Chemical Forums
Chemistry Forums for Students => Undergraduate General Chemistry Forum => Topic started by: ch3mgirl on February 14, 2011, 05:17:57 PM
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There were 8.35 x 1023 molecules of carbon dioxide produced when a sample of octane was burned in air. If the combustion process is known to produce an 86.7% yield, how many grams of octane were in the same?
The equation I have is:
2C8H18 + 25O2 :rarrow: 16CO2 + 18H2O
I'm not sure if I even went about this the right way.
I converted molecules of CO2 to moles then to grams and came out with 6.10 g of CO2.
Since it's in a 1:8 ratio, I divided 6.10g by 8 and got 0.7625.
Then I took 0.7625 x 86.7% and got 0.661 g of C8H18.
Could someone let me know if I went about this in the wrong way? Thanks so much.
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the 1 in 8 ratio is for moles of course... remember what the coefficients in the reaction equation mean?
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the 1 in 8 ratio is for moles of course... remember what the coefficients in the reaction equation mean?
The number of molecules?
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The reaction tells you that for every molecule of octane combusted, you theoretically get 8 molecules of CO2. This is different from saying that for every gram of octane combusted, you theoretically get 8 grams of CO2.
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There were 8.35 x 1023 molecules of carbon dioxide produced when a sample of octane was burned in air. If the combustion process is known to produce an 86.7% yield, how many grams of octane were in the same?
The equation I have is:
2C8H18 + 25O2 :rarrow: 16CO2 + 18H2O
So far so good.
I converted molecules of CO2 to moles then to grams and came out with 6.10 g of CO2.
This step is not required and is wrong as are all the steps after this
Since it's in a 1:8 ratio, I divided 6.10g by 8 and got 0.7625.
As DeraDevil has pointed out the 1:8 ratio is moles not mass so you can not simply divide the weight of carbon dioxide to get the answer.
Then I took 0.7625 x 86.7% and got 0.661 g of C8H18.
This step is wrong. If the yield is 86.7% would more or less octane be required to give an ammount of CO2?
2C8H18 + 25O2 :rarrow: 16CO2 + 18H2O
There were 8.35 x 1023 molecules of carbon dioxide produced
1) Is 8.35 x 1023 more or less than Avogadro's number?
2) So how many moles of carbon dioxide were formed?
If for example 8 moles of CO2 were formed from the complete combustion that would mean 1 mole of octane was burnt.
3) Now using your numbers how many moles of Octane buring completely would give the CO2 formed and calculated in 2?
As the yield is 86.7% would that mean more or less octane is need to give this amount of CO2?
4) So how many moles of octane were needed to produce the amount of CO2 formed when the yield was 86.7%?
5) Now convert the number of moles in part 4 to a weight using the molecular weight of octane.
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There were 8.35 x 1023 molecules of carbon dioxide produced when a sample of octane was burned in air. If the combustion process is known to produce an 86.7% yield, how many grams of octane were in the same?
The equation I have is:
2C8H18 + 25O2 :rarrow: 16CO2 + 18H2O
So far so good.
I converted molecules of CO2 to moles then to grams and came out with 6.10 g of CO2.
This step is not required and is wrong as are all the steps after this
Since it's in a 1:8 ratio, I divided 6.10g by 8 and got 0.7625.
As DeraDevil has pointed out the 1:8 ratio is moles not mass so you can not simply divide the weight of carbon dioxide to get the answer.
Then I took 0.7625 x 86.7% and got 0.661 g of C8H18.
This step is wrong. If the yield is 86.7% would more or less octane be required to give an ammount of CO2?
2C8H18 + 25O2 :rarrow: 16CO2 + 18H2O
There were 8.35 x 1023 molecules of carbon dioxide produced
1) Is 8.35 x 1023 more or less than Avogadro's number?
2) So how many moles of carbon dioxide were formed?
If for example 8 moles of CO2 were formed from the complete combustion that would mean 1 mole of octane was burnt.
3) Now using your numbers how many moles of Octane buring completely would give the CO2 formed and calculated in 2?
As the yield is 86.7% would that mean more or less octane is need to give this amount of CO2?
4) So how many moles of octane were needed to produce the amount of CO2 formed when the yield was 86.7%?
5) Now convert the number of moles in part 4 to a weight using the molecular weight of octane.
1) It is more than Avogadro's number.
2) I got 1.39 mol of CO2. Am I going on the right track so far?
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1) It is more than Avogadro's number.
2) I got 1.39 mol of CO2. Am I going on the right track so far?
Yes and Yes
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1) It is more than Avogadro's number.
2) I got 1.39 mol of CO2. Am I going on the right track so far?
Yes and Yes
3) Now using your numbers how many moles of Octane buring completely would give the CO2 formed and calculated in 2?
0.174 mol C8H18?
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1) It is more than Avogadro's number.
2) I got 1.39 mol of CO2. Am I going on the right track so far?
Yes and Yes
3) Now using your numbers how many moles of Octane buring completely would give the CO2 formed and calculated in 2?
0.174 mol C8H18?
Okay, so I then converted that into grams and then multiplied by the percent yield and got:
17.2 g of C8H18
Is this correct?
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almost right. One small thing left is to have a look at the yield.
Let us see:
Yield% = actual moles CO2 produced / theoretical moles of CO2 that could be produced given the amount of reactant
so, what does this lead us to:
you calculate the number of moles of CO2 that are produced correctly, now use them to calculate the amount of moles that could have been formed at full combustion. Then use that number to determine the number of moles (and from there mass) of octane that combusted.
You applied the yield the wrong way around.
one last remark: in one of your first posts you calculated the mass of CO2 incorrectly. 1.39 mol of CO2 = 61g, not 6.1!
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almost right. One small thing left is to have a look at the yield.
Let us see:
Yield% = actual moles CO2 produced / theoretical moles of CO2 that could be produced given the amount of reactant
so, what does this lead us to:
you calculate the number of moles of CO2 that are produced correctly, now use them to calculate the amount of moles that could have been formed at full combustion. Then use that number to determine the number of moles (and from there mass) of octane that combusted.
You applied the yield the wrong way around.
one last remark: in one of your first posts you calculated the mass of CO2 incorrectly. 1.39 mol of CO2 = 61g, not 6.1!
Eeek, hold on I just got confused! Okay, so I went back and took my actual moles of CO2 which was 1.39 and divided by the % yield to get 1.60 mol CO2. I took 1.60 mol CO2 and divided by 8. I took my answer and then converted to grams and got 22.8 g of C8H18?
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correct!
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correct!
Oh wow.... thanks so much!
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correct!
Oh wow.... thanks so much!
It turns out it was supposed to be 2.28g of octane.
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If it was actually 8.23 x 1022 molecules and 86.7% yield or 8.23 x 1023 molecules but an 8.67% yield then 2.28g is correct but for the question you posted 22.8g is correct.
If you transcribed the question correctly then you should contact the person who says 2.28g is correct and ask them to prove it.
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If it was actually 8.23 x 1022 molecules and 86.7% yield or 8.23 x 1023 molecules but an 8.67% yield then 2.28g is correct but for the question you posted 22.8g is correct.
If you transcribed the question correctly then you should contact the person who says 2.28g is correct and ask them to prove it.
The way he did it was:
(8.35 x 1023 molecules/0.867) x (1 mol CO2/6.02 x 1023 molecules) x (2 mol octane/16 mol CO2) x (114.2 g octane/1 mole octane
Yeah, I checked it and I did type the question correctly. I'll ask my professor about it, thank you!
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type that calculation into your calculator and tell me what the answer is ;)
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type that calculation into your calculator and tell me what the answer is ;)
I've tried 2 times and it keeps coming out to 2.28... could I be typing it wrong?
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The way he did it was:
(8.35 x 1023 molecules/0.867) x (1 mol CO2/6.02 x 1023 molecules) x (2 mol octane/16 mol CO2) x (114.2 g octane/1 mole octane
The answer to the calculation shown above is 22.8. Your professor is a fuckwit it is very simple to work out in your head that the final answer should be ~ 1/5 of the molar mass of octane so the answer has to be ~23.
8 and a bit divide 6 and a bit divide 8 is approx 1 and a bit over 6 factor in the 86.7% yield and the answer is ~1/5th of octanes molecular weight.
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type that calculation into your calculator and tell me what the answer is ;)
I've tried 2 times and it keeps coming out to 2.28... could I be typing it wrong?
the power of 10 will fall away above and below the divider (1023/1023 = 1)
so you get: ( ( 8.35 / 0.867 ) / 6.02 ) * 1/8 * 114.2 = 8.35 * 114.2 / (0.867 * 6.02 * 8 ) = 953.57 / 41.75 = 22.84