So I came across a problem in my chemistry textbook that brought up another question. The problem had to do with calculating the mass% composition of chlorine in a certain molecule.

So to go about solving this, you could calculate the mass in grams of 1 mole of the molecule and divide the mass in grams of the corresponding amount of moles of chlorine by that number and multiply that by 100 to get a percent.

The molecule had 3 Cl molecules, so I would look and see that the molar mass of 1 mole of Cl is 35.45g. So I would multiply that by 3 to get 106.35g of Cl, but I would have to round that to 4 significant figures since 3 has infinite significant figures, and the mass of Cl was given in 4. Which means the mass of 3 moles of Cl is 106.4g.

Which poses my question... Isn’t multiplying the molar mass of Cl by 3 the same as adding the molar mass of Cl 3 times? And if so, wouldn’t that mean that if you were to add the mass 3 times, then the significant figure rules for adding would tell us to round to the number in the problem with the least decimal places, which is 2 decimal places? So when you add the mass of Cl 3 times you get 106.35g so then you round to 2 decimal places and get 106.35g.

Why does this give a different answer? How come when you multiply by a number with infinite significant figures, the number of significant figures in your rounded answer can be different than if you were to add that something to itself that same amount of times?

Can’t you be certain to the tenths place in this example, and have the hundredths place by your uncertain digit since when you add, you don’t have to round any numbers off? If not, why not?

  • 5
    $\begingroup$ TLDR - With significant figures you should only round the final answer. For intermediate calculations you should carry extra figures to minimize round-off errors. I'd recommend two extra figures if you need to round intermediate results. $\endgroup$
    – MaxW
    Oct 19, 2019 at 23:40
  • 1
    $\begingroup$ I suggest to make the question better structured for easier reading. $\endgroup$
    – Poutnik
    Oct 20, 2019 at 4:45
  • $\begingroup$ If you multiply by an integer, it is clear that you are talking about the same quantity. When you are adding, it could be three independent measurements that happen to have the same value. In the latter case, some of the errors might cancel out (the combined error can be taken as root mean square). Thus, even in a more sophisticated error propagation treatment, the results might be different. $\endgroup$
    – Karsten Theis
    Oct 21, 2019 at 11:09

2 Answers 2


Bear in mind the significant digit rules are just simplifications of the rules of absolute (+, -) and relative (*,/) variance propagation.

The rule of significant digit numbers for summing is for 2 statistically independent variables. Their correlation coefficient = 0 and the error is $$ s_\mathrm{C}= \sqrt{(s_\mathrm{A})^2+(s_\mathrm{B})^2}$$ and therefore about at the same order as for less precisely known variable.

Multiplying a variable by number means summing the same variable $N$ times, with the correlation coefficient 1 and the error is growing linearly by $N$. As here applies the multiplicative rule for propagation of relative variance.

$$ s_\mathrm{C,r}= \sqrt{(s_\mathrm{A,r})^2+(s_\mathrm{B,r})^2}\\=\sqrt{(s_\mathrm{A,r})^2+(s_\mathrm{3})^2}=s_\mathrm{A,r}$$

I agree with the comment to follow the rule of significant digits on the final result only, to avoid cummulative rounding errors.

On the other side, it can be hard to track the digits but for simple formulas. Applying the variance propagation rules may be easier way to go.


Yes, there is an inconsistency in how significant figures work. As you noted when using significant figures $$35.45 \times 3 = 106.4$$ but $$35.45 + 35.45 + 35.45 = 106.35$$ As a final answer, I'd go with 106.4 but for intermediate calculations I'd use 106.35.

Significant figures are really a sloppy way of doing error propagation. The gist is that significant figures prevents stupid mistakes like equating 10.0/3 with 3.3333333333... which has infinite exactness.

To go in somewhat more detail, consider that $35.45$ is taken to mean $35.45 \pm 0.005$. In percent this is $0.005/35.45 = 0.014\% $.

$106.4 \pm 0.05 \ce{->} 0.047\%$
$106.35 \pm 0.005 \ce{->} 0.0047\%$

So an error of 0.047% is a little worse than the "real" error of 0.014%, but 0.0047% is an error which is better than the "real" error of 0.014%.


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