[integral0]

The density of water is determined by measuring the mass of a known volume of water.  If you determined the density of a 25-mL sample of water, how many significant figures should you report if you used the following instruments?

a) an analytical balance for measuring mass and a 25 mL pipet for the volume measurement?

An analytical balance has a typical uncertainty of + or - 0.0001g  and the uncertainty of a 25 mL pipet is + or - 0.02 mL .  ..  so I thought since this is density... it would be mass over volume...of whatever.... which would give you

0.0001 g / 0.02 mL  ..  . . which is two sig figs (or at least I thought, my teacher thought otherwise)

WHere is my reasoning going wrong?

Thx

[jdurg]

Your reasoning went wrong when you were going over the accuracy of the pipette.  Based upon the data you've provided, the pipette is accurate to one decimal place.  After that it is uncertain.  So if you have 20 mL in the pipette, your measurement would be 20.0 mL.  That would be 3 significant figures there.

[jdurg]

While these numbers aren't true for the density of water, they give a good approximation.  Let's say you measured out 19 mL on the pipette.  Based on its uncertainty, your answer is accurate out to 19.0 mL.  (Since your inaccuracy is that next decimal place).  Your volume has 3 significant figures.  Now you go and weight the water on the tared balance and you get a reading of 18.004 grams.  This is as accurate as you can get because the uncertainty lies in the next decimal place.  So your mass has 5 significant figures.  When you do your density calculation, you'll have 18.004/19.0 = 0.94757894..  Since the volume measurement only has 3 sig-figs, your final answer would be 94.8x10^-2 g/mL.

[integral0]

so jdurg, if you have the same analytical balance that has an uncertainty of + or - 0.0001g  and a graduated cylinder of + or - 0.1mL

wouldn't it be two sig figs?

however, I was wrong for this question for two sig figs.... i think i'm still missing some logic here....

any help appreciated!

[savoy7]

according to your question -

when multiplying or dividing numbers - the product is limited by the least number of sig figs.

You where able to estimate the analytical balance to several sig figs, but your pipet is limited by the pipet type.

I don't know your data - but let's throw in 24.9987 grams - that's 6 sig figs
Your pipet  - if volumetric was - 25.00 ml - that's 4 sig figs
if graduated - you estimate one past the graduation - maybe 25.0 ml - that's 3 sig figs

so, when you divide you are limited to either 4 or 3 sig figs depending on the pipet type.

That's why jdurg wrote that it will have 3 sig figs with his answer of 9.48 x 10-1 g/ml in his answer, because he estimated your pipet to have 3 sig figs 19.0 ml.

savoy out