Question

a composition of a organic compound found to contain 39.9%of carbon67%of hydrogen and remaining being oxygen the molar mass of the compound is 180 determine its empirical formula

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Answer 1

Correct question :-

A composition of a organic compound found to contain 39.9% of carbon, 6.7% of hydrogen  and remaining being oxygen. The molar mass of the compound is 180. Determine its empirical formula and molecular formula.

Solution :-

Given :-

Percentage composition of carbon (C) = 39.9%

Percentage composition of hydrogen (H) = 6.7%

Percentage composition of oxygen (O) = 100 - ( 39.9 + 6.7 )

                                                                 = 100 - 46.6

                                                                 = 53.4%$\bullet \sf \ No \ of \ moles \ of \ carbon \ (C) = \dfrac{39.9}{12} = 3.325 \\\\\bullet \ No \ of \ moles \ of \ hydrogen \ (H) = \dfrac{6.7} {1} = 6.7 \\\\\bullet \ No \ of \ moles \ of \ oxygen \ (O) = \dfrac{53.4}{16} = 3.338$

Simplest ratio :-

$\longrightarrow \sf \dfrac{3.325}{3.325} \ : \ \dfrac{6.7}{3.325} \ : \ \dfrac{3.338}{3.325} \\\\\longrightarrow 1 \ : \ 2 \ : 1$

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

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

                                       = 12 + 2 + 16

                                       = 30

We know :-

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

Here,

$\longrightarrow \sf n = \dfrac{Molar \ mass }{Empirical \ formula \ mass} \\\\\longrightarrow n = \dfrac{180}{30} \\\\\longrightarrow n = 6$

Molecular formula = $\sf 6 \times (CH_2O)$

$\boxed{\bf Molecular \ formula = C_6H_{12}O_6}$