Need some help on these chem calculations

[Asterix]

What weight of the following soluts is required to produce a 1 molar solution of each.

CaCL2
KCL
MGCL2
C12H22O11

Ca= 40, CL= 35.5, K=39, Mg=24, C=12, O=16, H=1

do you just use moles=mass/mr?

[Chosen1]

A 1 molar solution is equivalent to the molecular weight of the constituent ions.

I.e. For CaCl2 it would be 40 + 35.5 + 35.5 = 111g

Thus a 1 molar solution of CaCl2 will have 111g of CaCl2

It's as simple as that.

You could use the equation as mentioned:

Moles = Mass/Mr
1 = ?/111
And ? = 111

[johnep]

REmember material should be dried to constant weight.
johnep

[Pharmanaut]

REmember material should be dried to constant weight.
johnep

If you know it, you can account for the water of crystallisation.

Apologies to Admin for advertising, but as I don't have a financial interest here goes...

You need introduction to pharmaceutical calculations

Amazon.co.uk: Introduction to Pharmaceutical Calculations: Books: Judith A. Rees,Ian Smith,Brian Smith

There is also a work book.

If you are still at Uni they should have it in the library.

If you are a pre-reg you can borrow books by "mail order" from the RPSGB library.
RPSGB.org: Using the library

Make sure you send them back though - they check if you have any outstanding or overdue books when you try to register!

Alternatively, try looking around the internet its got loads of 'science and stuff' on it <g>

[admin]

Apologies to Admin for advertising, but as I don't have a financial interest here goes...

There's no need to apologise for this type of reference (or advertising as you've put it) my friend.......

The forums "policy" on advertising is all about common sense really. If you were blatently advertising your own website (selling drugs, recruitment, whatever else!) then that's unacceptable, and people often message me to complain before I've even seen it. Either that or one of the moderators just deletes it.

This is a case of you pointing out a useful book to attempt to help a member, which is obviously fine. We do live in a commercial world, and often need to refer to books or companies to forward the discussion.

I believe we all know when somebody is honestly trying to help somebody else, and when they are trying to help themselves by attempting to sneak in adverts. Once the post is removed, and they get a gentle warning e-mail, they invariably never come back.

The book you link to looks good. I was actually thinking of starting a section for book reviews. The NPA section is not used any more, and the person who requested it has not made a post since May 2007, so the interest has obviously gone there. I was thinking of changing it to a book review section, where we could have for example a section for pharmacy books, and another for all type of books to keep it interesting. I thought the pharmacy book part would be useful to us all (especially the students) and the "any other" books might kick off some debates and keep it interesting.

I think we'll give it a go and see if it's popular. If it turns out to be another NPA section we can always change it again.

[pharout]

If you know it, you can account for the water of crystallisation.

Apologies to Admin for advertising, but as I don't have a financial interest here goes...

You need introduction to pharmaceutical calculations

Amazon.co.uk: Introduction to Pharmaceutical Calculations: Books: Judith A. Rees,Ian Smith,Brian Smith

There is also a work book.

If you are still at Uni they should have it in the library.

If you are a pre-reg you can borrow books by "mail order" from the RPSGB library.
RPSGB.org: Using the library

Make sure you send them back though - they check if you have any outstanding or overdue books when you try to register!

Alternatively, try looking around the internet its got loads of 'science and stuff' on it <g>

There is a book here in the U.S. that back in the day helped both me and my wife tremendously. Quite a bit devoted to pharmaceutical calculations and may be worth a look as sometimes different perspectives/presentations can make things easier to learn. It's called: "Comprehensive Pharmacy Review" by Leon Shargel, Alan H. Mutnick, Paul F. Souney and Larry N. Swanson. Last edition was the 5th that we have, but may be another out now. I found it to be a good general overview of pharmacy principles and it is the most common text used to date for passing the NAPLEX exam in the U.S.

[Asterix]

ignore thanks

[Jeff]

1. How much solid drug is required to produce 250ml of a solution when diluted 40 fold to produce a 1 in 9000 diluted solution?

The NPSA should rapidly recommend a standard way of describing dilutions.
It's been so long since I had to know exactly what it meant that I no longer know with any certainty what "diluted 40 fold " means

2. Given 2 batches of syrup (batch x= 1.32g/ml and batch y=1.28g/ml) what volumes of each are required to produce 400ml of a syrup of 1.305g/ml.

I said I'd try and talk through my thought process.
(and it's been years since I did any of this so be patient)

So....
First thought
An equal mixture of each would give a syrup of 1.30g/ml
So we need a little bit more of batch x - the more concentrated.

We need Xml of syrup x and Yml of syrup y
X + Y = 400

(X * 1.32 ) + (Y*1.28) = 400 * 1.305

Then it's just simultaneous equations (see GCSE maths)
The answer should have X slightly more than 200 and Y slightly less than 200

Jeff

[admin]

Thank you to all who have helped me, I will look into those books. I have some other calculations which I would be grateful if someone showed me how to do. I do apologise for posting these questions but we actually get no support at uni for calculations and as I dont have a level maths I am not sure how I am going to cope on this course. Anyway here are the questions.

1. How much solid drug is required to produce 250ml of a solution when diluted 40 fold to produce a 1 in 9000 diluted solution?
2. Given 2 batches of syrup (batch x= 1.32g/ml and batch y=1.28g/ml) what volumes of each are required to produce 400ml of a syrup of 1.305g/ml.

1. (250 x 40) /9000 = 1.111g

so take 1.111g and put it in 250mls

to work it back: -

1.111/250 = 0.00444g/ml or 4.44mg/ml

0.00444 x 9000 = 40 = dilution factor

2. If we assume we want equal quantities of actual syrup from each batch, split in two initially: -

400 / 2 = 200mls

Then calculate how much syrup is needed to give 200mls at each strength: -

(strength required/strength we have) x final volume required = volume needed of batch

(1.305/1.32) x 200 = 197.73mls (less then 200 as is stronger then solution required)

(1.305/1.28) x 200 = 203.90mls (more then 200 as is weaker then solution required)

adding these two together gives us more then 400mls (1.97.73 + 203.90) = 401.63mls

so we have to reduce by a factor of (400/401.63) = 0.99594

so just multiply each volume by this factor

197.73 x 0.99594 = 196.92

203.90 x 0.99594 = 203.07

then add together 196.92 + 203.07 = 400 to check it equals the volume we want

so we can use 196.92mls of solution X and 203.07mls of solution Y = 400mls in total

I hope this is right - I'm tired!

No 2 is a bit long but that's how my brain works I'm afraid! I tried to write the steps down. In reality you can do this a lot quicker then it looks. It's also easier if you stick to your calculator as you can use the memory (and thus the same number of decimal places) for each calculation.

[admin]

The answer should have X slightly more than 200 and Y slightly less than 200Jeff

I make it the other way round - X is stronger then we require, so we need less. Y is weaker then we require, so we need more.

A simpler way to do it would be: -

in 200mls of a 1.305g/ml solution there is (1.305/2) x 200 = 130.5g of syrup

for solution x we thus need (130.5/1.32) x 2 = 197.73mls

for solution Y we thus need (130.5/1.28) x 2 = 203.90mls

These add up to 401.63 so we still need to reduce by 400/401.63 to get a final volume of 400mls

The same but maybe a bit shorter........