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

the only liquid elements at standard temperature and pressure are bromine (br) and mercury (hg). although, elements caesium (cs), rubidium (rb), francium (fr) and gallium (ga) become liquid at or just above room temperature.

explanation: