Question

A sample of an oxide of vanadium weighing \(4.589 \mathrm{~g}\) was heated with hydrogen gas to form water and another oxide of vanadium weighing \(3.782 \mathrm{~g} .\) The second oxide was treated further with hydrogen until only \(2.573 \mathrm{~g}\) of vanadium metal remained. (a) What are the simplest formulas of the two oxides? (b) What is the total mass of water formed in the successive reactions?

Short answer

Answer: The empirical formulas of the two vanadium oxides are V2O5 and VO. The total mass of water produced during the reactions is 3.6314 g.

Step-by-step solution

Calculate the moles of vanadium in the final product

We need to find the moles of vanadium in the final product, which is in the form of vanadium metal weighing 2.573 g. To find the moles, we'll use the formula: Moles = mass (g) / molar mass (g/mol) The molar mass of vanadium (V) is 50.94 g/mol. Moles of vanadium = \(\frac{2.573}{50.94}\) = 0.0505 mol

Determine moles of vanadium and oxygen in each oxide

Since both oxides contain the same amount of vanadium, we have 0.0505 mol of vanadium in each oxide. Subtracting the mass of vanadium (2.573 g) from the mass of each oxide, we will find the mass of oxygen in each oxide: Mass of oxygen in first oxide = 4.589 g - 2.573 g = 2.016 g Mass of oxygen in second oxide = 3.782 g - 2.573 g = 1.209 g Now, we'll find the moles of oxygen in each oxide. The molar mass of oxygen (O) is 16.00 g/mol. Moles of oxygen in first oxide = \(\frac{2.016}{16.00}\) = 0.126 mol Moles of oxygen in second oxide = \(\frac{1.209}{16.00}\) = 0.0756 mol

Determine the empirical formulas of the two oxides

To find the empirical formula for each oxide, we'll find the simplest whole number ratio between the moles of vanadium and the moles of oxygen in each. Ratio for first oxide = \(\frac{0.0505}{0.126}\) = \(\frac{1}{2.49}\). Since we need a whole number ratio, we'll multiply both numbers by 2. Empirical formula of the first oxide = V2O5 Ratio for second oxide = \(\frac{0.0505}{0.0756}\) = \(\frac{1}{1.497}\). Since we need a whole number ratio, the ratio is approximately 1:1. Empirical formula of the second oxide = VO

Calculate the total mass of water formed in the reactions

We know the mass of oxygen that reacted with hydrogen to form water in both steps. Summing up these values, we get: Total mass of oxygen in water = 2.016 g (first reaction) + 1.209 g (second reaction) = 3.225 g To find the total mass of water, we need to add the mass of hydrogen that reacted with oxygen. We'll first find the moles of oxygen that reacted: Moles of oxygen = 0.126 mol (first reaction) + 0.0756 mol (second reaction) = 0.2016 mol Since the ratio of hydrogen to oxygen in water is 2:1, we have twice as many moles of hydrogen reacting: Moles of hydrogen = 2 x 0.2016 mol = 0.4032 mol Now we'll convert moles of hydrogen to mass. The molar mass of hydrogen (H) is 1.008 g/mol. Mass of hydrogen = 0.4032 mol x 1.008 g/mol = 0.4064 g Finally, we find the total mass of water: Total mass of water = Mass of oxygen + Mass of hydrogen = 3.225 g + 0.4064 g = 3.6314 g (a) The simplest formulas of the two vanadium oxides are V2O5 and VO. (b) The total mass of water formed in the successive reactions is 3.6314 g.

Key concepts

Stoichiometry

Stoichiometry is a branch of chemistry that deals with the quantitative relationship between reactants and products in a chemical reaction. It is based on the law of conservation of mass, which states that in a chemical reaction, matter is neither created nor destroyed. Stoichiometry utilizes balanced chemical equations to calculate the amounts of substances consumed and produced in a reaction.

For example, in the reaction between vanadium oxides and hydrogen, stoichiometry allows us to determine the ratio of vanadium to oxygen in the products, and thus, infer the empirical formula of the vanadium oxides. By knowing the moles of vanadium, and then calculating the moles of oxygen based on the mass difference after reactions with hydrogen, we establish a stoichiometric relationship that helps identify the simplest formulas for the vanadium oxides.

Mole Concept

The mole concept is a fundamental idea in chemistry that relates the mass of substances to the number of particles, atoms, molecules, or ions they contain. One mole is defined as the amount of a substance that contains the same number of particles as there are atoms in 12 grams of carbon-12. This number, known as Avogadro's number, is approximately equal to \(6.022 \times 10^{23}\).

The significance of the mole is seen in how it interconnects the macroscopic and atomic scale. For instance, in determining the empirical formulas of the vanadium oxides, we first used the mole concept to convert the mass of each oxide to moles, providing a bridge between measurable quantities and chemical formulas.

Reaction of Oxides with Hydrogen

The reaction of oxides with hydrogen is a type of redox reaction where hydrogen reduces the oxide, often producing water as a byproduct. In the cases of vanadium oxides, heating them with hydrogen gas causes a stepwise reduction that results in a change in mass, attributable to the loss of oxygen in the form of water.

The reaction progression—from vanadium oxide to another oxide of vanadium, and eventually to vanadium metal—involves decreasing oxidation states of vanadium as oxygen is removed. The total mass of water formed can be deduced by tracking the amount of oxygen that has been removed from the vanadium oxides, as was done in the step-by-step solution.

Determining Empirical Formulas

Determining the empirical formula of a compound involves finding the simplest whole number ratio of atoms of each element in the compound. To ascertain the empirical formula, one must first determine the moles of each constituent element. The resulting mole ratios can then be simplified to yield the empirical formula.

In our vanadium oxide example, after obtaining the moles of vanadium and oxygen through their respective masses, we calculated their ratios to determine the empirical formulas of the two oxides. The empirical formula is a crucial aspect of understanding a compound's composition and provides foundational information for further studies in reactivity and molecular structure.