Question

Heating \(2.40 \mathrm{~g}\) of the oxide of metal \(\mathrm{X}\) (molar mass of \(\mathrm{X}=55.9 \mathrm{~g} / \mathrm{mol}\) ) in carbon monoxide (CO) yields the pure metal and carbon dioxide. The mass of the metal product is \(1.68 \mathrm{~g}\). From the data given, show that the simplest formula of the oxide is \(\mathrm{X}_{2} \mathrm{O}_{3}\) and write a balanced equation for the reaction.

Short answer

The simplest formula of the oxide is \( \mathrm{X}_2\mathrm{O}_3 \). The balanced equation is \( \mathrm{X}_2\mathrm{O}_3 + 3\mathrm{CO} \rightarrow 2\mathrm{X} + 3\mathrm{CO}_2 \).

Step-by-step solution

Determine the mass of oxygen in the oxide

First, find the mass of oxygen by subtracting the mass of the metal product from the mass of the oxide:\[ \text{Mass of oxygen} = 2.40\, \text{g} - 1.68\, \text{g} = 0.72\, \text{g} \]

Convert the masses to moles

Calculate the moles of metal \( \mathrm{X} \) and oxygen. The molar mass of \( \mathrm{X} \) is given as \( 55.9\, \text{g/mol} \) and that of oxygen \( \mathrm{O} \) is \( 16.0\, \text{g/mol} \).- Moles of \( \mathrm{X} \):\[ \text{Moles of } \mathrm{X} = \frac{1.68\, \text{g}}{55.9\, \text{g/mol}} = 0.0300\, \text{mol} \]- Moles of \( \mathrm{O} \):\[ \text{Moles of } \mathrm{O} = \frac{0.72\, \text{g}}{16.0\, \text{g/mol}} = 0.0450\, \text{mol} \]

Calculate the mole ratio of X:O

To find the simplest ratio, divide the moles of each by the smallest number of moles:- Ratio of \( \mathrm{X} \):\[ \frac{0.0300\, \text{mol}}{0.0300\, \text{mol}} = 1 \]- Ratio of \( \mathrm{O} \):\[ \frac{0.0450\, \text{mol}}{0.0300\, \text{mol}} = 1.5 \]Thus, the simplest ratio is 2:3, leading to the empirical formula \( \mathrm{X}_2\mathrm{O}_3 \).

Write the balanced chemical equation

Based on the reaction in which the oxide \( \mathrm{X}_2\mathrm{O}_3 \) reacts with CO to give \( \mathrm{X} \) and \( \mathrm{CO}_2 \), the balanced equation is:\[ \mathrm{X}_2\mathrm{O}_3 + 3\mathrm{CO} \rightarrow 2\mathrm{X} + 3\mathrm{CO}_2 \]

Key concepts

Molar Mass

Molar mass is the mass of one mole of a substance, usually expressed in grams per mole (g/mol). It represents the combined atomic masses of all atoms in the molecular or empirical formula of the compound. Knowing the molar mass is crucial to convert between the mass of a substance and the amount in moles. To calculate the molar mass of an element, you simply use its atomic mass from the periodic table. For example, the element X in our exercise has a molar mass of 55.9 g/mol, which means that one mole of element X weighs 55.9 grams. For compounds, such as water (H₂O), the molar mass is calculated by adding up the atomic masses of hydrogen and oxygen: - Hydrogen: 1.0 g/mol - Oxygen: 16.0 g/mol Adding these gives water a molar mass of 18.0 g/mol. Having the molar mass allows us to determine how many moles of a substance we have when given the mass. In this exercise, the metal X and the oxygen are converted into moles by dividing their mass in grams by their respective molar masses.

Balanced Chemical Equation

A balanced chemical equation represents a reaction with equal numbers of each type of atom on both the reactant and the product side. This balance is necessary to satisfy the law of conservation of mass, which states that matter cannot be created or destroyed in a closed system.In our example, the reaction involves the oxide X₂O₃ and carbon monoxide (CO), yielding pure metal X and carbon dioxide (CO₂). Balancing the equation requires ensuring the same number of X and O atoms are present before and after the reaction. The balanced chemical equation given in the exercise is:\[\mathrm{X}_2\mathrm{O}_3 + 3\mathrm{CO} \rightarrow 2\mathrm{X} + 3\mathrm{CO}_2\]Here's how we achieved this: - On the reactant side, we start with 2 X atoms and 3 O atoms in X₂O₃, and 3 O atoms in CO.- On the product side, 2 X atoms are maintained in 2X, and 3 O atoms appear in 3 CO₂.Balancing chemical equations is essential for predicting the quantities of reactants and products involved in a chemical reaction.

Mole Ratio Calculation

Mole ratios are derived from the coefficients of a balanced chemical equation. They serve as a quantitative bridge between reactants and products. In a given reaction, the mole ratio dictates how many moles of one substance react with a certain number of moles of another.In this exercise, to find the simplest whole number ratio of X to O, we first calculated the moles of each element. With 0.0300 mol of X and 0.0450 mol of O, we identified the smallest number of moles, which is 0.0300 mol for element X. By dividing both by this smallest number:- For X: \[\frac{0.0300\, \text{mol}}{0.0300\, \text{mol}} = 1\]- For O: \[\frac{0.0450\, \text{mol}}{0.0300\, \text{mol}} = 1.5\]Since we prefer whole numbers for empirical formulas, we multiply each coefficient by 2 to obtain the simplest ratio, resulting in the formula X₂O₃. These calculations enable us to deduce not only the empirical formula of the compound but also confirm the balanced nature of the chemical equation.