Sciencemadness Discussion Board

Help needed with maths please.

CHRIS25 - 15-7-2012 at 08:07

Yes yes, I know, but I simply can not compute my brain with maths, and I find it infuriating.

I want to make a 5% concentration of acetic acid from a 40% concentration into 1 litre. Yes stupid isn't it, but I simply can not work it out - I have 20 pages in two chapters of two chemistry books that deal with everything moles and mass and more, but not one clue on how to work this one out.

I reasoned that 100% acetic is 17.4 mol. Therefore 40% of this would be 6.96 mol. 5% of this would be 0.87 mol. But I don't have the concentrated so taking 5% of my 40% would be giving me around 2% strength because there is 60% water. So juggling maths and logic and getting all screwed up because I can not work out the maths to do this accurately I humbly admit my stupidity and ask for the formula please.

I did reason out that 5/40 x 100/1 = 12.5% and wondered if this was correct. 12.5 ml in 1 litre from the 40% would give me 5% strength I am looking for.

Thankyou.

[Edited on 15-7-2012 by CHRIS25]

Fossil - 15-7-2012 at 10:16

Give this a try
http://www.sigmaaldrich.com/chemistry/stockroom-reagents/lea...

If you scroll down a little, the formula will be listed as well as a step by step walkthrough.

CHRIS25 - 15-7-2012 at 13:40

Quote: Originally posted by Fossil  
Give this a try
http://www.sigmaaldrich.com/chemistry/stockroom-reagents/lea...

If you scroll down a little, the formula will be listed as well as a step by step walkthrough.


Well, thankyou fossil, that site is a great site for me. Much appreciated.

But how do I calculate the percentage of my result? When working with concentrated solutions that is not a problem, but after calculating 1 mole from 40% on that site's calculator I find my maths was spot on regarding moles, but can not figure out percentages when the solutions are already not concentrated.
The 1 mole is 143 mL in a 1000 mL water. Is this 14.3%?

[Edited on 15-7-2012 by CHRIS25]

kavu - 15-7-2012 at 14:07

No need to bother with molarities. It can be solved with the good old c1V1 = c2V2. Percentages are directly proportional with volumes. You want to have 1 L of 5% acetic acid, so V1 = 1L and c1 = 5%. V2 is the volume you need of the 40% solution and c2 = 40%. Solve for V2 and plug in the values:

V2 = c1/c2 * V1 = 0.125 * 1 L = 125 mL

So measure out 125 mL of 40% solution and dilute it to to 1 liters.

Funkerman23 - 15-7-2012 at 14:40

Quote: Originally posted by kavu  
No need to bother with molarities. It can be solved with the good old c1V1 = c2V2. Percentages are directly proportional with volumes. You want to have 1 L of 5% acetic acid, so V1 = 1L and c1 = 5%. V2 is the volume you need of the 40% solution and c2 = 40%. Solve for V2 and plug in the values:

V2 = c1/c2 * V1 = 0.125 * 1 L = 125 mL

So measure out 125 mL of 40% solution and dilute it to to 1 liters.
That from Vogel's bychance?

kavu - 15-7-2012 at 15:34

Nope, that's found in any introductory chemistry textbook that has ever existed! It should also be found in every chemist's head.

[Edited on 15-7-2012 by kavu]

CHRIS25 - 15-7-2012 at 16:22

Quote: Originally posted by kavu  
No need to bother with molarities. It can be solved with the good old c1V1 = c2V2. Percentages are directly proportional with volumes. You want to have 1 L of 5% acetic acid, so V1 = 1L and c1 = 5%. V2 is the volume you need of the 40% solution and c2 = 40%. Solve for V2 and plug in the values:

V2 = c1/c2 * V1 = 0.125 * 1 L = 125 mL

So measure out 125 mL of 40% solution and dilute it to to 1 liters.


Thanks, great help. Though it would have been easier if I had simply seen the equation 5/40 x 1000/1. Now That I can comprehend, that makes sense. Going through the chemistry section I see this formula, but yeh, they complicate matters by throwing out c's and v's c1's and v2's without a simple straight-forward explanation of the mathematical principle at work. That annoys me, had I have seen just good old fashioned figures written in a good old fashioned way then I would have grasped it and no need to post on the forum.