Phosphoric acid, one of the acids used in some cola drinks, is produced by the reaction of phosphorus(V) oxide, an acidic oxide, with water. Phosphorus(V) oxide is prepared by the combustion of phosphorus. (a) Write the empirical formula of phosphorus(V) oxide. (b) What is the molecular formula of phosphorus(V) oxide if the molar mass is about 280 . (c) Write balanced equations for the production of phosphorus( \(\mathrm{V}\) ) oxide and phosphoric acid. (d) Determine the mass of phosphorus required to make \(1.00 \times 10^{4} \mathrm{kg}\) of phosphoric acid, assuming a yield of 98.85\%.

Understanding the empirical formula is a foundational concept in chemistry, often leading students to insights into a compound’s composition. An empirical formula represents the simplest whole-number ratio of the elements in a compound. When we talk about phosphorus(V) oxide as mentioned in the exercise, we find that it has an empirical formula of P₂O₅. This indicates that for every two atoms of phosphorus, there are five atoms of oxygen in the simplest form of the compound. Though the empirical formula conveys the ratio, it does not necessarily reflect the actual number of atoms in the molecule, which is where the molecular formula comes into play.

To deduce the empirical formula from a given chemical, like in the exercise, you need to know the valency of the elements involved. Here, phosphorus has a valency of +5 and oxygen -2. By balancing the positive and negative charges, we can derive the empirical formula — a crucial step for students to master before moving on to more complex chemical equations.

In contrast to the empirical formula, the molecular formula gives us the exact number of each type of atom in a molecule. In the case of phosphorus(V) oxide, its empirical formula is P₂O₅, but its molecular formula is P₄O₁₀, which tells us that in a single molecule of the compound, there are actually four phosphorus atoms and ten oxygen atoms. The process to ascertain the molecular formula involves determining the empirical formula’s molar mass and comparing it with the compound’s actual molar mass - which in this exercise is given as approximately 280 g/mol.

It’s essential for students to grasp that while the empirical and molecular formulas can be the same, they often differ for compounds with higher molar masses, such as phosphorus(V) oxide. Taking these steps ensures accuracy when determining chemical compositions and later, when calculating to complete chemical reactions.

When representing chemical reactions, the balanced chemical equations play a critical role. These equations ensure the law of conservation of mass is followed, meaning the number of atoms for each element should be the same on both the reactants and the products side. The production of phosphorus(V) oxide from phosphorus and the subsequent formation of phosphoric acid from phosphorus(V) oxide are both illustrated with balanced equations in the original problem.

For instance, the formation of phosphorus(V) oxide is depicted by the reaction: 4P + 5O₂ -> P₄O₁₀, ensuring equal numbers of phosphorus and oxygen atoms on either side. For students, mastering the balancing of equations is vital as it lays the groundwork for understanding stoichiometry in chemical reactions.

The concept of stoichiometry describes the quantitative relationship between reactants and products in a chemical reaction. It is the calculation core of chemistry that allows students to predict the amount of substances consumed and produced. In our exercise scenario, the stoichiometry of the reaction is used to ascertain how much phosphorus is needed to make a certain quantity of phosphoric acid.

Understanding stoichiometry involves calculating the molar mass of the compounds involved, using balanced chemical equations to find the mole-to-mole ratios, and applying these ratios to find the quantities of reactants needed for a desired product. For real-world applications and experimental chemistry, stoichiometric calculations must also account for the reaction yield, giving a more realistic view of the actual product that can be obtained.

The chemical reaction yield is an essential concept for understanding the efficiency of chemical processes. In the provided problem, the yield is given as 98.85%, indicating that not all reactants are converted into the desired product in practical scenarios, a crucial piece of information for both laboratory and industrial chemistry practices. The yield is calculated by comparing the actual amount of product obtained to the theoretical amount that could be produced according to the stoichiometric calculations.

When students calculate the mass of phosphorus required to produce phosphoric acid, they must not only use the stoichiometry from the balanced equations but also adjust for the reaction yield, as complete conversion is rarely achieved due to factors like side reactions, incomplete reactions, or practical losses. By incorporating the percent yield into calculations, students learn to adjust theoretical outcomes to align with real, observed results.