# 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

#### Explanation:

**!! LONG ANSWER !!**

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,

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 **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**

#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 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

#"C"_5"H"_7"N"_1#

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")))#

Therefore,

#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)#