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I was finding the molecular weight of $\ce{Br2}$ the other day and noticed that if I add the weight of $\ce{Br}$ to itself, $79.90+79.90=159.80$, that answer for the weight has $5$ significant figures. But if I just multiply the weight by $2$, $79.90 \times 2 = 159.8$, the weight has $4$ significant figures. If adding something to itself is the same as multiplying it by $2$ mathematically, which accuracy is more justifiable for my calculations in cases like this?

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  • $\begingroup$ You forgot 2 * 0 = 0 . The multiplied answer contains 5 significant figs too $\endgroup$
    – slhulk
    Oct 10 '15 at 3:43
  • $\begingroup$ but you have to round it down to 4 because 79.90 only has 4 sig figs... sigfigscalculator.com $\endgroup$ Oct 10 '15 at 3:46
  • $\begingroup$ According to your argument the added answer should also be rounded off to 4 significant values. It should not be 159.80 $\endgroup$
    – slhulk
    Oct 10 '15 at 3:50
  • $\begingroup$ no, the rules for addition are that the lowest decimal place is kept, so two decimal places. Again, you can verify this with any online sig. fig. calculator $\endgroup$ Oct 10 '15 at 3:58
  • $\begingroup$ If this so, your argument "but you have to round it down to 4 because 79.90 only has 4 sig figs..." fails. $\endgroup$
    – slhulk
    Oct 10 '15 at 4:03
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Your calculations are in accordance with the rules of the so-called significance arithmetic even though they yield different results.

When multiplying numerical values, according to the rules of significance arithmetic, the result is rounded to the number of significant digits in the factor with the least significant figures. When you multiply the value of the molar mass $M_{\ce{Br}}=79.90\ \mathrm{g/mol}$ by 2, the value of the molar mass $M_{\ce{Br}}$ has four significant digits, whereas the factor 2 is assumed to be exact. Accordingly, the result is rounded to four significant digits:

$$M_{\ce{Br2}}=2M_{\ce{Br}}=2\times79.90\ \mathrm{g/mol}=159.8\ \mathrm{g/mol}\tag1$$

When adding numerical values, according to the rules of significance arithmetic, the result is rounded to the position of the least significant digit in the most uncertain of the numbers being summed. When you add the molar mass $M_{\ce{Br}}=79.90\ \mathrm{g/mol}$ to itself, the value of the molar mass $M_{\ce{Br}}$ is significant to the hundredth place. Accordingly, the result is rounded to the hundredth place:

$$M_{\ce{Br2}}=M_{\ce{Br}}+M_{\ce{Br}}=79.90\ \mathrm{g/mol}+79.90\ \mathrm{g/mol}=159.80\ \mathrm{g/mol}\tag2$$

However, this concept of significant digits is just an oversimplified form of the concept of uncertainty as a quantifiable attribute. In your question, you are testing the limits of the applicability of this simplification, which is certainly a good idea.

When a number is given without any further information, it is generally interpreted so that the last digit is rounded. Thus, the molar mass of $M_{\ce{Br}}=79.90\ \mathrm{g/mol}$ is generally assumed to represent a value between $79.895\ \mathrm{g/mol}$ and $79.905\ \mathrm{g/mol}$, or $M_{\ce{Br}}=\left(79.900\pm0.005\right)\ \mathrm{g/mol}$.

Therefore, the assumed uncertainty of the molar mass $M_{\ce{Br}}$ is $u(M_{\ce{Br}})=0.005\ \mathrm{g/mol}$. (Note that, normally, the uncertainty corresponds to the estimated standard deviation; in this case, however, the uncertainty just indicates an interval corresponding to a high level of confidence.)

When you multiply the value of the molar mass $M_{\ce{Br}}$ by 2, you also have to multiply its uncertainty by 2 (the factor 2 is assumed to be exact):

$$M_{\ce{Br2}}=2M_{\ce{Br}}=2\times79.90\ \mathrm{g/mol}=159.80\ \mathrm{g/mol}\tag{3a}$$

$$u(M_{\ce{Br2}})=u(2M_{\ce{Br}})=2u(M_{\ce{Br}})=2\times0.005\ \mathrm{g/mol}=0.010\ \mathrm{g/mol}\tag{3b}$$

When you add the value of the molar mass $M_{\ce{Br}}$ to itself, you also have to add the uncertainty to itself:

$$M_{\ce{Br2}}=M_{\ce{Br}}+M_{\ce{Br}}=79.90\ \mathrm{g/mol}+79.90\ \mathrm{g/mol}=159.80\ \mathrm{g/mol}\tag{4a}$$

$$\begin{align}u(M_{\ce{Br2}})&=u(M_{\ce{Br}}+M_{\ce{Br}})=u(M_{\ce{Br}})+u(M_{\ce{Br}})=0.005\ \mathrm{g/mol}+0.005\ \mathrm{g/mol}\\&=0.010\ \mathrm{g/mol}\tag{4b}\end{align}$$

(Note that such a simple addition of uncertainties is not correct when the errors of the added values are uncorrelated, for example when calculating $M_{\ce{HBr}}=M_{\ce{H}}+M_{\ce{Br}}$.)

Both calculations yield the same results for the molar mass $M_{\ce{Br2}}$ ($\text{(3a)}$ and $\text{(4a)}$) as well for its uncertainty ($\text{(3b)}$ and $\text{(4b)}$). Thus, the molar mass of $M_{\ce{Br2}}$ is estimated to represent a value between $159.79\ \mathrm{g/mol}$ and $159.81\ \mathrm{g/mol}$, or $$M_{\ce{Br2}}=\left(159.80\pm0.01\right)\ \mathrm{g/mol}$$

By way of comparison, the result of the multiplication according to the rules of significance arithmetic $\text{(1)}$, $M_{\ce{Br2}}=159.8\ \mathrm{g/mol}$, represents a value between $159.75\ \mathrm{g/mol}$ and $159.85\ \mathrm{g/mol}$, or $$M_{\ce{Br2}}=\left(159.80\pm0.05\right)\ \mathrm{g/mol}$$ Thus, this approach overestimates the uncertainty of the result.

The result of the addition according to the rules of significance arithmetic $\text{(2)}$, $M_{\ce{Br2}}=159.80\ \mathrm{g/mol}$, represents a value between $159.795\ \mathrm{g/mol}$ and $159.805\ \mathrm{g/mol}$, or $$M_{\ce{Br2}}=\left(159.800\pm0.005\right)\ \mathrm{g/mol}$$ Thus, this approach underestimates the uncertainty of the result.

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If the $0$ in $79.90$ is important, the when you multiply by $2$ also, you get $159.80$ so there is no discrepancy. However, you do realise that suppose the weight was $79.904$ (in fact, it is !), writing 2 decimals only will be $79.90$ and times $2$ will produce $159.80$. However, if you had written $79.904$, you would have got $159.808$, which up to $2$ decimals is $159.81$

In multiplication, the simple rule is, the number of significant digits of the answer is that of the multiplicand with the least number of sinificant digits.

Here is one such guide which you can go through.

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  • $\begingroup$ When you multiply 2 (exact) times 79.90 (4 sig figs) it equals 159.80, but because you have to round to 4 sig figs, it's 159.8, right? $\endgroup$ Oct 9 '15 at 19:45
  • $\begingroup$ Try sigfigscalculator.com to see what I mean (enter "2.000000*79.90") $\endgroup$ Oct 9 '15 at 21:56

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