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anonymous
 3 years ago
I found this question in the chemistry section of a test.
I need help with number 17 I have no idea where to start.
http://gyazo.com/8e8407570f9ec590c8485d47cf429beb
(PS: the answer for 16 is B, confirmed with answer manual.)
anonymous
 3 years ago
I found this question in the chemistry section of a test. I need help with number 17 I have no idea where to start. http://gyazo.com/8e8407570f9ec590c8485d47cf429beb (PS: the answer for 16 is B, confirmed with answer manual.)

This Question is Closed

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0good question! lots going on here.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Well, since you have the formula already from question 16, this is simple. The molar weight of CH is just (approx) 12+1=13 g/mol

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0The hard question is #16...

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0I only found the empirical formula for the gas, remember. The molecular mass depends on the MOLECULAR formula, which may or may not be CH. the gas could be CH, C2H2, C3H3, etc. so actually, I don't see how your solution is correct...

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0oh...ic what you mean. I misread that part. I'll have another look.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Ok....Ihave a solution. The mass in chamber 1 is all H20. THus, we can conclude that there was .9 grams of H20 originally. Converting this to moles gives approximately .05 moles of H20. We do the same to the material in Chamber 2 since that was all CO2, and we get .1 moles of CO2. By conservation of mass, we must conclude that there were .05*2 = .1 moles of H and .1 moles of C that were on the product side. Since the problem said the hydrocarbon only containted C and H, we can say that the original sample hydrocarbon contained .1 moles of C and .1 moles of H. From here, we can get the empirical formula by computing the ratio of C to H in the hydrocarbon. Since it is a 1:1 ratio, the empirical formula is CH!.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0for 17, we can compute the molar mass of the hydrocarbon by using this formula:\[Molar mass = density*R*T/P\]. Doing so for the hydrocarbon yields approximately 26 g/mol.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Thanks, your solution for number 16 is similar to mine. The way you answer 17 is strange to me, though. Where do you derive that formula from?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Ideal Gas Law: \[PV = nRT\]\[Density = mass/Volume.\] But in this case Mass = molar mass * n (number of moles). Thus n = Mass/molar mass. Solving for molar mass, we get\[Molar mass = Mass*R*T/(P*Volume)\]. But as established above, Mass = Mass/Volume. Using this arrives to the equation I gave.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0wow, that's really useful! I wonder why I never encountered that in class before... Thank you so much!
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