Question

Complete combustion of 6.30 g of a hydrocarbon produced 20.5 g of CO2 and 6.29 g of H2O. What is the empirical formula for the hydrocarbon?

Answer 1

In order to determine the emperical formular for the hydrocarbon, we will follow the steps below:

Step 1:Determine moles of CO2, H2O and hydrogen

• Moles of CO2- given that mass of CO2 = 20.5g

-Molecular Mass of CO2 = 44.01g/mole

Therefore, moles(n) = m/Mol.Mass= 20.5g/44.01g/mol = 0.466moles

• Moles of H2O - given that mass of H2O = 6.29g

-Molecula Mass of H2O = 18,01528 g/mol

Therefore,moles (n) = m/Mol.Mass = 6.29g/18.01528g/mol = 0.349moles

• Moles of Hydogen = ,0.349moles, H2O * 2Moles H /1mol H2O

=0.698 moles of hydogen

Step 2 : Determine ratio of Carbon and hydrogen molecules.

For C = 0.466moles/0.466moles = 1 * 4 = 4

For H = 0.698 moles/0.466moles = 1.49* 4= 5.99 = 6

Ratio is therefore C : H

4 : 6

is the same as 2: 3

Therefore , the emperical formula for the hydrocarbon = C2H3