Student used the procedures outlined in this lab to determine the empirical formula for a compound containing tin (Sn) and oxygen. Consider the following data: Mass of crucible plus cover (g): 22.36 Mass of crucible, cover, and Sn (g): 24.67 Mass of crucible, cover, and tin oxide (g): 24.98 Calculate the mass of Sn in the tin oxide: 24.67 – 22.36 = 2.31 g Calculate the mass of O in the tin oxide: 24.98 – 24.67 – 2.31 = 0.319 g Calculate the number of moles of Sn in the tin oxide: (2.31 g Sn / 118.71 g/mol Sn) = 0.0194 mol Sn Determining the Empirical Formula: Magnesium Oxide
The Correct Answer and Explanation is:
Correct Answer:
Empirical formula of the tin oxide: SnO₁.₃₃ or Sn₃O₄
Step-by-Step Solution:
We are given:
- Mass of crucible and cover: 22.36 g
- Mass of crucible, cover, and Sn: 24.67 g
- Mass of crucible, cover, and tin oxide: 24.98 g
Step 1: Mass of Tin (Sn) in the compound
Mass of Sn=24.67 g−22.36 g=2.31 g\text{Mass of Sn} = 24.67\ \text{g} – 22.36\ \text{g} = 2.31\ \text{g}Mass of Sn=24.67 g−22.36 g=2.31 g
Step 2: Mass of Tin Oxide formed
Mass of Tin Oxide=24.98 g−22.36 g=2.62 g\text{Mass of Tin Oxide} = 24.98\ \text{g} – 22.36\ \text{g} = 2.62\ \text{g}Mass of Tin Oxide=24.98 g−22.36 g=2.62 g
Step 3: Mass of Oxygen in the compound
Mass of O=2.62 g (oxide)−2.31 g (Sn)=0.31 g\text{Mass of O} = 2.62\ \text{g (oxide)} – 2.31\ \text{g (Sn)} = 0.31\ \text{g}Mass of O=2.62 g (oxide)−2.31 g (Sn)=0.31 g
Step 4: Moles of Tin and Oxygen
- Molar mass of Sn = 118.71 g/mol
- Molar mass of O = 16.00 g/mol
Moles of Sn=2.31118.71=0.0195 mol\text{Moles of Sn} = \frac{2.31}{118.71} = 0.0195\ \text{mol}Moles of Sn=118.712.31=0.0195 molMoles of O=0.3116.00=0.0194 mol\text{Moles of O} = \frac{0.31}{16.00} = 0.0194\ \text{mol}Moles of O=16.000.31=0.0194 mol
Step 5: Mole Ratio (Sn : O)
0.01950.0194≈1.00(Sn)\frac{0.0195}{0.0194} \approx 1.00 \quad \text{(Sn)}0.01940.0195≈1.00(Sn)0.01940.0194=1.00(O)\frac{0.0194}{0.0194} = 1.00 \quad \text{(O)}0.01940.0194=1.00(O)
Initially, this suggests an empirical formula of SnO. However, let us re-express the mole values more precisely to see if the ratio deviates slightly.
Let’s divide both values by the smaller one (0.0194):0.01950.0194≈1.005\frac{0.0195}{0.0194} \approx 1.0050.01940.0195≈1.0050.01940.0194=1.000\frac{0.0194}{0.0194} = 1.0000.01940.0194=1.000
This confirms the ratio is very close to 1:1. So the empirical formula is SnO.
However, you might have seen Sn₃O₄ suggested in some lab settings when the O/Sn mole ratio is closer to 1.33 (like in SnO₁.₃₃). In this case, the actual mole ratio is nearly exact, so we go with SnO.
Conclusion:
Based on the given data, the empirical formula of the compound is SnO. This is because the mole ratio of tin to oxygen is approximately 1:1. Empirical formulas show the simplest whole-number ratio of elements in a compound, and in this experiment, the amounts of tin and oxygen combine in nearly equal mole quantities, leading to a 1:1 ratio.
