Question

The reaction of equal molar amounts of benzene, C6H6, and chlorine, Cl2, carried out under special conditions, yields a gas and a clear liquid. Analysis of the liquid shows that it contains 64.03% carbon, 4.48% hydrogen, and 31.49% chlorine by mass and that is has a molar mass of 112.5 g/mol. The molecular formula will be determined. First, determine the number of moles of carbon in a 100 g sample.

Answer 1

Answer : The molecular formula of a compound is, $C_6H_5Cl_1$

Solution : Given,

Mass of C = 64.03 g

Mass of H = 4.48 g

Mass of Cl = 31.49 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 = $\frac{\text{ given mass of C}}{\text{ molar mass of C}}= \frac{64.03g}{12g/mole}=5.34moles$

Moles of H = $\frac{\text{ given mass of H}}{\text{ molar mass of H}}= \frac{4.48g}{1g/mole}=4.48moles$

Moles of Cl = $\frac{\text{ given mass of Cl}}{\text{ molar mass of Cl}}= \frac{31.49g}{35.5g/mole}=0.887moles$

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

For C = $\frac{5.34}{0.887}=6.02\approx 6$

For H = $\frac{4.48}{0.887}=5.05\approx 5$

For Cl = $\frac{0.887}{0.887}=1$

The ratio of C : H : Cl = 6 : 5 : 1

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

The Empirical formula = $C_6H_5Cl_1$

The empirical formula weight = 6(12) + 5(1) + 1(35.5) = 112.5 gram/eq

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

Formula used :

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

$n=\frac{112.5}{112.5}=1$

Molecular formula = $(C_6H_5Cl_1)_n=(C_6H_5Cl_1)_1=C_6H_5Cl_1$

Therefore, the molecular of the compound is, $C_6H_5Cl_1$