# 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.

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