# The mass fractions of carbon, hydrogen and oxygen in aldehydes are 66.67%, 11.11% and 22.22%, respectively.

The mass fractions of carbon, hydrogen and oxygen in aldehydes are 66.67%, 11.11% and 22.22%, respectively. How much hydrogen should be used to hydrogenate this 216g aldehyde? to the corresponding alcohol?

Given:
ω (C) = 66.67%
ω (H) = 11.11%
ω (O) = 22.22%
m (CxHyOz) = 216 g

To find:
CxHyOz -?
V (H2) -?

1) m (C) = ω (C) * m (CxHyOz) / 100% = 66.67% * 216/100% = 144.007 g;
2) n (C) = m / M = 144.007 / 12 = 12.001 mol;
3) m (H) = ω (H) * m (CxHyOz) / 100% = 11.11% * 216/100% = 23.998 g;
4) n (H) = m / M = 23.998 / 1 = 23.998 mol;
5) m (O) = ω (O) * m (CxHyOz) / 100% = 22.22% * 216/100% = 47.995 g;
6) n (O) = m / M = 47.995 / 16 = 3 mol;
7) x: y: z = n (C): n (H): n (O) = 12.001: 23.998: 3 = 4: 8: 1;
8) C4H8O + H2 => C4H10O;
9) n (C4H8O) = m / M = 216/72 = 3 mol;
10) n (H2) = n (C4H8O) = 3 mol;
11) V (H2) = n * Vm = 3 * 22.4 = 67.2 liters.

Answer: Unknown substance – C4H8O; volume H2 – 67.2 liters.

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