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A cereal sample was analysed for Calcium content as follows: A 5g cereal sample was digested with HCl and the resulting mixture made up to 50ml in a volumetric flask. The absorbance of this solution was found to be 0.351. A 7ppm calcium solution gave an absorbance of 0.254. Assuming linearity between absorbance and concentration, calculate the calcium content of the cereal in mg Ca/100g cereal. (8 marks)

Seems so simple but is confusing me as when I take into account calcium content of the cereal in mg Ca/100g cereal.

A 5g cereal sample was digested with HCl and the resulting mixture made up to 50ml in a volumetric flask.

The absorbance of this solution was found to be 0.351.

A 7ppm calcium solution gave an absorbance of 0.254

7 ppm Ca solution/ absorbance 1 x absorbance 2 =

step 1: 7 ppm/ (0.351 x 0.254) = 78.515 ppm ca in final dilute sample.

step 2: 78.5158/50 ml (of initial volume sample)= 1.57 ppm ca in the first solution.

Supposing the density of the first solution to be near that of water:

step 3: (1000g) x (1.57 x 10^-6) = 1564 g Ca

step 4: (1564 g ca / 5g) = 312.8 g/100 g

answer: 3.128x10^5 mg/100 g would be the initial answer I receive.

My second attempt looking at it a different way...

feeling that I have done step 1 correctly. Starting at step 2

78.5158/(50 ml x 2) (multiplying initial volume by 2 to achieve 100 ml) = .785158 ppm ca in 100 ml.

step 3: (1000g x 7.85x10^-4g)=7846 g ca /100 ml

7846000 mg ca / 100 ml

answer: 7.846 x 10^6 mg ca/ 100 ml

third attempt:

Seeing that the initial sample is 5 g and the question is looking for calcium content in a 100 g sample...

5g sample x 20 = 100 g

therefore other components are multiplied by 20

V= 50ml x 20 = 1000 ml

step 1: step 1: 7 ppm/ (0.351 x 0.254) = 78.515 ppm ca in final dilute sample.

step 2: 78.5158/1000 ml (of initial volume sample x20)= 0.0785 ppm ca in the first solution.

step 3: (1000g) x (.0785 x 10^-6g) = 72.5 g Ca

step 4: (72.5 g ca / 100g) = .725 g Ca

answer: 725 mg Ca content/ 100 g cerial

I feel that I am missing a step in calculations. I would prefer to read a resource of working this out myself than finding the answer.
I feel more confident in my third attempt of finding a correct answer.

This question is from a past paper.

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  • $\begingroup$ Good job writing your attempts in detail. It makes pinpointing errors easier. $\endgroup$ – sixtytrees Jul 20 '16 at 20:53
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The absorbance of this solution was found to be 0.351. A 7ppm calcium solution gave an absorbance of 0.254

You have slightly (~1.5 times) higher absorbance (.35) than a solution of 7 ppm calcium (0.24)

step 1: 7 ppm/ (0.351 x 0.254) = 78.515 ppm ca in final dilute sample.

This is not correct. You should have proportion
7/0.254 = x/0.351 leading to x = 7*0.351/0.254

You get 10 times higher value. This is the first error. You should have done 7*(0.351/0.254) = 9.67 ppm.

In 50 mL it is A, but when it was in 5 mL it was 50/5 = 10 times higher. 7*(0.351/0.254)*(50/5)=96.7 ppm.

This means you have 96.7 mg calcium per 1 kG of cereal (ppm is one millionth, and thus, mg per kG). Then one tenth of that amount of cereal (0.1 kg/1 kg) contains 1/10 of the calcium. that is 9.67 mg.

Once again, you should have had these steps: So, you should have these steps:

(1) Since the final sample absorbs .35/.254 ~1.38 times more it is 1.38 times more concentrated than 7 ppm. That is 9.67 ppm.

(2) It was diluted 50/5= 10 times, so it was 10 times more concentrated before. That is 96.7 ppm.

(3) For 100 g of cereal: 100 000 mg * 96.7 /1 000 000 = 96.7 mg /10 = 9.67 mg.

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  • $\begingroup$ Thank you very much sixtytrees. I Appreciate it. Are there any resources that you recommend? $\endgroup$ – Mike Jul 21 '16 at 3:00
  • $\begingroup$ I have made another question. I really hope i've wrapped my head around it. [link] (chemistry.stackexchange.com/questions/55335/…) @sixtytrees $\endgroup$ – Mike Jul 21 '16 at 4:09
  • $\begingroup$ @sixytrees I'm just wondering where 5 ml is coming from in the above workings. $\endgroup$ – Mike Jul 21 '16 at 6:19

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