0.07g of a hydride of carbon occupies 56.0cm\(^3\) at S.T.P. When vaporized and contains 14.29% by mass of hydrogen. The formula of the hydrocarbon is(C = 12, H = 1)
Hydrocarbon means the compound comprises Hydrogen and Carbon atoms.
% of H\(_2\) = 14.29
% of C = 100 - 14.29 = 85.71
Solving the empirical formula for the Hydrocarbon,
Atom of element C H
% by mass 85.71 14.29
Divide by their Atomic masses \(\frac{85.71}{12}\) \(\frac{14.29}{1}\)
Divide through by the smaller number \(\frac{7.1425}{7.1425}\) \(\frac{14.29}{7.1425}\)
Ratio 1 : 2
Empirical formula: CH\(_2\)
Recall that (Empirical formula)x n = Molecular mass
(CH\(_2\))n = Molecular mass
We need to get the molecular mass.To do that, we were given the:
mass, m = 0.07g and V = 56 cm\(^3\)
But Number of moles = \(\frac{volume}{Molar Volume}\) = \(\frac{56}{22,400}\) = 0.0025 mole
N.B : At stp, Molar volume of all gases = 22.4dm\(^3\) or 22,400cm\(^3\)
Thus, n = \(\frac{m}{M}\)
M = \(\frac{m}{n}\) = \(\frac{0.07}{0.0025}\) = 28.
(CH\(_2\))n = Molecular mass
(12 + (1 x 2))n = 28
14n = 28
n = \(\frac{28}{14}\) = 2
The formula for the hydrocarbon = (CH\(_2\))n = (CH\(_2\))\(_2\) = C\(_2\)H\(_4\)
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