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Topic: Trouble with sig figs...  (Read 5365 times)

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Offline Korokian

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Trouble with sig figs...
« on: September 08, 2006, 05:46:23 PM »
in a problem like this... (0.00058x1.311)-0.00004
what is the answer and would you do the multiplication sig figs or the addition sig figs?

Offline Borek

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Re: Trouble with sig figs...
« Reply #1 on: September 09, 2006, 05:54:51 PM »
0.00058x1.311 = 0.00076038

2 significant figures at most, so it is 0.00076, no doubt about it.

Now comes the tricky part.

0.00076 - 0.00004 = 0.00072

I would go for two significant figures here, even if one of the numbers contains only one digit.

Why?

Well, let's look at much more obvious case:

0.12345 - 0.00000000000000000001

Assuming all digits are precise, final and accurate result is 0.12345, not 0.1.

Note that significant digits are ill defined. While they make some sense when presenting results (don't write down 10 digits displayed by your calculator when you started with 2 digits of input value) they should be not treated as a way of displaying accuracy - you should give your results (especially experimental) as 1.2345+/-0.0032 whenever possible. In the case of calculations that can be challenging.
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Offline Korokian

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Re: Trouble with sig figs...
« Reply #2 on: September 09, 2006, 06:36:55 PM »
thanks!

Offline brettphillips

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Re: Trouble with sig figs...
« Reply #3 on: January 12, 2007, 05:37:24 PM »

I feel like such a dunce.  I have the following problem.

(0.0045 x 20020.0) + (2844 x 12) = ???

I have perfomred the following calculations in the following order.

1) .0045 x 20020.0 = 90.09
2) 2844 x 12 = 34,128

therfore...
??? = 90.09 + 34,128 = 34,218.09

My problem is that I can't figure out how to determine the number of significant digits.  I know the basic rules:
 * for multiplication/division, same as the number of sig figs as measurement with fewest sig figs
  * for addition/subratcion, same as number with fewest number of decimal places. 

What I don't get is what to do when I have a complex equation that involves both multiplication/division AND addition/subtraction.

Thanks,

Brett :)

Offline teclord

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Re: Trouble with sig figs...
« Reply #4 on: January 13, 2007, 12:57:51 AM »
I also have not been able to figure this out. I am curntly looking. I have worked some equatioons and would say that they require fenese.. spelled wrong of course.

Offline english

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Re: Trouble with sig figs...
« Reply #5 on: January 13, 2007, 02:04:40 AM »
General rule for addition and subtraction to keep the right amount of certainty in our results is to record your result in the same number of decimal places as the number with the fewest decimal places.

For example,

.001 + .00003 = .001

But this is a convenient way of skipping the more important aspects of analytical chemistry, as we are ignoring our reliability here.

Generally speaking, this should be enough for your case.


When you have multiple operations, such as multiplication followed by say addition, you follow operation rules and determine significance in that order.

For example,

2.34 x 3.423 + 2.000 = 10.00

Your multiplication should only carry 3 digits, of course.  Being careful not to round off and carry at least 2 more digits than 3, for example 8.0098 instead of 8.01, add this to 2.000 keeping in mind that your previous calculation had 3 digits, with 2 decimal places

After you have multiplied your 3 significant digits does not mean that after adding to 2.000 you have an answer with 3 significant digits.  Because now you are adding, the rule is to record the number with the fewer number of decimal places, which is 2.

If you rounded through, it would be easier to see, but I warn you not to round until the very end and carry some extra digits throughout.  This is just for clarification:

Rounding through
2.34 x 3.423 = 8.01 (3 sig figs)

8.01 + 2.000 = 10.01 (2 decimal places)

Notice that the final answer is 10.01 because we rounded in the middle of our calculation.  Do not do this.  Only round at the very end.  This, as calculated above, should give you 10.00.  You have to carry a couple of extra digits (2 or 3 should be enough) and keep in mind how many significant ones you need throughout the calculation.


Essentially, I have restated in a different way what Borek has said.  Whichever way is easier for you to understand.   ::)

« Last Edit: January 13, 2007, 02:42:13 AM by k.V. »

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