Problem 109 - Chemistry Matter and Change

Question

When 9.59 g of a certain vanadium oxide is heated in the presence of hydrogen, water and a new oxide of vanadium are formed. This new vanadium oxide has a mass of 8.76 g. When the second vanadium oxide undergoes additional heating in the presence of hydrogen, 5.38 g of vanadium metal forms. \begin{equation} \begin{array}{l}{\text { a. Determine the empirical formulas for the two }} \\\ {\text { vanadium oxides. }} \\ {\text { b. Write balanced equations for the steps of the reaction. }} \\ {\text { c. Determine the mass of hydrogen needed to complete }} \\ {\text { the steps of this reaction. }}\end{array} \end{equation}

Step-by-Step Solution

Answer

Empirical formulas: \( V_2O \) and \( VO_2 \); hydrogen mass needed: \( 1.056 \, \text{g} \).

Step 1: Determine Moles of Vanadium in First Reaction

Calculate the moles of vanadium produced from the data provided. Since the second vanadium oxide contains 5.38 g of pure vanadium, we can calculate the moles of vanadium:\[ \text{Moles of V} = \frac{5.38 \, \text{g}}{50.94 \, \text{g/mol}} \approx 0.1056 \, \text{mol} \]

Step 2: Find Mass of Oxygen in First Oxide

To find the mass of oxygen in the initial vanadium oxide (9.59 g), we subtract the mass of vanadium (8.76 g):\[ \text{Mass of O} = 9.59 \, \text{g} - 8.76 \, \text{g} = 0.83 \, \text{g} \]

Step 3: Calculate Moles of Oxygen in First Oxide

Use the mass of oxygen to find the moles of oxygen:\[ \text{Moles of O} = \frac{0.83 \, \text{g}}{16.00 \, \text{g/mol}} \approx 0.0519 \, \text{mol} \]

Step 4: Find Empirical Formula of First Vanadium Oxide

Using the molar ratio of vanadium to oxygen from Steps 1 and 3, determine the empirical formula. The ratio \( \frac{0.1056}{0.0519} \approx 2 \) suggests \( V_2O \) as the empirical formula for the first oxide.

Step 5: Determine Moles of Vanadium in Second Reaction

The mass of the second vanadium oxide is 8.76 g which contains 0.1056 mol V. Therefore, it contains the same moles of vanadium.

Step 6: Find Mass and Moles of Oxygen in Second Oxide

Calculate mass of oxygen:\[ \text{Mass of O in second oxide} = 8.76 \, \text{g} - 5.38 \, \text{g} = 3.38 \, \text{g} \]Calculate moles of oxygen:\[ \text{Moles of O} = \frac{3.38 \, \text{g}}{16.00 \, \text{g/mol}} \approx 0.2112 \, \text{mol} \]

Step 7: Find Empirical Formula of Second Vanadium Oxide

Using the molar ratio from vanadium to oxygen (0.1056 to 0.2112) in the second oxide, which is approximately 1:2, the empirical formula is \( VO_2 \).

Step 8: Write Balanced Equations for Reactions

Identified reactions are:First Reaction: \[ V_2O + H_2 \rightarrow VO_2 + H_2O \]Second Reaction: \[ VO_2 + H_2 \rightarrow V + H_2O \]

Step 9: Determine Total Mass of Hydrogen Needed

In the first reaction, each mole of \( VO_2 \) releases 0.0519 mol of oxygen; thus, it requires \( 0.1056 \times 2 = 0.1056 \times 16/2 = 0.8448 \, \text{g} \) of hydrogen.In the second reaction, the equivalent of \( 0.1056 \, \text{mol} \) of oxygen requires \( 0.2112 \, \text{g} \) of hydrogen.Total hydrogen needed:\[ 0.8448 \, \text{g} + 0.2112 \, \text{g} = 1.056 \, \text{g} \]

Step 10: Solution Summary

The empirical formulas are \( V_2O \) and \( VO_2 \); balanced reactions are \( V_2O + H_2 \rightarrow VO_2 + H_2O \) and \( VO_2 + H_2 \rightarrow V + H_2O \). The total hydrogen needed is \( 1.056 \, \text{g} \).

Key Concepts

Vanadium Oxides

Vanadium oxides are compounds composed of vanadium and oxygen, showcasing various oxidation states of vanadium, leading to different forms. Common examples include VO, V\(_2\)O\(_3\), and VO\(_2\). The exercise explores two vanadium oxides, catalyzed under different conditions: the first has an empirical formula of V\(_2\)O and the second VO\(_2\). Each oxide form represents unique properties and is part of industrial and chemical processes due to their stability and reactivity.
Both of these oxides differ in the ratio of vanadium to oxygen, indicating different chemical properties and applications. Precisely determining the empirical formulas helps in understanding and employing these substances appropriately in practical scenarios, vital for chemistry students and researchers alike. Using mass and conversion to moles, empirical formulas can be found by establishing the simplest whole number ratio between vanadium and oxygen atoms in the compounds formed.

Balanced Chemical Equations

Balanced chemical equations are essential in chemistry to reflect the conservation of mass. They show the stoichiometry of a reaction, ensuring that atoms of each element are equal on both sides of the equation. In the given problem, two reactions occur involving vanadium oxides, which need balanced equations to represent correctly.
In the first reaction, vanadium oxide (V\(_2\)O) reacts with hydrogen to form VO\(_2\) and water:

• \(V_2O + H_2 \rightarrow VO_2 + H_2O\)

In the second, VO\(_2\) further reacts with hydrogen to yield pure vanadium and water:

• \(VO_2 + H_2 \rightarrow V + H_2O\)

These equations highlight how vanadium oxides undergo transformations via oxidation-reduction reactions, showcasing vanadium's change in oxidation states. Balancing these equations is crucial for correctly predicting and measuring reactants and products.

Stoichiometry

Stoichiometry is the calculation of reactants and products in chemical reactions and is a central concept in chemistry. It ensures the precise measurement of the substances involved in chemical transformations. For students tackling the vanadium oxide exercise, stoichiometry connects macroscopic and microscopic perspectives.
By knowing the molar masses of the substances involved, you can compute the number of moles and use that to find out how much of each substance is required or produced in a reaction. In the exercise, the stoichiometric relationships helped determine the amount of hydrogen gas required to drive the reactions fully and predict the vanadium and products formed. This involved converting between grams and moles using molar masses, an essential step in solving real-world chemical problems.

Mass and Moles Calculation

Mass and moles calculation are foundational tools for understanding chemical reactions and determining the amount of substances involved. To convert mass to moles, you divide the mass by the molar mass of the substance, which is derived from the periodic table.
For the vanadium oxide reactions, calculating the moles of vanadium and oxygen helped in formulating the empirical formulas of the oxides. For example, using the data:

• Vanadium's molar mass: 50.94 g/mol
• Oxygen's molar mass: 16.00 g/mol

These calculations help determine how much hydrogen is needed for the reactions to proceed, a critical aspect of balancing the original equations. Understanding these principles allows chemists to convert laboratory data into useful information about chemical compounds and reactions.