So, in our experiment, we measured the optical rotation of a compound with a polarimeter, and then we have to calculate the specific rotation and the error in calculating the specific rotation.

I am using the formula $$\mathrm{[\alpha]=\frac{\alpha}{lc}}$$ where l is the path length and c is the concentration of the solution.

I am propagating the error using$$\mathrm{(\delta[\alpha]/[\alpha])^2=(\delta\alpha/\alpha)^2+(\delta l/l)^2+(\delta c/c)^2}$$

However, one of the compounds is a racemic mixture and the measured value of $\alpha$ is $0$. So, how do I calculate the error in this case, because clearly, $\delta\alpha/\alpha=\infty$ ?

There are some posts on various maths forums about calculating relative error when the magnitude is zero (they suggest avoiding the use of relative error), but I am not sure that I can apply that for error propagation.

[N.B.- The optical rotation is measured using a digital polarimeter, so we are considering that the error in the measurement is half of the last decimal point]


1 Answer 1


You can always propagate uncertainty without referring to relative quantities. In general, if you have a function $F(x_1,x_2,\ldots,x_N)$ where the $x_i$'s are all independent, then $$\delta F = \sqrt{\sum_{i=1}^N \left(\frac{\partial F}{\partial x_i} \cdot \delta x_i\right) ^2}$$

In this case, we have $F(\alpha, l, c)=\frac{\alpha}{lc}$, so $$\frac{\partial F}{\partial \alpha} = \frac{1}{lc}$$ $$\frac{\partial F}{\partial l} = -\frac{\alpha}{l^2 c}$$ $$\frac{\partial F}{\partial c} = -\frac{\alpha}{lc^2}$$

and so $$\delta F = \sqrt{\left(\frac{1}{lc} \delta \alpha \right)^2 + \left(\frac{\alpha}{l^2 c} \delta l\right)^2 + \left(\frac{\alpha}{lc^2} \delta c\right)^2}$$

If your measured value of $\alpha$ is zero, then the second and third terms vanish and your uncertainty reduces to

$$\delta F = \frac{\delta \alpha}{lc}$$

If $\alpha,l,c\neq 0$, then you can obtain the standard formula for the relative uncertainty $\frac{\delta F}{F}$ by dividing that big square root by $\frac{\alpha}{lc}$.

Again, this assumes that all of your errors can be approximated as uncorrelated, normally-distributed random variables. If your errors are correlated, or if they are sufficiently large that the assumption of normality is a very bad one, then all of this is out the window. You can find more information on error propagation in the wiki entry here.

  • 2
    $\begingroup$ This will be the same result if the OP had multiplied by $[\alpha]^2$ to normalise the infinities. $\endgroup$ Commented Dec 17, 2019 at 1:51
  • $\begingroup$ The OP has already made this assumtion in their choice of equation. It is good though to be made aware of. $\endgroup$ Commented Dec 17, 2019 at 1:52

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