What is the empirical formula of vanadium oxide, if \(2.74 \mathrm{~g}\) of the metal oxide contains \(1.53 \mathrm{~g}\) of metal ? (a) \(\mathrm{V}_{2} \mathrm{O}_{3}\) (b) VO (c) \(\mathrm{V}_{2} \mathrm{O}_{5}\) (d) \(\mathrm{V}_{2} \mathrm{O}_{7}\)

Stoichiometry is a branch of chemistry that deals with the quantitative relationships between reactants and products in a chemical reaction. It's the calculation of the relative quantities of substances consumed and produced in chemical reactions. Stoichiometry is based on conservation of mass where the total mass of the reactants equals the total mass of the products.

In the case of calculating the empirical formula of vanadium oxide, stoichiometry involves understanding the mass ratio of vanadium to oxygen in the compound, and then using that information to deduce the simplest ratio of atoms within a molecule. The process requires balancing the mass and the number of moles of each element, which leads us to the next essential concepts of molar mass and chemical composition.

Molar mass is defined as the mass in grams of one mole of any substance. The molar mass is numerically equal to the atomic or molecular weight of the substance in atomic mass units (amu). For instance, the atomic molar mass of vanadium (V) is approximately 50.94 g/mol and that of oxygen (O) is 16.00 g/mol.

When calculating an empirical formula, molar mass allows us to convert the mass of each element from grams to moles, which is a necessary step since chemical formulas are expressed in terms of the number of atoms (moles), not mass.

Chemical composition refers to the identity and ratio of the elements that make up a compound. It can be expressed qualitatively by the empirical or molecular formula and quantitatively by percent composition. An empirical formula represents the simplest whole number ratio of the types of atoms in a substance. For example, the empirical formula for water is always H2O, indicating a fixed ratio of 2 hydrogen atoms to every oxygen atom regardless of the sample size.

The chemical composition of the vanadium oxide in our exercise dictates that the masses of vanadium and oxygen must be converted into moles. By finding the molar ratio, you obtain the empirical formula that best represents the composition of the potentially complex vanadium oxide compound.

The concept of moles to grams conversion is pivotal in stoichiometry for translating the mass of substances to a common unit that can be used to understand the stoichiometry of chemical compounds. The conversion between moles and grams involves using the molar mass as the conversion factor.

To convert moles to grams, you multiply the number of moles by the molar mass of the substance. Conversely, to convert grams to moles, divide the mass by the molar mass. This conversion is exemplified in the exercise, where the masses of vanadium and oxygen are divided by their respective molar masses to find the number of moles, allowing for the determination of the empirical formula for vanadium oxide.