Determine the molecular formula of a hydrocarbon containing 93.75% carbon, the relative density in air is 4.41.

Given: W (C) = 93.75%
D (air) = 4.41;
M (air) = 29 g / mol;
Find: ShNu -?
Solution:
1. Calculate the molecular weight of the hydrocarbon by the formula:
M (CxHy) = D (air) * M (air) = 4.41 * 29 = 127.8 g / mol.
2. Determine the number of moles of carbon:
Y (C) = m / M = 93.75 / 12 = 7.8 mol.
3. Let’s calculate the mass of carbon, hydrogen:
m (C) = Y * M = 7.8 * 12 = 93.6 g;
m (H) = 127.8 – 93.6 = 34.2;
Y (H) = m / M = 34.2 / 1 = 34.2 mol.
4. Ratio:
X: Y = C: H = 7.8: 34.2 = 1: 4 (divide by the smallest number 7.4);
CH4 is the simplest formula, which means that the hydrocarbon belongs to the class of alkanes.
We will make calculations according to the general formula: CnH2n + 2.
5. Let’s establish the molecular formula:
CnH2n + 2;
C9H20;
Checking: M (C9H20) = 12 * 9 + 20 = 128 g / mol.
Answer: С9Н20 – nonane.