How do I find this empirical formula and determine amounts of a substance in a reaction?

Question

CO and CO2 form when hydrocarbons are burned in little air. When .450 g of a certain hydrocarbon was burned in air, .467 g of CO, .733 g of CO2, and .450 g of H2O were formed.

a) What's the empirical formula of the compound?

b) How many grams of O2 were used in the reaction?

c) How many grams would have been required for complete combustion?

Answer 1

a) All the carbons and hydrogens from you hydrocarbon are present in the CO, CO2 and H2O. So first work out the number of moles of C and H present in the products.

use moles = mass / molecular weight

CO2.

molecular weigth CO2 = 12.01 + (2 x 16.00) = 44.01 g/mol

moles CO2 = 0.733 g / 44.01 g/mol

= 0.016655 moles of CO2

CO2 contains only 1 carbon atom, therefore the number of moles of Carbon in the CO2 is also 0.016655 moles.

CO

molecular weigth = 12.01 + 16.00 = 28.01 g/mol

moles CO = 0.467 g / 28.01 g/mol

= 0.016673 moles of CO

CO also has only 1 Carbon, therefore moles of C in CO = 0.016673 moles

Total moles of Carbon in sample = 0.016673 + 0.016655

= 0.033328 moles of Carbon

hydrogen

All the H from your hydrogen is in the water.

molecular weigth H2O = 16.00 + (1.008 x 2) = 18.016 g/mol

moles of H2O = mass / molecular weight

= 0.450 g / 18.016 g/mol

= 0.024978 moles H2O

Each 1 H2O has 2 hydrogen, therefore there are 2 x 0.024978 moles of H in total sample

= 0.049956 moles of H in original hydrocarbon

So in your Hydrocarbon you had

C : H

0.033328 moles : 0.049956 moles

Now you need to get the ratio into whole numbers. Divide both numbers by the lowest value

C : H

0.033328 / 0.033328 : 0.049956 / 0.033328

= 1 : 1.5  (this is still not a whole ratio - x both by 2

= 2 : 3

So empirical formula is

C2H3

b)

We have already worked out the moles of CO and CO2. These contain all the O2 that was used in the reaction.

moles CO2 = 0.016655 moles

CO2 has 2 O, therefore moles O = (2 x 0.016655)

= 0.03331 moles O in CO2

moles CO = 0.016673 moles

moles O = moles CO = 0.016673

So total O = 0.03331 + 0.016673

= 0.049983 moles of O

mass O = moles x molecular weight

= 0.049983 mol x 16.00 g/mol

= 0.800 g of oxygen