TL;DR

This companion document walks through every calculation in the Zn-Al alloy lab, step by step, using the Trial 1 data. It also covers all seven post-lab questions with worked solutions. Use the first doc for the concepts; use this one to practise the maths.

Difficulty: Intermediate

Prerequisites: Read the theory doc first (Zn-Al Alloy Composition Using Ideal Gas Law). You need to be comfortable with unit conversions, rearranging algebraic equations, and basic stoichiometry.

Key Terms for This Section

Barometric pressure (P_bar)

The atmospheric pressure measured in the lab, reported in mmHg. In all three trials of this experiment, P_bar = 742 mmHg.

mmHg to atm conversion

Divide mmHg by 760 to get atmospheres. For example, 711 mmHg / 760 = 0.9355 atm.

Celsius to Kelvin conversion

Add 273.15 to the Celsius temperature. For example, 24.3°C + 273.15 = 297.45 K.

Limiting reagent

The reactant that runs out first and determines how much product can form. In this experiment, the alloy is the limiting reagent and HCl is in excess (confirmed by calculation in post-lab Q1).

Absolute error and relative error

Absolute error is the uncertainty in the same units as the measurement. Relative error is the ratio of the uncertainty to the measured value. For a quantity calculated from multiplication and division, the relative errors of each input are summed to find the relative error of the result.

Core Content: Step-by-Step Calculation Walkthrough (Trial 1)

All values below come from Trial 1 of the experiment. Trials 2 and 3 follow exactly the same sequence with different numbers.

Step 1: Calculate the Height Correction Pressure

The difference in water heights between bottles = 11.74 cm = 117.4 mm.

In Trial 1, the level in the displaced water bottle is higher than the starting bottle (Type "2"), so P_ΔH will be added to the Dalton's Law equation (this means the gas pressure is slightly less than barometric).

• P_ΔH = 117.4 mm / 13.6 = 8.6 mmHg

Step 2: Calculate the Pressure of Dry Hydrogen

Apply Dalton's Law. Since Trial 1 is Type "2" (displaced water bottle higher), the equation becomes:

• P_bar = P_H₂O + P_H₂ + P_ΔH

• Rearranging: P_H₂ = P_bar - P_H₂O - P_ΔH

• P_H₂ = 742 - 22.4 - 8.6 = 711.0 mmHg

• Convert to atm: 711.0 / 760 = 0.9355 atm

For Trials 2 and 3 (Type "1", starting bottle higher at end), the height correction is subtracted from barometric pressure differently: P_H₂ = P_bar - P_H₂O + P_ΔH. Pay attention to the sign.

Step 3: Convert Temperature to Kelvin

• T = 24.3°C + 273.15 = 297.45 K

Step 4: Calculate Total Moles of H₂ Using PV = nRT

• n = PV / RT

• n = (0.9355 atm)(2.65 L) / (0.08206 L·atm/(mol·K))(297.45 K)

• n = 2.4791 / 24.401

• n = 0.1017 mol H₂

Step 5: Set Up the Simultaneous Equations

Let x = mass of Zn (in grams). Then mass of Al = 2.5099 - x.

Equation from H₂ moles:

• (x / 65.38)(1) + ((2.5099 - x) / 26.98)(3/2) = 0.1017

Note: the lab report uses 63.38 for Zn molar mass. Follow whichever value your course provides. The working here uses the lab report's value (63.38) to match their answers.

• (x / 63.38)(1) + ((2.5099 - x) / 26.98)(1.5) = 0.1017

Expanding:

• 0.01578x + (2.5099 - x)(0.05560) = 0.1017

• 0.01578x + 0.13955 - 0.05560x = 0.1017

• -0.03982x = 0.1017 - 0.13955

• -0.03982x = -0.03785

• x = 0.95 g (mass of Zn)

Step 6: Calculate Percent Composition

• Mass Zn = 0.95 g

• Mass Al = 2.5099 - 0.95 = 1.56 g

• % Zn = (0.95 / 2.5099) x 100 = 37.82%

• % Al = (1.56 / 2.5099) x 100 = 62.18%

Step 7: Verify with Individual H₂ Contributions

• Moles H₂ from Zn = (0.95 / 63.38)(1) = 0.0150 mol

• Moles H₂ from Al = (1.56 / 26.98)(1.5) = 0.0867 mol

• Total = 0.0150 + 0.0867 = 0.1017 mol (matches Step 4)

Summary of All Three Trials

Common Misconceptions in Calculations

• Students sometimes apply the height correction in the wrong direction. Remember: if the displaced-water side is higher, the extra water column is pushing down on the gas, so the gas pressure is less than barometric minus water vapour alone. The sign depends on which bottle has the higher water level at the end.

• Forgetting to convert cm to mm before dividing by 13.6 is a frequent arithmetic slip. The height difference is often recorded in cm but the conversion factor 13.6 expects mm.

• When solving the simultaneous equation, students occasionally forget the 3/2 factor on the aluminium term and use 1 : 1 for both metals. This will give incorrect masses.

• Rounding too early in intermediate steps compounds errors. Carry at least four significant figures through the calculation and round only at the final answer.

Why It Matters / Exam Flags

• The entire pressure correction chain (barometric to dry hydrogen) is a favourite multi-step exam problem. Practise it until you can do it without looking at notes.

• "What if" questions (gas leak, no equilibration time, soluble gas instead of H₂) test your understanding of how each measured variable feeds into the final answer. Know the direction of the error, not just that there is one.

• Error propagation (post-lab Q7) may appear as a standalone problem on the exam. The pattern is always the same: for multiplication/division, sum the relative errors.

• The limiting reagent question (post-lab Q1) is straightforward but easy to lose marks on if you forget to multiply by the stoichiometric coefficient for H⁺.

Practice Q&A: Post-Lab Questions with Worked Answers

Q: (Post-lab Q1) Why was it unnecessary to discard and refill the HCl between trials?

A: 75 mL of 6 M HCl provides 0.450 mol H⁺ (n = M x V = 6 x 0.075). The total alloy across three trials is 5.205 g. The total Zn used (1.917 g) needs (1.917 / 63.38) x 2 = 0.0605 mol H⁺. The total Al used (3.288 g) needs (3.288 / 26.98) x 3 = 0.3656 mol H⁺. Total H⁺ required = 0.426 mol, which is less than 0.450 mol available. The acid was in excess throughout.

Q: (Post-lab Q2) What problems arise if the reaction produces H₂S or NH₃ instead of H₂?

A: Both H₂S and NH₃ are polar molecules that dissolve readily in water. The water displacement method relies on the gas being insoluble so it pushes water out of the collection bottle. A soluble gas would dissolve instead of displacing water, so the measured volume would be far less than the actual amount of gas produced, making the method unreliable.

Q: (Post-lab Q3) Why is the amount of air in the generator flask at the start not a concern?

A: Both the generator flask and the collection bottle start saturated with air at atmospheric pressure. The air is already at equilibrium throughout the system, so it does not contribute to any net gas displacement. Only the newly produced hydrogen gas creates additional pressure that pushes water out.

Q: (Post-lab Q4) If 150 mL of H₂ leaked, would % Zn be higher, lower, or the same?

A: Higher. With a leaked volume, the measured volume drops from 2.65 L to 2.50 L. Recalculating: n = (0.9355)(2.50) / (0.08206)(297.45) = 0.09581 mol. Solving the simultaneous equation with 0.09581 mol instead of 0.1017 mol gives mass_Zn = 1.097 g and % Zn = 43.71% (up from 37.82%). The shift happens because aluminium contributes more H₂ per gram, so when total H₂ is underestimated, the equations assign more of the alloy mass to the lower-yield metal (zinc).

Q: (Post-lab Q5) What temperature assumptions are made, and are they valid?

A: The experiment assumes constant temperature and pressure in both the generator flask and the collection bottle. This is reasonable because the system is allowed to equilibrate for about 10 minutes, and the temperature readings across the gas and water in the collection bottle are very close (within 0.1 to 0.2°C), indicating thermal equilibrium was essentially reached.

Q: (Post-lab Q6) What happens if the generator flask does not fully equilibrate?

A: If the reaction is still producing gas when you measure, the volume of displaced water will be less than the true total, because not all the hydrogen has been released yet. This gives an undercount of moles, which (as in the gas-leak scenario) inflates % Zn and deflates % Al.

Q: (Post-lab Q7) Find the molar mass and absolute error for a gas with m = 1.545 +/- 0.001 g, T = 298.5 +/- 0.1 K, P = 715 +/- 1 mmHg, V = 0.260 +/- 0.001 L.

A: First, convert P to atm: 715 / 760 = 0.94079 atm. Then MM = mRT / PV = (1.545)(0.08206)(298.5) / (0.94079)(0.260) = 154.72 g/mol. For the error: relative error = (0.001/1.545) + (0.1/298.5) + (1/715) + (0.001/0.260) = 0.000647 + 0.000335 + 0.001399 + 0.003846 = 0.006227. Absolute error = 154.72 x 0.006227 = 0.963, which rounds to 1. Final answer: 154.72 +/- 1 g/mol.

Quick Self-Test

1. Fill in the blank: To convert mmHg to atm, divide by ______. (760)

1. True or False: In Trial 1, the height correction pressure is subtracted from barometric pressure when calculating P_H₂. (False, in Trial 1 the displaced bottle is higher, so P_ΔH is effectively subtracted from P_bar through the rearranged equation P_H₂ = P_bar - P_H₂O - P_ΔH. The sign depends on the water level arrangement.)

1. Fill in the blank: If 75 mL of 6 M HCl is used, the total moles of H⁺ available are ______. (0.450 mol)

1. True or False: Rounding intermediate calculation steps to two significant figures is acceptable. (False, carry at least four significant figures through and round only at the end.)

1. Fill in the blank: For error propagation in a product/quotient, you sum the ______ errors of each input. (relative)

Related Terms / Search Tags

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