Question

The empirical formulae and molar masses of three compounds, $A$ , $B$ and $C$ , are shown in table. Calculate the molecular formula of each of these compounds. ( $2$ values: $C = 12.0, Cl = 35.5, H = 1.0$ ).

Compound	Empirical formula	$M_{r}$
$A$	$C_{3} H_{5}$	$82$
$B$	${CCl}_{3}$	$237$
$C$	${CH}_{2}$	$112$

Compound	Empirical formula	$M_{r}$
$A$	$C_{3} H_{5}$	$82$
$B$	${CCl}_{3}$	$237$
$C$	${CH}_{2}$	$112$

Accepted Answer

Molecular formulae show the simple numbers of each type of atom in a molecule.

Steps to find molecular formula:

First divide the molar mass of the given compound by the empirical formula mass, we will get the results those should be a whole number or very close to a whole number.

Now, multiply all the subscripts in the empirical formula by the whole number found in step in the first step. Then resulted formula is the molecular formula. The molar mass of the molecular formula must be match with the compound mass.

$A$ .

We have given empirical formula, $C_{3} H_{5}$ and, the molar mass, $82$

Empirical formula mass of $C_{3} H_{5} = (3 \times 12) + (5 \times 1) = 41$

Divide molar mass by empirical formula mass, $\frac{82}{41} = 2$

Multiply all the subscripts in the empirical formula by the whole number found in step in the first step i.e with $C_{3} H_{5}$

$C_{3} H_{5} \times 2 = C_{6} H_{10}$

So, the molecular for given empirical formula, $C_{3} H_{5}$ is ${CH}_{2}$ (The molar mass of the molecular formula matches with the compound mass)

$B$ .

We have given empirical formula, ${CCl}_{3}$ and, the molar mass, $237$

Empirical formula mass of ${CCl}_{3} = (2 \times 12) + (3 \times 35.5) = 130.5$

Divide molar mass by empirical formula mass, $\frac{237}{130.5} = 1.82 ≃ 2.0$

Multiply all the subscripts in the empirical formula by the whole number found in step in the first step i.e with $2$

${CCl}_{3} \times 2 = C_{2} {Cl}_{6}$

So, the molecular for given empirical formula, ${CCl}_{3}$ is $C_{2} {Cl}_{6}$ (The molar mass of the molecular formula matches with the compound mass)

$C$ .

We have given empirical formula, ${CH}_{2}$ and, the molar mass, $112$

Empirical formula mass of ${CH}_{2} = (1 \times 12) + (2 \times 1) = 14$

Divide molar mass by empirical formula mass, $\frac{112}{14} = 8$

Multiply all the subscripts in the empirical formula by the whole number found in step in the first step i.e with $8$

${CH}_{2} \times 8 = C_{8} H_{16}$

So, the molecular for given empirical formula, ${CH}_{2}$ is $C_{8} H_{16}$ (The molar mass of the molecular formula matches with the compound mass)

Lawrie Ryan and Roger Norris Solutions for Chapter: Atoms, Molecules and Stoichiometry, Exercise 14: Question

Lawrie Ryan Chemistry Solutions for Exercise - Lawrie Ryan and Roger Norris Solutions for Chapter: Atoms, Molecules and Stoichiometry, Exercise 14: Question

Questions from Lawrie Ryan and Roger Norris Solutions for Chapter: Atoms, Molecules and Stoichiometry, Exercise 14: Question with Hints & Solutions

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