Question

A compound contains 4.07% hydrogen, 24.27% carbon and 71.65% chlorine. Its molar mass is 98.96 g. What are its empirical and molecular formulas?

Answer 1

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$\star \: \bf\small{Step\: 1.}$ Conversion of mass per cent to grams :

Since we are having mass per cent, it is convenient to use 100 g of the compound as the starting material. Thus, in the  100 g sample of the above compound, 4.07 g hydrogen, 24.27 g carbon and 71.65 g chlorine are present.

$\star\: \bf\small{Step \:2.}$ Convert into number moles of each element :

Divide the masses obtained above by respective atomic masses of various elements. This gives the number of moles of constituent elements in the compound.

• Moles of Cl = $\dfrac{\sf 71.65\:g}{\sf 35.435\:g}$ = 2.021

• Moles of H = $\dfrac{\sf 4.07\:g}{\sf 1.008\:g}$ = 4.04

• Moles of C = $\dfrac{\sf 24.27\:g}{\sf 12.01\:g}$ = 2.021

$\star \: \bf\small{Step \:3.}$ Divide each of the mole values obtained above by the smallest number amongst them :

Since 2.021 is smallest value, division by it gives a ratio of 2:1:1 for H:C:Cl

In case the ratios are not whole numbers, then they may be converted into whole number by multiplying by the suitable coefficient.

$\star \: \bf\small{Step \:4.}$ Write down the empirical formula by mentioning the numbers after writing the symbols of respective elements :

CH₂Cl is the Empirical formula of the above compound.

$\star \: \bf\small{Step \:5.}$ Writing molecular formula :

(a) Determine empirical formula mass by  adding the atomic masses of various atoms present in the empirical formula -

For CH₂Cl, Empirical formula mass is :

12.01 + (2 × 1.008) + 35.453  = 49.48 g

(b) Divide Molar mass by empirical formula mass -

$\dfrac{\sf Molecular \:mass}{\sf Empirical \:formula\: mass} = \dfrac{\sf 98.96\:g}{\sf 49.48\:g}$

(c) Multiply empirical formula by n obtained above to get the molecular formula -

Empirical formula = CH₂Cl, n = 2.

Hence molecular formula is C₂H₄Cl₂

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Answer 2

Answer:

C₂H₄Cl₂

Explanation:

Conversion of mass per cent to grams :

Since we are having mass per cent, it is convenient to use 100 g of the compound as the starting material. Thus, in the  100 g sample of the above compound, 4.07 g hydrogen, 24.27 g carbon and 71.65 g chlorine are present.

Convert into number moles of each element :

Divide the masses obtained above by respective atomic masses of various elements. This gives the number of moles of constituent elements in the compound.

• Moles of Cl =  = 2.021

• Moles of H =  = 4.04

• Moles of C =  = 2.021

Divide each of the mole values obtained above by the smallest number amongst them :

Since 2.021 is smallest value, division by it gives a ratio of 2:1:1 for H:C:Cl

In case the ratios are not whole numbers, then they may be converted into whole number by multiplying by the suitable coefficient.

Write down the empirical formula by mentioning the numbers after writing the symbols of respective elements :

CH₂Cl is the Empirical formula of the above compound.

Writing molecular formula :

(a) Determine empirical formula mass by  adding the atomic masses of various atoms present in the empirical formula -

For CH₂Cl, Empirical formula mass is :

12.01 + (2 × 1.008) + 35.453  = 49.48 g

(b) Divide Molar mass by empirical formula mass -

(c) Multiply empirical formula by n obtained above to get the molecular formula -

Empirical formula = CH₂Cl, n = 2.

Hence molecular formula is C₂H₄Cl₂