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