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
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.