Question

An organic compound contain carbon =40 percent, hydrogen =6.67 percent and remain oxygen. If the molecular mass of the compound is 30,find its empirical formula and molecular formulae by ​

Answer 1

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Answer 2

Solution :-

Given :-

Percentage composition of carbon (C) = 40%

Percentage composition of hydrogen (H) = 6.67%

Percentage composition of oxygen (O) = 100 - ( 40 + 6.67 )

                                                                 = 100 - 46.67

                                                                 = 53.33%

We know :-

Atomic mass of carbon (C) = 12g

Atomic mass of hydrogen (H) = 1g

Atomic mass of oxygen (O) = 16g

$\bullet \sf \ No \ of \ moles \ of \ carbon \ (C) = \dfrac{40}{12} = 3.33 \\\\\bullet \ No \ of \ moles \ of \ hydrogen \ (H) = \dfrac{6.67}{1} = 6.67 \\\\\bullet \ No \ of \ moles \ of \ oxygen \ (O) = \dfrac{53.33}{16}= 3 .33$

Simplest ratio :-

$\longrightarrow \sf \dfrac{3.33}{3.33} \ : \ \dfrac{6.67}{3.33} \ : \ \dfrac{3.33}{3.33} \\\\\longrightarrow 1 \ : \ 2 : \ 1$

Empirical formula = $\sf CH_2O$

Empirical formula mass =  12 + 2 + 16

                                       = 30g

We know :-

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

Here :-

$\longrightarrow\sf n=\dfrac{Molar \ mass }{Empirical \ formula \ mass } \\\\\longrightarrow n = \dfrac{30}{30} \\\\\longrightarrow \bf n= 1$

$\underline{\boxed{\bf Molecular \ formula = CH_2O}}$