[ch3mgirl]

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'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.

[DevaDevil]

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

[ch3mgirl]

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

The number of molecules?

[rabolisk]

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.

[DrCMS]

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.

[ch3mgirl]

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?

[DrCMS]

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

[ch3mgirl]

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?

[ch3mgirl]

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?

[DevaDevil]

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!

[ch3mgirl]

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?

[DevaDevil]

correct!

[ch3mgirl]

correct!

Oh wow.... thanks so much!

[ch3mgirl]

correct!

Oh wow.... thanks so much!

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

[DrCMS]

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.