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Q.
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:
$ % $ 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 $ .