C(65.5%), H(5.5%), O(29%)
Molar weights of the elements are:
m(C) = 12, m(H) = 1, m(O) = 16.
65.5 : 5.5 : 29 = 12 x : 1 y : 16 z
65.5/5.5 = 12x/y, x/y = 0.99 x = y
29/5.5 = 16z/y, z/y = 0.33 z = 1/3 y
65/29 = 12x / 16z, x/z = 3 x = 3z
Thus we can find that if z = 1, x = y = 3, the empirical formula would be C3H3O.
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