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Topic: Stoichiometry  (Read 1850 times)

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Offline purplemonkey123

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Stoichiometry
« on: November 17, 2015, 12:26:32 PM »
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

Offline mjc123

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Re: Stoichiometry
« Reply #1 on: November 17, 2015, 12:51:47 PM »
Quote
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.

Offline purplemonkey123

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Re: Stoichiometry
« Reply #2 on: November 17, 2015, 04:27:05 PM »
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?

« Last Edit: November 17, 2015, 04:42:11 PM by purplemonkey123 »

Offline mjc123

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Re: Stoichiometry
« Reply #3 on: November 18, 2015, 04:46:54 AM »
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.

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