Question

For a research project, a student decided to test the effect of the lead(II) ion \(\left(\mathrm{Pb}^{2+}\right)\) on the ability of salmon eggs to hatch. This ion was obtainable from the water-soluble salt, lead(II) nitrate, \(\mathrm{Pb}\left(\mathrm{NO}_{3}\right)_{2},\) which the student decided to make by the following reaction. (The desired product was to be isolated by the slow evaporation of the water.) \(\mathrm{PbO}(s)+2 \mathrm{HNO}_{3}(a q) \longrightarrow \mathrm{Pb}\left(\mathrm{NO}_{3}\right)_{2}(a q)+\mathrm{H}_{2} \mathrm{O}\) Losses of product for various reasons were expected, and a yield of \(86.0 \%\) was expected. In order to have \(5.00 \mathrm{~g}\) of product at this yield, how many grams of \(\mathrm{PbO}\) should be taken? (Assume that sufficient nitric acid, \(\mathrm{HNO}_{3}\), would be used.

Short answer

To obtain 5.00 g of \(\mathrm{Pb}\left(\mathrm{NO}_{3}\right)_{2}\) at an 86% yield, approximately 5.814 g of \(\mathrm{PbO}\) is needed.

Step-by-step solution

Calculate the theoretical yield

The theoretical yield is the maximum amount of product that can be produced from a given amount of reactants. Since the expected yield is 86%, the theoretical yield of \(\mathrm{Pb}\left(\mathrm{NO}_{3}\right)_{2}\) to get 5.00 g of the product is calculated by the equation: \(\text{theoretical yield} = \frac{\text{actual yield}}{\text{percent yield}}\times 100\).

Calculate the mass of PbO required

We now use the molar masses of \(\mathrm{PbO}\) and \(\mathrm{Pb}\left(\mathrm{NO}_{3}\right)_{2}\) to find the mass of \(\mathrm{PbO}\) needed. This involves using stoichiometry based on the balanced chemical equation \(\mathrm{PbO}(s) + 2 \mathrm{HNO}_{3}(aq) \rightarrow \mathrm{Pb}\left(\mathrm{NO}_{3}\right)_{2}(aq) + \mathrm{H}_{2}\mathrm{O}\), which indicates a 1:1 mole ratio between \(\mathrm{PbO}\) and \(\mathrm{Pb}\left(\mathrm{NO}_{3}\right)_{2}\).

Perform the stoichiometric calculation

The mass of \(\mathrm{PbO}\) required can be calculated by first determining the moles of \(\mathrm{Pb}\left(\mathrm{NO}_{3}\right)_{2}\) needed for the theoretical yield, and then converting those moles to grams of \(\mathrm{PbO}\) using the molar mass of \(\mathrm{PbO}\).

Use the percentage yield to calculate the required \(\mathrm{PbO}\) mass

The final calculation is to find the actual mass of \(\mathrm{PbO}\) that should be used based on the expected percentage yield. Multiply the theoretical mass of \(\mathrm{PbO}\) by the percentage yield to find the required starting mass.

Key concepts

Theoretical Yield

Theoretical yield is a foundational concept in stoichiometry, describing the maximum quantity of product that could be generated from complete conversion of reactants in a chemical reaction. It's determined by the balanced chemical equation and the amount of limiting reagent available.

Imagine baking cookies with exactly enough dough for 12 cookies—the dough represents the limiting reagent, and the 12 cookies are the theoretical yield. Even if the oven's capacity or your cookie sheet only allows you to bake 10 at a time, the potential to make 12 exists given adequate resources.

To calculate the theoretical yield:

• Identify the limiting reactant, which is the substance that will run out first during the reaction.
• Use stoichiometry to convert this reactant's mass to moles, based on its molar mass.
• Apply the mole ratio from the balanced equation to find the number of moles of product formed.
• Convert the moles of product back to mass, using its molar mass, for the theoretical yield in grams.

Percent Yield

Percent yield is a measure of the efficiency of a chemical reaction, indicating how much product was actually obtained compared to the theoretical yield. In reality, achieving 100% yield is rare due to side reactions, incomplete reactions, or loss of product during the process.

Returning to our cookie analogy, if you planned for 12 cookies (theoretical yield), but only produced 10 because some dough was left on the spoon or stuck to the bowl, your percent yield is the actual number of cookies baked divided by the potential number of cookies multiplied by 100: \[\begin{equation}Percent Yield = \left(\frac{Actual Yield}{Theoretical Yield}\right) \times 100\%ewlineewlineewlineewlineewlineewlineewlineewline\end{equation}\]

• First, find the actual yield of the product from the reaction.
• Then, calculate the theoretical yield using stoichiometry.
• Last, divide the actual yield by the theoretical yield and multiply by 100 to get the percent yield.

It provides insight into the reaction's success and potential areas for process optimization.

Chemical Reaction

A chemical reaction is a process where substances, known as reactants, transform into different substances called products. It's similar to a culinary recipe, where ingredients are combined and altered by cooking to create a final dish. Each reaction is governed by a balanced chemical equation that preserves the number of atoms for each element.

In these equations, reactant molecules and product molecules are represented with chemical formulas. For instance, in our exercise, \(\mathrm{PbO}\) and \(\mathrm{HNO}_{3}\) are reactants, and \(\mathrm{Pb}\left(\mathrm{NO}_{3}\right)_{2}\) and \(\mathrm{H}_{2}\mathrm{O}\) are products. The reaction's stoichiometry ensures that for each mole of \(\mathrm{PbO}\) used, one mole of \(\mathrm{Pb}\left(\mathrm{NO}_{3}\right)_{2}\) is produced.

Understanding the nature of chemical reactions is essential, as it enables us to predict products, calculate yields, and control reaction conditions effectively.

Molar Mass

Molar mass is the weight of one mole (6.022 x 10²³ particles) of a given substance. It's the molecular weight of a compound expressed in grams per mole (g/mol), and it's fundamental for converting between the mass of a substance and the number of moles present.

To calculate it, sum up the atomic masses of all atoms in the molecule as listed on the periodic table. For instance, the molar mass of water \(H_2O\) is approximately 18 g/mol, adding the mass of two hydrogen atoms (about 1 g/mol each) with one oxygen atom (roughly 16 g/mol).

This value is crucial in stoichiometry for solving problems that involve mass-to-mole conversions or vice versa, as seen in our exercise where the molar mass of \(\mathrm{PbO}\) helps connect the mass of the PbO needed to the moles required by the balanced equation.