When 2.686 grams of a hydrocarbon, CxHy, were burned in a combustion analysis apparatus, 9.224 grams of CO2 and 1.511 grams of H2O were produced.
In a separate experiment, the molar mass of the compound was found to be 128.2 g/mol. Determine the empirical formula and the molecular formula of the hydrocarbon.
Enter the elements in the order presented in the qustion.
In a separate experiment, the molar mass of the compound was found to be 128.2 g/mol. Determine the empirical formula and the molecular formula of the hydrocarbon.
Enter the elements in the order presented in the qustion.
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To solve this, you first need to know that combustion in this question means the burning of the hydrocarbon in Oxygen gas to produce Carbon dioxide + Water. Or in terms of an equation
CxHy + O2 ---> CO2 + H2O
Obviously the above is not balanced and you are not given an empirical formula for the hydrocarbon. You are however given a molar mass and some mass values for your hydrocarbon and your two products. You can therefore find the mols of each of those and write out the full equation.
I'll start with the hydrocarbon.
You have 2.686 grams of it and you know the Molar Mass (MM) is 128.2 g/mol.
2.686g/128.2 g/mol = .0209 mols hydrocarbon. Save this number now.
Now pick one of your products. I did CO2 first. Do the same as above to get the mols of CO2 produced. To get MM of CO2 simply add up their atomic weights. 12.0107 + (2 x 1.00794). You can hopefully do the math to get the mols.
You should get mols CO2 = .2096 mols
Now to find out the molar ratio between CO2 and the hydrocarbon, simply take the ratio of the mols.
.2096 mol CO2 / .0209 mols hydrocarbon = ~10. This tells you that for every mol of hydrocarbon used in the reaction, 10 mols of CO2 were produced.
Now do the same thing with water. Find mols water produced and take ratio. You should get 4.
You can now write out your balanced equation.
C10H8 + 12O2 -----> 10CO2 + 4H2O
Because C10H8 also has an MM of 128.2 you are done. It is both the empirical and the molecular formula.
CxHy + O2 ---> CO2 + H2O
Obviously the above is not balanced and you are not given an empirical formula for the hydrocarbon. You are however given a molar mass and some mass values for your hydrocarbon and your two products. You can therefore find the mols of each of those and write out the full equation.
I'll start with the hydrocarbon.
You have 2.686 grams of it and you know the Molar Mass (MM) is 128.2 g/mol.
2.686g/128.2 g/mol = .0209 mols hydrocarbon. Save this number now.
Now pick one of your products. I did CO2 first. Do the same as above to get the mols of CO2 produced. To get MM of CO2 simply add up their atomic weights. 12.0107 + (2 x 1.00794). You can hopefully do the math to get the mols.
You should get mols CO2 = .2096 mols
Now to find out the molar ratio between CO2 and the hydrocarbon, simply take the ratio of the mols.
.2096 mol CO2 / .0209 mols hydrocarbon = ~10. This tells you that for every mol of hydrocarbon used in the reaction, 10 mols of CO2 were produced.
Now do the same thing with water. Find mols water produced and take ratio. You should get 4.
You can now write out your balanced equation.
C10H8 + 12O2 -----> 10CO2 + 4H2O
Because C10H8 also has an MM of 128.2 you are done. It is both the empirical and the molecular formula.