"nicotine contains 74.03%C,8.70%H,and 17.27%N by mass. the molar mass of nicotine...

D

DaRealist1

New member
"nicotine contains 74.03%C,8.70%H,and 17.27%N by mass. the molar mass of nicotine...

...is 16.23g/mol. What is the? empirical formula and molecular formular of nicotrine?
 
A

ashishsinha

New member
carbon--
percentage = 74.03
atomic mass = 12
number of moles = 6.17

hydrogen--
percentage = 8.70
atomic mass = 1
number of moles = 8.70/1 = 8.70

nitrogen--
percentage = 17.27
atomic mass = 14
number of moles = 1.23

simples molar ratio of carbon = 6.17/1.23 = 5.01 or 5
simplest molar ratio of hydrogen = 8.70/1.23 = 7.07 or 7
simples molar ratio of nitrogen = 1.23/1.23 = 1

so empirical formula is C5H7N
empirical mass = 12 X 5 + 7 X 1 + 14 = 81 g/mole
molecular mass of Nicotine = 162 g/mole
molecular mass / empirical mass = 162/81 = 2
so molecular formula = (C5H7N)2 = C10H14N2

see this link--
http://chemistry.about.com/od/workedchemistryproblems/a/empirical.htm
 
G

geekeeeee

New member
method use to determine empirical formula is to assume that percent given is in grams, find mole, divide by the smallest mole, and arrived at empirical formula. Then find molar mass of empirical formula and divide by g/mole given. I believe your given g/mole should be bigger as empirical is the lowest ratio structure. For example, 160.23 g/mole given.

Step 1: 74.03 g C (mol C/12.01 g C) = 6.164 mol
8.70 g H (mol H/1.01 g H) = 8.614 mol
17.27 g N (mol N/14.01 g N) = 1.233 mol

Step 2: C =6.164/1.233= 4.99 round up 5
H=8.614/1.233= 6.98 round up 7
N=1.233/1.233= 1

Step 3: C5H7N empirical formula to find molecular use given mass 160.23 g/mole/81.13 g/mole (this is MM of C5H7N)= 2

Step 4: 2 x the number of atoms C5H7N = C10H14N2 molecular formula
 
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