The empirical formula and the molecular formula are the same : C₁₇H₁₉NO₃
<h3>Further explanation</h3>
Given
3.22 g of carbon, 0.300 g of hydrogen, 0.221 g of nitrogen, and the rest is oxygen.
Required
The empirical formula and then the molecular formula.
Solution
Mass of Oxygen = 4.5 - (3.22+0.3+0.221)=0.759 g
Mol ratio of elements C : H : N : O =
3.22/12 : 0.3/1 : 0.221/14 : 0.759/16 = 0.268 : 0.3 : 0.016 : 0.047
Divide by 0.016 :
16.75 : 18.75 : 1 : 2.9375 = 17 : 19 : 1 : 3
(C₁₇H₁₉NO₃)n = 285
(12.17+19+14+3.16)n = 285
(285)n = 285
n = 1
Then the empirical formula and the molecular formula are the same
