3.1g of an organic compound containing carbon hydrogen and oxygen only produced 4.4g of carbon 4 oxide and 2.0g water on complete combustion

a)determine its empirical formula
b)determine its molecular formula if its formula mass is 62

1 answer

a) To determine the empirical formula, we need to find the mole ratio of the elements present in the compound.

First, we calculate the moles of carbon dioxide produced:

moles of CO2 = mass / molar mass = 4.4 g / 44.01 g/mol = 0.1 mol

Next, we calculate the moles of water produced:

moles of H2O = mass / molar mass = 2.0 g / 18.015 g/mol = 0.111 mol

Since the combustion reaction is balanced, the mole ratio of C : H : O in the compound should be the same as the mole ratio in the products.

Mole ratio of C : H : O = 0.1 mol : (0.111 mol/2) : (0.111 mol/2) = 1 : 1 : 1

Therefore, the empirical formula is CH2O.

b) To determine the molecular formula, we need to know the molecular weight of the compound. From the empirical formula, the molecular weight would be:

molecular weight = (12.01 g/mol x 1) + (1.01 g/mol x 2) + (16.00 g/mol x 1) = 30.03 g/mol

The formula mass given is 62 g/mol, which is approximately double the empirical formula mass. Therefore, we need to multiply the empirical formula by a factor to get the molecular formula.

molecular formula = (CH2O)n

molecular weight = 30.03 g/mol x n

30.03 g/mol x n = 62 g/mol

n = 2

Therefore, the molecular formula of the organic compound is C2H4O2.