Heating \(2.40 \mathrm{~g}\) of the oxide of metal \(\mathrm{X}\) (molar mass of \(X=55.9 \mathrm{~g} / \mathrm{mol}\) ) in carbon monoxide \((\mathrm{CO})\) 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.

Understanding molar mass is crucial to mastering stoichiometry, which is the calculation of quantities in chemical reactions. The molar mass of a substance is the mass of one mole of that substance. It serves as a bridge between the microscopic world of atoms and molecules and the macroscopic world we observe.

For instance, in the given exercise, the molar mass of metal X is specified as 55.9 g/mol. This means each mole of metal X weighs 55.9 grams. By dividing the mass of the element by its molar mass, we find how many moles we have, which is the first step in many stoichiometry calculations. This molar mass led us to calculate the moles of metal X present in the sample, further allowing us to delve into the stoichiometric conversions involved in determining the empirical formula.

Balancing chemical reactions is a fundamental skill in chemistry that ensures the law of conservation of mass is respected. In other words, the number of atoms of each element must be the same on both sides of the reaction equation. To balance a reaction, coefficients are added in front of the chemical formulas to adjust the number of atoms or molecules.

In the solution provided, after determining the empirical formula of the oxide of metal X, the reaction is then balanced. The final step illustrates that for every one mole of the compound \( \mathrm{X}_{2}\mathrm{O}_{3} \), three moles of carbon monoxide \( \mathrm{CO} \) react to produce two moles of the metal X and three moles of carbon dioxide \( \mathrm{CO}_{2} \). This balanced equation provides a macroscopic view of the proportions in which reactants combine to form products.

The mole-to-mole ratio is derived from a balanced chemical equation and indicates the proportion of reactants that react and the products formed. It is a critical aspect of stoichiometry as it allows for the conversion between moles of different substances in a chemical reaction.

In the given exercise, once the moles of metal X and oxygen are determined, we used the mole-to-mole ratio to find the composition of the metal oxide. By simplifying the ratio of the moles of X to O (1.5:1), we identified the empirical formula, revealing that two moles of metal X react with three moles of oxygen atoms to form the oxide \( \mathrm{X}_{2}\mathrm{O}_{3} \). Therefore, we can infer the stoichiometric relationship in the compound and the reacting gases for the balanced equation.

The empirical formula represents the simplest whole-number ratio of the elements in a compound. Calculating the empirical formula involves determining the molar ratio of the elements present, using the mass and molar mass to find the number of moles.

In the context of the exercise, once we have the number of moles of metal X and oxygen, the empirical formula of the oxide is deduced by finding the simplest whole-number ratio of moles of metal X to moles of oxygen. The calculated moles lead to a mole ratio of 1.5:1 (or simplified to 3:2), indicating the empirical formula \( \mathrm{X}_{2}\mathrm{O}_{3} \). This process provides the foundation for stoichiometric calculations in chemistry, allowing the prediction of quantities required and produced in a given reaction.