[suhjata]

63.2mg of pure L-cysteine hydrochloride is required to prepare 100ml of a 4mM (mmol.l-1) solution of L-cysteine.
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.l-1) L-cysteine.

Express your answer in mg to 1 decimal place. Do not include the units.

Molar mass (molecular weight): L-cysteine hydrochloride = 158 g mol-1.

please help me!!!!!!!

[hmx9123]

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., and you'll have your total weight.

[jdurg]

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., and you'll have your total weight.

Don't you have to take into account the fact that the source is L-cysteine hydrochloride and the final solution is in terms of pure L-cysteine?

[hmx9123]

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.

[AWK]

1mole of L-cysteine hydrochloride contains 1 mole of L-cysteine.
So we can calculate number of moles of L-Cys, and for weight calculation use MM of L-Cys^.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.

[hmx9123]

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