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

Seems so simple but is confusing me.

A steel sample ($\pu{0.3500 g}$) was dissolved in nitric acid and the resulting solution made up to $\pu{500 ml}$.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.

In the final diluted sample, $$\frac{\pu{6 ppm}\ \ce{Ni}}{0.410\times0.522}=\pu{28.03476311 ppm}\ \ce{Ni} \tag{1 }$$

In the first solution $$\frac{\pu{28.03476311 ppm}\ \ce{Ni}}{\pu{500 ml}}=\pu{0.05606952622 mg}\ \ce{Ni} \tag{2}$$

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

$$\pu{1000 g}\times\pu{0.05606952622\times10^{-6} g}=\pu{5.60695\times10^{-5} g} \tag{3}$$

In the steel sample:

$$\frac{\pu{5.60695\times10^{-5} g}}{\pu{0.35 g}}=0.000160198=0.016\ \%\ \ce{Ni}\tag{4}$$

$2$nd attempt:

\begin{align} \frac{\pu{6 ppm}}{0.522}&=\frac x{0.410}\\ \implies \frac{\pu{6 ppm}\times0.410}{0.522}& =x\\ x&=\pu{4.7126ppm}\ \ce{Ni} \tag{6} \end{align}

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

Since it was marked up to $\pu{500 ml}$

In the steel sample: \begin{align} \pu{0.47126 mg}\times\frac{500}{50}&=\pu{4.7126 mg ml-1}\\ \frac{4.7126}{100}&=0.047126\\ \frac{\pu{0.047126 g}}{\pu{0.35 g}}&=0.134645\\ &=13.46\ \%\ \ce{Ni} \tag{7} \end{align}

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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    $\begingroup$ 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?

The 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. $5$ people $\cdot$($14$ eyes/$10$ eyes) $ = 5$ people $ \times 1.4$ (unitless ratio) $= 7$ people. Here, the units make sense. This is correct.

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

First you find the amount of $\ce{Ni}$ in the final sample.

Your sample absorbs $0.410$ which is slightly less than $0.522$ ($\pu{6 ppm }$sample which). So your sample has $(0.410/0.522) \times \pu{6 ppm} = \pu{4.713 ppm}.$

Water solution has density of $\pu{1 kg/L}$, so $\pu{500 mL} \to \pu{500 g}$. $\pu{4.713 ppm}$ in $\frac{\pu{500 g}}{1\,000\,000} = \pu{0.5 mg}$. $\pu{0.5 mg} \cdot 4.713 = \pu{2.356 mg}$.

The total weight of the sample was $\pu{0.35 g} = \pu{350 mg}$

$$100\ \% \cdot \frac{\pu{2.356 mg}}{\pu{350mg}} = 1.003\ \%$$

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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