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KungKemi

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Challenging Stoichiometry Question
« on: October 01, 2016, 06:46:02 AM »
So, I came across the following stoichiometry question in an RSC journal:

A mixture of CH4O, C6H6 and C7H6O weighting 44.37 g gives the elemental analysis: C = 68.74%; H = 8.905%; O = 22.355%. How many grams of C6H6 are contained in the mixture?

I get an answer of 2.80 g, however, although I have reworked my solution and found that it fits, I have no external means of proving or disproving this validity. Any help would be appreciated.

Thank you,
KungKemi

Borek

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Re: Challenging Stoichiometry Question
« Reply #1 on: October 01, 2016, 07:39:39 AM »
That's not what I got. I have not checked, but there is a slight chance that the solution is highly dependent on the atomic masses used. Please show how you got your result.
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KungKemi

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Re: Challenging Stoichiometry Question
« Reply #2 on: October 01, 2016, 08:52:53 AM »
Actual mass of mixture = 44.32 g
1 mol H = 1.008 g
1 mol C = 12.01 g
1 mol CH4O = 32.042 g
1 mol C6H6 = 78.108 g
1 mol C7H6O = 106.118 g

Let mixture = 100 g

CH4O + C6H6 + C7H6O = 100 g
O = 22.355 g
C = 68.74 g
H = 8.905 g

CH4O & C7H6O: Mol O = Mol C

16 g/22.335 g = 12.01 g/xC
xC = 16.765 g C

CH4O & C7H6O: Mol O = 4 Mol H

16 g/22.335 g = 4.032 g/x4H
x4H = 5.628 g H

Unknown: C6H6 + C6H2 = 100 g - 22.335 g - 5.628 g - 16.765 g = 55.272 g ...
Remaining carbon: 68.74 g - 16.765 g = 51.975 g
Remaining hydrogen: 8.905 g - 5.628 g = 3.277 g

C6H6 & C6H2: Mol 6C = Mol 2H

72.06 g/51.975 g = 2.016 g/x2H
x2H = 1.454 g

Remaining: H4 of C6H6 ...
Remaining hydrogen: 3.277 g - 1.454 g = 1.823 g...

Find percentage of C6H6 in mixture ...

% = 1.823 g / (4.032 g/78.108 g) = 35.32 %

Mass C6H6 = 44.32 g × 0.3532 = 15.65 g

KungKemi

« Last Edit: October 01, 2016, 09:57:19 AM by KungKemi »

KungKemi

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Re: Challenging Stoichiometry Question
« Reply #3 on: October 01, 2016, 08:54:08 AM »
Haha, I see where I went wrong. I feel that this answer is correct now, is it not? Just out of curiosity, Borek, I would quite like to see what working out you would have done in order to solve such a question. That would be greatly appreciated.

Thank you, KungKemi.
« Last Edit: October 01, 2016, 09:46:26 AM by KungKemi »

Borek

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Re: Challenging Stoichiometry Question
« Reply #4 on: October 01, 2016, 03:23:24 PM »
CH4O & C7H6O: Mol O = Mol C

No idea what you mean by that.

Quote
Mass C6H6 = 44.32 g × 0.3532 = 15.65 g

And this is still not the result I got.
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KungKemi

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Re: Challenging Stoichiometry Question
« Reply #5 on: October 01, 2016, 07:55:50 PM »
Curious, what answer did you get?

And what I mean by that is that in a compound of, say, C6H12O6, the moles of carbon and oxygen are equal, however, the mass is not the same.

Okay, so I reworked my method because I did see some stupid discrepancies in my last. I don't know what I was thinking...

1 mol H = 1.008 g
1 mol C = 12.01 g
1 mol O = 16.00 g
1 mol CH4O = 32.042 g
1 mol C6H6 = 78.108 g
1 mol C7H6O = 106.118 g

Let mixture = 100 g instead of 100% ...

Then,
O = 22.355 g
C = 68.740 g
H = 08.905 g

The moles of oxygen in C7H6O is equivalent to moles of oxygen in CH4O, which is equivalent to moles of carbon in both substances (which is a 1:1 mole ratio)...

molar mass O/grams O = molar mass C/grams C

16.00/22.355 = 12.01/x g C

mass carbon = 16.780 g...

Remaining carbon = 68.740 g - 16.780 g = 51.960 g...
Remaining oxygen = 00.000 g...
Remaining hydrogen = 08.905 g...

Non-worked components: H4 (of CH4O) + C6H6 + C6H6 (of C7H6O)...

The moles of carbon (6) in C6H6 is now equivalent to the moles of carbon (6) in C7H6O which is further equivalent to the moles of hydrogen (6) in C6H6 and the moles of hydrogen (6) in C7H6O...

molar mass C/grams C = molar mass H/grams H

72.060/51.960 = 6.048/x g H

mass hydrogen = 04.361 g...

Remaining carbon = 00.000 g...
Remaining oxygen = 00.000 g...
Remaining hydrogen = 08.905 g - 04.361 g = 04.544 g...

Non-worked components: H4 (of CH4O)

Therefore the mass of hydrogen in CH4O is 04.544 g...

Find the mass of CH4O...

04.544/(04.032/32.042) = 36.111 g or 36.111 %

Find the mass of oxygen in CH4O...

36.111 g × (16.000/32.042) = 18.032 g...

Oxygen in C7H6O = 22.355 g - 18.032 g = 04.323 g...

Find the mass of C7H6O...

04.323/(16.000/106.118) = 28.672 g or 28.672 %

Find the mass of C6H6...

C6H6 = 100 g - 28.672 g - 36.111 g = 35.217 g

Proof

Find the mass of hydrogen in C6H6, C7H6O, and CH4O...

C6H6: 35.217 g × (6.048/78.108) = 2.727 g
C7H6O: 28.672 g × (6.048/106.118) = 1.633 g
CH4O: 36.111 g × (4.032/32.042) = 4.544 g

Total hydrogen = 2.727 + 1.633 + 4.544 = 8.904 g ≈ original hydrogen mass...

Find the mass of oxygen in C7H6O, and CH4O...

C7H6O: 28.672 g × (16.000/106.118) = 4.323 g
CH4O: 36.111 g × (16.000/32.042) = 18.032 g

Total oxygen = 4.323 + 18.032 = 22.355 g = original oxygen mass...

Find the mass of carbon in C6H6, C7H6O, and CH4O...

C6H6: 35.217 g × (72.06/78.108) = 32.490 g
C7H6O: 28.672 g × (84.07/106.118) = 22.715 g
CH4O: 36.111 g × (12.01/32.042) = 13.535 g

Total carbon = 32.490 + 22.715 + 13.535 = 68.740 g = original carbon mass...

Find actual mass of C6H6 in mixture...

Answer: 44.32 g × 0.35217 = 15.608 g

Would this now be correct?
« Last Edit: October 01, 2016, 11:26:58 PM by KungKemi »

AWK

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Re: Challenging Stoichiometry Question
« Reply #6 on: October 01, 2016, 11:49:40 PM »
I got a very close result.
But why so complicated calculations?
Using moles of your compounds in 100 g of mixture you can set of 3 equations for moles C, H and O in elemental analysis.
eg: a+c = 1.3972 (22.355/16) for O and so on.
AWK

KungKemi

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Re: Challenging Stoichiometry Question
« Reply #7 on: October 02, 2016, 12:14:55 AM »
I got a very close result.
But why so complicated calculations?
Using moles of your compounds in 100 g of mixture you can set of 3 equations for moles C, H and O in elemental analysis.
eg: a+c = 1.3972 (22.355/16) for O and so on.

Hmm...Thank you AWK. I've used such a method as the one you stated for different stoichiometry algebra questions, however, I wasn't quite sure of how to set it up for this question. I definitely see what you are getting at, although.

Thanks for the help.
KungKemi

n.b. For those interested, I found the following question along with others at the following source:
"Problem Solving in Stoichiometry"

Borek

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Re: Challenging Stoichiometry Question
« Reply #8 on: October 02, 2016, 03:51:06 AM »
And what I mean by that is that in a compound of, say, C6H12O6, the moles of carbon and oxygen are equal

Which is not true for the C7H6O, so it can't lead to a correct result.

Quote
Find actual mass of C6H6 in mixture...

Answer: 44.32 g × 0.35217 = 15.608 g

Would this now be correct?

Close.

I did more or less what AWK suggests. We can easily calculate percentages of each element in all three compounds. Then, for example for hydrogen, we can write following mass balance:

0.1258*x+0.0774*y+0.057*z=0.08905*44.37

where x, y, z are respectively masses of all three compounds, 0.1258, 0.0774, 0.0507 are mass fractions of hydrogen in each compound and 0.08905 is the mass fraction of hydrogen in the mixture. Mass fraction is almost the same thing percentage is, just not multiplied by 100.

Write identical equations for carbon and oxygen, solve.

Actually we have four equations, as for obvious reasons x+y+z=1. Choose any three to get the answer. In reality x+y+z=1 is the only exact equation, all others are approximate (as the numbers are rounded down), so each selection of equations can give a slightly different result (but the differences are pretty small).
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KungKemi

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Re: Challenging Stoichiometry Question
« Reply #9 on: October 02, 2016, 04:13:47 AM »
And what I mean by that is that in a compound of, say, C6H12O6, the moles of carbon and oxygen are equal

Which is not true for the C7H6O, so it can't lead to a correct result.

Quote
Find actual mass of C6H6 in mixture...

Answer: 44.32 g × 0.35217 = 15.608 g

Would this now be correct?

Close.

I did more or less what AWK suggests. We can easily calculate percentages of each element in all three compounds. Then, for example for hydrogen, we can write following mass balance:

0.1258*x+0.0774*y+0.057*z=0.08905*44.37

where x, y, z are respectively masses of all three compounds, 0.1258, 0.0774, 0.0507 are mass fractions of hydrogen in each compound and 0.08905 is the mass fraction of hydrogen in the mixture. Mass fraction is almost the same thing percentage is, just not multiplied by 100.

Write identical equations for carbon and oxygen, solve.

Actually we have four equations, as for obvious reasons x+y+z=1. Choose any three to get the answer. In reality x+y+z=1 is the only exact equation, all others are approximate (as the numbers are rounded down), so each selection of equations can give a slightly different result (but the differences are pretty small).

As for the first part of my post which you quoted, Borek, as you would have seen I did not strictly use that philosophy and say that 7 mol C = 6 mol H, however, I used such a philosophy across the compounds (such as saying that 1 mol O = 1 mol C, for example).

As for the second part, yes, I now very easily see what you did. Thank you for the suggestion, I will definitely keep that in mind for future questions!

Cheers,
KungKemi

AWK

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Re: Challenging Stoichiometry Question
« Reply #10 on: October 02, 2016, 04:36:44 AM »
My attempt is different, but as in stoichiometry happens, if equations are correctly set, then give the same result.
I take 100 g of mixture that contains a (or x) moles CH4O, b moles C6H6 (or y) and c moles of C7H6O (or z).
Using molecular formulas I can set a system of equations:
(C) a+6b+7c=68.74/12.01 (moles of C corresponding to 68.74 g of C in 100 g of mixture)
(H) 4a+6b+6c=...
(O) a+c=...
solving for needed b, changing moles b to mass of benzene in 100 g of mixture and rescaling to 44.37 g of mixture one can get answer.
AWK

KungKemi

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Re: Challenging Stoichiometry Question
« Reply #11 on: October 02, 2016, 04:45:41 AM »
My attempt is different, but as in stoichiometry happens, if equations are correctly set, then give the same result.
I take 100 g of mixture that contains a (or x) moles CH4O, b moles C6H6 (or y) and c moles of C7H6O (or z).
Using molecular formulas I can set a system of equations:
(C) a+6b+7c=68.74/12.01 (moles of C corresponding to 68.74 g of C in 100 g of mixture)
(H) 4a+6b+6c=...
(O) a+c=...
solving for needed b, changing moles b to mass of benzene in 100 g of mixture and rescaling to 44.37 g of mixture one can get answer.

Ingenious. That ought to be the simplest method. I guess that there is more than one way to skin a cat.

KungKemi

Vidya

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Re: Challenging Stoichiometry Question
« Reply #12 on: October 02, 2016, 05:24:30 AM »
So, I came across the following stoichiometry question in an RSC journal:

A mixture of CH4O, C6H6 and C7H6O weighting 44.37 g gives the elemental analysis: C = 68.74%; H = 8.905%; O = 22.355%. How many grams of C6H6 are contained in the mixture?

I get an answer of 2.80 g, however, although I have reworked my solution and found that it fits, I have no external means of proving or disproving this validity. Any help would be appreciated.

Thank you,
KungKemi
This is more of algebra kind of problem
If we assume x moles  of CH4O, y moles of  C6H6 and and Z moles of C7H6O
Then we get three simple algebric equations and you can solve three variables with three equations
I am getting these three equations as
from C moles and mass in each compound
12*1*x + 12*6*y+12*7*z= 68.74*44.37/100-------(1)
From H
I am taking approx molar masses
1*4*x+1*6*y+1*6*z= 8.905*44.37/100--------(2)
For O
16*1*x+16*1*z= 22.355*44.37/100-----------(3)
solve for y and you will get moles of C6H6
Multiply moles with molar mass to know mass of C6H6 in the sample