SUSB-003

1. Diameter of Sphere and Computation of its Area and Volume

The area of a sphere of radius, r, is given by A = 4 π r2. The volume, by (4/3) π r3 . Measuring a radius is difficult, so we measure the more easily determined diameter. Since the diameter, d, is twice the radius, ( d = 2 r )we can rewrite the above as:        A = π d2  and  V = (1/6) π d3

Suppose our sphere is a basketball and we measure its diameter with a ruler whose smallest division is 1 inch. (Such a ruler is shown below.) Using our rule of thumb, we can read the ruler to the nearest 0.2 in.). We measure the diameter to be 9.4 in. Our understanding is that our diameter may have an error of ±0.2 in. due to our visual estimation which corresponds to a percent error of 100 X 0.2 / 9.4 = 2.1%. We seek to find out what the impact of such an error will be on the area and volume of the ball.

A simple way is to:

• calculate the area and volume of the sphere from the value of our measured diameter, and then
• to add and subtract the uncertainty in the diameter and recalculate the area and volume.
 Diameter (in) Area (in2) Volume (in3) We write the Area and Volume in scientific notation because of the number of signficant figures (2) justified in their values 9.4 2.8 X 102 4.4 X 102 9.4 - 0.2 = 9.2 2.7 X 102 4.1X 102 9.4 + 0.2 = 9.6 2.9 X 102 4.6 X 102

The Range of the calculated volume (the difference between the maximum and minimum values) is
4.6 X 102 - 4.1 X 102 = 5 X 101 (I.e., ~50 in3 separate the minimum and maximum values!).
Note that while the percent error in the diameter is 100 X 0.2 / 9.4 = 2%, the percent error in the area is 100 X 0.1 / 2.8 = 4%, and that in the volume is 100 X 0.3 / 4.4 = 7%.

In SUSB-003, as in all future exercises, distances are measured in cm (or mm), weights in grams (or mg).
and volumes in L (or mL)

2a. Weight of the Sphere

Suppose we weigh the above basketball on a scale whose smallest division is 1 oz. Again, by our rule of thumb, we should be able to estimate the weight to within the nearest 0.2 oz. Assume our weighing produces 21.0 oz. Our rule of thumb tells us that our best estimate of the weight of the ball is that it lies between 20.8 oz and 21.2 oz (i.e. 21.0 ± 0.2 oz) which corresponds to a percent error of 100 X 0.2/21.0 = 1.0 %.

2b. Calculation of the Density of the Sphere

What uncertainty shall we associate with the density calculated from our experimental measurements? The largest value of the calculated density, dmax, will come from using the largest weight (numerator) and the smallest volume (denominator). I.e.,

dmax = 21.2 oz / 4.1 X 102 in3 = 0.052 oz/in3

Similarly, the smallest value of the density will obtain by using the smallest weight and the largest volume, i.e.,

dmin = 20.8 oz / 4.6 X 102 in3 = 0.045 oz/in3

The value obtained by using our "raw" measurements is

draw = 21.0 oz / 4.4 X 102 in3 = 0.048 oz/in3

We can summarize the result by writing our density as 0.048 ± 0.004 oz/in3.. Note that an error of 0.004 in the number 0.048 corresponds to 100 X 0.004 / 0.048 = 8 % error.

Errors of 2% in the diameter and 1% in the weight have produced an error of 8 % in the computed density!

3a. The transfer Pipet

The intrinsic precision of the analytical balance makes it possible to determine weights to high precision. If the relationship between weight and other quantities (e.g., volume) is also known to high precision, we can use measurements of weight as indirect measures of the other quantity (e.g. volume) to high precision. This is the basis for our using the analytical balance to measure the precision with which the transfer pipet delivers its nominal volume (given that the experimenter using the pipet uses it properly).

The quantity that relates the weight of water to its volume is the density of water, d. For a substance such as water, the density is known to high precision. Not surprisingly, that value is a function of temperature. Tables of the density of water as a function of temperature are given in most references. (A section of such a table is posted in the laboratory and is shown below.)

DENSITY OF WATER AS A FUNCTION OF TEMPERATURE
(from the Handbook of Chemistry and Physics, 76th Edition)
 T (oC) Density (g/cm3) T (oC) Density (g/cm3) T (oC) Density (g/cm3) 0 0.99984 10 0.99970 20 0.99820 1 0.99990 11 0.99961 21 0.99799 2 0.99994 12 0.99950 22 0.99777 3 0.99997 13 0.99938 23 0.99754 4 0.99997 14 0.99925 24 0.99730 5 0.99997 15 0.99910 25 0.99705 6 0.99994 16 0.99895 26 0.99678 7 0.99990 17 0.99878 27 0.99652 8 0.99985 18 0.99860 28 0.99624 9 0.99978 19 0.99841 29 0.99595

Using the table, we can convert the weight of a sample of water into its volume if we know the temperature.
Example:

The weight of water (at 23.3 oC) delivered by a 10 mL transfer pipet is 9.9424 g. What volume of water was delivered by the pipet?

 Interpolation: The table does not list the temperature in question, 23.3oC,but it does provide the density for 23oC (0.99754 g/cm3) and 24oC (0.99730 g/cm3). Over a one degree range of temperature, we assume that the density changes linearly with density, so the value we should use is: 0.99754 + 0.3 (0.99730 - 0.99754) = 0.99754 - 0.00004 = 0.99750 g/cm3 The process of finding values between listed values in a table, or between marks on the axis of a graph, is called interpolation.

Density is defined as weight / volume (d = w/v), so, volume is given in terms of weight by v = w/d. The requested volume then is:

v = 9.9424 g / 0.99750 g/cm3 = 9.9673 cm3    (Note that all the numbers are known to 5 sig figs)

If the manufacturer's tolerance for a 10 mL pipet is 0.02 cm3. I.e., when properly used, the pipet will deliver 10.00 ± 0.02 cm3. The above result (rounded to two decimals) is 9.97 which departs from 10.00 by more than the manufacturer's specification.

6. The Buret

The calculations for the buret are similar to those for the pipet. The analogous volumes of water are determined by buret readings instead of a nominal value. Since we will generally use the buret to deliver 20 - 30 mL, the exercise uses samples in that range.

7. Preparation of a Solution of Known Concentration (This part may be omitted from the exercise)

The procedure asks that you prepare 250.0 mL of a solution of sodium chloride (NaCl) with a precisely known concentration of 4 g/L where the result should be within 20% of the specified value. (A 20% range will apply, unless otherwise specified, whenever weights or concentrations are specified in the lab manual.)

20% X 4 g/L = 0.8 g/L, so the specification requires the final solution to have a concentration of:

4.0 ± 0.8 g/L

i.e., between 3.2 g/L and 4.8 g/L

Since we will prepare only 250.0 mL, we will need only 250.0 / 1000.0 = 0.2500 times as much, or

0.2500 X 1.0000 L X ( 4.0 ± 0.8 ) g/L = 1.0 ± 0.2 g

Suppose we actually weigh out a sample of NaCl ( BY DIFFERENCE! ) of 1.1736 g into our volumetric flask.

That weight is within the specified range ( 1.0 ± 0.2 g ), so we add water until the flask is filled to its mark. The volume will then be 250.0 mL and the concentration of our solution is:

1.1736 g / 250.0 mL = 4.69

which is within the specified range