Chemistry Question on Laws of Chemical Combinations

Question

Complete combustion of 0.858 g of compound $X$ gives 2.63 g of $C{{O}_{2}}$ and 1.28 g of ${{H}_{2}}O$ . The lowest molecular weight which $X$ can have, is:

Answer 1

Solution

$%$ of $C=\frac{12}{44}\times \frac{2.63}{0.858}\times 100=83.5%$ $%$ if $H=\frac{2}{18}\times \frac{1.28}{0.858}\times 100=16.5%$
Element
%
Relative no. of atoms
Simplest ration
C
83.5
83.5/12=7
7/7=1
H
16.5
16.5/1=16.5
16.5/1 =2.35
$C:H=1:2.35$ Multiply by 6 to make whole number $6:14$ . Hence, empirical formula $={{C}_{6}}{{H}_{14}}$ The minimum value of $n=1$ Hence, its molecular formula is ${{C}_{6}}{{H}_{14}}$ and molecular weight $=86\text{ }g$ .