The combustion of \(20.63 \mathrm{mg}\) of compound \(\mathrm{Y},\) which contains only \(\mathrm{C}, \mathrm{H},\) and \(\mathrm{O},\) with excess ox ygen gave \(57.94 \mathrm{mg}\) of \(\mathrm{CO}_{2}\) and \(11.85 \mathrm{mg}\) of \(\mathrm{H}_{2} \mathrm{O}\). (a) Calculate how many milligrams of \(\mathrm{C}, \mathrm{H},\) and \(\mathrm{O}\) were present in the original sample of \(Y\). (b) Derive the empirical formula of Y. (c) Suggest a plausible structure for Y if the empirical formula is the same as the molecular formula.

Combustion analysis is a type of chemical analysis used to determine the elemental composition of a compound, particularly organic compounds composed of elements such as carbon (C), hydrogen (H), and oxygen (O). In this process, a known mass of a compound is burned in excess oxygen to produce carbon dioxide (CO2) and water (H2O). The masses of these combustion products are then converted to moles using their known molar masses.

For instance, in our exercise, the combustion of 20.63 mg of compound Y resulted in 57.94 mg of CO2 and 11.85 mg of H2O. By determining the moles of CO2 and H2O produced, we can back-calculate to find the moles of the individual elements present in the original compound. This is achieved using stoichiometry - the molar relationship between the reactant and products in a chemical reaction.

Stoichiometry is the section of chemistry that deals with the quantities of substances involved in chemical reactions. It is based on the conservation of mass where the quantity of each element must be the same at the start and end of a chemical reaction. This allows us to relate the masses of different compounds in a reaction.

In our exercise, once the moles of CO2 (which represent carbon atoms) and H2O (which represent hydrogen atoms) are known, stoichiometry is used to calculate the moles, and thus the mass, of carbon and hydrogen in compound Y. Because the original compound also contains oxygen, we can find the mass of oxygen by subtracting the mass of carbon and hydrogen from the total mass of compound Y. Stoichiometry is crucial for balancing chemical reactions and determining the quantity of reactants or products.

The molar mass is the mass of one mole of a substance, usually expressed in grams per mole (g/mol). It is used to convert between moles and grams, enabling chemists to count molecules by weighing them.

During the combustion analysis in our exercise, we use the molar masses of CO2 and H2O to convert the mass of the combustion products into moles. For example, the molar mass of CO2 (approximately 44.01 g/mol) and H2O (approximately 18.02 g/mol) are used to determine the number of moles from the measured masses of CO2 and H2O. Similarly, to find the mass of carbon and hydrogen in the original compound, we use their respective molar masses, 12.01 g/mol for carbon and 1.01 g/mol for hydrogen.

The molecular structure defines the three-dimensional arrangement of atoms within a molecule. It not only includes the molecular formula but also shows the connectivity and spatial orientation of the atoms. The empirical formula, derived from the stoichiometric calculation, gives the simplest whole-number ratio of atoms in the compound, but it doesn't provide the specific structure.

In our exercise, once we determine the empirical formula of compound Y as C6H6O, we can propose potential molecular structures. Because the empirical formula could be the same as the molecular formula, a plausible structure could be that of phenol, which possesses a six-carbon aromatic ring (benzene) with one hydrogen atom replaced by a hydroxyl (-OH) group. It is important for students to understand that multiple molecular structures can have the same empirical formula, which is why information about the molecular formula and structure is essential.