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:

• A. 43 g
• B. 86 g
• C. 129 g
• D. 172 g
• E. 22 g

Answer 1

Answer: (B)

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 $=86g$ .