(D) To find the empirical formula,we calculate the mole ratio of each element:$1$. Moles of $C = \frac{24 \ g}{12 \ g/mol} = 2 \ mol$$2$. Moles of $H

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(D) To find the empirical formula,we calculate the mole ratio of each element:$1$. Moles of $C = \frac{24 \ g}{12 \ g/mol} = 2 \ mol$$2$. Moles of $H = \frac{4 \ g}{1 \ g/mol} = 4 \ mol$$3$. Moles of $O = \frac{32 \ g}{16 \ g/mol} = 2 \ mol$Divide each by the smallest number of moles $(2)$:$C = \frac{2}{2} = 1$$H = \frac{4}{2} = 2$$O = \frac{2}{2} = 1$Thus,the empirical formula is $CH_2O$.

Class 11 Chemistry Some Basic Concepts of Chemistry Percentage composition and Molecular formula

Easy

$60 \ g$ of a compound on analysis gave $C = 24 \ g$,$H = 4 \ g$ and $O = 32 \ g$. Its empirical formula is

A
$C_2H_4O_2$
B
$C_2H_2O$
C
$CH_2O_2$
D
$CH_2O$

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Percentage composition and Molecular formula

$0.30 \ g$ of an organic compound containing $C$,$H$ and $O$ on combustion yields $0.44 \ g \ CO_2$ and $0.18 \ g \ H_2O$. If $1 \ mol$ of the compound weighs $60 \ g$,then the molecular formula of the compound is:

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$A$ $400 \ mg$ iron capsule contains $100 \ mg$ of ferrous fumarate,$(C_2H_2O_4)Fe$. The percentage of iron present in it is approximately $.......... \%$.

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In $CH_3CH_2Br$,the percentage of $Br$ is:

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Define empirical formula and molecular formula.

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The empirical formula of a compound is $CH_2$. If the molar mass of the compound is $42 \ g/mol$,what is its molecular formula?

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$60 \ g$ of a compound on analysis gave $C = 24 \ g$,$H = 4 \ g$ and $O = 32 \ g$. Its empirical formula is

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$0.30 \ g$ of an organic compound containing $C$,$H$ and $O$ on combustion yields $0.44 \ g \ CO_2$ and $0.18 \ g \ H_2O$. If $1 \ mol$ of the compound weighs $60 \ g$,then the molecular formula of the compound is:

$A$ $400 \ mg$ iron capsule contains $100 \ mg$ of ferrous fumarate,$(C_2H_2O_4)Fe$. The percentage of iron present in it is approximately $.......... \%$.

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