Enter the true formula of a substance with a mass fraction of carbon 80% and hydrogen 20%.

Enter the true formula of a substance with a mass fraction of carbon 80% and hydrogen 20%. Under normal conditions, 1 liter of this gas has a mass of 1.34 grams.

Let’s write down the condition of the problem:
Given: RxO hydrocarbon;
W (C) = 80%;
W (H) = 20%;
1 l (RxOy) = 1.34 g;
V (m) = 22.4 L / mol;
Determine: RxOу;
Solution:
1. Let’s make calculations using the formula:
1 mol of gas at n. y- 22.4 l
X mol RxOy -1 l from here, X mol (RxOy) = 1 * 1 / 22.4 = 0.0446 mol;

Y (RxOy) = m / M;
M (RxOy) = m / Y = 1.34 / 0.0446 = 30 g / mol;
2. Let’s calculate the number of moles:
Y (C) = 80/12 = 6.6 mol;
Y (H) = 20/1 = 20 mol;
3. We take into account the ratio X: Y = C: H = 6.6: 20 = 1: 3;
The simplest formula of the substance: CH3;
4. General formula of the hydrocarbon: CnH2n + 2;
5. M (CnH2n + 2) = 30 g / mol, find the value – n;
6.12n + 2 + 2 = 30;
12n = 26;
n = 2;
M (C2H6) = 2 * 12 + 1 * 6 = 30 g / mol.
Answer: C2H6 (ethane).