In an experiment 3.10grams of carbon hydrogen and oxygen produced 4.4grams of carbon4 oxide and 2.7of water on complete combustion. Determine the empirical formula of the compound
To determine the empirical formula of the compound, we need to find the mole ratios of carbon, hydrogen, and oxygen in the compound.
First, let's determine the moles of CO2 produced:
Molar mass of CO2 = 12.01 g/mol + (2 * 16.00 g/mol) = 44.01 g/mol
moles of CO2 = 4.4 g / 44.01 g/mol = 0.0999 mol (approximately)
Next, let's determine the moles of H2O produced:
Molar mass of H2O = 2 * 1.01 g/mol + 16.00 g/mol = 18.02 g/mol
moles of H2O = 2.7 g / 18.02 g/mol = 0.1498 mol (approximately)
Now, let's calculate the mole ratio of carbon to carbon dioxide and hydrogen to water:
moles of C = moles of CO2 = 0.0999 mol
moles of H = 2 * moles of H2O = 2 * 0.1498 mol = 0.2996 mol
Finally, let's calculate the mole ratio of oxygen:
moles of O = (moles of CO2 * 2) + (moles of H2O * 1)
= (0.0999 mol * 2) + (0.1498 mol * 1)
= 0.3496 mol
Now, let's convert the moles to whole numbers by dividing by the smallest moles (0.0999 mol):
moles of C ≈ 0.0999 mol / 0.0999 mol = 1
moles of H ≈ 0.2996 mol / 0.0999 mol ≈ 3
moles of O ≈ 0.3496 mol / 0.0999 mol ≈ 3.5
The empirical formula of the compound is CH3.5O (or CH7O2 in the lowest whole number ratio).