# 16.9: Correcting Mass for the Buoyancy of Air

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Calibrating a balance does not eliminate all sources of determinate error that might affect the signal. Because of the buoyancy of air, an object always weighs less in air than it does in a vacuum. If there is a difference between the object’s density and the density of the weights used to calibrate the balance, then we can make a correction for buoyancy [Battino, R.; Williamson, A. G. J. Chem. Educ. 1984, 61, 51–52]. An object’s true weight in vacuo, Wv, is related to its weight in air, Wa, by the equation

$W_v = W_a \times \left[ 1 + \left( \frac {1} {D_o} - \frac {1} {D_w} \right) \times 0.0012 \right] \label{16.1}$

where Do is the object’s density, Dw is the density of the calibration weight, and 0.0012 is the density of air under normal laboratory conditions (all densities are in units of g/cm3). The greater the difference between Do and Dw the more serious the error in the object’s measured weight.

The buoyancy correction for a solid is small and frequently ignored. The correction may be significant, however, for low density liquids and gases. This is particularly important when calibrating glassware. For example, we can calibrate a volumetric pipet by carefully filling the pipet with water to its calibration mark, dispensing the water into a tared beaker, and determining the water’s mass. After correcting for the buoyancy of air, we use the water’s density to calculate the volume dispensed by the pipet.

Example $$\PageIndex{1}$$

A 10-mL volumetric pipet is calibrated following the procedure outlined above, using a balance calibrated with brass weights with a density of 8.40 g/cm3. At 25oC the pipet dispenses 9.9736 g of water. What is the actual volume dispensed by the pipet and what is the determinate error in this volume if we ignore the buoyancy correction? At 25oC the density of water is 0.997 05 g/cm3.

Solution

Using Equation \ref{16.1} the water’s true weight is

$W_v = 9.9736 \text{ g} \times \left[ 1 + \left( \frac {1} {0.99705} - \frac {1} {8.40} \right) \times 0.0012 \right] = 9.9842 \text{ g} \nonumber$

and the actual volume of water dispensed by the pipet is

$\frac {9.9842 \text{ g}} {0.99705 \text{ g/cm}^{3}} = 10.014 \text{ cm}^{3} \nonumber$

If we ignore the buoyancy correction, then we report the pipet’s volume as

$\frac {9.9736 \text{ g}} {0.99705 \text{ g/cm}^{3}} = 10.003 \text{ cm}^{3} \nonumber$

introducing a negative determinate error of –0.11%.

Exercise $$\PageIndex{1}$$

To calibrate a 10-mL pipet a measured volume of water is transferred to a tared flask and weighed, yielding a mass of 9.9814 grams. (a) Calculate, with and without correcting for buoyancy, the volume of water delivered by the pipet. Assume the density of water is 0.99707 g/cm3 and that the density of the weights is 8.40 g/cm3. (b) What is the absolute error and the relative error introduced if we fail to account for the effect of buoyancy? Is this a significant source of determinate error for the calibration of a pipet? Explain.

For (a), without accounting for buoyancy, the volume of water is

$\frac {9.9814 \text{ g}} {0.99707 \text{ g/cm}^3} = 10.011 \text{ cm}^3 = 10.011 \text{ mL} \nonumber$

When we correct for buoyancy, however, the volume is

$W_v = 9.9814 \text{ g} \times \left[ 1 + \left( \frac {1} {0.99707 \text{ g/cm}^3} - \frac {1} {8.40 \text{ g/cm}^3} \right) \times 0.0012 \text{ g/cm}^3 \right] = 9.920 \text{ g} \nonumber$

For (b), the absolute and relative errors in the mass are

$10.011 \text{ mL} - 10.021 \text{ mL} = -0.010 \text{ mL} \nonumber$

$\frac {- 0.010 \text{ mL}} {10.021 \text{ mL}} \times 100 = -0.10\% \nonumber$

Table 4.2.8 shows us that the standard deviation for the calibration of a 10-mL pipet is on the order of ±0.006 mL. Failing to correct for the effect of buoyancy gives a determinate error of –0.010 mL that is slightly larger than ±0.006 mL, suggesting that it introduces a small, but significant determinate error.

Exercise $$\PageIndex{2}$$

Repeat the questions in Exercise $$\PageIndex{1}$$ for the case where a mass of 0.2500 g is measured for a solid that has a density of 2.50 g/cm3.

The sample’s true weight is

$W_v = 0.2500 \text{ g} \times \left[ 1 + \left( \frac {1} {2.50 \text{ g/cm}^3} - \frac {1} {8.40 \text{ g/cm}^3} \right) \times 0.0012 \text{ g/cm}^3 \right] = 0.2501 \text{ g} \nonumber$

In this case the absolute and relative errors in mass are –0.0001 g and –0.040%.

Exercise $$\PageIndex{3}$$

Is the failure to correct for buoyancy a constant or proportional source of determinate error?

The true weight is the product of the weight measured in air and the buoyancy correction factor, which makes this a proportional error. The percentage error introduced when we ignore the buoyancy correction is independent of mass and a function only of the difference between the density of the object being weighed and the density of the calibration weights.

Exercise $$\PageIndex{4}$$

What is the minimum density of a substance necessary to keep the buoyancy correction to less than 0.01% when using brass calibration weights with a density of 8.40 g/cm3?

To determine the minimum density, we note that the buoyancy correction factor equals 1.00 if the density of the calibration weights and the density of the sample are the same. The correction factor is greater than 1.00 if Do is smaller than Dw; thus, the following inequality applies

$\left( \frac {1} {D_o} - \frac {1} {8.40} \right) \times 0.0012 \le (1.00)(0.0001) \nonumber$

Solving for Do shows that the sample’s density must be greater than 4.94 g/cm3 to ensure an error of less than 0.01%.

This page titled 16.9: Correcting Mass for the Buoyancy of Air is shared under a CC BY-NC-SA license and was authored, remixed, and/or curated by David Harvey.