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#### ch3mgirl

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##### How many grams of octane were in the sample?
« on: February 14, 2011, 11:17:57 AM »

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   16CO2 + 18H2O

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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##### Re: How many grams of octane were in the sample?
« Reply #1 on: February 14, 2011, 11:58:55 AM »

the 1 in 8 ratio is for moles of course... remember what the coefficients in the reaction equation mean?
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#### ch3mgirl

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##### Re: How many grams of octane were in the sample?
« Reply #2 on: February 14, 2011, 12:12:22 PM »

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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#### rabolisk

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##### Re: How many grams of octane were in the sample?
« Reply #3 on: February 14, 2011, 01:55:06 PM »

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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#### DrCMS

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##### Re: How many grams of octane were in the sample?
« Reply #4 on: February 15, 2011, 12:20:42 AM »

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  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  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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#### ch3mgirl

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##### Re: How many grams of octane were in the sample?
« Reply #5 on: February 15, 2011, 02:45:59 AM »

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  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  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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#### DrCMS

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##### Re: How many grams of octane were in the sample?
« Reply #6 on: February 15, 2011, 04:06:20 AM »

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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#### ch3mgirl

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##### Re: How many grams of octane were in the sample?
« Reply #7 on: February 15, 2011, 04:12:52 AM »

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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#### ch3mgirl

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##### Re: How many grams of octane were in the sample?
« Reply #8 on: February 15, 2011, 06:19:10 AM »

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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##### Re: How many grams of octane were in the sample?
« Reply #9 on: February 15, 2011, 06:52:59 AM »

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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#### ch3mgirl

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##### Re: How many grams of octane were in the sample?
« Reply #10 on: February 15, 2011, 07:28:48 AM »

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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##### Re: How many grams of octane were in the sample?
« Reply #11 on: February 15, 2011, 07:49:56 AM »

correct!
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#### ch3mgirl

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##### Re: How many grams of octane were in the sample?
« Reply #12 on: February 15, 2011, 07:54:09 AM »

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#### ch3mgirl

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##### Re: How many grams of octane were in the sample?
« Reply #13 on: February 21, 2011, 03:11:22 AM »

correct!
Oh wow.... thanks so much!

It turns out it was supposed to be 2.28g of octane.
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#### DrCMS

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##### Re: How many grams of octane were in the sample?
« Reply #14 on: February 21, 2011, 05:19:14 AM »

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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