Calibration Drift and the Magnesium Ribbon
A student determines the percentage of magnesium in a ribbon sample by reacting it with excess 1.00 M HCl and collecting the hydrogen gas over water. The student uses a 50.0 mL burette to measure the HCl, but the burette was not calibrated: its markings are 0.30 mL too high at every reading (i.e., the actual delivered volume is 0.30 mL less than the reading change). The student reads an initial volume of 0.50 mL and a final volume of 25.50 mL. The gas is collected at 22.0 °C and 748 mmHg; the vapor pressure of water at 22.0 °C is 19.8 mmHg. The student weighs 0.0523 g of magnesium ribbon and calculates the percentage of Mg using the ideal gas law. The true percentage of Mg in the ribbon is 92.0%. The student's calculated percentage is 89.5%. The student is asked to explain the discrepancy and propose corrections.
1. Identify the most likely source of the systematic error in the calculated percentage of Mg, and show quantitatively how the burette calibration affects the result.
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The burette error is the most likely systematic error. The student reads a volume change of 25.50 − 0.50 = 25.00 mL. Because the markings are 0.30 mL too high at every reading, the actual volume delivered is 25.00 − 0.30 = 24.70 mL. This is a 1.2% relative error in the volume of HCl, but the HCl is in excess, so the volume of HCl does not directly affect the amount of H2 produced. The amount of H2 depends on the mass of Mg, not on the volume of acid, as long as the acid is in excess. Therefore, the burette error alone cannot explain the 2.5 percentage-point discrepancy. The student must look for another systematic error, such as incomplete reaction, gas leakage, or incorrect pressure correction. The burette error would matter if the student used the volume of HCl to calculate the moles of HCl consumed, but that is not the case here.
2. The student's calculated percentage is lower than the true value. Propose two plausible systematic errors that would cause a low result, and explain how each would affect the calculated moles of H2.
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Two plausible systematic errors that would cause a low calculated percentage of Mg are: (1) The magnesium ribbon is not fully reacted. If some Mg remains unreacted, less H2 is produced, so the calculated moles of Mg (from H2) are lower than the actual moles of Mg in the sample. This would make the calculated percentage lower. (2) The gas collection system leaks, allowing some H2 to escape before the volume is measured. A leak would reduce the measured volume of H2, leading to fewer calculated moles of Mg and a lower percentage. Another possibility is that the student did not subtract the water vapor pressure correctly; if the student forgot to subtract the vapor pressure, the calculated pressure of H2 would be too high, which would increase the calculated moles of H2 and give a high result, not low. If the student subtracted too much (e.g., used a higher vapor pressure), the calculated H2 pressure would be too low, giving a low result. However, the most likely errors are incomplete reaction or leakage.
3. Calculate the theoretical volume of H2 gas (in mL) that should be collected at 22.0 °C and 748 mmHg if the ribbon is 92.0% Mg and the sample mass is 0.0523 g. Use this to assess whether the student's measured volume is consistent with the stated errors.
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Mass of Mg in sample = 0.0523 g × 0.920 = 0.04812 g. Moles of Mg = 0.04812 g / 24.305 g/mol = 1.980 × 10^-3 mol. From the balanced equation Mg + 2HCl → MgCl2 + H2, moles of H2 = moles of Mg = 1.980 × 10^-3 mol. The pressure of H2 = 748 mmHg − 19.8 mmHg = 728.2 mmHg = 728.2/760 atm = 0.9582 atm. Temperature = 22.0 + 273.15 = 295.15 K. Using PV = nRT, V = nRT/P = (1.980 × 10^-3 mol)(0.08206 L atm/mol K)(295.15 K) / 0.9582 atm = 0.0501 L = 50.1 mL. The student's measured volume can be estimated from the calculated percentage: if the student calculated 89.5% Mg, then the calculated moles of Mg = (0.0523 g × 0.895)/24.305 = 1.927 × 10^-3 mol, corresponding to V = 48.8 mL. The difference between 50.1 mL and 48.8 mL is 1.3 mL, which is about 2.6% of the expected volume. This small difference could be due to incomplete reaction or a small leak, but it is larger than the burette error alone. The student should check for leaks and ensure complete reaction.