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Topic: Question on purity  (Read 1230 times)

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Offline Kunshirostorm

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Question on purity
« on: January 10, 2020, 12:27:38 PM »
we whant to make S2 buy in the protocole there is something implicit...help me plz...
"to give a white solide S2 (530 mg, 63% yield)"
they say in a formula, what does it mean ?

1) is S2 = 63% of 530mg and the rest is impurities ?
like 63% of 530 is S2 = 333mg

2)or is 63% of the white solid actually 530mg of S2 ?
like
solid = 841mg and
63% x 841mg =  530mg = S2
and 311mg of impurities

so after recrystallisation, do i end up with 331mg of pure S2 crystal ?
or do i end up with 530mg of pure crystal of S2?

Offline chenbeier

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Re: Question on purity
« Reply #1 on: January 10, 2020, 12:52:38 PM »
The second one is the case.

Offline hollytara

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Re: Question on purity
« Reply #2 on: January 10, 2020, 06:46:55 PM »
The theoretical yield (if the reaction went 100%) was 841 mg.  They only got 530 mg (63% of what they could have obtained). 

They assume the S2 is 100% pure but that might not be the case!

Offline MNIO

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Re: Question on purity
« Reply #3 on: January 10, 2020, 08:12:49 PM »
the key word here is "yield"

by definition,
  theoretical yield = maximum amount possible if the reaction goes to completion
  actual yield = amount recovered
  % yield = (actual yield / theoretical yield) * 100%

or if it helps, the % yield is a measure of the efficiency of your overall reaction, collection, purification processes.

********
you were given this information
  530mg of solid (recovered)... this is "actual yield"
  63% yield
so
  theoretical yield = 530 / 0.630 = 841mg

meaning
  IF the reaction went to completion and IF you isolated and purified the solid S2
  THEN you could have recovered 841mg of S2
  but you didn't.  You only recovered 530mg of S2... whatever that is.
  so your overall "efficiency" was (530 / 841) * 100% = 63%

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