Derive the formula for a substance, the mass fraction of carbon in which is 82.74%
Derive the formula for a substance, the mass fraction of carbon in which is 82.74%, and hydrogen is 17.25%, the vapor density of the substance for hydrogen is 29.
Given:
w% (C) = 82.74%
w% (H) = 17.26%
D (H2) = 29
To find:
Formula -?
Decision:
1) Find the molecular weight of the hydrocarbon by its hydrogen density:
D (H2) = Mr (UV): Mr (H2)
Mr (HC) = D (H2) * Mr (H2) = 29 * 2 = 58.
2) Knowing the mass fraction of carbon, we find its molecular weight in the composition and the number of atoms:
58 – 100%
x – 82.74%,
then x = 82.74% * 58: 100% = 48
N (C) = 48: Mr (C) = 48: 12 = 4 atoms.
3) Then the molecular weight of hydrogen in the composition is 58 – 48 = 10.
N (H) = 10: Mr (H) = 10: 1 = 10 atoms.
We got a substance with the formula C4H10 – butane.
Answer: C4H10 – butane.