[Guest]

Groan........ I'm on my last limb here.... I've gotta figure this out if I'm going to sleep tonight....

I think that I understand the concept of Significant Figures. It's been a long time, but I do know that when multiplying or dividing two numbers together, the number of sig figs in the answer need to equal the number in the question that has the least amount of numbers. So, in other words, the answer to the problem 25.2 * 6.1 would have 2 sig figs in it (the answer being 1.5*10²,) and the answer to the problem 252 / 3.44623 would have three sig figs in it (the answer being 7.31.)

Also, I know that when adding and subtracting sig figs, when there's a decimal involved, you need the same number of digits in the answer as the number with the fewest decimal place numbers in the question. So, in other words, the problem 3.14159 + 25.2 would have 3 sig figs (the answer being 28.3,) and the problem 33.14159 - 22.04 would have four sig figs in it (the answer being 11.10.)

So, here's my problem: how many sig figs would be in the answer of problems such as this?

[(345.9*10⁶ )+(1234-10^-4)]/2346.25

Now, please, don't torture yourselves with solving this monster. This isn't an actual problem I have to do, but it's much like what I do have to do. I can get the answer easily enough (that's what calculators are for, after all,) but I do need to know how you would determine how many sig figs would be in the answer. So.... Help? Anyone?

[Guest]

So, here's my problem: how many sig figs would be in the answer of problems such as this?

[(345.9*10⁶ )+(1234-10^-4)]/2346.25

If the bit in blue is supposed to read (1234*10^-4) then you should already know that it is 4 significant figures.

If it is supposed to be (1234-10^-4) then it should only be 2 significant figures (based on 10).

To further explain the importance of what you are studying a joke is in order:

A family was looking around a dinosaur exhibit in a museum when a security guard joined them while they were looking at one of the skeletons. The security guard started a conversation explaining how he was in awe about how accurate the curators are. "Take this skeleton for example." the security guard explained. "The curators know that this died 65 million and 4 years ago."

"How can you be so accurate?" the family asked.

"Well, when I started this job they said that it died 65 million years ago. That was 4 years ago."

Not a great joke I admit, but it gets the point across I hope.

[Guest]

I understand that much.... However, what I don't get is how many sig figs I should wind up with in the first place. Would I just do each part of the problem one at a time and decide how many sig figs is in the answer from that, would do I determine the number of sig figs I need from the problem now and solve, then shrink the solution to fit the sig figs needed, or would I do something else?

[plasmid]

Just handle each operation one-at-a-time, in the usual order of precedence. For the example above, I would first re-write it in standard scientific notation (none of this ###.## x 10^x BS) to get

[ (3.459*10⁸) + (0.1234) ] / 2.34625*10³

So the first operation you do is to evaluate is the thing in the brackets. In this case, that would simply be

(3.459*10⁸) / 2.34625*10³

because the second term is so small that no sig figs from the first term are affected. Then do the division knowing that the numerator has 4 sig figs and the denominator has 6 sig figs, so the final answer has 4.

So in general, for your actual problem, just do one step at a time (and maybe convert everything to standard scientific notation if that helps) and you should be able to handle whatever they throw at you.

[Guest]

Even though many calculators have a 10 or 12 digit display, they usually work with two more significant figures than their display. This is to ensure accuracy.

1/3 + 1/3 + 1/3 = 1

0.3 + 0.3 + 0.3 = 1 or 0.9?

It all comes down to how confident can you be in your answer. I try to involve any and all numbers that I know during the calculation and then round off at the end (if using a calculator).

Another example:

The relative molecular mass of a powder is 253.4

There is a 2kg pile of the powder.

You have been asked to work out how many molecules there are.

Avogadro's constant can be defined to over 10 decimal places but you only have one significant figure for the mass. Using as accurate a figure for Avogadro as possible (as most graphical calculators will have it stored already) you find that it is approx 4.75307166 x 10²¹ I would include this answer in any math related test I do and then show to round off to 5 x 10²¹. Admitted that in this example it doesn't show too well but if I chose 300 as the relative molecular mass (rounded to one significant figure) then the answer would be 4 x 10²¹

Does this answer the question? If not, we're still here.