Dilution factor?

[hakke]

The question is:

Iron is essential to our health. To determine the iron content in a flavoured milk drink, a 5.0 mL sample was diluted to 50.0 mL. The absorbance of the diluted solution and of several standard solutions was measured using AAS.

a) Plot a graph of abs against C of iron. (I've done this, and found C(iron) to be 2.4ppm)
b) Calculate the concentration of iron, in ppm, in the undiluted drink.

In the past ive always calculated the dilution factor like this: put the volume of what you want to find over the volume of what you have and then multiply that
in this case: c(iron)undiluted = c(iron)diluted x 5/50 = 2.4 x 0.1 = 0.24 ppm, which is wrong because the dilution factor is supposed to be 50/5... could anyone please explain the dilution factor concept to me? Have i been doing it wrong? that's how i worked it out during titration questions and it was fine..

thanks!11

[Zien]

In this case, the original concentration has been diluted 50/5 times, i.e. a tenfold dilution factor. From the graph, c(Fe) in the diluted sample is 2.4ppm. From logic, since we know this is diluted, the ppm should be higher than 2.4.

n(Fe)original = c(Fe)diluted x 50/5 = 2.4 x 10 = 24ppm.

What you've done is that you've diluted the concentration of iron even more, rather than 'undiluting' or tracing backwards to find the original concentration. You basically found the concentration of iron diluted twice, each time by a factor of 10 (so your answer is 1/100 of the original concentration).

Hope this helps? :p

[hakke]

In this case, the original concentration has been diluted 50/5 times, i.e. a tenfold dilution factor. From the graph, c(Fe) in the diluted sample is 2.4ppm. From logic, since we know this is diluted, the ppm should be higher than 2.4.

n(Fe)original = c(Fe)diluted x 50/5 = 2.4 x 10 = 24ppm.

What you've done is that you've diluted the concentration of iron even more, rather than 'undiluting' or tracing backwards to find the original concentration. You basically found the concentration of iron diluted twice, each time by a factor of 10 (so your answer is 1/100 of the original concentration).

Hope this helps? :p

yea If i think about it in a logical sense that it works, but for questions like this (also in titrations) i've just used my own 'method'
like for example a if a 20ml aliquot is removed from 100 ml of a 1.25 M solution of HCl, then the aliquot will contain:
n(aliquot) = v(aliquot)/v(solution) x n(HCl) = 20/100 x 0.125 = 0.025 mol of HCl in the aliquot

can i still think like that for dilutions and such? i've found it works before but hmmm.. arghhhh

edit: i think like this when calculating mole ratios as well, and i've found it always works

[Zien]

Ah. What you've done there isn't a dilution/undilution factor. You just found a fraction of the total amount of HCl present in 100ml. Basically you've said "Out of 20 apples, I have 2/20 = 2 apples". And if you look at the units, you divided a volume by a volume which cancels all units and makes it into a ratio. Since you already found the total amount of HCl present in 100ml (i.e. 0.125mol), the ratio you found multiplied by that total gives you 0.025mol. This is a correct way of doing this, but this isn't a dilution factor. :p

A dilution factor is something like, "A 20.00mL sample of wine is diluted up to a 750.0mL volumetric flask with distilled water. An aliquot of 20.00mL of this diluted solution was titrated against 1.25M of HCl solution and an average titre of 5.85mL was found. Find the original concentration of ethanoic acid in wine." This is a basic titration question which involves a dilution factor of 750.0mL/20.00mL = 37.5. This means the concentration you find for the aliquot is 1/37.5 of the original concentration.

A good way I like to use is to think that, to 'undilute' something, the bigger number is the numerator and the denominator is the smaller number. And to dilute something, the smaller number is the numerator and the denominator is the bigger number. <-- Not sure if this always works out like this though. :p

[hakke]

Ah. What you've done there isn't a dilution/undilution factor. You just found a fraction of the total amount of HCl present in 100ml. Basically you've said "Out of 20 apples, I have 2/20 = 2 apples". And if you look at the units, you divided a volume by a volume which cancels all units and makes it into a ratio. Since you already found the total amount of HCl present in 100ml (i.e. 0.125mol), the ratio you found multiplied by that total gives you 0.025mol. This is a correct way of doing this, but this isn't a dilution factor. :p

A dilution factor is something like, "A 20.00mL sample of wine is diluted up to a 750.0mL volumetric flask with distilled water. An aliquot of 20.00mL of this diluted solution was titrated against 1.25M of HCl solution and an average titre of 5.85mL was found. Find the original concentration of ethanoic acid in wine." This is a basic titration question which involves a dilution factor of 750.0mL/20.00mL = 37.5. This means the concentration you find for the aliquot is 1/37.5 of the original concentration.

A good way I like to use is to think that, to 'undilute' something, the bigger number is the numerator and the denominator is the smaller number. And to dilute something, the smaller number is the numerator and the denominator is the bigger number. <-- Not sure if this always works out like this though. :p

thanks! I dont quite get what you mean here thoug?

"This is a basic titration question which involves a dilution factor of 750.0mL/20.00mL = 37.5. This means the concentration you find for the aliquot is 1/37.5 of the original concentration."

:s

[Zien]

So the amount present in the original solution is the amount present in the diluted solution right, regardless of the volume it's in. e.g. n(Ethanoic acid)before dilution = n(Ethanoic acid) after dilution. But if you consider the volume after dilution, the concentration (governed by c = n/v) of the diluted solution is 37.5 times less than the original concentration. Same amount (n) but different volumes.

[hakke]

So the amount present in the original solution is the amount present in the diluted solution right, regardless of the volume it's in. e.g. n(Ethanoic acid)before dilution = n(Ethanoic acid) after dilution. But if you consider the volume after dilution, the concentration (governed by c = n/v) of the diluted solution is 37.5 times more than the original concentration. Same amount (n) but different volumes.

Shouldn't it be the C(diluted solution) is 37.5 times less than C(original solution)?

For your example, "A 20.00mL sample of wine is diluted up to a 750.0mL volumetric flask with distilled water."

Let us assign C(20.00ml sample of wine) = x
So then wouldn't C(ethanoic acid in 750 mL after dilution) = x * 20/750?

Is that right? :S

[Zien]

Ah, my bad. It should be 37.5 less (I wondered why I put more..). Sorry about that! = ='' I'll edit my post up there.

Now this makes me wonder how many other errors I posted. : /

[hakke]

Ah, my bad. It should be 37.5 less (I wondered why I put more..). Sorry about that! = ='' I'll edit my post up there.

Now this makes me wonder how many other errors I posted. : /

phewwww, you scared me i thought i was having it wrong the entire time LOL
so it'd involve a dilution factor of 20/375 instead of 375/20?

so for the usual titration questions, do you usually just think like this? "A good way I like to use is to think that, to 'undilute' something, the bigger number is the numerator and the denominator is the smaller number. And to dilute something, the smaller number is the numerator and the denominator is the bigger number." (i.e. what you said before), or would you think about it logically?

btw, do you think my 'fraction/ratio-ing' of the moles and volume etc would work in pretty much any scenario?

[Zien]

I think the reason why I put more instead of less is because I was confusing the terminology. As long as you understand the concept then it's fine.

So 20/375 when you want to want find its concentration as a diluted solution, and 375/20 when you want to find its concentration as a undiluted (i.e. original) solution.

I would usually think about it logically I think. If you have a diluted sample, then to find the original concentration which would be higher, the factor to multiply it by would be >1 (bigger number / smaller number). If you have a concentrated sample, then to find the diluted concentrated which would be lower, the factor to multiply it by would be <1 (smaller number / bigger number). At least, that's what I think anyway.

[Zebra]

theres 1ml of acid. you add 9ml of water.
is the dilution factor 10?

[The Detective]

yeah it should be diluted by a factor of 10
50/5 = 10

sorry this is in response to original qns