Question

an organic compound contains carbon 53.1 8% hydrogen 15.55% and Nitrogen 31.27% % molar mass equal to 45 find empirical formula and molecular formula?​

Answer 1

GIVEN :-

• Mass percentage of carbon ( C ) = 53.18 %

• Mass percentage of hydrogen ( H ) = 15.55 %

• Mass percentage of nitrogen ( N ) = 31.27 %

• Molar mass = 45

TO FIND :-

• Empirical formula

• Molecular formula

SOLUTION :-

We know that,

Atomic mass of carbon ( C ) = 12 g

Atomic mass of hydrogen ( H ) = 1 g

Atomic mass of nitrogen ( N ) = 14 g

$\bullet \sf \ No \ of \ moles \ of \ carbon = \dfrac{53.18}{12}= 4.432 \\\\\bullet \sf \ No \ of \ moles \ of \ hydrogen =\dfrac{15.55}{1}= 15.55\\\\\bullet \sf \ No \ of \ moles \ of \ nitrogen = \dfrac{31.27}{14}=2.234$

Simplest ratio,

$\longrightarrow \sf \ \ \ \ C \ \ \ \ \ \ \ \ : \ \ \ \ H \ \ \ \ \ \ \ : \ \ \ \ \ \ \ N \\\\\longrightarrow \ \ \ \dfrac{4.432}{2.234} \ \ \ \ : \ \ \ \dfrac{15.55}{2.234}\ \ : \ \ \ \ \ \dfrac{2.234}{2.234} \\\\\longrightarrow \ \ \ \ \ 2 \ \ \ \ \ \ \ \ : \ \ \ \ 7 \ \ \ \ \ \ \ : \ \ \ \ \ 1$

$\boxed{\bf Empirical \ formula = C_{2}H_{7}N}$

Molecular mass of $\sf C_{2}H_{7}N$ = ( 12 × 2 ) + ( 1 × 7 ) ( 14 × 1 )

                                              = 24 + 7 + 14

                                              = 45

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

Here,

$\bf n = \dfrac{Molar \ mass }{Molecular \ mass}$

$\longrightarrow\sf Molecular \ formula = (C_{2}H_{7}N)\times \dfrac{45}{45}$

                                  $= \sf (C_{2}H_{7}N)\times 1$

                                   $=\boxed{ \bf C_{2}H_{7}N}$