One of my exam questions is about lipstick and calculating the amount of lithium in the lipstick by AAS. The determination took place by making standard solutions with known concentrations of lithium. For the stock solution $653.3~\mathrm{mg}$ lithium chloride, $\ce{LiCl}$, was dissolved in $100~\mathrm{ml}$.

$M(\ce{Li}) = 6.49~\mathrm{g/mol}$
$M(\ce{LiCl}) = 42.4~\mathrm{g/mol}$

From the stock solution the following volumes were pipetted into $100~\mathrm{ml}$ flasks: $3$, $5$, $8$ and $10~\mathrm{ml}$. Also $1~\mathrm{ml}$ $\ce{KCl}$ solution, $1.0~\mathrm{M}$, was added to prevent ionisation.

I calculated the concentration Lithium in the stock as follows:

The mass of Lithium in the weighed $\ce{LiCl}$ is equal to the molar mass of lithium divided by the molar mass of lithium chloride multiplied by the mass that was weighed. So in the form of an equation:

$$m(Li)=\frac{M(\ce{Li})}{M(\ce{LiCl})}\cdot m(\text{total})=99.9~\mathrm{mg}$$

Therefore the concentration of the stock solution is equal to the mass of lithium divided by the volume of the flask:

$$C(\text{Li, stock}) = \frac{m(\ce{Li})}{V(\ce{Li})}=\frac{99.9}{0.1}=999~\mathrm{mg/l}$$

The concentration of the standard solutions therefore would be:

$$C(\text{standard}) = C(\text{stock}) \times \frac{V(\text{stock})}{V(\text{flask volume})}$$

With the concentration of the standard in $\mathrm{mg/L}$ and the volumes of both stock and flash volume in $\mathrm{mL}$.

This would give concentrations of $29.9$; $49.9$; $79.9$ and $99.9~\mathrm{mg/L}$ to the standard solutions where $3$, $5$, $8$ and $10~\mathrm{ml}$ were pipetted into. The answer sheet however provides us with $300$, $500$, $800$ and $1000~\mathrm{\mu g/L}$. I suspect I've made a mistake in unit conversion somewhere, due to the factor 100 mistake in the answer, although the units don't correspond either.

  • $\begingroup$ I don't see anything wrong with the math. I calculate the same numbers you have. $\endgroup$
    – LDC3
    Commented Apr 19, 2015 at 16:02
  • $\begingroup$ @LDC3 Which is strange, because the answer sheet provides me with a different answer O_O. I'll send an email to my teacher. $\endgroup$
    – Shaggy
    Commented Apr 20, 2015 at 12:22

1 Answer 1


Well, I'll start by saying I redid your math, and you have rounding errors. Using correct sig figs, your stock solution has 100. mg of Li in it, giving a concentration of 1.00E3 mg/L. Those dilutions therefore should be 30.0 mg/L, 50.0 mg/L, 80.0 mg/L, and 100. mg/L.

You're correct aside from the rounding errors, and the answer key is wrong. The question as you've written can't have those answers, as 300 μg/L is 0.3 mg/L, which is not the answer.


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