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A steel sample (0.3500 g) was dissolved in nitric acid and the resulting solution made up to 500 ml. The absorbance of this solution measured by atomic absorption was found to be 0.410. A standard 6 ppm solution of Ni gave an absorbance of 0.522 under the same conditions. Assuming linearity between absorbance and concentration, calculate the % Ni in the steel sample. (8 marks)

Seems so simple but is confusing me.

A steel sample (0.3500 g) was dissolved in nitric acid and the resulting solution made up to 500 ml.The absorbance of this solution measured by atomic absorption was found to be 0.410.

A standard 6 ppm solution of Ni gave an absorbance of 0.522 under the same conditions.

step 1: $\displaystyle\frac{6\ \mathrm{ppm}\ \ce{Ni}}{0.410\times0.522}=28.03476311\ \mathrm{ppm}\ \ce{Ni}$ in the final diluted sample

step 2: $\displaystyle\frac{28.03476311\ \mathrm{ppm}\ \ce{Ni}}{500\ \mathrm{ml}}=0.05606952622\ \mathrm{mg}\ \ce{Ni}$ in the first solution

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

step 3: $\displaystyle1000\ \mathrm g\times0.05606952622\times10^{-6}\ \mathrm g=5.60695\times10^{-5}\ \mathrm g$

step 4: $\displaystyle\frac{5.60695\times10^{-5}\ \mathrm g}{0.35\ \mathrm g}=0.000160198=0.016\ \%\ \ce{Ni}$ in the steel sample.

2nd attempt:

$\begin{align}\displaystyle\frac{6\ \mathrm{ppm}}{0.522}&=\frac x{0.410}\\[6pt] \frac{6\ \mathrm{ppm}\times0.410}{0.522}&=x\\[6pt] x&=4.7126\ \mathrm{ppm}\ \ce{Ni}\end{align}$

$4.7126\ \mathrm{ppm}$

This means you have $4.7126\ \mathrm{mg}$ nickel per $1\ \mathrm{kg}$ of steel (ppm is one millionth, and thus, mg per kg). Then one tenth of that amount of steel $(0.1\ \mathrm{kg}/1\ \mathrm{kg})$ contains 1/10 of the $\ce{Ni}$. that is $0.47126\ \mathrm{mg}$

since it was marked up to $500\ \mathrm{ml}$

$\begin{align}\displaystyle0.47126\ \mathrm{mg}\times\frac{500}{50}&=4.7126\ \mathrm{mg/ml}\\[6pt] \displaystyle\frac{4.7126}{100}&=0.047126\\[6pt] \displaystyle\frac{0.047126\ \mathrm g}{0.35\ \mathrm g}&=0.134645\\[6pt] &=13.46\ \%\ \ce{Ni}\end{align}$
in the steel sample.

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 second attempt of finding a correct answer.

This question is from a past paper.

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  • 2
    $\begingroup$ Your first attempt goes wrong in your first step; this is corrected in your second attempt, but then you choose to work with 50 mL, which comes out of nowhere. It may help if you first -- ignoring the mathematics -- explain to yourself what you're trying to accomplish in every step and make sure you understand the relevant concepts (proportions and unit conversion). $\endgroup$ – a-cyclohexane-molecule Jul 21 '16 at 13:41
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Your first step is incorrect. Think about it this way:

5 people have 10 eyes. This group of people has 14 eyes. How many people in the group?

Correct solution is "this group has 14/10 ~1.4 times more eyes than the known group. Thus they have 1.4 times more people. 5people *(14 eyes/10 eyes) 5 people *1.4 (unitless ratio) = 7 people". Here units make sense. This is correct.

In your calculation you do (5 people)/(10 eyes * 14 eyes)= 1 people/(28 eyes$^2). This unit people/eyes$^2$ makes no sense. This is incorrect.

First you find the amount of Ni in the final sample.
Your sample absorbs 0.410 which is slightly less than 0.522 (6 ppm sample which). So your sample has (0.410/0.522) * 6ppm = 4.713 ppm.
Water solution has density of 1 kg/L, so 500 mL -> 500g. 4.713 ppm in 500g/1 000 000 = 0.5 mg. 0.5 * 4.713 = 2.356 mg.
The total weight of the sample was .35 g = 350 mg.
100% * 2.356mg/350mg = 1.003%.
You should round it down to 2 digits, because that was the accuracy of other data. The answer is 1.00%.

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