Question

A compound contains 4.07% hydrogen 24.27% carbon and 71.65% chlorine. Its molar mase is 98.96g.what are its empirical molecular formulas?​

Answer 1

Answer:

step 1

Divide the given percentages of atoms with their molecular masses

H---4.07/1 =4.07

C---24.27/12 =2.02

Cl--71.65/35.5=2.01

step 2

divide all values with the lowest value obtained.

H---4.07/2.01=2

C---2.02/2.01=1

Cl---2.01/2.01=1

therefore the empirical formula is

CH

2

Cl

WEIGHT OF EMPIRICAL FORMULA=12+2+35.5=49.5

Given molecular weight =98.96 which is double of empirical weight .therefore molecular formula is C

2

H

4

Cl

2

Explanation:

i hope this will help you

Answer 2

Solution :-

Given :-

Percentage composition of carbon (C) = 24.27%

Percentage composition of hydrogen (H) = 4.07%

Percentage composition of chlorine (Cl) = 71.65%

We know :-

Atomic mass of carbon (C) = 12g

Atomic mass of hydrogen (H) = 1g

Atomic mass of chlorine (Cl) = 35.5g

$\bullet \sf \ No \ of \ moles \ of \ carbon \ (C) = \dfrac{24.27}{12} =2.022 \\\\\bullet \ No \ of \ moles \ of \ hydrogen \ (H) = \dfrac{4.07}{1} = 4.07 \\\\\bullet \ No \ of \ moles \ of \ chlorine \ (Cl) = \dfrac{71.65}{35.5}= 2.018$

Simplest ratio :-

$\longrightarrow \sf \dfrac{2.022}{2.018} \ : \ \dfrac{4.07}{2.018} \ : \ \dfrac{2.018}{2.018} \\\\\longrightarrow 1 \ : \ 2 \ : \ 1$

$\boxed{\bf Empirical \ formula = CH_2Cl}$

Empirical formula mass = 12 + 2 + 35.5

                                       = 49.5

We know :-

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

Here,

$\longrightarrow \sf n = \dfrac{Molar \ mass }{Empirical \ formula \ mass } \\\\\longrightarrow n = \dfrac{98.96}{49.5} \\\\\longrightarrow n = 1.99 \approx 2$

Molecular formula = $\sf 2 \times (CH_2Cl)$

$\boxed{\bf Molecular \ formula = C_2H_4Cl_2}$