an organic compound CxHyO which occupied 112ml at ntp and on combustion gave 0.44g co2. The ratio of x to y in the compound (answer- 1:2)

Answer 1

0.22g of an organic compounds CxHyO which occupied 112ml at ntp.

so, number of moles of CxHyO = given volume/22400 mL

= 112mL/22400mL = 1/200 = 0.005

or, number of mole of CxHyO = given weight /molecular weight

or, 0.005 = 0.22/(12x + y + 16)

or, 12x + y + 16 = 220/5 = 44

or, 12x + y = 28 .......(1)

again according to question, on combustion gave 0.44g of CO2

combustion reaction is ....

$C_xH_yO+(x+\frac{y}{4}-1)O_2\rightarrow xCO_2+\frac{y}{2}H_2O$

here it is clear that 1 mole of CxHyO gave x mole of CO2

given, weight of CO2 = 0.44g

so, mole of CO2 = 0.44/44 = 0.01

so, 0.01 mole of CO2 is given by, 0.01/x mole of CxHyO.

so, 0.01/x = 0.22/(12x + y + 16)

or, 1/x = 22/(12x + y + 16)

or, 12x + y + 16 = 22x

or, y + 16 = 10x .......(2)

from equations (1) and (2),

x = 2 and y = 4

so, x : y = 1 : 2