# Would the density and radius of your metal appear to be higher or lower if you forgot to tap the flask to release air bubbles?

For this lab we had to measure out copper in flasks, but tap them out before so no air bubbles would affect the weight. We have this question at the end of the lab and so I wasn't sure if the density/radius would be higher or lower if I didn't tap them out and why.

We were trying to figure out the size on an atom of copper. To find the density of the sample, we had to add about 1/3 of copper to an empty flask and weigh it with the metal. Then after we had to add enough water to just cover the metal, then gently tap and rotate the volumetric flask to remove any air bubbles. Then fill to the calibration mark with water and reweigh the volumetric flask containing the water solid mixture.

• In short, you were measuring the mass and volume of copper. Which of these measurements would be affected by the air bubbles, and how? Sep 14 '15 at 21:02
• Yes. We are trying to figure out if those and density would be higher or lower based on if I hadn't tapped out the bubbles.
– Zoe
Sep 14 '15 at 21:04

## 1 Answer

In this experiment, you were determining the average volume that a copper atom occupied in your sample of copper. In order to do this, you had to know the number of copper atoms and the volume that number of copper atoms occupied.

The moles of copper was measured by gravimetrically determining the mass of the copper (e.g. by weighing it). From that you determined the number of moles of copper using it's atomic weight, and from there you calculated the number of copper atoms using Avogadro's number.

The volume occupied by your copper sample was determined by measuring how much water the copper displaced. So, if the volume of water increased by 10 $\mathrm{cm^3}$, then you knew your volume of copper was $\mathrm{cm^3}$. But air bubbles occupy space also. If, for example, 1 $\mathrm{cm^3}$ of the total displaced volume were actually due to air bubbles, then you would actually only have 9 $\mathrm{cm^3}$ of copper, but you would mistakenly have used a value of 10 $\mathrm{cm^3}$.

So, the bottom line is that because you were calculating the molecular volume as the volume occupied by the copper per molar quantity of copper, your calculated molar volume would have been proportionally higher than the true value.