# When organic matter with a mass of 23 grams is burned, 44 grams of carbon dioxide and 27

**When organic matter with a mass of 23 grams is burned, 44 grams of carbon dioxide and 27 g of water are formed, the relative density of hydrogen 23 determine the formula**

Let’s find the molecular weight of the unknown substance:

M (CxHy) = D (H2) • n = 23 • 2 = 46.

Find the number of moles of carbon and hydrogen:

n (C) = m (CO2) / Mr (CO2) = 44/44 = 1 mol.

n (H) = m (H2O) / Mr (H2O) = 27/18 = 1.5 mol.

Since water contains 2 hydrogen atoms, there will be more of it in the compound:

n (H) = 2 x 1.5 = 3 mol.

Let’s calculate the masses of the elements:

m (C) = n • Mr = 1 • 12 = 12 g.

m (H) = 3 • 1 = 3 g.

Determine the amount of oxygen in the compound:

m (O) = m (in-va) – m (C) – m (H) = 23 – 12 – 3 = 8 g.

Mole oxygen:

n (O) = m / Mr = 8/16 = 0.5 mol.

We get the ratio:

n (C): n (H): n (O) = 1: 3: 0.5 = 2: 6: 1.

So the true formula C2H6O or C2H5OH is ethanol.