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A \(1.50-\mathrm{g}\) sample of sodium nitrate is decomposed by heating. If the resulting sodium nitrite has a mass of \(1.29 \mathrm{~g}\) and the calculated yield is \(1.22 \mathrm{~g},\) what is the percent yield?

Short Answer

Expert verified
The percent yield is approximately 105.74%.

Step by step solution

01

Understand the Concept of Percent Yield

Percent yield is a measure of how much product is obtained compared to the maximum amount that could be obtained, calculated by the formula: \( \text{Percent Yield} = \left( \frac{\text{Actual Yield}}{\text{Theoretical Yield}} \right) \times 100\% \).
02

Identify the Given Values

The given values are: the actual yield (the mass of sodium nitrite obtained) is \( 1.29 \text{ g} \), and the theoretical yield (the calculated mass of sodium nitrite) is \( 1.22 \text{ g} \).
03

Substitute the Values into the Formula

Substitute the given values into the percent yield formula: \( \text{Percent Yield} = \left( \frac{1.29}{1.22} \right) \times 100\% \).
04

Calculate the Percent Yield

Perform the division followed by the multiplication: \( \text{Percent Yield} = \left( \frac{1.29}{1.22} \right) \times 100\% \approx 105.74\% \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Theoretical Yield
Theoretical yield is the maximum amount of product that can be formed in a chemical reaction. It is calculated based on the stoichiometric coefficients from the balanced chemical equation. In this case, before beginning the actual reaction, you calculate how much sodium nitrite should be produced from the decomposition of sodium nitrate if everything reacts perfectly.
  • To find the theoretical yield, you generally start with the balanced chemical equation.

  • Using the molar masses of the reactants and products, you determine the mole relationships to calculate the amount of product that should form if the reaction were to proceed perfectly.

  • Sodium nitrate decomposes into sodium nitrite, so you would use the weights and mole ratios to predict the theoretical amount of sodium nitrite formed.

In the given problem, the theoretical yield is provided as 1.22 g of sodium nitrite, which is the amount expected if the reaction had gone without any losses due to byproducts, incomplete reactions, or measurement inaccuracies.
Actual Yield
The actual yield is the amount of product that is actually obtained from a chemical reaction. This differs from the theoretical yield due to practical losses during the experiment or incomplete reactions. In a lab experiment, these losses can come from various factors such as side reactions, errors in measurement, or inefficiencies in the procedure followed.
  • The actual yield is usually obtained by measuring the mass of the products formed at the end of the reaction. In our scenario, this is the 1.29 g of sodium nitrite achieved from the decomposition of sodium nitrate.

  • A higher or lower actual yield compared to the theoretical yield can indicate the efficiency of the reaction or the presence of impurities.

It is important in laboratory settings to carefully measure and record the actual yield to accurately calculate the reaction's efficiency and understand any discrepancies from the theoretical prediction.
Chemical Decomposition
Chemical decomposition is a type of chemical reaction where one compound breaks down into two or more simpler substances. This reaction type is crucial in understanding and calculating theoretical and actual yields as it directly impacts how products are formed.
  • In our example, sodium nitrate is subjected to heat, leading it to chemically decompose into sodium nitrite and possibly other byproducts, if any.

  • Decomposition reactions generally require some form of energy input, like heat, light, or electricity, to break chemical bonds in the reactants.

  • Knowing how a compound decomposes can help predict the theoretical yield by allowing for the calculation of what products are expected and in what amounts.

Understanding chemical decomposition helps in designing experiments with better efficiency, minimizing side reactions, and reaching higher actual yields compared to theoretical expectations.

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