Difficulty: Intermediate | Prerequisites: Stoichiometry and Mass Relationships study notes (Part 1)
These are the full worked calculations from Lab 8, covering the Fe + S → FeS system and the Fe + Cl₂ → FeCl₃ system. Each example walks through mole conversions, limiting reagent identification, theoretical yield, excess reagent calculation, and percent yield. The graphical analysis section explains what the lab's plots demonstrate about the law of definite proportions.
Balanced equation: Fe + S → FeS (1:1 mole ratio)
Given: 5 g Fe, 2 g S
Moles of Fe = 5 g / 55.85 g/mol = 8.95 × 10⁻² mol
Moles of S = 2 g / 32.06 g/mol = 6.24 × 10⁻² mol
The mole ratio is 1:1. S has fewer moles (0.0624 vs 0.0895), so S is the limiting reagent.
Fe consumed = moles of S × molar mass of Fe = 0.0624 × 55.85 = 3.48 g (this follows from the 1:1 ratio: moles of Fe consumed equals moles of S).
Theoretical excess Fe = 5 g - 3.48 g = 1.52 g.
Note: the lab document writes 5 - 1.2 = 3.48 g for mass of Fe reacted. The 1.2 appears to be a rounded or experimental value for excess Fe; the calculated value is 1.52 g.
Mass of Fe reacted = 5 g - 1.52 g = 3.48 g.
Moles of Fe reacted = 3.48 / 55.85 = 6.23 × 10⁻² mol.
Mass of Fe + S combined = 3.48 + 2 = 5.48 g.
Theoretical mass of FeS = moles of S × molar mass of FeS = 0.0624 × 87.91 = 5.49 g.
Mass ratio (Fe/S) = 3.48 / 2 = 1.74
Mole ratio (Fe/S) = 6.23 × 10⁻² / 6.23 × 10⁻² = ~1.00 (confirming the 1:1 stoichiometry)
Fe: 55.85 g/mol
S: 32.06 g/mol
FeS: 87.91 g/mol
Balanced equation: 2 Fe + 3 Cl₂ → 2 FeCl₃
Given: 10.00 g Fe, 0.50 g Cl₂
Moles of Fe = 10.00 / 55.85 = 0.1791 mol
Moles of Cl₂ = 0.50 / 70.90 = 0.0071 mol
Divide each by its coefficient in the balanced equation:
Fe: 0.1791 / 2 = 0.08955
Cl₂: 0.0071 / 3 = 0.00237
Cl₂ gives the smaller value, so Cl₂ is the limiting reagent. Iron is in large excess.
From the balanced equation, 3 mol Cl₂ produces 2 mol FeCl₃.
Moles of FeCl₃ = 0.0071 × (2/3) = 0.00473 mol.
Mass of FeCl₃ = 0.00473 × 162.2 = 0.77 g.
This is the theoretical yield.
Given: Actual yield = 0.65 g FeCl₃; theoretical yield = 0.77 g FeCl₃ (from Example 2)
Percent yield = (actual yield / theoretical yield) × 100
Percent yield = (0.65 / 0.77) × 100 = 84.4%
The reaction was 84.4% efficient. The remaining 15.6% of the expected product was lost. Common reasons for a yield below 100% include:
Incomplete reaction (not all limiting reagent reacted)
Transfer losses (product left on glassware, filter paper, or spatula)
Side reactions producing unwanted byproducts
Measurement errors in weighing
Product loss during filtration or evaporation
The lab produced several plots. Each one tests or illustrates a specific principle.
Best-fit line: y = 0.775x - 0.4116. This plot demonstrates the law of definite proportions. Because Fe and S combine in a fixed mass ratio (approximately 1.74:1), plotting one against the other should give a straight line. The slope is close to the theoretical mass ratio of Fe to S (55.85/32.06 = 1.74). A non-zero y-intercept reflects experimental error.
Best-fit line: y = 1.6981x - 0.1419. The slope here reflects the ratio of the mass of Fe consumed to the mass of FeS produced. Theoretically, the mass fraction of Fe in FeS is 55.85/87.91 = 0.635, so the inverse (mass of FeS per gram of Fe reacted) should be about 1.57. The experimental slope of ~1.70 is in the right range.
Best-fit line: y = 1.9431x + 1.1321. The theoretical ratio of FeS mass to S mass is 87.91/32.06 = 2.74. The experimental slope of ~1.94 is lower, suggesting systematic error or data scatter.
Best-fit line: y = 1.0022x - 8 × 10⁻⁵. This is the cleanest demonstration of the 1:1 mole ratio. The slope of ~1.00 confirms that Fe and S react in equal moles, consistent with Fe + S → FeS.
Best-fit line: y = -0.0192x + 1.0009. The mole ratio should be constant at 1.00 regardless of how much iron is used. The near-horizontal line with a y-intercept of ~1.00 confirms this. The slight negative slope is within experimental scatter.
Best-fit line: y = 1.0789x - 0.1432. If mass is conserved, the mass of reactants consumed should equal the mass of product formed, giving a slope of 1. The experimental slope of ~1.08 is close, consistent with conservation of mass within experimental error.
Using mass instead of moles to identify the limiting reagent. 10 g of Fe vs. 0.50 g of Cl₂ makes iron look dominant by mass, but Cl₂ is still limiting because the balanced equation demands 1.5 mol Cl₂ per mol Fe.
Forgetting to divide by the stoichiometric coefficient. Computing moles alone is not enough; you must divide by the coefficient before comparing.
Mixing up the mole ratio direction. When converting from Cl₂ to FeCl₃, the ratio is 2 mol FeCl₃ per 3 mol Cl₂, not 3/2. Writing it upside down doubles or halves your answer.
Rounding too early. Carry at least four significant figures through intermediate steps and round only at the end. Early rounding in multi-step stoichiometry can shift the final answer noticeably.
Confusing theoretical yield with actual yield when computing percent yield. The theoretical number goes in the denominator.
Q: 3.00 g of Fe reacts with 3.00 g of S. Which is the limiting reagent, and how many grams of FeS form?
A: Moles of Fe = 3.00/55.85 = 0.0537 mol. Moles of S = 3.00/32.06 = 0.0936 mol. The ratio is 1:1, so Fe is limiting (fewer moles). Theoretical FeS = 0.0537 × 87.91 = 4.72 g.
Q: For the reaction 2 Fe + 3 Cl₂ → 2 FeCl₃, if you start with 2.00 g of Fe and 5.00 g of Cl₂, what is the theoretical yield of FeCl₃?
A: Moles of Fe = 2.00/55.85 = 0.0358 mol. Moles of Cl₂ = 5.00/70.90 = 0.0705 mol. Divide by coefficients: Fe: 0.0358/2 = 0.0179. Cl₂: 0.0705/3 = 0.0235. Fe is limiting. Moles of FeCl₃ = 0.0358 × (2/2) = 0.0358 mol. Mass = 0.0358 × 162.2 = 5.81 g.
Q: A student collects 4.50 g of FeCl₃ from the reaction above. What is the percent yield?
A: Percent yield = (4.50/5.81) × 100 = 77.5%.
Q: In the Fe + S → FeS reaction, what does a plot of moles of S vs. moles of Fe with a slope of 1.00 confirm?
A: It confirms the 1:1 mole ratio predicted by the balanced equation, which is evidence for the law of definite proportions.
Q: Why does a graph of m(Fe + S) vs. m(FeS) with a slope near 1.0 support the law of conservation of mass?
A: If the total mass of reactants consumed equals the mass of product formed, the slope is 1. A slope near 1.0 means mass is neither gained nor lost, consistent with conservation of mass.
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