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1 November, 16:04

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?

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  1. 1 November, 18:17
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    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
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