Question

the percentage composition of organic compound is found to contain 39.9are carbon 6.97 of hydrogen the rest is oxygen.if the molecular mass is 60. find the molecular formula of the compound

Answer 1

Solution :-

Given :-

Percentage composition of carbon (C) = 39.9%

Percentage composition of hydrogen (H) = 6.97%

Percentage composition of oxygen (O) = 100 - ( 39.9 + 6.97 )

                                                                 = 100 - 46.87

                                                                 = 53.13

We know :-

Atomic mass of carbon (C) = 12g

Atomic mass of hydrogen (H) = 1g

Atomic mass of oxygen (O) = 16g

$\bullet \sf \ No \ of \ moles \ of \ carbon \ (C) = \dfrac{39.9}{12} = 3.325 \\\\\bullet \ No \ of \ moles \ of \ hydrogen \ (H) = \dfrac{6.97}{1} = 6.97 \\\\\bullet \ No \ of \ moles \ of \ oxygen \ (O) = \dfrac{53.13}{16} = 3.32$

Simplest ratio :-

$\longrightarrow \sf \dfrac{3.325}{3.32} \ : \ \dfrac{6.97}{3.32} \ : \ \dfrac{3.32}{3.32} \\\\\longrightarrow 1 \ : \ 2 \ : \ 1$

Empirical formula = $\sf CH_2O$

Empirical formula mass = 12 + 2 + 16

                                       = 30

We know :-

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

Here :-

$\longrightarrow \sf n = \dfrac{Molecular \ mass }{Empirical \ formula \ mass } \\\\\longrightarrow n = \dfrac{60}{30} \\\\\longrightarrow \bf n = 2$

Molecular formula = $\sf 2\times (CH_2O)$

$\boxed{\bf Molecular \ formula = C_2H_4O_2}$

Answer 2

Answer:

Given :-

Percentage composition of carbon (C) = 39.9%

Percentage composition of hydrogen (H) = 6.97%

Percentage composition of oxygen (O) = 100 - ( 39.9 + 6.97 )

= 100 - 46.87

= 53.13

We know :-

Atomic mass of carbon (C) = 12g

Atomic mass of hydrogen (H) = 1g

Atomic mass of oxygen (O) = 16g

\begin{gathered}\bullet \sf \ No \ of \ moles \ of \ carbon \ (C) = \dfrac{39.9}{12} = 3.325 \\\\\bullet \ No \ of \ moles \ of \ hydrogen \ (H) = \dfrac{6.97}{1} = 6.97 \\\\\bullet \ No \ of \ moles \ of \ oxygen \ (O) = \dfrac{53.13}{16} = 3.32\end{gathered}

∙ No of moles of carbon (C)=

12

39.9

=3.325

∙ No of moles of hydrogen (H)=

1

6.97

=6.97

∙ No of moles of oxygen (O)=

16

53.13

=3.32

Simplest ratio :-

\begin{gathered}\longrightarrow \sf \dfrac{3.325}{3.32} \ : \ \dfrac{6.97}{3.32} \ : \ \dfrac{3.32}{3.32} \\\\\longrightarrow 1 \ : \ 2 \ : \ 1\end{gathered}

⟶

3.32

3.325

:

3.32

6.97

:

3.32

3.32

⟶1 : 2 : 1

Empirical formula = \sf CH_2OCH

2

O

Empirical formula mass = 12 + 2 + 16

= 30

We know :-

\bf Molecular \ formula = n \times Empirical \ formulaMolecular formula=n×Empirical formula

Here :-

\begin{gathered}\longrightarrow \sf n = \dfrac{Molecular \ mass }{Empirical \ formula \ mass } \\\\\longrightarrow n = \dfrac{60}{30} \\\\\longrightarrow \bf n = 2\end{gathered}

⟶n=

Empirical formula mass

Molecular mass

⟶n=

30

60

⟶n=2

Molecular formula = \sf 2\times (CH_2O)2×(CH

2

O)

\boxed{\bf Molecular \ formula = C_2H_4O_2}

Molecular formula=C

2

H

4

O

2