Question #61897

1 Answer
Mar 3, 2017

Answer:

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)))#

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