The oxidizing density of the hydrocarbon for oxygen is 1.25. Mass fraction of carbon in it is 90%

The oxidizing density of the hydrocarbon for oxygen is 1.25. Mass fraction of carbon in it is 90%. Derive the molecular formula of the hydrocarbon.

Given:
D O2 (CxHy) = 1.25
ω (C) = 90%

To find:
CxHy -?

1) Let m (CxHy) = 100 g;
2) m (C) = ω (C) * m (CxHy) / 100% = 90% * 100/100% = 90 g;
3) n (C) = m (C) / M (C) = 90/12 = 7.5 mol;
4) m (H) = m (CxHy) – m (C) = 100 – 90 = 10 g;
5) n (H) = m (H) / M (H) = 10/1 = 10 mol;
6) x: y = n (C): n (H) = 7.5: 10 = 1: 1.333 = 3: 4;
The simplest formula for an unknown substance is C3H4;
7) M (CxHy) = D O2 (CxHy) * M (O2) = 1.25 * 32 = 40 g / mol;
8) M (C3H4) = Mr (C3H4) = Ar (C) * N (C) + Ar (H) * N (H) = 12 * 3 + 1 * 4 = 40 g / mol;
9) M (CxHy) = M (C3H4);
Unknown substance – C3H4 – propadiene.

Answer: Unknown substance – C3H4 – propadiene.