[BME_Dev]

I haven't taken a Chem. course since fall 2015 so I'm having trouble with remembering a few things. One of which is a combustion problem, if anyone could help me figure it out I would greatly appreciate.
PROBLEM: An 4.64 g sample of a compound with the formula C₃H₆O₂ was combusted in oxygen producing CO₂ and H₂O What mass of H₂O will be formed?

Thanks in advance.

[AWK]

Start from a balanced reaction.

[BME_Dev]

That's the only part of the problem I remember to do, everything after that is really what I'm lost on. Tried searching google, but no luck. Thank you

[BME_Dev]

H₂O mass = 3.88 grams.

[Arkcon]

OK, lets see that balanced reaction.  Do you know how you will use it?  Do you know the units of a balanced chemical equation, and how to convert units you have to units you need.

Sorry to answer a question with a question, but that's what we do here.  That's detailed in  the  Forum Rules{click}.  You agreed to follow these rules when you signed up.

If you need, "Just everything" you'll find it in your textbook.  If you know some bits, but want to tell us one post at a time, this will be a tedious thread.  If you just want to assure us you know something, and not show it to us, this will be a completely pointless thread.

[AWK]

H₂O mass = 3.88 grams.

Too much

[BME_Dev]

Well, here's my work so maybe this will help clarify what I asking. The reason I put 3.88 grams was because the problem stated to give three significant figures. To better state my question(s): I know the answer is correct because I got full credit for it, but I didn't feel comfortable with how I achieved it as I had to rearrange the labeling multiple times. With that being said, I was wondering if my set up is correct? Furthermore, I feel as if the time it took to actually solve the problem from start to beginning was incredibly longer than it should've been. Would anyone recommend a way I can shorten any part of the process to achieve my answer?

NOTE: What's in the top right corner is a table I use to keep track of elements when balancing. My work starts in top left corner and is going in order. I apologize in advance for my sloppy 2^nd grade hand writing. As well, feedback would be greatly appreciated.


http://m.imgur.com/a/LtLOd

[AWK]

Only balancing OK
2C₃H₆O₂ + 7O₂ = 6CO₂ + 6H₂O
Now simplify extracting only balancing hydrogen atoms
C₃H₆O₂  3H₂O
74.0784  54.0444
Without calculation you can estimate result as ~5/7 ≈ ~70 % of your sample

[Borek]

Several problems with what you did.

You tried to apply a unit conversion scheme, but you were not consistent with the units. Mol of C₃H₆O₂ doesn't cancel out with mol of H₂O.

You have never calculated molar mass of the substance that was burnt, and that should be the next step after balancing.

108.1g/mol is not a molar mass of 6 moles of water, no such thing. Molar mass is always mass of a single mole. 108.1 g is a MASS of 6 MOLES of water.

I guess you answer was accepted accidentally - the correct one is very similar.

[BME_Dev]

Didn't notice that I put it only per 1 mol. AWK, your explanation worked as well and was put in a simpler way, thank you. Borek, what answer did you get and what did you do differently than I did to reach it? Not saying you're wrong, just curious.

[Borek]

3.38 g is a correct answer, and I have not bothered with manual calculations.

There are many ways to calculate the final answer. The one you selected (unit conversion) is as good as any if applied correctly. I told you what is wrong with your approach. Your answer is not 3.88 g of water, but [itex]3.88 \frac {mol C_3H_6O_2}{mol H_2O}[/itex] (whatever it means).

Basically you should start with finding number of moles of the burnt substance, convert it to the number of moles of water, then convert number of moles of water to the mass of water.

AWK's approach is a convenient shortcut which shouldn't be used blindly before you become fluent in a standard approach (at which stage the shortcut will be obvious).

[BME_Dev]

Okay, thanks for your help man. Greatly appreciated.

[AWK]

4.64*3*H2O/C3H6O2 = 3.385(14352361822)

[Borek]

H2O/C3H6O2

You could at least explain you mean ratio of molar masses here, unless you are trying to win "the most cryptic answer" competition.

[AWK]

I explained it previously

Now simplify extracting only balancing hydrogen atoms
C3H6O2  3H2O
74.0784  54.0444

I used calculator that directly replace formulas by masses (it use the most exact atom masses by IUPAC)
http://www.periodni.com/scientific_calculator.html
This is a copy from notebook of calculator (history of calculations) with brackets inserted by me.
4.64*3*H2O/C3H6O2 = 3.385(14352361822)