Question

A compound was found to contain 10.06% carbon, 89.10% chlorine, and 0.84% hydrogen, by mass. If the molar mass of the compound was found to be 119.6 g/mol, its molecular formula will be _________.

Answer 1

Answer : The molecular formula of a compound is,
`CHCl_3`

Solution :

If percentage are given then we are taking total mass is 100 grams.

So, the mass of each element is equal to the percentage given.

Mass of C = 0.1006 g

Mass of H = 0.0084 g

Mass of Cl = 0.8910 g

Molar mass of C = 12 g/mole

Molar mass of H = 1 g/mole

Molar mass of Cl = 35.5 g/mole

Step 1 : convert given masses into moles.

Moles of C =
`\frac{\text{ given mass of C}}{\text{ molar mass of C}}= (0.1006g)/(12g/mole)=0.0084moles`

Moles of H =
`\frac{\text{ given mass of H}}{\text{ molar mass of H}}= (0.0084g)/(1g/mole)=0.0084moles`

Moles of Cl =
`\frac{\text{ given mass of Cl}}{\text{ molar mass of Cl}}= (0.8910g)/(35.5g/mole)=0.0251moles`

Step 2 : For the mole ratio, divide each value of moles by the smallest number of moles calculated.

For C =
`(0.0084)/(0.0084)=1`

For H =
`(0.0084)/(0.0084)=1`

For Cl =
`(0.0251)/(0.0084)=2.98\approx 3`

The ratio of C : H : Cl = 1 : 1 : 3

The mole ratio of the element is represented by subscripts in empirical formula.

The Empirical formula =
`C_1H_1Cl_3=CHCl_3`

The empirical formula weight = 1(12) + 1(1) + 3(35.5) = 119.5 gram/eq

Now we have to calculate the molecular formula of the compound.

Formula used :

`n=\frac{\text{Molecular formula}}{\text{Empirical formula weight}}`

`n=(119.6)/(119.5)=1`

Molecular formula =
`(CHCl_3)_n=(CHCl_3)_1=CHCl_3`

Therefore, the molecular of the compound is,
`CHCl_3`