first we need to find the empirical formula of nicotine
empirical formula is the simplest ratio of whole numbers of elements making up a compound
the percentage compositions for each element has been given. So we can calculate for 100 g of the compound.
masses of elements and the number of moles
C - 74.1 g - 74.1 g/12 g/mol = 6.17 mol
H - 8.6 g - 8.6 g / 1 g/mol = 8.6 mol
N - 17.3 g - 17.3 g / 14 g/mol = 1.23
divide all by the least number of moles
C - 6.17 / 1.23 = 5.01
H - 8.6 / 1.23 = 6.99
N - 1.23 / 1.23 = 1.00
when the atoms are rounded off to the nearest whole numbers
C - 5
H - 7
N - 1
empirical formula is C₅H₇N
we have to find what the mass of 1 empirical unit is
mass - 5 x 12 g/mol + 7 x 1 g/mol + 14 g/mol = 81 g
molecular mass is 162.26 g/mol
we have to find how many empirical units make up 1 molecule
number of empirical units = molecular mass / mass of 1 empirical unit
= 162.26 g/mol / 81 g = 2.00
there are 2 empirical units
molecular formula is - 2 (C₅H₇N)
molecular formula - C₁₀H₁₄N₂
