Question #61897
1 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
In other words, the mass of
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)))#