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To solve any problem in chemistry, understanding how to calculate moles is essential. The mole is a fundamental unit in chemistry used to measure the amount of substance. One mole corresponds to Avogadro's number, which is approximately \(6.022 \times 10^{23}\) particles of that substance.

When converting mass to moles, use the formula: \text{Moles} = \frac{\text{Mass of substance (g)}}{\text{Molar mass (g/mol)}}\.

Let's consider the above problem, where 104 grams of \(\text{O}_2\) are provided. Given that the molar mass of \(\text{O}_2\) is 32 g/mol, the calculation will be:

\text{Moles of } \text{O}_2 = \frac{104 \text{ g}}{32 \text{ g/mol}} = 3.25 \text{ moles}\.

By performing this calculation, we are able to know how much \(\text{O}_2\) is present in terms of moles, making further calculations more manageable.

Stoichiometry involves understanding the quantitative relationships between reactants and products in a chemical reaction. It is often compared to a recipe that tells you how much of each ingredient you need to make a certain amount of a dish.

In our provided exercise, we need to find the empirical formula of compound \(\text{X}\). The empirical formula is the simplest whole number ratio of the elements in a compound. By identifying the moles of each element present in the products, we can derive this ratio.

Step by step, we determine:

• 2 moles of \(\text{CO}_2\) yield 2 moles of Carbon (C).
• 2.5 moles of \(\text{H}_2\text{O}\) yield 5 moles of Hydrogen (H).
• Combining the results: 2 moles of C, 5 moles of H, and according to the provided oxygen data, we need 6.5 moles of O in total from products.

Subtracting and adjusting by whole numbers allows us to form the ratio and therefore the empirical formula: \(\text{CH}_{1.25}\text{O}\), which simplifies as we consider the nearest whole numbers to \(C_4H_5O\).

Molar mass is a critical concept used in determining the molecular formula of a compound. The molar mass is the mass of one mole of a substance (usually in g/mol). Knowing the molar mass, along with the empirical formula, helps to determine the actual molecular formula of the compound.

In the exercise, the empirical formula of compound \(\text{X}\) was found to be \(\text{C}_4\text{H}_5\text{O}\). To convert this to a molecular formula, one would need additional information about the molar mass of compound \(\text{X}\).

Suppose the molar mass of compound \(\text{X}\) is known. We compare it with the empirical formula's molar mass. For instance, if the given molar mass of \(\text{X}\) is twice the empirical formula mass, then the molecular formula is double the empirical formula units, resulting in \(\text{C}_8\text{H}_{10}\text{O}_2\).

Therefore, understanding molar mass is essential for converting empirical formulas into the molecular formulas.