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

altitude, density, pressure

air gets less dense the higher the altitude and also the air pressure is less as altitude rises.

so here we go

atomic mass

239 + 1 = y + 134 + 3 combine like terms

240 = y + 137             subtract 137 from both terms

240 - 137 = y

103 = y

atomic number

94 = z +54 + 0

94 = z + 54               subtract 54 from both sides.

40 = z

answer