Question

An unknown compound with a molar mass of 223.94 g/mol consists of 32.18% c, 4.50% h, and 63.32% cl. find the molecular formula for the compound.

Answer 1

The actual number of atoms of each element present in the molecule of the compound is represented by the formula known as molecular formula.

Molar mass of the unknown compound = 223.94 g/mol (given)

Mass of each element present in the unknown compound is determined as:

• Mass of carbon, $C$:

$\frac{32.18}{100}\times 223.94 = 72.06 g$

• Mass of hydrogen, $H$:

$\frac{4.5}{100}\times 223.94 = 10.08 g$

• Mass of chlorine, $Cl$:

$\frac{63.32}{100}\times 223.94 = 141.79 g$

Now, the number of each element in the unknown compound is determined by the formula:

$number of moles = \frac{given mass}{molar mass}$

• Number of moles of $C$:

$number of moles = \frac{72.06}{12} = 6.005 mole\simeq 6 mole$

• Number of moles of $H$:

$number of moles = \frac{10.08}{1} = 10.08 mole\simeq 10 mole$

• Number of moles of $Cl$

$number of moles = \frac{141.79}{35.5} = 3.99 mole\simeq 4 mole$

Dividing each mole with the smallest number of mole, to determine the empirical formula:

$C_{\frac{6}{4}}H_{\frac{10}{4}}Cl_{\frac{10}{4}}$

$C_{1.5}H_{2.5}Cl_{1}$

Multiplying with 2 to convert the numbers in formula into a whole number:

So, the empirical formula is $C_{3}H_{5}Cl_{2}$.

Empirical mass = $12\times 3+1\times 5+2\times 35.5 = 112 g/mol$

In order to determine the molecular formula:

n = $\frac{molar mass}{empirical mass}$

n = $\frac{223.94}{112} = 1.99 \simeq 2$

So, the molecular formula is:

$2\times C_{3}H_{5}Cl_{2} = C_{6}H_{10}Cl_{4}$