High School Chemistry Significant Figures [closed]

This problem appeared on my AP Chemistry summer work form, and I do not understand how to use significant figures to do this.

Here is the question word-for-word. Convert these from +- (plus-minus sign) to Significant Figures Notation

a) 2.0646 +- 0.050

b) 5.04 +- 0.12

c) 12.675 +- 0.20

d) 24.81 +- 1.0

I understand the basics of significant figures (how they work, what do do when adding, subtracting, multiplying, dividing). But the given information does not mention anything about how to do variability

Any help would be appreaciated.

• @ACR Actually, the error in the first example has three decimal places. Extra zeros are still considered significant figures (such as the difference between 2.00 and 2; the precision for the first measurement is higher than the precision for the second, regardless of the fact that the decimals are 0). Commented Jul 17 at 7:51
• Revised comment, I did not want to solve the actual example but added one extra zero. My mistake! "There are no hard and fast rules for that. The message in this question is do not use too many significant figures in the main number, if it is not known with a higher certainity. For example in a similar example as (a), the uncertainty, 0.05, has two decimal places. The main value should be rounded to two decimal places: 2.06±0.05. In the same way, you can solve (a) wher you have three decimal places." –
– ACR
Commented Jul 17 at 13:06
• The most significant figure of the variability becomes the least significant figure of the value.
– Karsten
Commented Jul 17 at 22:58
• I’m voting to close this question because significant figures are the topic of numerical analysis, which is mathematics, not chemistry, and the question lacks any use of significant figures in a chemical context. Commented Jul 19 at 15:53

A rule of thumb is that the last significant figure in any stated answer should be of the same order of magnitude, i.e. in the same decimal position, as the uncertainty. For example,

$$\displaystyle 92.81 \pm 0.3 \text{ should be reported as } 92.8\pm 0.3\\ 92.81\pm 3 \to 93\pm 3\\ 92.81\pm 30 \to 90 \pm 30$$

In the last case, the rounded result is a little smaller than the result but the error is so large that this is of no consequence. In other words, the $$92.81$$ is only one of many results that could have been obtained had the experiment been repeated many more times and values from at least $$60 \to 120$$ are to be expected. In cases such as $$92.5\pm 0.35$$, you might not want to round up either to $$93$$ or down to $$92$$. In this case the rounded number could be reported as $$(0.92_5 \pm 0.04) \times 10^2$$.

Illustrating the error of $$\pm 30$$ on the number $$-6051.78$$. The red dot is the original number, the blue dot the rounded one, and the red line the error.