# A steel sample was analysed for nickel

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.

• 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). Commented Jul 21, 2016 at 13:41

$$5$$ 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 $$1.00\ \%$$.