Question

Nicotine is 74.1% carbon, 8.6% hydrogen, and 17.3% nitrogen by mass. What is its molecular formula if its molar mass is 162.26 g/mol

Answer 1

So the C:H:N ratio is 5:7:1 and the empirical formula for nicotine is C5H7N.

Answer 2

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