Mass fraction of carbon 54.54 Mass fraction of hydrogen 9.09 Mass fraction of oxygen 36.36

Mass fraction of carbon 54.54 Mass fraction of hydrogen 9.09 Mass fraction of oxygen 36.36 Oxygen density 1.345 Find the formula.

Let’s implement the solution:
1. Let’s designate the substance: CxHyOz;
2. Let’s calculate the molar mass of a substance, if the density for oxygen is known,
M (O2) = 32 g / mol;
M (CxHyOz) = D (O2) * M (O2) = 32 * 1.345 = 43 g / mol;
3. Determine the amount of moles of carbon, hydrogen, oxygen:
Y (C) = 54.54 / 12 = 4.5 mol;
Y (H) = 9.09 / 1 = 9.09 mol;
Y (O) = 36.36 / 16 = 2.27 mol.

4. The ratio X: Y: Z = C: H: O = 4.5: 9.09: 2.27 = 2: 4: 1;
5. Molecular formula: C2H4O, M (C2H4O) = 2 * 12 + 4 + 16 = 44 g / mol.
Answer: C2H4O is ethanal.