[ch3mgirl]

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!

[DevaDevil]

type that calculation into your calculator and tell me what the answer is

[ch3mgirl]

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?

[DrCMS]

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.

[DevaDevil]

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