Fundamental M.25

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Hyein Cha 2I
Posts: 103
Joined: Fri Sep 29, 2017 7:05 am
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Fundamental M.25

Postby Hyein Cha 2I » Tue Oct 03, 2017 4:39 pm

So the question is:

In addition to determining the elemental composition of pure unknown compounds, combustion analysis can be used to determine the purity of known compounds. A sample of 2-naphthol, C10H7OH, which is used to prepare antioxidants to incorporate into synthetic rubber, was found to be contaminated with a small amount of LiBr. The combustion analysis of this sample gave the following results: 77.48% C and 5.20% H. Assuming that the only species present are 2-naphthol and LiBr, calculate the percentage purity by mass of the sample.

I looked at the solutions Manuel, but I still do not understand how to solve this question.. :( Please, can anyone explain to me?

Thank you!

Wenjie Dong 2E
Posts: 53
Joined: Fri Jun 23, 2017 11:40 am

Re: Fundamental M.25

Postby Wenjie Dong 2E » Tue Oct 03, 2017 9:29 pm

1. Use the molecule formula C10H7OH to calculate mass percentage composition of C, H, O
2. Get the simplest ratio of mass of C, H, O in C10H7OH (C:H:O=15:1:2)
3. Get the percentage of O in C10H7OH in the sample using the ratio above (H : O = 5.20% : ? = 1:2 --> ? = 10.40%)
4. Add up the percentage of C, H, O (77.48% + 5.20% + 10.40%) to get the purity

That's how I do that. I don't know if it is right.

Andrea Grigsby 1I
Posts: 60
Joined: Fri Sep 29, 2017 7:03 am

Re: Fundamental M.25

Postby Andrea Grigsby 1I » Tue Oct 03, 2017 10:30 pm

@Hyein Cha 1K

I do it the same way

Hyein Cha 2I
Posts: 103
Joined: Fri Sep 29, 2017 7:05 am
Been upvoted: 1 time

Re: Fundamental M.25

Postby Hyein Cha 2I » Thu Oct 05, 2017 4:45 pm

Wenjie Dong 1K wrote:1. Use the molecule formula C10H7OH to calculate mass percentage composition of C, H, O
2. Get the simplest ratio of mass of C, H, O in C10H7OH (C:H:O=15:1:2)
3. Get the percentage of O in C10H7OH in the sample using the ratio above (H : O = 5.20% : ? = 1:2 --> ? = 10.40%)
4. Add up the percentage of C, H, O (77.48% + 5.20% + 10.40%) to get the purity

That's how I do that. I don't know if it is right.


Thank you for your help!


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