Question #18ab0

1 Answer
Oct 2, 2015

Answer:

#"C"_4"H"_8#

Explanation:

The idea here is that you need to use the amount of carbon dioxide and water produced by the combustion reaction to figure out how much carbon and how much hydrogen was initially found in the hydrocarbon.

You can assume that all the carbon that was initially a part of the #"X"# is now a part of #"CO"_2#. Carbon's percent composition in carbon dioxide is

#(12.011 color(red)(cancel(color(black)("g/mol"))))/(44.01color(red)(cancel(color(black)("g/mol")))) xx 100 = 27.29%#

This means that your sample of carbon dioxide will contain

#88color(red)(cancel(color(black)("g CO"""_2))) * "27.29 g C"/(100color(red)(cancel(color(black)("g CO"""_2)))) = "24.02 g C"#

Do the same for hydrogen's percent composition in water - keep in mind that one mole of water contains two moles of hydrogen!

#(2 xx 1.008color(red)(cancel(color(black)("g/mol"))))/(18.015color(red)(cancel(color(black)("g/mol")))) xx 100 = 11.19%#

This means that the water produced by the reaction will contain

#36color(red)(cancel(color(black)("g H"_2"O"))) * "11.19 g H"/(100color(red)(cancel(color(black)("g H"_2"O")))) = "4.028 g H"#

This is how much carbon and how much hydrogen was originally a part of #"X"#.

Use carbon and hydrogen's respective molar masses to find how many moles of each were present in the hydrocarbon

#24.02color(red)(cancel(color(black)("g C"))) * "1 mole C"/(12.011color(red)(cancel(color(black)("g C")))) = 1.9998 ~~ "2 moles C"#

and

#4.028color(red)(cancel(color(black)("g H"))) * "1 mole H"/(1.008color(red)(cancel(color(black)("g H")))) = 3.996 ~~ "4 moles H"#

Now, this is how many moles of carbon and hydrogen were present in 0.5 moles of #"X"#. This means that one mole of #"X"# will actually contain twice as many moles of carbon and hydrogen.

The molecular formula of #"X"# will thus be

#("C"_2"H"_4)_2 = color(green)("C"_4"H"_8)#

You can test the result by finding the total mass of 0.5 moles of #"X"#

#m = m_"C" + m_"H"#

#m = "24.02 g C" + "4.028 g H" = "28.048 g"#

This means that the molar mass of #"X"# is

#M_M = m/n = "28.048 g"/"0.5 moles" = "56.096 g/mol"#

Therefore, you have

#(2 xx 12.011 + 4 xx 1.008) xx color(blue)(n) = 56.096#

#color(blue)(n) = 56.096/28.054 = 1.9996 ~~ 2#

The molecular formula will once again be

#("C"_2"H"_4)_2 = color(green)("C"_4"H"_8)#