Enter the molecular formula of the hydrocarbon, where the mass fraction of carbon and hydrogen

Enter the molecular formula of the hydrocarbon, where the mass fraction of carbon and hydrogen are 85.71% and 14.29%, respectively

1. Let’s find the chemical amount of carbon and hydrogen in 100 g of the required substance:

m (C) = m (in-va) * ω (C) / 100% = 100 g * 85.71% / 100% = 85.71 g.

n (C) = m (C) / M (C) = 85.71 g / 12 g / mol = 7.14 mol.

m (H) = m (in-va) * ω (H) / 100% = 100 g * 14.29% / 100% = 14.29 g.

n (H) = m (H) / M (H) = 14.29 g / 1 g / mol = 14.29 mol.

2. Let’s compose the ratio of the quantities of these elements:

n (C): n (H) = 7.14 mol: 14.29 mol = 1: 2, hence the initial formula of the substance CH2, multiply it by 2 and get C2H4, that is, ethylene.

Answer: C2H4.