Problem 139
Question
A certain lead ore contains the compound \(\mathrm{PbCO}_{3} .\) A sample of the ore weighing \(1.526 \mathrm{~g}\) was treated with nitric acid, which dissolved the \(\mathrm{PbCO}_{3}\). The resulting solution was filtered from undissolved rock and required \(29.22 \mathrm{~mL}\) of \(0.122 \mathrm{M} \mathrm{Na}_{2} \mathrm{SO}_{4}\) to completely precipitate all of the lead as \(\mathrm{PbSO}_{4}\). (a) How many moles of lead were in the ore sample? (b) How many grams of lead were in the ore sample? (c) What is the percentage by mass of lead in the ore? (d) Would you expect the same results if the solid \(\mathrm{PbSO}_{4}\) was collected, washed, dried, and weighed and the final mass was used to answer this question?
Step-by-Step Solution
Verified Answer
The moles of lead were 0.01413 mol, the mass of lead in the sample was 3.115 grams, and the percentage by mass of lead in the sample was 204.15%. The theoretical calculation may differ from the experimental result due to impurities and experimental errors.
1Step 1: Determine moles of Na2SO4 used
Using the concentration and volume of the Na2SO4 solution, calculate the number of moles used in the reaction. The concentration of Na2SO4 is given as 0.122 M (moles per liter), and the volume used is 29.22 mL (which is 0.02922 L after converting from mL to L). The number of moles is found by multiplying the concentration by the volume (in liters).
2Step 2: Determine moles of PbCO3 reacted
Each mole of Na2SO4 reacts with one mole of PbCO3 to form PbSO4. Using the stoichiometry of the reaction, the moles of PbCO3 that reacted will be equal to the moles of Na2SO4 used in the reaction, as determined in step 1.
3Step 3: Calculate the mass of lead (Pb)
(a) The atomic mass of lead (Pb) is necessary to convert moles of lead to grams. (b) Multiply the number of moles of Pb by the atomic mass of Pb to obtain the mass in grams.
4Step 4: Calculate the percentage by mass of lead in the ore
The percentage by mass of lead in the ore is found by dividing the mass of Pb (from step 3) by the original mass of the ore sample and multiplying by 100%.
5Step 5: Consider the measurement method for theoretical calculation
In an actual experiment, weighing the precipitate would include not only the desired PbSO4 but also any impurities or additional substances that might be present. The theoretical calculations assume pure substances and complete reactions without side products or impurities. Therefore, different results should be expected since discrepancies might arise from experimental errors or impurities present in the sample or reagents.
Key Concepts
Molar ConcentrationPrecipitation ReactionsPercent CompositionQuantitative Analysis
Molar Concentration
Understanding molar concentration is foundational when studying chemistry, especially in quantitative analysis. It's a measure of the amount of a substance, in moles, that's present in one liter of solution. It's denoted as molarity (M) and is calculated by the formula:
\[ \text{Molarity (M)} = \frac{\text{moles of solute}}{\text{liters of solution}} \]
In the given exercise, molar concentration of a sodium sulfate (\( \mathrm{Na}_{2}\mathrm{SO}_{4} \)) solution was used to find the moles of solute. The concentration is already provided, meaning we can use it directly with the volume of the solution to find the number of moles. Here's a sample calculation using the figures given:
\[ \text{Moles of } \mathrm{Na}_{2}\mathrm{SO}_{4} = 0.122 \, M \times 0.02922 \, \text{L} \]
This calculation translates the abstract concept of molar concentration into a tangible quantity of substance, which then plays a vital role in stoichiometric calculations for the reaction.
\[ \text{Molarity (M)} = \frac{\text{moles of solute}}{\text{liters of solution}} \]
In the given exercise, molar concentration of a sodium sulfate (\( \mathrm{Na}_{2}\mathrm{SO}_{4} \)) solution was used to find the moles of solute. The concentration is already provided, meaning we can use it directly with the volume of the solution to find the number of moles. Here's a sample calculation using the figures given:
\[ \text{Moles of } \mathrm{Na}_{2}\mathrm{SO}_{4} = 0.122 \, M \times 0.02922 \, \text{L} \]
This calculation translates the abstract concept of molar concentration into a tangible quantity of substance, which then plays a vital role in stoichiometric calculations for the reaction.
Precipitation Reactions
Moving onto precipitation reactions, these are chemical reactions where two solutions react to form an insoluble solid, known as the precipitate. The reaction between the nitric acid-treated lead compound from the ore (\( \mathrm{PbCO}_{3} \)) and sodium sulfate (\( \mathrm{Na}_{2}\mathrm{SO}_{4} \)) results in the formation of lead sulfate (\( \mathrm{PbSO}_{4} \)), which is a precipitate.
In these reactions, key to identify the insoluble product. The solubilities of compounds in water are often presented in tables and solubility rules, which state that most sulfate compounds are soluble, but with important exceptions like lead sulfate.
Improving the understanding of precipitation reactions involves identifying the reactants and predicting the products, using the stoichiometry of the balanced chemical equation to relate moles of reactants and products, and also recognizing that not all ionic compounds form precipitates.
In these reactions, key to identify the insoluble product. The solubilities of compounds in water are often presented in tables and solubility rules, which state that most sulfate compounds are soluble, but with important exceptions like lead sulfate.
Improving the understanding of precipitation reactions involves identifying the reactants and predicting the products, using the stoichiometry of the balanced chemical equation to relate moles of reactants and products, and also recognizing that not all ionic compounds form precipitates.
Percent Composition
Percent composition is critical in determining the purity of a compound or the elemental makeup of a substance. It is the percentage by mass of each element in a compound. This concept is of great importance in quantitative analysis and material characterization.
Calculating percent composition involves knowing the mass of the individual elements in the sample, and the total mass of the sample itself. The calculation for the percent of lead in the lead carbonate ore looks like this:
\[ \text{Percent composition} = \frac{\text{mass of element in sample}}{\text{total mass of sample}} \times 100 \% \]
The percent composition enables us to relate the mass of the element to the mass of the whole sample, thus giving a percentage that indicates the proportion of that particular element in the sample.
Calculating percent composition involves knowing the mass of the individual elements in the sample, and the total mass of the sample itself. The calculation for the percent of lead in the lead carbonate ore looks like this:
\[ \text{Percent composition} = \frac{\text{mass of element in sample}}{\text{total mass of sample}} \times 100 \% \]
The percent composition enables us to relate the mass of the element to the mass of the whole sample, thus giving a percentage that indicates the proportion of that particular element in the sample.
Quantitative Analysis
Finally, quantitative analysis in chemistry refers to the determination of the quantity or concentration of a substance in a sample. It involves the use of various techniques, from simple titrations to advanced spectroscopy, to ascertain the amount of a specific chemical component.
The exercise presents a scenario where quantitative analysis is used to calculate the amount of lead in a sample. This encompasses various steps: solubilizing the sample, precipitating out a pure compound containing the element of interest, and then using stoichiometry to back-calculate to the original substance.
In the classroom and lab, understanding each step's purpose and the methods utilized aids in not just solving problems but also in developing practical lab skills. Real-world samples might contain impurities, and part of quantitative analysis is considering and compensating for potential errors these might introduce.
The exercise presents a scenario where quantitative analysis is used to calculate the amount of lead in a sample. This encompasses various steps: solubilizing the sample, precipitating out a pure compound containing the element of interest, and then using stoichiometry to back-calculate to the original substance.
In the classroom and lab, understanding each step's purpose and the methods utilized aids in not just solving problems but also in developing practical lab skills. Real-world samples might contain impurities, and part of quantitative analysis is considering and compensating for potential errors these might introduce.
Other exercises in this chapter
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