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Chapter 4: Problem 112

A mixture contains only NaCl and \(\mathrm{Fe}\left(\mathrm{NO}_{3}\right)_{3}\) . A \(0.456-\mathrm{g}\) sample of the mixture is dissolved in water, and an excess of NaOH is added, producing a precipitate of \(\mathrm{Fe}(\mathrm{OH})_{3}\) . The precipitate is filtered, dried, and weighed. Its mass is 0.107 \(\mathrm{g}\) . Calculate the following. a. the mass of iron in the sample b. the mass of Fe(NO \(_{3} )_{3}\) in the sample c. the mass percent of \(\mathrm{Fe}\left(\mathrm{NO}_{3}\right)_{3}\) in the sample

Short Answer

Expert verified
The mass of iron in the sample is 0.0560 g, the mass of Fe(NO₃)₃ in the sample is 0.2428 g, and the mass percent of Fe(NO₃)₃ in the sample is 53.25%.

Step by step solution

01

Write the balanced chemical equation

The reaction between iron(III) nitrate and sodium hydroxide forms a precipitate of iron(III) hydroxide and sodium nitrate. The balanced equation is: \(Fe(NO_3)_3 + 3NaOH \rightarrow Fe(OH)_3 + 3NaNO_3\)
02

Calculate the moles of \(\mathrm{Fe}\left(\mathrm{OH}\right)_{3}\) precipitate

We are given the mass of the precipitate, which is 0.107g. To calculate the moles, use the molar mass of Fe(OH)₃ (106.87 g/mol): Moles of \(Fe(OH)_3 = \frac{mass}{molar\,mass} = \frac{0.107\,g}{106.87\,g/mol} = 1.002\,×10^{-3}\,mol\)
03

Calculate the moles of \(\mathrm{Fe}\left(\mathrm{NO}_{3}\right)_{3}\)

From the balanced equation, we can see that 1 mole of Fe(NO₃)₃ produces 1 mole of Fe(OH)₃. Therefore, moles of Fe(NO₃)₃ are the same as moles of Fe(OH)₃: Moles of \(Fe(NO_3)_3 = 1.002 × 10^{-3}\,mol\)
04

Calculate the mass of Fe in the sample

Now, we can calculate the mass of iron in the sample: Mass of Fe = (moles of \(Fe(NO_3)_3) * (molar\,mass\,of\,Fe)\) Mass of Fe = \(1.002 × 10^{-3} mol * 55.85\,g/mol = 0.0560\,g\)
05

Calculate the mass of \(Fe(NO_3)_3\) in the sample

Next, we can calculate the mass of Fe(NO₃)₃ in the sample: Mass of \(Fe(NO_3)_3\) = (moles of \(Fe(NO_3)_3) * (molar\,mass\,of\,Fe(NO_3)_3)\) Mass of \(Fe(NO_3)_3 = 1.002 × 10^{-3}\,mol * 241.86\,g/mol = 0.2428\,g\)
06

Calculate the mass percent of \(\mathrm{Fe}\left(\mathrm{NO}_{3}\right)_{3}\) in the sample

Finally, we can determine the mass percent of Fe(NO₃)₃ in the sample: Mass percent of \(Fe(NO_3)_3 = \frac{mass\,of\,Fe(NO_3)_3}{mass\,of\,sample} × 100\%\) Mass percent of \(Fe(NO_3)_3 = \frac{0.2428\,g}{0.456\,g} × 100\% = 53.25\%\) To sum up, the mass of iron in the sample is 0.0560 g, the mass of Fe(NO₃)₃ in the sample is 0.2428 g, and the mass percent of Fe(NO₃)₃ in the sample is 53.25%.

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Key Concepts

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

Chemical Reactions
A chemical reaction is a process in which substances, known as reactants, are transformed into other substances, called products. In the given exercise, the reaction involves iron(III) nitrate \((\mathrm{Fe(NO}_3)_3)\) and sodium hydroxide \((\mathrm{NaOH})\) reacting to form iron(III) hydroxide \((\mathrm{Fe(OH)}_3)\) as a precipitate and sodium nitrate \((\mathrm{NaNO}_3)\). This transformation is described by a balanced chemical equation:\[ \mathrm{Fe(NO_3)_3 + 3NaOH \rightarrow Fe(OH)_3 + 3NaNO_3} \]Balancing chemical equations is crucial because it ensures that the same number of each type of atom is present on both sides of the equation. It reflects the Law of Conservation of Mass, stating that matter cannot be created or destroyed during a chemical reaction. This principle helps accurately determine the amounts of reactants and products involved, enabling further calculations in stoichiometry.
Molar Mass Calculation
Molar mass is an essential concept in stoichiometry that relates the mass of a substance to the number of moles. It allows us to convert between grams and moles, facilitating the calculation of reactants and products in a chemical reaction. For instance, the molar mass of iron(III) hydroxide \((\mathrm{Fe(OH)}_3)\) is 106.87 g/mol. To find it, sum up the atomic masses of each element in its chemical formula:- Fe = 55.85 g/mol- O = 16.00 g/mol (for 3 oxygen atoms, multiply by 3)- H = 1.01 g/mol (for 3 hydrogen atoms, multiply by 3)By adding these values, the molar mass of \(\mathrm{Fe(OH)}_3\) becomes:\[ \text{Molar Mass of } \mathrm{Fe(OH)}_3 = 55.85 + (3 \times 16.00) + (3 \times 1.01) = 106.87 \text{ g/mol} \]Molar mass aids in calculating the amount of substance, or moles, using the formula:\[ \text{Moles} = \frac{\text{Mass}}{\text{Molar Mass}} \]This calculation is crucial for interpreting chemical reactions quantitatively.
Precipitation Reaction
Precipitation reactions are a type of chemical reaction where an insoluble solid, called a precipitate, forms when two solutions are mixed. In the exercise example, iron(III) nitrate \((\mathrm{Fe(NO}_3)_3)\) reacts with sodium hydroxide \((\mathrm{NaOH})\), resulting in the formation of a precipitate, iron(III) hydroxide \((\mathrm{Fe(OH)}_3)\). The reaction can be represented by the following equation:\[ \mathrm{Fe(NO_3)_3 + 3NaOH \rightarrow Fe(OH)_3 + 3NaNO_3} \]Here, \(\mathrm{Fe(OH)}_3\) is the precipitate, which can be observed as a solid that separates from the solution. This type of reaction is used to isolate specific compounds from mixtures, as the precipitate can be filtered and weighed. It plays a vital role in various applications, including chemical analysis and environmental science, where unwanted contaminants are removed as solids.
Mass Percent Calculation
Mass percent is a measure of the concentration of a component in a mixture and is expressed as a percentage. It is calculated by dividing the mass of the component by the total mass of the sample and then multiplying by 100%.To find the mass percent of \(\mathrm{Fe(NO}_3)_3\) in the sample, follow these steps:1. Calculate the mass of \(\mathrm{Fe(NO}_3)_3\), which from the solution is 0.2428 g.2. Divide this value by the total mass of the sample, which is 0.456 g.3. Multiply the result by 100 to convert it to a percentage.Mathematically, this is shown as:\[ \text{Mass percent of } \mathrm{Fe(NO}_3)_3 = \left( \frac{0.2428 \text{ g}}{0.456 \text{ g}} \right) \times 100\% = 53.25\% \]Understanding mass percent is crucial for determining the composition of a mixture and is widely used in fields like chemistry, pharmacy, and metallurgy. This calculation helps in analyzing the purity of substances and the formulation of chemical solutions.

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Most popular questions from this chapter

The blood alcohol \(\left(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}\right)\) level can be determined by tirrating a sample of blood plasma with an acidic potassium dichromate solution, resulting in the production of \(\mathrm{Cr}^{3+}(a q)\) and carbon dioxide. The reaction can be monitored because the dichromate ion \(\left(\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}\right)\) is orange in solution, and the \(\mathrm{Cr}^{3+}\) ion is green. The unbalanced redox equation is $$\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}(a q)+\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}(a q) \rightarrow \mathrm{Cr}^{3+}(a q)+\mathrm{CO}_{2}(g)$$ If 31.05 \(\mathrm{mL}\) of 0.0600\(M\) potassium dichromate solution is required to titrate 30.0 \(\mathrm{g}\) of blood plasma, determine the mass percent of alcohol in the blood.

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