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Calculate how many grams of p-nitrophenol are required to prepare $\pu{250 mL}$ of a $\pu{11 mM}$ solution. (Answer to 3 significant figures.)

I worked it out as

\begin{align} \frac{250}{1000}\,\mathrm{L} \times (\pu{11E-3 M}) \times \pu{139.11 g//mol} &= \pu{2.75E-3 mol} \times \pu{139.11 g//mol} \\ &= \pu{0.3825525 g} \end{align}

Is that $\pu{0.383 g}$ or $\pu{0.382 g}$ when approximated to 3 significant figures? I think it is $\pu{0.383 g}$, but that's only going by the general rule of "up" approximating a five.

Martin - マーチン
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Edward
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  • I'll point out that you made a mistake when multiplying 250 ml times 11. The result is 2.750 not 2.75. 250 ml implies +/- 0.5 ml. If it was +/- 10 ml then it should have be written as $2.5 * 10^2$ ml. – MaxW Nov 03 '15 at 02:44
  • The 11 is the "limiting" precision. It is 11 +/- 0.5. – MaxW Nov 03 '15 at 02:47

1 Answers1

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In rounding with significant figures, you want to avoid undue precision, yes, but it's also important to minimize error. It's this error minimization which is the primary goal of all "round to nearest" approaches.

So in short:

\begin{align} 0.383 - 0.3825525 &= 0.0004475\\ 0.3825525 - 0.382 &= 0.0005525 \end{align}

$0.0004475$ is less than $0.0005525$, so you should round up to $0.383$, as that's the nearest value, and will minimize any error you introduce by rounding.

Even & Odd?

So what's this about the even/odd rule? Is there any validity there?

This technique is called "Banker's rounding" and is also attempting to minimize error.

When a number falls exactly in the middle of the two values to be rounded to (that is, when the digit to round is "5" exactly), each of the two difference is equally far away. So to a first order, both will result in an equal amount of error.

But picking a consistent direction (e.g. "always round up") introduces a subtle bias in the sums over the long run. This is because the errors are biased.

.     |  Always  |  Up   | Banker's | Banker's |
Value | Round Up | Error | Rounding | Error |
------+----------+-------+---------+------+
2.0   |     2    |  0.0  |    2    |  0.0  |
2.1   |     2    | -0.1  |    2    | -0.1  |
2.2   |     2    | -0.2  |    2    | -0.2  |
2.3   |     2    | -0.3  |    2    | -0.3  |
2.4   |     2    | -0.4  |    2    | -0.4  |
2.5   |     3    | +0.5  |    2    | -0.5  |   <<<<<
2.6   |     3    | +0.4  |    3    | +0.4  |
2.7   |     3    | +0.3  |    3    | +0.3  |
2.8   |     3    | +0.2  |    3    | +0.2  |
2.9   |     3    | +0.1  |    3    | +0.1  |
3.0   |     3    |  0.0  |    3    |  0.0  |
3.1   |     3    | -0.1  |    3    | -0.1  |
3.2   |     3    | -0.2  |    3    | -0.2  |
3.3   |     3    | -0.3  |    3    | -0.3  |
3.4   |     3    | -0.4  |    3    | -0.4  |
3.5   |     4    | +0.5  |    4    | +0.5  |   <<<<<
3.6   |     4    | +0.4  |    4    | +0.4  |
3.7   |     4    | +0.3  |    4    | +0.3  |
3.8   |     4    | +0.2  |    4    | +0.2  |
3.9   |     4    | +0.1  |    4    | +0.1  |
4.0   |     4    |  0.0  |    4    |  0.0  |
------+----------+-------+---------+-------+
63.0  |    64    | +1.0  |    63   |  0.0  |   - Totals

With "always round up" you get a consistent $+0.5$ bias per decade, and this is consistent even if you extend the range to more decimal places. You can balance the errors for $0.00...01$ to $0.49...99$ with those for $0.99...99$ to $0.50...01$, but there's that dead center $0.50...00$ (aka $0.5$) which is the odd man out - consistent rounding up (or down) will introduce a bias because of this entry.

So what are you to do? Well, introduce a rule which rounds that entry up half the time and rounds down the other half. Bankers' rounding is the most common such rule. This rule says that if the digits to be rounded off are $5$ exactly, then round such that the last digit of the number after rounding is even. (Note that this is equivalent to a rule to round up if the digit to the left of the $5$ is odd, but round down if the digit to the left is even - two different ways of saying the same thing.)

0.255 -> 0.26 ... because 6 is even (or round up because 5 is odd)
0.265 -> 0.26 ... because 6 is even (or round down because 6 is even) 
0.275 -> 0.28 ... because 8 is even (or round up because 7 is odd)
0.285 -> 0.28 ... because 8 is even (or round down because 8 is even)

But again, this is the rule only if the number to be rounded falls exactly between the two possibilities. If the number is closer to one or the other, then round to the nearest.

0.2453   -> 0.25 (Round to nearest)
0.2553   -> 0.26 (Round to nearest)
0.2651   -> 0.27 (Round to nearest)
0.275000 -> 0.28 (Falls exactly between 0.27 and 0.28, apply even/odd rules)
0.275378 -> 0.28 (Round to nearest)
0.285000 -> 0.28 (Falls exactly between 0.28 and 0.29, apply even/odd rules)
0.285023 -> 0.29 (Round to nearest)

Note also that you should only round once, and don't cascade your rounding:

0.14555 -> 0.1    (Round to nearest)
0.14555 -> 0.1456 -> 0.146 -> 0.15 -> 0.2  (WRONG!)
Martin - マーチン
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R.M.
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  • I agree with your objection but you have the wrong examples. The first table should be for 0.5, 1.5, 2.5 and so on. Over 10 possibilities rounding with even odd "balances." //The second table is wrong. You compare to left digit. // The last table should compare rounding a 4 digit number to 3 digits or to 2 digits. So 53 always rounds up since it is more than 50. So for 0.2550 you'd have to apply the odd/even rounding rule when rounding to two digits, but you would not need the odd/even rule to round 0.2553 to two digits. – MaxW Nov 02 '15 at 22:46
  • @MaxW The first table is just to demonstrate that there's a bias, not how Bankers' rounding fixes it - I'll extend it to help clarify. On the second table, I think you're using the same rule, but with different phrasing - I'll clarify. On the third I'll add additional examples. – R.M. Nov 02 '15 at 23:20
  • In second table you have 0.2X5. Rounding up/down depends on if the X is even or odd. // In third table 2.553 rounds to 2.55 if we rounding to 2 decimals and 2.6 if we're rounding to one decimal. But if we rounding to two decimals then 53 is bigger than 50 and it is always up. // Think of it this way. 2.549 rounds to two decimals as 2.5. You don't round to 2.55 then to 2.6. – MaxW Nov 02 '15 at 23:29
  • In your bankers example 2.5 round to 2, so the "error" averages out. That is why you don't have to flip a coin at 5 (which is wrong in the accepted answer). – MaxW Nov 02 '15 at 23:35
  • Would it be ok for me just to edit your answer? If you don't like my edits you can flip it back. – MaxW Nov 02 '15 at 23:46
  • I agree; when there are numbers to the right of 5, it does not stay just 5, it becomes a quantity bigger than 5, thus it makes sense to round 5 up in this case. – Rok Apr 16 '16 at 01:07
  • My teacher gave a rationale for rounding to the even number, even if the ending is not EXACTLY 5. For example 0.2533 would round to 0.2. His rationale was: due to imprecision the digits after the 5 must be ignored because they are insignificant. Could you explain why this is wrong? – Hanry Hu Mar 12 '17 at 03:27
  • Random rounding rules do, in general, assume that digits are largely independent. However, for example, in the realm of commerce and price marketing, prices are not likely composed of random digits. So, depending on how the underlying numbers were created, perhaps one should think twice before assuming on aggregation some perceived benefit. – AJKOER Feb 04 '20 at 15:03