Question 910ee

Oct 4, 2016

Given 0.44g of sample on combustion produces 0.88g $C {O}_{2}$
We know molar mass of $C {O}_{2} = 44 \text{ g/mol}$ which contains 12g carbon.

So 0.88g $C {O}_{2}$ will contain

$\frac{0.88 \times 12}{44} g = 0.24 g \text{ } C$

Similarly molar mass of ${H}_{2} O = 18 \text{ g/mol}$ which contains 2g hydrogen.

So 0.36g ${H}_{2} O$ will contain

$\frac{2 \times 0.36}{18} g = 0.04 g \text{ } H$

So the remaing amount in 0.88g of sample $\left(0.88 - 0.24 - 0.04\right) g = 0.16 g$ must be Oxygen.

Given molar mass of sample $132 \text{ g/mol}$

Hence 1 mole or 132g of the sample will contain

C->(0.24xx132)/0.44g=72g=(72g)/(12g/"mol")=6" mol"

H->(0.04xx132)/0.44g=12g=(12g)/(1g/"mol")=12" mol"

O->(0.16xx132)/0.44g=48g=(48g)/(16g/"mol")=3" mol"#

Hence the molecular formula of the sample is-$\text{ } {C}_{6} {H}_{12} {O}_{3}$