What is the true volume?

Question

I completed a homework assignment similar to the actual experiment, which read: A student calibrated a eudiometer by first recording the mass of the empty eudiometer followed by filling the eudiometer to the 50.00 mL line with deionized water at 22°C and recorded the mass. The mass of water was 49.012 g. A reading from the eudiometer of 46.78 mL corresponds to what true volume? I obtained 49.96mL.

I repeated this in lab, measured about 50mL (came to 50.21mL), and weighed (49.771g) @ 22 Celsius. For some reason, I'm having difficulty determining the true volume for the real lab. Help?

Answer 1

Hi Lily S.,

So, a eudiometer is essentially a pycnometer, but ultimately for measurement of gas volumes, not liquid densities, and a buret, rather than a closed vessel. But the principle is the same. You have a glass vessel or tube, with calibration mark or marks, and you fill it to a calibration mark, or otherwise use the calibration marks to determine a volume.

Now, initally someone (you?) filled the eudiometer to the 50.00 mL mark with deionized water at 22C and recorded the mass at 49.012 g. Fine. From standard tables of water density vs. temperature, obtain the theoretical density of the water at 22C (0.997799 g/mL). Therefore, your water had actual volume 49.012g/(0.997799 g/mL) = 49.120 mL.

At this point you must make a decision: is the volume of the eudiometer proportional (i.e. is the error % constant across all volumes), or is the eudiometer simply off by a constant 50-49.12 = 0.88 mL across the board, so to speak? I don't know your experimental setup, so I can't be sure -- but I suspect the latter is true -- in other words, the calibration marks on the eudiometer tube are at proper intervals, but there is a constant volume error due to uncontrolled volume at the end of the eudiometer tube. If that is in fact true, you would have to add 0.88 mL to any reading, to obtain the true volume. In other words, your 46.78 mL reading is corrected to 46.78 + 0.88 = 47.66 mL. Don't know where you got 49.96 mL??

So when you did your second trial, 49.771g of water have volume (use same density as above, water is water, in this case):** see comments below!

=49.771g/(0.997799g/mL) = 49.881 mL

So your eudiometer error on this second trial is 50.21 - 49.881 = 0.329 or 0.33mL

The difference between those two values for correction -- 0.88 mL and 0.33 mL -- is rather large! But, you are a learning student, so there is plenty of opportunity for error to creep in. Air bubbles , miniscuses read at an angle or without a black/white backdrop card, a little water clinging to the outside of the eudiometer or present as a droplet where it shouldn't be -- all these are potential error sources. As you continue your education, you will get better and better at ferreting out sources of potential error.

** Note on "water is just water" above: well yes, there's water (tap), you use it in lab for running aspirators and rinsing glassware before sending for glasswash, and drinking -- you need the minerals!; there's de-ionized water (good for making analytical solutions); there's Milli-Q water (18 M-ohm), used for making liquid chromatography eluents, where great purity is required, and then there's freshly-boiled Milli-Q water, used occasionally where CO2-free water is required, for pH measurements on essentially uncharged molecules in solution. Pure water picks up CO2 like mad from the air, which lowers its pH. The US Pharmacopoeia (USP), which is the testing "Bible" for the prescription drug industry, requires this material to be prepared sometimes. However, a more reliable way of making water CO2-free is purging (bubbling) continuously with N2 gas. A little lab secret for you!

Let me know (via a comment) if that was what your lab analysis needed, or not --

-- Cheers, --Mr. d.