Derive the molecular formula of the chlorine derivative of alkane, if its mass fractions are

Derive the molecular formula of the chlorine derivative of alkane, if its mass fractions are: carbon-45.85%, hydrogen-8.92%, chlorine-45.22%.

Given:
ω (C) = 45.85%
ω (H) = 8.92%
ω (Cl) = 45.22%

Find:
CxHyClz -?

Solution:
1) Let m (CxHyClz) = 100 g;
2) m (C) = ω (C) * m (CxHyClz) / 100% = 45.85% * 100/100% = 45.85 g;
3) n (C) = m (C) / M (C) = 45.85 / 12 = 3.821 mol;
4) m (H) = ω (H) * m (CxHyClz) / 100% = 8.92% * 100/100% = 8.92 g;
5) n (H) = m (H) / M (H) = 8.92 / 1 = 8.92 mol;
6) m (Cl) = ω (Cl) * m (CxHyClz) / 100% = 45.22% * 100/100% = 45.22 g;
7) n (Cl) = m (Cl) / M (Cl) = 45.22 / 35.5 = 1.274 mol;
8) x: y: z = n (C): n (H): n (Cl) = 3.821: 8.92: 1.274 = 3: 7: 1;
9) C3H7Cl.

Answer: Unknown substance – C3H7Cl.