Chemical Forums
Chemistry Forums for Students => High School Chemistry Forum => Topic started by: purplemonkey123 on November 17, 2015, 12:26:32 PM
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What is the percentage yield of indigo?
Mass of indigo obtained in the experiment: 0.17g
My work:
2C3H6O + 2C7H5NO3 C16H10N2O2 + 2C2H4O2 + 2H2O
MC7H5NO3 = 151.13 g/mol
MC16H10N2O2 = 262.28 g/mol
nC7H5NO3 = 0.25 g C7H5NO3 ((2 mol)/(151.13g))
= 0.0033084 mol C7H5NO3
nC16H10N2O2= 0.0033084 mol C7H5NO3 ((1 mol )/(2 mol))
= 0.0016542 mol C16H10N2O2
mC16H10N2O2= 0.0016542 mol C16H10N2O2 ((262.28 g C16H10N2O2)/(1 mol C16H10N2O2))
= 0.43 g
Percentage yield = (actual yield)/(theoretical yield) ×100
= (0.17 g)/(0.43 g) ×100
= 39%
My friend got 73%? So I dont know if this is right
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nC7H5NO3 = 0.25 g C7H5NO3 ((2 mol)/(151.13g))
Why multiply by 2 mol? You divide by 2 in the next step to get the stoichiometry right.
You actually have 0.0016542 mol, so you expect 0.0008271 mol indigo.
Your answer is out by a factor of 2.
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I'm confused... so you're saying the first step should be:
nC7H5NO3 = 0.25 g C7H5NO3 (1 mol C7H5NO3/151.13 g C7H5NO3) ?
Why is that? It has a coefficient of 2. I thought this means that there is 2 moles?
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At this stage you're just working out how many moles of compound you have, irrespective of the reaction stoichiometry. MW = 151.13 g/mol, so
nC7H5NO3 = 0.25 g C7H5NO3 (1 mol C7H5NO3/151.13 g C7H5NO3)
is correct.
The stoichiometry is taken care of by the factor (1 mol/2 mol) in your next equation - i.e. you expect 1 mol indigo for every 2 mol of precursor.