Answer:
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
