Question

an organic compound contain 39.5%of carbon 6.4%of hydrogen and rest of oxygen and the molar mass is 60 g / mol what art it's empirical and molecular formulas?​

Answer 1

Solution :-

Given :-

Percentage composition of carbon ( C ) = 39.5 %

Percentage composition of hydrogen ( H ) = 6.4 %

Percentage composition of oxygen ( O )

                      = 100 - ( 39.5 + 6.4 )

                      = 100 - 45.9

                      = 54.1 %

We know :-

Atomic mass of carbon = 12 g

Atomic mass of hydrogen = 1 g

Atomic mass of oxygen = 16 g

$\bullet \sf \ No \ of \ mol \ of \ carbon \ (C)= \dfrac{39.5}{12}= 3.29 \\\\\bullet \ No \ of \ mol \ of \ hydrogen \ (H)= \dfrac{6.4}{1}= 6.4 \\\\\bullet \ No \ of \ mol \ of \ oxygen \ (O) = \dfrac{54.1}{16}= 3 . 38$

Simplest ratio :-

$\longrightarrow \sf \dfrac{3.29}{3.29} \ : \ \dfrac{6.4}{3.29} \ : \ \dfrac{3.38}{3.29} \\\\\longrightarrow 1 \ : \ 2 \ : \ 1$

$\boxed{\bf Empirical \ formula = CH_2O}$

→ Empirical formula mass = 12 + ( 2 × 1 ) + 16

                                           = 12 + 2 + 16

                                           = 30

We know :-

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

Here,

$\sf n = \dfrac{Molar \ mass}{Empirical \ formula \ mass}$

$\longrightarrow \sf n= \dfrac{60}{30} \\\\\longrightarrow \sf n = 2$

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

                               = $\sf C_2H_4O_2$

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

Answer 2

Answer:

Solution :-

Given :-

Percentage composition of carbon ( C ) = 39.5 %

Percentage composition of hydrogen ( H ) = 6.4 %

Percentage composition of oxygen ( O )

= 100 - ( 39.5 + 6.4 )

= 100 - 45.9

= 54.1 %

We know :-

Atomic mass of carbon = 12 g

Atomic mass of hydrogen = 1 g

Atomic mass of oxygen = 16 g

\begin{gathered}\bullet \sf \ No \ of \ mol \ of \ carbon \ (C)= \dfrac{39.5}{12}= 3.29 \\\\\bullet \ No \ of \ mol \ of \ hydrogen \ (H)= \dfrac{6.4}{1}= 6.4 \\\\\bullet \ No \ of \ mol \ of \ oxygen \ (O) = \dfrac{54.1}{16}= 3 . 38\end{gathered}

∙ No of mol of carbon (C)=

12

39.5

=3.29

∙ No of mol of hydrogen (H)=

1

6.4

=6.4

∙ No of mol of oxygen (O)=

16

54.1

=3.38

Simplest ratio :-

\begin{gathered}\longrightarrow \sf \dfrac{3.29}{3.29} \ : \ \dfrac{6.4}{3.29} \ : \ \dfrac{3.38}{3.29} \\\\\longrightarrow 1 \ : \ 2 \ : \ 1\end{gathered}

⟶

3.29

3.29

:

3.29

6.4

:

3.29

3.38

⟶1 : 2 : 1

\boxed{\bf Empirical \ formula = CH_2O}

Empirical formula=CH

2

O

→ Empirical formula mass = 12 + ( 2 × 1 ) + 16

= 12 + 2 + 16

= 30

We know :-

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

Here,

\sf n = \dfrac{Molar \ mass}{Empirical \ formula \ mass}n=

Empirical formula mass

Molar mass

\begin{gathered}\longrightarrow \sf n= \dfrac{60}{30} \\\\\longrightarrow \sf n = 2\end{gathered}

⟶n=

30

60

⟶n=2

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

2

O)

= \sf C_2H_4O_2C

2

H

4

O

2

\boxed{\bf Molecular \ formula = C_2H_4O_2}

Molecular formula=C

2

H

4

O

2