Question

The mass composition of a compound that assists in the coagulation of blood is 76.71% carbon, 7.02% hydrogen, and 16.27% nitrogen. Determine the empirical formula of the compound and report the answer by specifying X, Y & Z in the format below:
C_X H_Y N_Z

Answer 1

Answer : The empirical of the compound is, $C_{11}H_{12}N_2$

Explanation :

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 = 76.71 g

Mass of H = 7.02 g

Mass of N = 16.27 g

Molar mass of C = 12 g/mole

Molar mass of H = 1 g/mole

Molar mass of N = 14 g/mole

Step 1 : convert given masses into moles.

Moles of C = $\frac{\text{ given mass of C}}{\text{ molar mass of C}}= \frac{76.71g}{12g/mole}=6.39moles$

Moles of H = $\frac{\text{ given mass of H}}{\text{ molar mass of H}}= \frac{7.02g}{1g/mole}=7.02moles$

Moles of N = $\frac{\text{ given mass of N}}{\text{ molar mass of N}}= \frac{16.27g}{14g/mole}=1.16moles$

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

For C = $\frac{6.39}{1.16}=5.5$

For H = $\frac{7.02}{1.16}=6.0\approx 6$

For N = $\frac{1.16}{1.16}=1$

The ratio of C : H : N = 5.5 : 6 : 1

To make in a whole number we are multiplying the ratio by 2, we get:

The ratio of C : H : N = 11 : 12 : 2

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

The Empirical formula = $C_{11}H_{12}N_2$

Therefore, the empirical of the compound is, $C_{11}H_{12}N_2$