ATAR Notes: Forum
VCE Stuff => VCE Science => VCE Mathematics/Science/Technology => VCE Subjects + Help => VCE Chemistry => Topic started by: bucket on May 05, 2008, 07:01:16 pm
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Calculate the percentage yield if 5.0g of ethanol is oxidised to produce 4.8g of ethanoic acid.
how on earth do you work this out?!!
edit**
looks like the answer I was given was not right haha.
thanks for all the help!
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Haha I asked coblin the exact same question and the answer isn't 92%! Apparently it's something around the 70s.
Anyway, this is how you do it
The percentage yield is defined as (the actual amount that you got)/(the amount that you would theoretically get) x 100.
The actual amount that you got would be 4.8g.
To figure out the amount you would theoretically get:
Basically one ethanol molecule would produce one ethanoic acid molecule. Thus the mole ratios of the two are the same. Thus you find out the mole of the ethanol
n(CH3CH2OH) = m/M = 5.0/[2(12)+6(1)+16] = 5.0/46 = 0.10869565mol
Therefore n(CH3COOH) = 0.10869565mol.
Theoretical mass of ethanoic acid that you would get = nxM
= 0.10869565 x (2(12)+4(1)+2(16))
=6.521739g
Therefore, percentage yield = 4.8/6.521739 x 100 = 73.6% = 74%
I don't know if the values are correct, I did them using my computer calculator and might have missed out a few digits! But that's basically how you go about doing it.
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Calculate the percentage yield if 5.0g of ethanol is oxidised to produce 4.8g of ethanoic acid.
(http://www.chemguide.co.uk/organicprops/acids/oxalceq4.gif)
assuming the two oxygens came from an oxidant during oxidation of ethanol.
THEORETICAL YIELD:
m(CH3CH2OH) = 5.0g
n(CH3CH2OH) = m/Mr = 5.0/46 = 0.10869 mol
n(CH3COOH) = n(CH3CH2OH) = 0.10869 mol
ACTUAL YIELD:
m(CH3COOH) = 4.8g
n(CH3COOH) = m/Mr = 4.8/60 = 0.08 mol
yeah i don't know how you got 92%
% yield = (actual yield)/(theoretical yield) * 100%
= 0.08 / 0.108695 * 100%
= 73.6% = 74%
oh my bad, someone's beat me to it.