Question

An unknown liquid is composed of 34.31% c, 5.28% h, and 60.41% i. The molecular weight is 210.06 amu. What is the molecular formula of the compound.

Answer 1

In an unknown liquid, the percentage composition with respect to carbon, hydrogen and iodine is 34.31%, 5.28% and 60.41% respectively.

Let the mass of liquid be 100 g thus, mass of carbon, hydrogen and oxygen will be 34.31 g, 5.28 g and 60.41 g respectively.

To calculate molecular formula of compound, convert mass into number of moles as follows:

$n=\frac{m}{M}$

Molar mass of carbon, hydrogen and iodine is 12 g/mol, 1 g/mol and 126.90 g/mol.

Taking the ratio:

$C:H:I=n_{C}:n_{H}:n_{I}$

Putting the values,

$C:H:I=\frac{34.31 g}{12 g/mol}:\frac{5.28 g}{1 g/mol}:\frac{60.41 g}{126.90 g/mol}=6:11:1$

Thus, molecular formula of compound will be $C_{6}H_{11}I$.