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### Question 7 on Module 1 Post-Assessment

Posted: **Thu Oct 04, 2018 4:50 pm**

by **Tyra Nguyen 4H**

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**

by **Nicholas Carpo 1L**

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**

by **Henry Krasner 1C**

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.