Question

A compound contains 37.5% of carbon,12.5% of hydrogen and rest of oxygen. calculate its empirical formula. if its molar mass is 32, find its molecular formula.
please solve this question ​

Answer 1

Solution :-

Given :-

Percentage composition of carbon (C) = 37.5%

Percentage composition of hydrogen (H) = 12.5%

Percentage composition of oxygen (O) = 100 -  ( 37.5 + 12.5 )

                                                                 = 100 - 50

                                                                 = 50%

We know :-

Atomic mass of carbon (C) = 12g

Atomic mass of hydrogen (H) = 1g

Atomic mass of oxygen (O) = 16g

$\bullet \sf \ No \ of \ moles \ of \ carbon \ (C) = \dfrac{37.5}{12} = 3.125 \\\\ \bullet \ No \ of \ moles \ of \ hydrogen \ (H) = \dfrac{12.5}{1} = 12.5 \\\\\bullet \ No \ of \ moles \ of \ oxygen \ (O) = \dfrac{50}{16} = 3.125$

Simplest ratio :-

$\longrightarrow \sf \dfrac{3.125}{3.125} \ : \ \dfrac{12.5}{3.125} \ : \ \dfrac{3.125}{3.125} \\\\\longrightarrow 1 \ : \ 4 \ : \ 1$

$\boxed{\bf Empirical \ formula= CH_4O}$

Empirical formula mass = 12 + ( 1 × 4 ) + 16

                                        = 12 + 4 + 16

                                         = 32

We know :-

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

Here,

$\longrightarrow \sf n =\dfrac {Molar \ mass}{Empirical \ formula \ mass} \\\\\longrightarrow n = \dfrac{32}{32} \\\\\longrightarrow n = 1$

Molecular formula = $\sf 1 \times (CH_4O)$

$\boxed{\bf Molecular \ formula = CH_4O}$

Answer 2

the compound contains 37.5% of carbon,12.5% of hydrogen and rest is oxygen.

empirical formula. its molar mass is 32, its molecular formula.