Question 7 on Module 1 Post-Assessment  [ENDORSED]

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Tyra Nguyen 4H
Posts: 74
Joined: Fri Sep 28, 2018 12:25 am

Question 7 on Module 1 Post-Assessment

Postby Tyra Nguyen 4H » 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?

Nicholas Carpo 1L
Posts: 36
Joined: Fri Sep 28, 2018 12:19 am

Re: Question 7 on Module 1 Post-Assessment  [ENDORSED]

Postby Nicholas Carpo 1L » 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.

Henry Krasner 1C
Posts: 64
Joined: Fri Sep 28, 2018 12:15 am

Re: Question 7 on Module 1 Post-Assessment

Postby Henry Krasner 1C » 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.

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