How to correctly solve density tolerance

Question

I want to know if my solution to a textbook problem has any major problems with it. Here is the problem:

Ethanol has a given density of 0.789 g/mL at 20 degrees Celsius and isopropanol has a given density of 0.785 g/mL at 20 degrees Celsius. A chemist analyzes a substance using a pipet that is accurate to $\pm 0.02$ mL and a balance accurate to $\pm 0.003$ g. Is this equipment precise enough to distinguish between ethanol and isopropanol?

And here is my solution:

We can calculate with tolerances in the same way we calculate measurements. The mass tolerance of $\pm 0.003$ g has three significant figures. The volume tolerance of $\pm 0.02$ mL has two significant figures. The density tolerance will therefore have two significant figures. $\pm 0.003 \text{ g} / \pm 0.02 \text{ mL} = \pm 0.15 \text{ g/mL}$.In order to distinguish between ethanol and isopropanol, whose densities differ by 0.789 - 0.785 g/ mL, or 0.004 g/ mL, we need a precision smaller than half the difference, or 0.002 g/mL. But we can only measure density to within 0.15 g/mL of the actual value. Therefore, this equipment is not precise enough to distinguish between ethanol and isopropanol.

But what I don't like or feel is right about dividing the tolerances like that is that having a smaller tolerance (more precise) for volume in the denominator blows up (bad) your density tolerance. Shouldn't higher precision (smaller tolerance) of either pipet OR balance result in higher precision (smaller tolerance) of density measurement?

Edit: I tried to give just as much information as is relevant to my problem, but I guess one detail from part (a) of the textbook problem (what I described was part (b)), which said the nominal sample volume was 15.00 mL, must carry over to part (b). So I think I assume the nominal volume of my sample is 15.00 mL plus or minus 0.02 mL.

Answer 1

Assuming your pipette has a volume of 15ml, the error in measuring the volume of your sample is 0.02 in 15 i.e. 0.133%.

The weight of 15cc of ethanol is 0.789 $\times$ 15 = 11.835g. The error in measuring the weight is 0.002 in 11.835 i.e. 0.025%.

Hedge physicists like me would immediately note that the error in the weight is a lot lower than the volume, so we can ignore it and just take the error in the density to be 0.13%. However, for the sake of the exercise let's do this thoroughly. The density is given by:

$$ \rho = \frac{m}{V} $$

so the percentage error in the density, $\sigma_\rho$ is given by:

$$ \sigma_\rho = \sqrt{\sigma_m^2 + \sigma_V^2} $$

where $\sigma_m$ and $\sigma_V$ are the percentage errors in the weight and volume respectively. This gives:

$$\sigma_\rho = 0.136\%$$

So the absolute error in the density of 0.789 would be 0.136% of 0.789 or 0.0011. There's one last step to do: to distinguish the two fluids we need to measure their densities (with an error of 0.0011) then subtract the measurements. Because we're subtracting two measured densities, each with an error of 0.0011, the error in the result is given by an expression similar to the one above:

$$ \sigma_{diff} = \sqrt{0.0011^2 + 0.0011^2} = 0.0016 $$

Note that this time we're using absolute errors not percentage errors. When you're multiplying or dividing you combine the percentage errors and when you're adding and subtracting you combine the absoluite errors.

Anyhow, the difference in the densities is 0.789 - 0.785 = 0.004, so our result would be 0.004 plus or minus 0.0011. Assuming the errors quoted are the 1$\sigma$ errors, the error in the result is 4$\sigma$, so we'd be 98% confident we could tell the difference.

Answer 2

I like to do my tolerance analysis the long hard way, which sometimes comes in very handy, will include it here just for the sake of it...

So $\rho = m/V$. We can then write $\rho$ as a Taylor series around $\rho_0=m_0/V_0$, and keeping only the linear terms, we would get,

$$\Delta \rho =\rho - \rho_0 = \frac{1}{V_0}\Delta m - \frac{m_0}{V_0^2}\Delta V + \ldots$$

IF we consider $\Delta \rho$, $\Delta m$ and $\Delta V$ random variables, we can compute their variance as

$$Var(\Delta \rho) = \sigma_\rho^2 = \frac{1}{V_0^2}\sigma_m^2+\frac{m_0^2}{V_0^4}\sigma_V^2$$.

These variances are absolute, not relative. By dividing both sides of the equation by $\rho_0^2 = m_0^2/V_0^2$, it is very easy to turn these into relative errors, getting John's formula for the percent error of the quotient of two variables.