Cyanogen is a gas which contains 46.2% C and 53.8% N by mass. At a temperature of 25°C and a pressure of 750 mm Hg, 1.50 g of cyanogen occupies 0.714 L. What is the molecular formula of cyanogen?

Molecular formula of cyanogen is C₂N₂

Explanation:

We apply the ideal gases law to find out the mole of cyanogen

P. V = n. R. T

Firstly let's convert the pressure in atm, for R

750 mmHg = 0.986 atm

25°C + 273 = 298K

0.986 atm. 0.714L = n. 0.082 L. atm/mol. K. 298K

(0.986 atm. 0.714L) / (0.082 L. atm/mol. K. 298K) = n

0.0288 mol = n

Molar mass of cyanogen = mass / mol

1.50 g / 0.0288 mol = 52.02 g/m

Let's apply the percent, to know the quantity of atoms

100 g of compound contain 46.2 g of C and 53.8 g of N

52.02 g of compound contain:

(52.02. 46.2) / 100 = 24 g → 2 atoms of C

(52.02. 53.8) / 100 =.28 g → 2 atoms of N