Question 1
If I complete the multiplication in each parentheses, I get $4440+308$ which is $4748$. Why does my answer have 4 sigfigs but the answer has 3?
Question 2
I got $12.9$ amu instead. Here's my work (significant digits are underlined):
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IWantToLearn
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In the first question you assume that $4.44\times10^3=4440$. But you don't know what the next digit would be. – Andrei Sep 18 '21 at 20:30
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Same principle. I can use more digits, and I get 12.8671428571. But I don't know more than 2 digits for the average and the other isotope, so I truncate my result to the same precision – Andrei Sep 18 '21 at 20:36
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@Andrei But don't you have to use the multiplication & addition rules for sigfigs (& round only at the end). That's how I got 12.9. – IWantToLearn Sep 18 '21 at 20:40
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Edit your answer to show how you did your calculation. – Andrei Sep 18 '21 at 20:45
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@Andrei K I did it. – IWantToLearn Sep 18 '21 at 21:12
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You are right. They have one more digit than they can reliably calculate – Andrei Sep 18 '21 at 21:27
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@Andrei Actually wait. Why can't I do 4.44×103=4440 when the zero is not significant. I am not saying that 4.44×103=4440. w/a decimal point after the zero – IWantToLearn Sep 18 '21 at 21:43
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How do you know that $4.44$ is $4.440$ and not $4.442$? – Andrei Sep 18 '21 at 21:52
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@Andrei But the zero in 4.440 is significant. The zero in 4440 is not. – IWantToLearn Sep 18 '21 at 22:05
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Let us continue this discussion in chat. – Andrei Sep 18 '21 at 22:10
1 Answers
1
- As Andrei has pointed out, you have interpreted $4.44 \times 10^3$ as being exactly $4440$. But we don't know that. It could be anywhere between $4435$ and $4445$. Similarly, $3.08 \times 10^2$ could be anything between $307.5$ and $308.5$. So the sum could be anywhere from $4742.5$ to $4753.5$. The former rounds to $4.74 \times 10^3$ while the latter to $4.75 \times 10^3$. So why is the correct answer only $4.75 \times 10^3$? Because the rules for significant figures are only an approximate approach to error analysis. They are sufficient for most things, but if you truly need accuracy in your error limits, they are not how you should be tracking error.
- The official answer and you are both wrong. The correct answer per the rules of significant figures is 13 amu. What you both overlooked is that $65\%$ and $35\%$ are not perfectly known integer values. But if we overlook that and pretend that $^{11}\mathrm M$ and $^{13}\mathrm M$ are somehow known to exist in nature in an exact $65\% - 35\%$ split, then you are correct and the official answer is still wrong. They simply used the rule "in calculations involving multiplications and divisions, the number of significant digits in the output is the least number of significant digits in any of the inputs". They saw the inputs (at least those that they recognized) had four digits each, and so they put four digits in the answer. But they failed to pay attention to the subtraction, which requires a different approach.
The particular problem encountered here - where simularly sized numbers are subtracted from each other, resulting in a difference with significantly more relative error than the original numbers - is called cancellation error, and it is by far the single biggest issue in numerical calculation. I have put together calculational tools where the 16 or so decimal places available normally were not sufficient to guarantee the 3 decimal places I needed in the result because of cancellation error.
Paul Sinclair
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