J.R. S. answered • 08/01/15
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If the purity is 93%, then there is a 7% impurity. To correct for this, you need to increase the mass of the standard by 7%. 7% x 6 mg = 0.07 x 6mg = 0.42 mg, so you need to weigh out 6.42 mg. Another way to approach this problem is to weigh out 1.07 x as much as originally needed. This amounts to 1.07 x 6 mg = 6.42 mg.
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Amy A.
How would you correct for a compound which has a salt vs free base molecular weight? Multiply by the difference?
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10/19/18
Micha C.
Taking into account the above standard correction, should I also use the standard purity in the formula for assay?
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03/13/19
J.R. S.
tutor
@Micha C. I don’t understand the question.
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03/13/19
Sam S.
This only gives a ballpark correction because the extra 7% that you're saying to add also has a 7% impurity. You should be multiplying your target mass by 100/%purity to get a more accurate answer. [Total mass = 6 mg * 100 / 93% = 6.45 mg.] This is derived by rearranging the equation: Purity(%) = 100 * mass of useful product / total mass.
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04/21/20
Rd L.
08/01/15