Answer to Question #166334 in Analytical Chemistry for Joel

Question

. You calibrate a 100 mL volumetric flask by weighing the water it contains. At the 

temperature of the measurement, the density of water is 0.997 906 ± 0.000 111 g/mL. 

(This uncertainty is due to random fluctuations in the ambient temperature.) The 

buoyancy correction factor mtrue/mread = 1.001 053; its uncertainty is negligible. Write all 

values and uncertainties in this problem with one guard digit. [Throughout this exam, 

you may type “+/-” for ± and use an underscore to indicate subscripts: “+/- 0.003_3” 

for 0.0033.]

__/3 a. After performing several replicate measurements, you determine that the average 

uncorrected mass of water contained by the flask is 99.7562 ± 0.0033 g. Calculate 

the true mass of the water. [Ignore uncertainties until part c.]

__/3 b. Now calculate the calibrated volume of the flask in mL.

__/3 c. Which quantity has the larger absolute uncertainty? Briefly explain your choice.

density cannot be compared [uncorrected] mass

__/3 d. Which quantity contributes more to the uncertainty in the volume? Report the 

numbers you compared to reach your answer.

density cannot be compared [uncorrected] mass

__/6 e. Propagate these uncertainties to determine the uncertainty in the calibrated volume, 

and report the volume of the flask (in mL) as you would an intermediate result. [If 

you could not answer part b use 99.999 999... mL instead.]

__/4 f. Rewrite your answer from e in L. Which numeric value should remain the same when 

you multiply or divide by an exact conversion factor, the relative uncertainty or the 

bsolute uncertainty?

__/2 g. What would the relative uncertainty of the volume be if you thermostatted the water 

to eliminate the temperature fluctuations and thus the uncertainty in its density?

[You should have calculated this value earlier in this question.]

Answer 1

This uncertainty is due to random fluctuations in the ambient temperature.) The 

buoyancy correction factor mtrue/mread = 1.001 053; its uncertainty is negligible. Write all 

values and uncertainties in this problem with one guard digit. [Throughout this exam, 

you may type “+/-” for ± and use an underscore to indicate subscripts: “+/- 0.003_3” 

for 0.0033.]