Chemical Forums
Chemistry Forums for Students => Undergraduate General Chemistry Forum => Topic started by: suhjata on April 27, 2004, 08:49:23 AM

63.2mg of pure Lcysteine hydrochloride is required to prepare 100ml of a 4mM (mmol.l1) solution of Lcysteine.
Chemical analysis of an old batch of the chemical has shown that it is only 80% pure.
Calculate the amount of the impure material that is required to prepare 100ml of 4mM (mmol.l1) Lcysteine.
Express your answer in mg to 1 decimal place. Do not include the units.
Molar mass (molecular weight): Lcysteine hydrochloride = 158 g mol1.
please help me!!!!!!!

Well, you're actually got more information than you need, as far as I can tell.
You are given that for a pure solution, you need 63.2mg; this means that you still need this amount of pure cysteine. However, now your weight is coming from a sample that is only 80% pure. This means that if you take 80% of the weight of the impure sample, you'll have to get 63.2mg. So, if 63.2mg if only 80% of the weight, how do you figure out what the full weight is? Just divide 63.2mg by 80% (0.8), and you'll have your total weight.

Well, you're actually got more information than you need, as far as I can tell.
You are given that for a pure solution, you need 63.2mg; this means that you still need this amount of pure cysteine. However, now your weight is coming from a sample that is only 80% pure. This means that if you take 80% of the weight of the impure sample, you'll have to get 63.2mg. So, if 63.2mg if only 80% of the weight, how do you figure out what the full weight is? Just divide 63.2mg by 80% (0.8), and you'll have your total weight.
Don't you have to take into account the fact that the source is Lcysteine hydrochloride and the final solution is in terms of pure Lcysteine?

s#*$, I read that wrong. Sorry about that... I thought the end product was the hydrochloride as well.... OK, that makes the question a little more complicated. I'll reply to this tonight when I have more time.

1mole of Lcysteine hydrochloride contains 1 mole of Lcysteine.
So we can calculate number of moles of LCys, and for weight calculation use MM of LCys^{.}HCl. Then we should multiply our result by 1.25 (which is equivalent to Hmx9123 "/80%").
So the first post of Hmx9123 is absolutely correct.

Man, I need to stop reading these when I'm halfasleep. AWK, thanks, you're right, the original answer is correct.