Question

An organic compound contains 40.65% carbon ,8.55% hydrogen & 23.77% nitrogen. Its vapour density is 29.5 Find its molecular formula.

Answer 1

molecular formula = C2H5N1O1

Answer 2

Answer : The molecular of the compound is, $C_2H_5NO$

Solution : Given,

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 = 40.65 g

Mass of H = 8.55 g

Mass of N = 23.77 g

Mass of O = 100 - 72.97 = 27.03 g

Molar mass of C = 12 g/mole

Molar mass of H = 1 g/mole

Molar mass of N = 14 g/mole

Molar mass of O = 16 g/mole

Step 1 : convert given masses into moles.

Moles of C = $\frac{\text{ given mass of C}}{\text{ molar mass of C}}= \frac{40.65g}{12g/mole}=3.387moles$

Moles of H = $\frac{\text{ given mass of H}}{\text{ molar mass of H}}= \frac{8.55g}{1g/mole}=8.55moles$

Moles of N = $\frac{\text{ given mass of N}}{\text{ molar mass of N}}= \frac{23.77g}{14g/mole}=1.69moles$

Moles of O = $\frac{\text{ given mass of O}}{\text{ molar mass of O}}= \frac{27.03g}{16g/mole}=1.69moles$

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

For C = $\frac{3.387}{1.69}=2.004\approx 2$

For H = $\frac{8.55}{1.69}=5.059\approx 5$

For N = $\frac{1.69}{1.69}=1$

For O = $\frac{1.69}{1.69}=1$

The ratio of C : H : N : O = 2 : 5 : 1 : 1

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

The Empirical formula = $C_2H_5N_1O_1$

The empirical formula weight = 12(2) + 5(1) + 1(14) + 1(16) = 59 gram/eq

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

Formula used :

$n=\frac{\text{Molecular formula}}{\text{Empirical formula weight}}\\\\n=\frac{2\times \text{Vapor density}}{\text{Empirical formula weight}}=\frac{2\times 29.5}{59}=1$

Molecular formula = $(C_2H_5N_1O_1)_n=(C_2H_5N_1O_1)_1=C_2H_5N_1O_1=C_2H_5NO$

Therefore, the molecular of the compound is, $C_2H_5NO$