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. a. Determine the empirical formulas for the two vanadium oxides. b. Write balanced equations for the steps of the reaction. c. Determine the mass of hydrogen needed to complete the steps of this reaction.

Step-by-Step Solution

Answer

The empirical formulas for the two vanadium oxides are V2O and VO2. The balanced equations for the reaction steps are: (a) V2O + H2 → VO2 + H2O, and (b) VO2 + H2 → V + H2O. The mass of hydrogen needed to complete the reaction steps is 0.158 g.

Step 1: Calculate the mass of vanadium and oxygen in the first oxide

We are given that 9.59 g of the first vanadium oxide undergoes a reaction. After the reaction, a new vanadium oxide is formed with a mass of 8.76 g. Therefore, the mass of oxygen removed from the first vanadium oxide is: Mass of oxygen = Mass of first vanadium oxide - Mass of second vanadium oxide Mass of oxygen = 9.59 g - 8.76 g Mass of oxygen = 0.83 g The mass of vanadium in the first oxide is equal to the mass of vanadium in the second oxide, which we will calculate now. ## Step 2: Calculation of moles of Vanadium and Oxygen ##

Step 2: Calculate moles of vanadium and oxygen

We are given the mass of vanadium (5.38 g). Therefore, we can calculate the moles of vanadium using the molar mass of vanadium, which is given by: Moles of vanadium = Mass of vanadium / Molar mass of vanadium Moles of vanadium = 5.38 g / 50.94 g/mol Moles of vanadium ≈ 0.106 mol Similarly, we can calculate the moles of oxygen in the first oxide using the molar mass of oxygen (16 g/mol): Moles of oxygen = Mass of oxygen / Molar mass of oxygen Moles of oxygen = 0.83 g / 16 g/mol Moles of oxygen ≈ 0.052 mol ## Step 3: Empirical formula of the first vanadium oxide ##

Step 3: Determine the empirical formula of the first vanadium oxide

To determine the empirical formula, we will divide the moles of vanadium and oxygen by the lowest number of moles: Lowest moles = 0.052 mol Divide the moles of vanadium and oxygen by the lowest moles: Ratio of vanadium to oxygen = 0.106 mol / 0.052 mol ≈ 2 Ratio of oxygen to vanadium = 0.052 mol / 0.052 mol ≈ 1 Therefore, the empirical formula of the first vanadium oxide is V2O1 or simply V2O. ## Step 4: Mass of Vanadium and Oxygen in the second vanadium oxide ##

Step 4: Calculate the mass of vanadium and oxygen in the second oxide

We already know the mass of vanadium in the second oxide (5.38 g), and the mass of the second vanadium oxide is given as 8.76 g. We can simply calculate the mass of oxygen in the second vanadium oxide as follows: Mass of oxygen = Mass of second vanadium oxide - Mass of vanadium Mass of oxygen = 8.76 g - 5.38 g Mass of oxygen = 3.38 g ## Step 5: Calculation of moles of Vanadium and Oxygen ##

Step 5: Calculate moles of vanadium and oxygen

Since we have already calculated the moles of vanadium (0.106 mol), we only need to calculate the moles of oxygen in the second oxide: Moles of oxygen = Mass of oxygen / Molar mass of oxygen Moles of oxygen = 3.38 g / 16 g/mol Moles of oxygen ≈ 0.211 mol ## Step 6: Empirical formula of the second vanadium oxide ##

Step 6: Determine the empirical formula of the second vanadium oxide

As we did for the first oxide, divide the moles of vanadium and oxygen in the second oxide by the lowest number of moles: Lowest moles = 0.106 mol Divide the moles of vanadium and oxygen by the lowest moles: Ratio of vanadium to oxygen = 0.106 mol / 0.106 mol ≈ 1 Ratio of oxygen to vanadium = 0.211 mol / 0.106 mol ≈ 2 Therefore, the empirical formula of the second vanadium oxide is V1O2 or simply VO2. ## Step 7: Balanced equations for the reactions ##

Step 7: Write balanced equations for the steps of the reaction

We can now write balanced equations for the reaction steps: a. V2O + H2 → VO2 + H2O b. VO2 + H2 → V + H2O ## Step 8: Mass of hydrogen needed to complete the reaction steps ##

Step 8: Determine the mass of hydrogen needed

To determine the mass of hydrogen needed to complete the reactions, we first need the moles of hydrogen consumed in both reactions. Moles of hydrogen in reaction a = Moles of oxygen in reaction a = 0.052 mol Moles of hydrogen in reaction b = Moles of vanadium in reaction b = 0.106 mol Total moles of hydrogen = 0.052 mol + 0.106 mol = 0.158 mol Now, we can calculate the mass of hydrogen using its molar mass (1 g/mol): Mass of hydrogen = Moles of hydrogen * Molar mass of hydrogen Mass of hydrogen = 0.158 mol * 1 g/mol Mass of hydrogen = 0.158 g Therefore, the mass of hydrogen needed to complete the reaction steps is 0.158 g.

Key Concepts

Vanadium Oxides

Vanadium oxides are compounds made of vanadium and oxygen. These oxides can form in different chemical states, making them quite interesting in chemistry. Such versatility means that vanadium can combine with oxygen in varying proportions to create different oxides.
For example, in this exercise, we have two types of vanadium oxides involved in a reaction sequence. The first oxide undergoes a transformation upon heating with hydrogen, leading to the formation of a second oxide and water.

• V2O: The first vanadium oxide has fewer oxygen atoms.
• VO2: In the second vanadium oxide, the ratio of oxygen to vanadium increases.

By studying these oxides, chemists can understand the changes that occur at a molecular level, such as the rearrangement of atoms and alteration in chemical formulas.

Moles Calculation

Moles calculation is crucial in understanding chemical reactions since it quantifies the amount of substance in a given sample. For vanadium and oxygen, moles are calculated using the atomic or molecular mass of the elements involved.
The mole, a fundamental concept in chemistry, allows scientists to "count" atoms by weighing them. This process makes it easier to work with elements and compounds even though they are too small to see.
In this exercise:

• We find the moles of vanadium by using its mass (5.38 g) and molar mass (50.94 g/mol). This gives us 0.106 moles.
• Similarly, the moles of oxygen in different stages are determined using its mass and molar mass (16 g/mol).
  For example, in the first oxide, we had 0.83 g of oxygen, which converted to 0.052 moles.

These calculations allow us to express the chemical reactions mathematically.

Chemical Reactions Balancing

Balancing chemical reactions is an exercise in showing both the reactants and products in a reaction are accounted for, following the law of conservation of mass.
This means the atoms of each element must be equal on both sides of a reaction equation, signifying that no atoms are lost or gained during the reaction.

• In the first reaction: Vanadium oxide (V2O) reacts with hydrogen (H2) to form a new vanadium oxide (VO2) and water (H2O).
• In the second reaction: The new oxide (VO2) further reacts with hydrogen to produce pure vanadium metal and water.

Understanding how to balance these reactions is critical to predicting how much of each product forms and ensuring accuracy in chemical experiments.

Mass and Molar Mass Calculation

Mass and molar mass calculations help convert between the mass of a substance and how many molecules or atoms it contains. This concept is vital in chemistry to understand the relationship between mass and reactant/product quantities in a reaction.
Here’s how it’s applied:

• Mass is the actual amount of substance you have, measured in grams.
• Molar mass, expressed in g/mol, is the mass of one mole of a given substance.
  For instance, hydrogen has a molar mass of approximately 1 g/mol, while oxygen's is 16 g/mol.

In our exercise, we calculated:

• The mass of the first vanadium oxide by subtracting the mass of oxygen removed from the total mass of the initial oxide.
• The molar masses to find moles of substances involved, assisting in determining empirical formulas for vanadium oxides.

This understanding helps us appreciate the underlying principles of reactions and predict outcomes based on reactant quantities.