October 15, 2019, 02:37:36 AM
Forum Rules: Read This Before Posting Topic: Calculate the concentration (v/v%)  (Read 372 times)

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• Mole Snacks: +0/-0 Calculate the concentration (v/v%)
« on: April 16, 2019, 08:40:19 AM »
Hey Guys!

I have a problem with task below, hope you can navigate/ help me with it In a 100 mL measuring flask 30 µL of butan-1-ol is placed by pipetting, The flask is filled to the mark with technical grade acetone,

The solution is analysed once and the following areas are found:

Area of
propan-2-ol   854
Area of
butan-1-ol           405

Calculate the concentration (v/v%) of propan-2-ol in the sample, Show the calculations:

What I did until now, I deviated Area of sample / Area of Internal standard (which is butan-1-ol, 0,03% v/v), I know that concentration I should find by drawing standard curve, but how if concentration is not given?

So formula should looks like that: c= (Area of sample/Area of IS - intercept)/slope . but, how to calculate concetration to draw std curve? Thank you in advance! Borek Re: Calculate the concentration (v/v%)
« Reply #1 on: April 16, 2019, 09:27:12 AM »
The only thing you can do it to assume signal to be directly proportional to the surface area.

This is equivalent to assuming at C=0 Area=0 - this way you have two points to build your curve on.
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• Mole Snacks: +0/-0 Re: Calculate the concentration (v/v%)
« Reply #2 on: April 16, 2019, 10:03:27 AM »
I'm not quite sure if assumec=0, please look below

the answer is given c= (854/405+0,0116)/35,529 = 0,0597%

this is what the Lecturer gave to us and I really do not how he got intercept, and slope for that, plus concentration Borek Re: Calculate the concentration (v/v%)
« Reply #3 on: April 16, 2019, 10:10:58 AM »
There is no way to derive the formula used from the numbers given, so either you are missing something (earlier part of the lecture, page in a book, whatever), or you were given incomplete data.
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« Reply #4 on: April 16, 2019, 12:07:59 PM »
All class got same as I, so I think I will write to Lecturer and confirm Thank you for reply!