[KungKemi]

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]

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.

[KungKemi]

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

[KungKemi]

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.

[Borek]

CH₄O & C₇H₆O: Mol O = Mol C

No idea what you mean by that.

Mass C6H6 = 44.32 g × 0.3532 = 15.65 g

And this is still not the result I got.

[KungKemi]

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?

[AWK]

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.

[KungKemi]

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]

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.

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).

[KungKemi]

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.

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]

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.

[KungKemi]

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]

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