I thought I had a handle on the concept, and maybe I do and the practice test I’m taking is just lousy (I’ve already spotted several errors).
The original problem is as follows:
"While students were practicing calculations with significant figures, the density of a chemical solution was found to be 2.567 g/l. The students accurately measured 6.710 liters of the solution and calculated the mass of the solute to be:
A. 23.93457 g
B. 23.935 g
C. 23.9 g
D. 24 g
The multiplication was easy, and looking at the original measurements, I decided that they were measuring to three places past the decimal. So, I chose B as my answer. The answer key says the correct answer is C, which has nothing past the decimal.
Now, I could see if the correct answer were 23.94, showing four significant figures like the original measurement, but that’s not offered.
What am I missing? How is C the correct answer (if indeed it is)?
Answers to arithmetic problems is only accurate to the degree of the least accurate measurement. In this case, the measured 6.710 liters is the measurement most in question. The problem states that the measurement is accurate at 6.710 leading me to believe that the trailing zero is a sig fig. Trailing zeroes are typically not considered sig figs, but can be indicated as such (often by underlining the last digit considered significant).
So, the measurement is accurate to either 3 or 4 figures, and if a four sig fig answer was available, I would chose that. The answers available preclude that, so you are left with selecting an answer with more or fewer sig figs. You should always pick the answer with an equal or lesser number of sig figs. In this case, C has three sig figs, and is the most accurate answer to the problem.
Given the significant figures as shown, the density could be as low as 3.5665 g/l. The volume of solution could be as low as 6.7095. Thus, the mass of the solute could be as low as 23.931. That’s why C. is wrong.
First, the number of significant figures has nothing to do with where the decimal happens to be; it’s all about the number of, uh, significant figures. For example, if you measure the weight of something to be 12.34 grams [four sig fig], then the weight of one hundred of them isn’t 1234.00 [six s.f.] grams, it’s 1234. [four s.f] grams.
Second, when you’re multiplying and dividing two measurements with limited accuracy, you need to drop a significant figure, because the errors in the two measurements will multiply. That’s why, when the two original measurements have four s.f.'s, and you multiply them, you end with a result that has three s.f.s.
Now, if you ask why you don’t drop a signifcant figure when multiplying 12.34 grams times 100 items, the answer is that the 100 items has no error at all (it’s exactly 100; it can’t be 99.97 items, right?) , so the significant figures are infinite (for significant figures purposes, think of it as 100.000000000000… items).
I thought trailing zeroes after a decimal point were always significant–after all, if they weren’t significant you wouldn’t have to show them.
Also, personal pet peeve, there is a difference between accuracy and precision. My graduated cylinder can be marked to the .001 liter so I can give you results like 6.710 but if the guy who marks the graduations was drunk that day I will have a very precise but inaccurate measurement.
After checking a number of sites, it become clear that you are correct about the trailing zeroes, CookingWithGas. So I went back to my own notes…now I wonder what else the physics department taught us wrong.
I agree with you about the accuracy/precision concept as well, but tend to feel that the distinction is less important in everyday speech.
Seems pretty clear to me. 3.567 (I assume that 2.567 was a typo) and 6.710 have 4 significant digits. So the final answer can only have at most 4 significant digits. From the choices, only C satisfiies this requirement.
When I was taught significant figures, we were told that the result of a multiplication should have the same number of sigfigs as the least precise multiplicand. There is actually some loss of precision, but it’s not so easily modeled as “lose one sigfig”. Suppose, for instance, the students were measuring at the very limit of their instruments, and had a density of 7 g/ml and a volume of 3 ml… You can express the answer with one sigfig as “20 g”, but how would you express it with zero sigfigs?
Also, the number of digits past the decimal point is relevant, but only if you’re adding or subtracting. If I start with 100.06392 grams of a substance, and lose 99.837 grams, it’s inappropriate for me to say that I have 0.22692 grams left, even though that’s the same number of sigfigs as 99.837. In that case, the fewest number of digits past the decimal point is 3, so I should only have 3 digits past the decimal point in my answer, or 0.227 grams.
This is why most high precision calculations are expressed either in scientific or engineering notation, i.e. 6.023x10[sup]23[/sup]. (Bonus points for everyone who knows the significance of that number.) This way, you know exactly how many figures are significant. There’s also that whole decimal/comma divide across the Atlantic, though I think everybody is pretty much sliding toward using a decimal point these days. And it does make the math between different magnitudes a lot easier.
Now, try figuring significant figures for allowable part tolerances in converting an engineering drawing from English to Metric. Grrr. I’ve wasted more time arguing over that interpretation than one needs to a hundred (1.00x10[sup]2[/sup]) lifetimes.
Yes, but if you have a figure in inches with 2 sigfigs, how many sigfigs are appropriate in the metric conversion? 1/10 of a centimeter implies a greater precision than does 1/10 of an inch.
Why? I have an engineering scale that is marked in tenths of an inch. Why does a scale marked in millimeters imply a greater precision than my engineering scale?
Learning how to “carry significant digits” is worthless. Actually, it’s worse than worthless, since it teaches you to do the wrong thing. (I’ve been a metrologist going on 10 years now, and I’ve never seen anyone do this. In fact, if anyone in my field did do this they’d be laughed at.)
How many significant digits to write down? As usual, it depends. If you’re recording data from an alphanumeric readout (e.g. digital voltmeter) then you would typically record all the digits, else the published uncertainty value for the device will no longer be valid. Furthermore, when you’re in the “middle steps” of performing calculations, you obviously want to use as many significant digits as possible. (This is what Excel does by default.)
But I suspect the OP is referring to this question: How many significant digits should be displayed for the final answer? To understand how to do this you need to know that, in the real world, every measurement value has an uncertainty value associated with it. If you have to make a bunch of calculations, each and every variable in the beginning, middle, and end has an uncertainty value associated with it. This is because uncertainties propagate through the calculations according to well-established rules. The number of significant digits to display for the final answer depends on the uncertainty value of the final answer. If, for example, the final answer is 763.4227543 ± 0.1, then you would display it as 763.4 ± 0.1 or perhaps 763.42 ± 0.1. If the final answer is 763.4227543 ± 0.001, then you would display it as 763.423 ± 0.001 or perhaps 763.4228 ± 0.001. If the final answer is 763.4227543 ± 0.0132, then you would display it as 763.4228 ± 0.0132 or perhaps 763.42275 ± 0.0132.
Of course, there’s also the question of, “How many significant digits should the final uncertainty value have?” I’m not aware of any rule for this but, due to random variability, there’s rarely a reason to have more than four. Two or three significant digits for the final uncertainty value is typical.
'Cause millimeters are smaller than tens of inches, of course.
On engineering drawings you have explicit and implied dimensional tolerances. Explict are stated as something like “1.00±.03 inch”. In converting from in to mm you just convert both the dimension and the tolerance by the scaled factor, e.g. 25.4±0.76mm.; no hu-hu. But then you have implied dimenions which are stated in the title block; usually something like:
x.x → ±0.1
x.xx → ±0.03
x.xxx → ±0.01
Well, now you have to figure out what that means in millimeters. You can’t just do a bulk conversion from x.xx in to yy.y mm, cause some x.xx in will be yy.yy and some will be yyy.y. So, the question is, do you slap tolerances on every dimension, or fiddle some kind of goofball conversion, or what?
Yes, these are the items of discussion that draw engineers into four hour long discussions on things that the normal person would dismiss in thirty seconds. :rolleyes:
I’ve taken to slapping explicit tolerances on any critical dimension, which pisses off manufacturing engineers and QA people because then they typically have to specifically check the tolerance (instead of just stating that the machining operation is “good for it”.) That’s okay; I like to torture them with GD&T, too. I figure if I’m going to hate my job there’s no reason why anyone else should have a good time.
Anyway, we could dispense with the whole bloody mess if we’d just go to one consistent system of measurement…but then we’d have to convince Ryerson to start selling metric sizes of steel. You can get away with that if you are Caterpillar or John Deere, but for those who don’t purchase mill run quantities of steel, we’re stuck with a measurement scale based upon the size of some long dead king’s finger-bone. :rolleyes:
Unfortunately, the girl in the front row taking my test today didn’t know it. She kept using as part of her molar mass: 1 mole / 6.023x10[sup]23[/sup] , giving gram answers. I kept crossing it out and handing it back to her, but I’m not sure she got it still.
Look at it this way: Say you want the area of a rectangle and you’ve measured the sides as 3.445 and 7.581. What this really means is that to the best of your measuring ability the sides are somewhere in the ranges of 3.4445 - 3.4455 and 7.5805 - 7.5815. Multiply it out and you know that the area is somewhere between 26.11103225 and 26.12205825. So you know that the area is 26.1 to the best of your ability to measure.
Er, well, I wouldn’t call it worthless. In these days of double precision variable types its all well and good to talk about carrying all figures to the end of the calculation, but back in the day (and way before my time) when slide rules were the calulating device of choice, you could only carry so many digits through any particular step in a calucluation. Four or five places was the limit, even with the top quality slapstick, so you needed to have an appreciation for when you could round your intermediate results and what that did to your calculations.
Well, it depends on your compound tolerances/uncertainties, but for any engineering purposes four significant places is about the extent for any operation. I would suspect that astronomers and particle physicists can probably use something on the order of eight or nine significant figures in their most extreme calculations, and of course mathematicians can use all they can get on the rare occasions they are interested in actual numbers, but in general, three to four significant figures is all you are going to get out of any real-world measurement without extreme effort.
Then again, from some things I’ve seen in manufacturing practice, you’re lucky to get a single significant figure. :smack:
if the density is less than 3 grams per litre, and there is less than 7 litres of the stuff, how can the mass be more than 3 x 7 = 21 grams?