Question

An organic compound, whose vapor density is 94, contains C=12.66%,H=2.13%and Br=85.11%.find the molecular formula.
(given, Atomic masses:C=12,H=1 and Br=80)​

Answer 1

Solution :-

Given :-

Percentage composition of carbon (C) = 12.66%

Percentage composition of hydrogen (H) = 2.13%

Percentage composition of bromine (Br) = 85.11%

We know :-

Atomic mass of carbon (C) = 12g

Atomic mass of hydrogen (H) = 1g

Atomic mass of bromine (Br) = 80g

$\bullet \sf \ No \ of \ moles \ of \ carbon \ (C) = \dfrac{12.66}{12}= 1.055\\\\\bullet \ No \ of \ moles \ of \ hydrogen \ (H) = \dfrac{2.13}{1} = 2.13 \\\\\bullet \ No \ o f \ moles \ of \ bromine \ (Br) = \dfrac{85.11}{80}= 1.063$

Simplest ratio :-

$\longrightarrow \sf \dfrac{1.055}{1.055} \ : \ \dfrac{2.13}{1.055} \ : \ \dfrac{1.063}{1.055} \\\\\longrightarrow 1 \ : \ 2 \ : \ 1$

Empirical formula = $\sf CH_2Br$

Empirical formula mass = 12 + 2 + 80

                                       = 94g

Molecular mass = 2 × Vapor density

                           = 2 × 94

                           = 188g

We know :-

$\bf Molecular \ formula = n \times Empirical \ formula$

Here,

$\longrightarrow \sf n = \dfrac{Molecular \ mass }{Empirical \ formula \ mass } \\\\\longrightarrow n = \dfrac{188}{94} \\\\\longrightarrow n = 2$

Molecular formula = $\sf 2 \times CH_2Br$

                              = $\sf C_2H_4Br_2$

$\boxed{\bf Molecular \ formula = C_2H_4Br_2 }$