Question

Question #61897

Answer 1

Use the compound's molar mass.

Explanation:

The idea here is that the molecular formula is always a multiple of the empirical formula

`color(blue)(ul(color(black)("molecular formula" = "empirical formula" xx n)))`

Here `n` must be a whole number.

In other words, the mass of `1` mole of molecular formulas of nicotine must be equal to the mass of `1` mole of empirical formulas of nicotine multiplied by a constant `n`.

You know that the empirical formula for nicotine is

`"C"_5"H"_7"N"`

use the molar masses of the elements that make up the empirical formula to find the molar mass of the empirical formula

`5 xx "12.01 g mol"^(-1) " "+`
`7 xx "1.008 g mol"^(-1)`
`1 xx "14.07 g mol"^(-1)`
`color(white)(aaaaaaaaaaaaaaa)/color(white)(a)`
`" 81.18 g mol"^(-1)`

Now look up the molar mass of nicotine, which you'll find listed as

`M_"M nicotine" = "162.23 g mol"^(-1)`

You can thus say that

`162.23 color(red)(cancel(color(black)("g mol"^(-1)))) = 81.18 color(red)(cancel(color(black)("g mol"^(-1)))) xx color(blue)(n)`

This will get you

`color(blue)(n) = 162.23/81.18 = 1.998 ~~ 2`

Therefore, you can say that the molecular formula for nicotine is

`color(darkgreen)(ul(color(black)(("C"_5"H"_7"N")_color(blue)(2) = "C"_10"H"_14"N"_2)))`