Gravimetric Analysis: Videos & Practice Problems

In gravimetric analysis, the mass of a product from a chemical reaction helps determine the original analyte's amount. For example, a 25 mL solution of bromide ion precipitated 0.7550 grams of lead(II) bromide. The molarity of bromide ion is calculated using the formula: $M=\frac{\mathrm{moles}}{\mathrm{liters}}$. Converting 25 mL to liters gives 0.025 L, and stoichiometry reveals 4.114 x 10^-3 moles of bromide ion, resulting in a molarity of 0.1646 M.

Electrogravimetric Analysis Example 1

Gravimetric analysis is a quantitative method used to determine the amount of an analyte based on the mass of a product formed in a chemical reaction. In this context, we can explore how to calculate the molarity of bromide ions in a solution using stoichiometry and the mass of a precipitate formed.

Consider a scenario where a 25 milliliter solution containing bromide ions is treated with excess lead(II) sulfate, resulting in the precipitation of 0.7550 grams of lead(II) bromide. To find the molarity of the bromide ions, we start with the formula for molarity:

Molarity (M) = \(\frac{\text{moles of solute}}{\text{liters of solution}}\)

First, we convert the volume of the solution from milliliters to liters. Since 1 milliliter is equal to \(10^{-3}\) liters, we have:

25 mL = 0.025 L

Next, we need to determine the moles of bromide ions. We know the mass of lead(II) bromide precipitated, which is 0.7550 grams. To find the moles of lead(II) bromide, we use its molar mass. The molar mass of lead(II) bromide (PbBr₂) can be calculated as follows:

Molar mass of PbBr₂ = Atomic mass of Pb + 2 × Atomic mass of Br

Using the periodic table, we find:

• Atomic mass of Pb ≈ 207.2 g/mol
• Atomic mass of Br ≈ 79.9 g/mol

Thus, the molar mass of PbBr₂ is approximately:

207.2 g/mol + 2 × 79.9 g/mol ≈ 367.008 g/mol

Now, we can calculate the moles of lead(II) bromide:

Moles of PbBr₂ = \(\frac{0.7550 \text{ g}}{367.008 \text{ g/mol}} \approx 0.002056 \text{ moles}\)

From the balanced chemical equation, we know that 1 mole of lead(II) bromide produces 2 moles of bromide ions. Therefore, the moles of bromide ions can be calculated as:

Moles of Br^- = 2 × Moles of PbBr₂ = 2 × 0.002056 moles ≈ 0.004112 moles

Now, we can substitute the moles of bromide ions into the molarity equation:

Molarity of Br^- = \(\frac{0.004112 \text{ moles}}{0.025 \text{ L}} \approx 0.1645 \text{ M}\)

Considering significant figures, since both the volume (25.00 mL) and the mass (0.7550 g) have four significant figures, the final answer for the molarity of bromide ions is:

0.1646 M

This example illustrates the application of stoichiometry in gravimetric analysis, allowing us to determine the concentration of an unknown analyte based on the mass of a precipitate formed during a chemical reaction.

Electrogravimetric Analysis Practice

To determine the weight percent of iron in a sample of iron ore, we start with the known mass of the ore, which is 1.1530 grams. The process involves precipitating the iron as iron(III) oxide and measuring the mass of the resulting product, which is 0.6310 grams of pure iron(III) oxide.

The first step is to calculate the moles of iron(III) oxide using its molar mass. The molar mass of iron(III) oxide (Fe₂O₃) is calculated as follows:

\[\text{Molar mass of Fe}_2\text{O}_3 = (2 \times 55.845 \, \text{g/mol}) + (3 \times 16.00 \, \text{g/mol}) = 159.687 \, \text{g/mol}\]

Next, we convert the mass of iron(III) oxide to moles:

\[\text{Moles of Fe}_2\text{O}_3 = \frac{0.6310 \, \text{g}}{159.687 \, \text{g/mol}} \approx 0.00395 \, \text{mol}\]

From the balanced chemical equation, we know that 1 mole of iron(III) oxide produces 2 moles of iron. Therefore, the moles of iron can be calculated as:

\[\text{Moles of Fe} = 0.00395 \, \text{mol Fe}_2\text{O}_3 \times 2 = 0.00790 \, \text{mol Fe}\]

Now, we convert the moles of iron to grams using its atomic mass:

\[\text{Mass of Fe} = 0.00790 \, \text{mol} \times 55.845 \, \text{g/mol} \approx 0.44134 \, \text{g}\]

Finally, to find the weight percent of iron in the ore, we use the formula:

\[\text{Weight percent of Fe} = \left( \frac{\text{grams of Fe}}{\text{grams of ore}} \right) \times 100 = \left( \frac{0.44134 \, \text{g}}{1.1530 \, \text{g}} \right) \times 100 \approx 38.3\%\]

This calculation illustrates how the mass of the product obtained from a chemical reaction can be used to infer the amount of the original analyte, in this case, the weight percent of iron in the iron ore sample.

Electrogravimetric Analysis Calculations Example 1

In the reaction between Piperazine and acetic acid, an adduct known as Piperazine diacetate is formed. The stoichiometry of the reaction indicates that 1 mole of Piperazine reacts with 2 moles of acetic acid to produce 1 mole of Piperazine diacetate. To determine the mass of Piperazine diacetate produced from a sample of impure Piperazine, we start with a 7.50 gram sample that contains 83.01% Piperazine.

First, we calculate the mass of pure Piperazine in the sample. The mass percent tells us that for every 100 grams of the solution, there are 83.01 grams of Piperazine. Thus, the mass of Piperazine in the 7.50 grams of the impure sample can be calculated as follows:

Mass of Piperazine = (7.50 g) × (83.01 g Piperazine / 100 g solution) = 6.22575 g Piperazine.

Next, we convert the mass of Piperazine to moles using its molar mass. The molar mass of Piperazine, which has the formula C₄H₁₀N₂, is calculated as:

Molar mass of Piperazine = (4 × 12.01 g/mol) + (10 × 1.008 g/mol) + (2 × 14.01 g/mol) = 86.136 g/mol.

Now, we can find the number of moles of Piperazine:

Moles of Piperazine = (6.22575 g) / (86.136 g/mol) ≈ 0.0723 moles.

According to the balanced equation, the molar ratio between Piperazine and Piperazine diacetate is 1:1. Therefore, the moles of Piperazine diacetate produced will also be approximately 0.0723 moles.

To find the mass of Piperazine diacetate, we need its molar mass. The molar mass of Piperazine diacetate, with the formula C₆H₁₀N₂O₄, is calculated as:

Molar mass of Piperazine diacetate = (6 × 12.01 g/mol) + (10 × 1.008 g/mol) + (2 × 14.01 g/mol) + (4 × 16.00 g/mol) = 206.236 g/mol.

Finally, we convert moles of Piperazine diacetate to grams:

Mass of Piperazine diacetate = (0.0723 moles) × (206.236 g/mol) ≈ 14.9 g.

In reporting the final answer, we consider significant figures. The initial mass of the impure Piperazine (7.50 g) has three significant figures, which dictates that our final answer should also be reported as 14.9 g of Piperazine diacetate.

This example illustrates the application of stoichiometry and dimensional analysis in determining the mass of a product from a given mass of an impure reactant, reinforcing the principles of gravimetric analysis.

Electrogravimetric Analysis Calculations Example 2

In this example, we explore the determination of iron content in an ore sample through an oxidation-reduction titration using potassium permanganate (KMnO₄) as the titrant. A 0.5600 gram sample of the ore is treated with acid, converting iron(III) ions to iron(II) ions. The titration process requires 39.82 milliliters of a 0.0315 M potassium permanganate solution to reach the endpoint.

To find the mass percent of iron(III) oxide (Fe₂O₃) in the sample, we start with the formula for mass percent:

Mass percent of iron oxide = \(\frac{\text{grams of iron oxide}}{\text{grams of sample}} \times 100\)

Here, the mass of the sample is already known to be 0.5600 grams. The next step involves calculating the moles of potassium permanganate used in the titration. We convert milliliters to liters:

39.82 mL = 0.03982 L

Using the molarity, we find the moles of KMnO₄:

\(\text{Moles of KMnO}_4 = 0.0315 \, \text{mol/L} \times 0.03982 \, \text{L} = 0.001253 \, \text{mol}\)

Since one mole of KMnO₄ corresponds to one mole of permanganate ion (MnO₄^-), we have:

\(\text{Moles of MnO}_4^{-} = 0.001253 \, \text{mol}\)

Next, we use the stoichiometry of the balanced redox reaction, which indicates that 1 mole of MnO₄^- reacts with 5 moles of Fe³⁺. Thus, the moles of Fe³⁺ can be calculated as:

\(\text{Moles of Fe}^{3+} = 5 \times 0.001253 \, \text{mol} = 0.006265 \, \text{mol}\)

To find the moles of iron(III) oxide, we note that 2 moles of Fe³⁺ correspond to 1 mole of Fe₂O₃:

\(\text{Moles of Fe}_2\text{O}_3 = \frac{0.006265 \, \text{mol}}{2} = 0.0031325 \, \text{mol}\)

Next, we calculate the mass of iron(III) oxide using its molar mass. The molar mass of Fe₂O₃ is calculated as follows:

Fe: 55.845 g/mol × 2 = 111.690 g/mol

O: 15.999 g/mol × 3 = 47.997 g/mol

Total molar mass of Fe₂O₃ = 111.690 g/mol + 47.997 g/mol = 159.687 g/mol

Now, we can find the grams of Fe₂O₃:

\(\text{Grams of Fe}_2\text{O}_3 = 0.0031325 \, \text{mol} \times 159.687 \, \text{g/mol} = 0.5000 \, \text{g}\)

Finally, we substitute this value into the mass percent formula:

Mass percent of iron oxide = \(\frac{0.5000 \, \text{g}}{0.5600 \, \text{g}} \times 100 = 89.42\%\)

Thus, the mass percent of iron(III) oxide in the ore sample is approximately 89.4%, rounded to three significant figures. This example illustrates the importance of careful stoichiometric calculations and the precision required in analytical chemistry, particularly in titration methods.

Here’s what students ask on this topic:

What is gravimetric analysis and how is it used to determine the amount of an analyte?

Gravimetric analysis is a method in analytical chemistry where the mass of a product from a chemical reaction is used to determine the amount of the original analyte. This technique involves precipitating the analyte as a solid, filtering, drying, and weighing the precipitate. The mass of the precipitate is then used to calculate the amount of the analyte based on stoichiometry. For example, if a solution containing bromide ions is treated with excess lead(II) sulfate to form lead(II) bromide, the mass of the lead(II) bromide precipitate can be used to determine the concentration of bromide ions in the original solution.

How do you calculate the molarity of an ion in a solution using gravimetric analysis?

To calculate the molarity of an ion in a solution using gravimetric analysis, follow these steps: First, convert the volume of the solution from milliliters to liters. Then, determine the mass of the precipitate formed. Use the molar mass of the precipitate to convert the mass to moles. Next, use the stoichiometry of the reaction to find the moles of the ion in the solution. Finally, divide the moles of the ion by the volume of the solution in liters to find the molarity. For example, if 0.7550 grams of lead(II) bromide is formed from a 25 mL solution, the molarity of bromide ions can be calculated as follows:

$\frac{4.114\times {10}^{-3} \mathrm{mol}}{0.025 L}=0.1646 M$

What are the steps involved in performing a gravimetric analysis?

The steps involved in performing a gravimetric analysis are as follows: 1) Prepare the sample solution. 2) Add a reagent to precipitate the analyte. 3) Filter the precipitate from the solution. 4) Wash the precipitate to remove impurities. 5) Dry the precipitate to a constant mass. 6) Weigh the dried precipitate. 7) Calculate the amount of analyte based on the mass of the precipitate and the stoichiometry of the reaction. These steps ensure accurate determination of the analyte's amount in the original sample.

What are some common applications of gravimetric analysis in real-world scenarios?

Gravimetric analysis is widely used in various real-world applications, including: 1) Determining the purity of a substance by measuring the amount of a specific component. 2) Analyzing the composition of ores and minerals in the mining industry. 3) Measuring the concentration of pollutants in environmental samples, such as water and air. 4) Quality control in the pharmaceutical industry to ensure the correct dosage of active ingredients. 5) Food industry for analyzing the content of specific nutrients or contaminants. These applications highlight the importance of gravimetric analysis in different fields.

What are the advantages and limitations of gravimetric analysis?

Gravimetric analysis offers several advantages, including high accuracy and precision, simplicity of the procedure, and the ability to determine the analyte without the need for expensive instrumentation. However, it also has limitations, such as being time-consuming, requiring careful handling to avoid loss of the precipitate, and being less suitable for very low concentrations of analytes. Additionally, the method relies on the formation of a stable and pure precipitate, which may not always be achievable for all analytes.