Formaldehyde formed during the oxidation of methyl alcohol weighing 48 g

Formaldehyde formed during the oxidation of methyl alcohol weighing 48 g was dissolved in a water mass of 120 g. Determine the mass fraction of methanal.

Given:
m (CH3OH) = 48 g
m (H2O) = 120 g

Find:
ω (HCHO) -?

Solution:
1) 2CH3OH + O2 => 2HCHO + 2H2O;
2) M (CH3OH) = Mr (CH3OH) = Ar (C) * N (C) + Ar (H) * N (H) + Ar (O) * N (O) = 12 * 1 + 1 * 4 + 16 * 1 = 32 g / mol;
M (HCHO) = Mr (HCHO) = Ar (H) * N (H) + Ar (C) * N (C) + Ar (O) * N (O) = 1 * 2 + 12 * 1 + 16 * 1 = 30 g / mol;
3) n (CH3OH) = m (CH3OH) / M (CH3OH) = 48/32 = 1.5 mol;
4) n (HCHO) = n (CH3OH) = 1.5 mol;
5) m (HCHO) = n (HCHO) * M (HCHO) = 1.5 * 30 = 45 g;
6) m solution (HCHO) = m (HCHO) + m (H2O) = 45 + 120 = 165 g;
7) ω (HCHO) = m (HCHO) * 100% / m solution (HCHO) = 45 * 100% / 165 = 27%.

Answer: Mass fraction of HCHO is 27%.