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Question 7 on Module 1 Post-Assessment
Posted: Thu Oct 04, 2018 4:50 pm
Question: Nicotine has mass percentage composition 74.03% C, 8.70% H, and 17.27% N, and a molar mass of 162.23 g/mol. Determine the molecular formula.
I'm struggling to answer this question, as I keep getting the ratio of C:H:N to be 5.7:7.9:1 and cannot solve for the empirical formula. None of the multiple choice options appeared to work for me either. Any idea where I'm going wrong?
Re: Question 7 on Module 1 Post-Assessment [ENDORSED]
Posted: Thu Oct 04, 2018 5:00 pm
You can assume a 100g sample since the percentages add up to 100g to first find the empirical formula.
C:74.03g/12.01g/mol = 6.164mol
H:8.70g/1.01g/mol = 8.614mol
N:17.27g/14.01g/mol = 1.233mol
Simplify the ratios by dividing by the smallest result and rounding up.
C:6.164/1.233 = 5
H:8.614/1.233 = 7
N:1.233/1.233 = 1
Use these results to find the empirical mass and compare to the molecular mass given.
(5x12.01g)(7x1.01g)(14.01g) = 81.13g
Since 162.23g/mol is around twice the empirical mass, you can find that the molecular formula is also twice the empirical formula.
So, the molecular formula is C10H14N2.
Re: Question 7 on Module 1 Post-Assessment
Posted: Thu Oct 04, 2018 5:16 pm
If you multiply the percentages (.7403, .8700, and .1727) by 162.23 g/mol and then divide by the molar mass of C, H and N respectively, you will receive 9.98:14.02:1.99 which is essentially 10:14:2 (the molecular formula). However, this can be simplified down to a 5:7:1 ratio, the empirical formula.