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Test 1 Compound X

Posted: Fri Oct 13, 2017 8:57 am
by Lorie Seuylemezian-2K
Can someone help me work through the compound X question on Test 1. It was the second to last question on the test?

Re: Test 1 Compound X  [ENDORSED]

Posted: Fri Oct 13, 2017 3:25 pm
by Cynthia Tsang
So you are given the mass of CO2 and H2O. You convert these both to moles. We can start with CO2. 7.48 grams of CO2 is 0.16996 moles of CO2. You multiple this value by (1 mol of C/1 mol of CO2) to isolate the moles of C. This will give you 0.16996 moles of C. You do the same for oxygen, multiplying 0.16996 moles of CO2 by (2 mol of O/1 mol of CO2). This gives you 0.33992 moles of O. You repeat this for H2O. In the end, you get 0.16996 moles of C, 0.59464 moles of O(make sure to add the 2 oxygen values from H2O and CO2), and 0.5094 moles of H. You divide the moles by the lowest value to get the ratios.

You are left with C2H6 and an unknown value of O. This is because in a combustion reaction, oxygen gas is involved, so you cannot be certain how much oxygen is in the original compound. You use the mass of compound X to calculate how much oxygen there is. C2H6 weighs about 30.08g. You subtract this from 46 g/mol(molar mass of compound x) to get 16g, the mass of 1 mole of oxygen. This results in C2H6O.

Re: Test 1 Compound X

Posted: Mon Oct 16, 2017 11:33 am
by Lorie Seuylemezian-2K
This helps so much! Thank you, Is there any combustion where we know the oxygen because in the homework there were sceniors where we did not approach the problem this way.

Re: Test 1 Compound X

Posted: Mon Oct 16, 2017 8:53 pm
by Peri Bingham 1G
Section M of the Fundamentals (specifically M.3) talks about this type of problem (combustion analysis). Problem M.19 is a good example of this type of problem if you want more practice with it for the midterm and final. Also, because it is an odd problem, the solution is in the workbook so that way you can check your work and make sure you understand it.

Re: Test 1 Compound X

Posted: Tue Nov 07, 2017 6:52 pm
by Paula Dowdell 1F
I'm confused by this question because when i divide 0.0594mol H/0.16996 I get 2.99 (approximately 3) and then when dividing 0.16996mol C/0.16996 I get 1. This leaves me with CH(sub)3 when I really should be getting C(sub)2 H(sub)6
What am I doing wrong?

Re: Test 1 Compound X

Posted: Thu Nov 09, 2017 4:30 pm
by Sarah Brauer
I also got CH3 and was wondering what I did wrong.