Identifying a Metal from a Density Measurement
A student in a Regents chemistry lab is given a small, irregular piece of metal taken from a recycling bin. The electronic balance reads 23.45 g. She places 15.0 mL of water in a 25-mL graduated cylinder, then slides the metal in. After the metal settles, the meniscus reads 17.6 mL. The room temperature is 21.5 °C. The instructor says the sample is probably aluminum (density 2.70 g/mL), copper (8.96 g/mL), or zinc (7.14 g/mL). The student computes a density and announces that the metal is aluminum because her answer is 'about 9.' A small air bubble was visible on the metal just before she recorded the final volume, but it broke free after the reading. The class must decide whether the identification is justified, report the density with the correct number of significant figures, and explain how the bubble and reading technique affect the result.
1. Calculate the density of the metal from the student's data. State the answer with the correct number of significant figures and explain how the volume subtraction limits the certainty.
Read the worked answer
Volume change: 17.6 mL − 15.0 mL = 2.6 mL.
Both volume readings are recorded to the tenths place, so the subtraction rule limits the result to the tenths place. That gives 2.6 mL, which has two significant figures.
Density: d = m/V = 23.45 g / 2.6 mL = 9.019... g/mL.
The mass has four significant figures, but the volume has only two, so the density must be reported to two significant figures: 9.0 g/mL.
Reporting 9.02 g/mL would overstate the certainty. The measurement cannot distinguish 8.9 g/mL from 9.1 g/mL because the volume change is known only to about ±0.1 mL.
2. The student concludes the metal is aluminum. Evaluate this conclusion using the accepted densities and the calculated density. Which metal is most consistent, and what limitation remains?
Read the worked answer
The calculated density is 9.0 g/mL.
Compare with the accepted values:
- Aluminum: 2.70 g/mL - far from 9.0 g/mL
- Zinc: 7.14 g/mL - different by about 1.9 g/mL
- Copper: 8.96 g/mL - rounds to 9.0 g/mL at two significant figures
The student's own result points to copper, not aluminum. Saying the answer is 'about 9' actually supports copper because 8.96 rounds to 9.0. Aluminum would require a density near 2.7 g/mL, which is not close even with poor technique.
The remaining limitation is that two-significant-figure data cannot prove the sample is pure copper. It could be a copper-rich alloy, or the volume reading could be off enough to shift the result. Among the three choices, however, copper is the only reasonable identification.
3. A small air bubble adhered to the metal when the final volume was read. Predict whether the calculated density would be too high, too low, or unaffected, and explain. Would the bubble change the identification among the three choices?
Read the worked answer
An air bubble takes up space in the cylinder, so the final volume reading is too high. That makes the calculated volume of the metal too large:
V_metal = V_final − V_initial
A larger V_metal in the denominator gives a density that is too low.
For example, if the bubble added about 0.2 mL, the true metal volume would be about 2.4 mL, giving d ≈ 23.45/2.4 ≈ 9.8 g/mL. That is still far from aluminum (2.70) and zinc (7.14); it remains closest to copper. A very large bubble would be needed to move the result toward zinc, and the observed bubble was small.
Reading the meniscus at eye level also matters. Looking from above or below would shift the volume reading and change the density in a predictable direction, but the identification as copper is still the best choice among the three.