Set the formula of the substance mass fraction of carbon 26.67% hydrogen 2.22% oxygen 71.11%

Set the formula of the substance mass fraction of carbon 26.67% hydrogen 2.22% oxygen 71.11% The molar mass of the substance is 90 g mol.

Given:
CxHyOz
ω (C) = 26.67%
ω (H) = 2.22%
ω (O) = 71.11%
M (CxHyOz) = 90 g / mol

To find:
CxHyOz -?

Solution:
1) Let m (CxHyOz) = 100 g;
2) m (C) = ω (C) * m (CxHyOz) / 100% = 26.67% * 100/100% = 26.67 g;
3) n (C) = m (C) / M (C) = 26.67 / 12 = 2.223 mol;
4) m (H) = ω (H) * m (CxHyOz) / 100% = 2.22% * 100/100% = 2.22 g;
5) n (H) = m (H) / M (H) = 2.22 / 1 = 2.22 mol;
6) m (O) = ω (O) * m (CxHyOz) / 100% = 71.11% * 100/100% = 71.11 g;
7) n (O) = m (O) / M (O) = 71.11 / 16 = 4.444 mol;
8) x: y: z = n (C): n (H): n (O) = 2.223: 2.22: 4.444 = 1: 1: 2;
The simplest formula is CHO2;
9) M (CHO2) = Mr (CHO2) = Ar (C) * N (C) + Ar (H) * N (H) + Ar (O) * N (O) = 12 * 1 + 1 * 1 + 16 * 2 = 45 g / mol;
10) M (CxHyOz) / M (CHO2) = 90/45 = 2;
Unknown substance – (CHO2) 2 – C2H2O4 – oxalic acid.

Answer: Unknown substance – C2H2O4 – oxalic acid.