When 2.5 g of matter was burned, 8.46 g of carbon dioxide and 1.73 g

When 2.5 g of matter was burned, 8.46 g of carbon dioxide and 1.73 g of water were released. the mass of the substance per 1 liter is 3.5 g, find the formula.

Given:
m (in-va) = 2.5 g
m (CO2) = 8.96 g
m (H2O) = 1.73 g
ρ (in-islands) = 3.5 g / l

Find:
in -?

Solution:
1) n (CO2) = m (CO2) / Mr (CO2) = 8.96 / 44 = 0.204 mol;
2) n (C in org.v-ve) = n (C in CO2) = n (CO2) = 0.204 mol;
3) m (C in org.v-ve) = n (C in org.v-ve) * Ar (C) = 0.204 * 12 = 2.448 g;
4) n (H2O) = m (H2O) / Mr (H2O) = 1.73 / 18 = 0.096 mol;
5) n (H in org.v-ve) = n (H in H2O) = n (H2O) * 2 = 0.096 * 2 = 0.192 mol;
6) m (H in org.v-ve) = n (H in org.v-ve) * Ar (H) = 0.192 * 1 = 0.192 g;
7) m (O in org.v-ve) = m (org.v-va) – m (C in org.v-ve) – m (H in org.v-ve) = 2.5 – 2.448 – 0.192 ≈ 0 g;
8) C (x) H (y)
x: y = m (C in org.v-ve): m (C in org.v-ve) = 0.204: 0.192 ≈ 1: 1;
CH is the simplest formula;
10) Mr (islands) = ρ (islands) * Vm = 3.5 * 22.4 = 78.4 ≈ 78 g / mol;
11) Mr (CH) = Ar (C) + Ar (H) = 12 + 1 = 13 g / mol;
12) Mr (in-va) = Mr (CH) * 6;
13) Unknown substance – C6H6 – benzene.

Answer: Unknown substance – C6H6 – benzene.