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For a lab exercise to calibrate volumetric glassware I was given the follow equation to correct weighings for the density of air:

$$m_\mathrm{c} = m_\mathrm{s}\,\frac{1 - \rho_\mathrm{a}/\rho_\mathrm{c}}{1 - \rho_\mathrm{a}/\rho_\mathrm{s}},$$

where:

  • $m_\mathrm{c}$ is the corrected mass of the sample;
  • $m_\mathrm{s}$ is the measured mass of the sample tared for the container;
  • $\rho_\mathrm{a}$ is the density of air;
  • $\rho_\mathrm{c}$ is the density of weights used to calibrate the balance, assumed brass;
  • $\rho_\mathrm{s}$ is the density of the sample.

Does $\rho_\mathrm{s}$ include the container for the sample? How is $\rho_\mathrm{s}$ determined without knowing $m_\mathrm{c}$?

If the tared container does not affect the calculation, $\rho_\mathrm{s}$ for water is straightforward. What about other material?

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  • $\begingroup$ You can determine mass and volume of the sample without using the buoyancy correction. There will be a secondary error (small error in density affecting the magnitude of the buoyancy correction), but it will be much smaller than the primary error (neglecting buoyancy difference between calibration standards and sample). If in doubt, you can recalculate the density with the correction and apply to the correction an iterative manner. It should converge in a single step, I bet. $\endgroup$
    – Karsten
    Commented Sep 20, 2022 at 18:43
  • $\begingroup$ If the container is tared, it does not figure in the measured mass. So, tare the container, and use the same one for the brass weights and your volumetric sample. $\endgroup$ Commented Sep 20, 2022 at 20:31
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    $\begingroup$ @Karsten also to consider is the volume of the standard weight vs sample volume as this is the difference of air mass displaced. Brass would be around 8x the density of water. $\endgroup$ Commented Sep 20, 2022 at 22:40
  • $\begingroup$ I wonder how all this compares to the %RSD of meniscus reading errors. $\endgroup$ Commented Sep 21, 2022 at 4:41

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