A 11.04g sample of a hydrocarbon produces 34.71g of CO2 and 14.20g H2O. What is empirical formula of hydrocarbon?
1 Answer
Explanation:
The key here is to realize that you're dealing with a hydrocarbon, that is, a compound that contains only carbon and hydrogen.
Notice that the products of this combustion reaction are carbon dioxide,
This tells you that all the carbon that was initially a part of the hydrocarbon will now be part of the carbon dioxide. Likewise, all the hydrogen that was initially a part of the hydrocarbon is now a part of the water.
This means that you can use the number of moles of water and carbon dioxide, respectively, to determine how many moles of carbon and of hydrogen were originally present in the hdyrocarbon.
So, for water you have
#14.20color(red)(cancel(color(black)("g"))) * ("1 mole H"_2"O")/(18.015color(red)(cancel(color(black)("g")))) = "0.78823 moles H"_2"O"#
and for carbon dioxide
#34.71color(red)(cancel(color(black)("g"))) * "1 mole CO"_2/(44.01color(red)(cancel(color(black)("g")))) = "0.78868 moles CO"_2#
Now, you know that every mole of water contains 2 moles of hydrogen and 1 mole of oxygen, which means that the reaction produced
#0.78823color(red)(cancel(color(black)("moles H"_2"O"))) * "2 moles H"/(1color(red)(cancel(color(black)("mole H"_2"O")))) = "1.5765 moles H"#
SInce every mole of carbon dioxide contains 1 mole of carbon and 2 moles of oxygen, it follows that the reaction also produced
#0.78868color(red)(cancel(color(black)("moles CO"_2))) * "1 mole C"/(1color(red)(cancel(color(black)("mole CO"_2)))) = "0.78868 moles C"#
Finally, to find the mole ratio that exists between carbon and hydrogen in the hydrocarbon, divide these values by the smallest one
#"For C: " (0.78868color(red)(cancel(color(black)("moles"))))/(0.78868color(red)(cancel(color(black)("moles")))) = 1#
#"For H: " (1.5765color(red)(cancel(color(black)("moles"))))/(0.78868color(red)(cancel(color(black)("moles")))) = 1.999 ~~ 2#
The empirical formula of the hydrocarbon will thus be
#"C"_1"H"_2 implies "CH"_2#
