Empirical Formula: SnO2
Mass of tin oxide formed: 1.2665g
The overall reaction is represented in the following unbalanced equation: aSn(s) + b HNO 3 (aq) ⟶ c Sn x Oy (s) + d H 2 O(g) + e NO 2 (g)
- Having determined the empirical formula of the tin oxide, re-write and balance the overall reaction representing the formation of the oxide.
- Based on the balanced chemical equation above, calculate the theoretical mass of tin oxide that could have been produced in each determination. Show your work.
- How do these theoretical amounts compare to the actual amounts obtained? Explain your answer.
- What is the mass percent of tin and oxygen in the synthesized product? Show your work.
- How would the loss of some of the metastannic acid during the evaporation of water affect the determination of the empirical formula of the product tin oxide? Explain your answer.
- What is the Theoretical Yield?
The Correct Answer and Explanation is :
Let’s address each question step by step.
1. Balanced Chemical Equation
The given empirical formula of tin oxide is ( \text{SnO}_2 ). Thus, the reaction is:
[ a \, \text{Sn}(s) + b \, \text{HNO}_3(aq) \rightarrow c \, \text{SnO}_2(s) + d \, \text{H}_2\text{O}(g) + e \, \text{NO}_2(g) ]
To balance the reaction:
- Tin ((\text{Sn})) is oxidized to (\text{SnO}_2), and (\text{HNO}_3) acts as the oxidizing agent.
- The nitrogen in (\text{HNO}_3) is reduced to (\text{NO}_2).
The balanced equation is:
[ \text{Sn}(s) + 4 \, \text{HNO}_3(aq) \rightarrow \text{SnO}_2(s) + 2 \, \text{H}_2\text{O}(g) + 4 \, \text{NO}_2(g) ]
2. Theoretical Mass of Tin Oxide
Given the mass of tin oxide (( \text{SnO}_2 )) formed is ( 1.2665 \, \text{g} ):
- Molar masses:
- (\text{Sn} = 118.71 \, \text{g/mol})
- (\text{O} = 16.00 \, \text{g/mol})
- (\text{SnO}_2 = 118.71 + 2(16.00) = 150.71 \, \text{g/mol})
1 mole of (\text{Sn}) produces 1 mole of (\text{SnO}_2). From the balanced equation:
[
\text{Mass of Sn required} = \frac{\text{Mass of SnO}_2}{\text{Molar mass of SnO}_2} \times \text{Molar mass of Sn}
]
For (1.2665 \, \text{g}) of (\text{SnO}_2):
[
\text{Mass of Sn required} = \frac{1.2665}{150.71} \times 118.71 = 0.9968 \, \text{g}
]
This indicates (0.9968 \, \text{g}) of tin theoretically produces (1.2665 \, \text{g}) of (\text{SnO}_2).
3. Comparing Theoretical and Actual Amounts
If the actual mass of tin oxide matches the calculated theoretical mass ((1.2665 \, \text{g})), the reaction proceeded with 100% efficiency. Any discrepancy between the theoretical and actual yield could be attributed to:
- Loss of product during handling.
- Side reactions or incomplete reaction.
- Impurities in the starting materials.
4. Mass Percent of Tin and Oxygen
Mass percent is calculated as:
[
\text{Mass percent of element} = \left( \frac{\text{Mass of element in 1 mole of compound}}{\text{Molar mass of compound}} \right) \times 100
]
- For tin in (\text{SnO}_2):
[
\text{Mass percent of Sn} = \left( \frac{118.71}{150.71} \right) \times 100 = 78.76\%
] - For oxygen in (\text{SnO}_2):
[
\text{Mass percent of O} = \left( \frac{2 \times 16.00}{150.71} \right) \times 100 = 21.24\%
]
5. Effect of Loss of Metastannic Acid
If metastannic acid ((\text{H}_2\text{SnO}_3)) is lost during evaporation:
- The calculated mass of tin oxide would be lower.
- This would affect the empirical formula calculation by showing a higher oxygen-to-tin ratio, leading to an incorrect empirical formula.
6. Theoretical Yield
The theoretical yield is the maximum mass of product that can be formed from the limiting reactant, calculated as (1.2665 \, \text{g}) in this case.
Explanation (300 Words):
The theoretical yield represents the ideal quantity of product formed if a chemical reaction proceeds perfectly without any losses or side reactions. For the reaction:
[
\text{Sn}(s) + 4 \, \text{HNO}_3(aq) \rightarrow \text{SnO}_2(s) + 2 \, \text{H}_2\text{O}(g) + 4 \, \text{NO}_2(g)
]
The theoretical yield of (\text{SnO}_2) depends on the amount of tin available, as it is the limiting reactant. Using stoichiometry and the molar masses of (\text{Sn}) and (\text{SnO}_2), we determine the theoretical yield as (1.2665 \, \text{g}). This assumes 100% conversion of tin to (\text{SnO}_2) without losses during the reaction, handling, or measurement.
In practical terms, achieving the theoretical yield is rare due to potential experimental errors, incomplete reactions, or competing side reactions. Comparing the actual yield with the theoretical yield provides the percentage yield, which reflects the reaction’s efficiency. If some metastannic acid is lost or impurities are present, the actual yield would be lower than the theoretical yield, affecting the reaction’s accuracy and efficiency.