How do you find the molecular formula of nicotine (CxHyNz), when 0.438 grams of nicotine burns to make 1.188 g CO2 and 0.341 g H2O? The molecular weight of the nicotine is 162.2g per mole.

1 Answer
Jan 8, 2016




Notice that despite the fact that nicotine is said to contain carbon, hydrogen, and nitrogen, the problem only provides you with information about how much carbon dioxide, #"CO"_2#, and water, #"H"_2"O"#, are produced by the reaction.

This means that you will have to determine the mass of nitrogen by subtracting the mass of carbon and the mass of hydrogen from the total mass of the sample.

So, the idea here is that after compound undergoes combustion, all the carbon that it originally contained will not be apart of the carbon dioxide.

Likewise, all the hydrogen that it originally contained will now be a part of the water.

Use the percent composition of carbon dioxide to determine how much carbon you'd get in that #"1.188-g"# sample of #"CO"_2# - keep in mind that every mole of carbon dioxide contains one mole of carbon

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

This tells you that every #"100 g"# of carbon dioxide will contain #"27.29 g"# of carbon. Your sample will thus contain

#1.188 color(red)(cancel(color(black)("g CO"_2))) * "27.29 g C"/(100color(red)(cancel(color(black)("g CO"_2)))) = "0.324 g C"#

Do the same for water, but do not forget that every mole of water contains two moles of hydrogen

#(2 xx 1.00794color(red)(cancel(color(black)("g/mol"))))/(18.015color(red)(cancel(color(black)("g/mol")))) xx 100 = "11.19% H"#

This means that your #"0.341-g"# sample of water will contain

#0.341 color(red)(cancel(color(black)("g H"_2"O"))) * "11.19 g H"/(100color(red)(cancel(color(black)("g H"_2"O")))) = "0.0382 g H"#

The mass of nitrogen originally contained in the nicotine sample will thus be

#m_"sample" = m_(N) + m_(H) + m_(C)#

#m_(N) = "0.438 g" - "0.324 g" - "0.0382 g"#

#m_(N) = "0.0758 g N"#

Next, focus on finding the compound's empirical formula, which as you know tells you the smallest whole number ratio that exists between the elements that make up the compound.

To do that, use the molar masses of carbon, hydrogen, and nitrogen, respectively, to find how many moles of each you get in the original sample

#"For C: " (0.324 color(red)(cancel(color(black)("g"))))/(12.0110.324 color(red)(cancel(color(black)("g")))/"mol") = "0.02698 moles C"#

#"For H: " (0.0382 color(red)(cancel(color(black)("g"))))/(1.00794color(red)(cancel(color(black)("g")))/"mol") = "0.03790 moles H"#

#"For N: " (0.0758 color(red)(cancel(color(black)("g"))))/(14.007color(red)(cancel(color(black)("g")))/"mol") = "0.005412 moles N"#

Divide these values by the smallest one to get

#"For C: " (0.02698color(red)(cancel(color(black)("moles"))))/(0.005412color(red)(cancel(color(black)("moles")))) = 4.985 ~~ 5#

#"For H: " (0.03790color(red)(cancel(color(black)("moles"))))/(0.005412color(red)(cancel(color(black)("moles")))) = 7.003 ~~ 7#

#"For N: " (0.005412color(red)(cancel(color(black)("moles"))))/(0.005412color(red)(cancel(color(black)("moles")))) = 1#

The empirical formula of nicotine will thus be


Now, the molecular formula, which tells you exactly how many atoms of each element are needed to form a molecule of your compound, will always be a whole number multiple of the empirical formula.

This means that if you find the molar mass of the empirical formula, you can use the molar mass of the compound to find the molecular formula.

The molar mass of the empirical formula will be

#5 xx "12.011 g/mol" + 7 xx "1.00794 g/mol" + 1 xx "14.007 g/mol" = "81.12 g/mol"#

This means that you have

#81.12 color(red)(cancel(color(black)("g/mol"))) xx color(blue)(n) = 162.2 color(red)(cancel(color(black)("g/mol")))#


#color(blue)(n) = 162.2/81.12 = 1.9995 ~~ 2#

The molecular formula of nicotine will thus be

#("C"_5"H"_7"N"_1)_color(blue)(2) implies color(green)("C"_10"H"_14"N"_2)#