Question

itm5 ; Combustion of 6.650 mg of an unknown sample yielded 6.648 mg water, and 16.24 mg carbon dioxide. Data analysis found that the sample was only composed of carbon, hydrogen, and oxygen.

Part A

How many moles of carbon were in the original compound?

Part B

How many moles of hydrogen were in the original compound?

Part C

Give the empirical formula of the unknown sample.

Answer 1

Part A. 1 moles of CO2 has 1 moles of C

moles of CO2 = g/molar mass = 0.01624/44.01 = 3.70 x 10^-4 mols

So, moles of carbon = 3.70 x 10^-4 mols

Part B. 1 moles of H2O has 2 moles of hydrogen

moles of H2O = 0.006648/18.015 = 3.70 x 10^-4 mols

moles of Hydrogen = 2 x 3.70 x 10^-4 = 7.38 x 10^-4 mols

Part C. divide mols of C and H by smallest number

C = 3.70 x 10^-4 mols/3.70 x 10^-4 mols = 1

H = 7.38 x 10^-4 mols/3.70 x 10^-4 mols = 2

empirical formaula thus becomes = CH2