\(P\) and \(Q\) are two elements which form \(P_{2} Q_{3}, P Q_{2}\) molecules. If \(0.15\) mole of \(P_{2} Q_{3}\) and \(P Q_{2}\) weighs \(15.9 \mathrm{~g}\) and \(9.3 \mathrm{~g}\), respectively, what are atomic weighs of \(P\) and \(Q ?\)

Understanding the basics of stoichiometry is crucial for delving into the realm of chemical reactions and composition. It involves calculations that revolve around the quantitative relationships of reactants and products in a chemical reaction. In the case of the textbook exercise, stoichiometric calculations help us determine the atomic weights of elements in a compound by using the given masses and the molar amounts of the compounds.

Applying stoichiometry to our problem, we first need to interpret the chemical formula of the compounds (like \(P_{2}Q_{3}\) and \(PQ_{2}\)) to represent elemental composition and quantity. This allows us to set up a mathematical representation of the chemical equation that can be translated into a system of linear equations, reflecting the conservation of mass. Through these steps, we conclude that stoichiometry is not only a tool for balancing chemical equations but also a fundamental method in determining unknown chemical quantities. It's like a detective's work, piecing together clues (mass and moles) to uncover the identity (atomic weights) of the elements involved.

The determination of atomic weight is a central task in chemistry, providing us with an understanding of the relative weight of different atoms. It is an essential parameter for the calculations in stoichiometry. Determining the atomic weights, as seen in the exercise, requires a careful analysis of the substance's molecular formula and integrating that with known quantities like mass and the mole concept.

To find the atomic weights of \(P\) and \(Q\) from the compounds \(P_{2}Q_{3}\) and \(PQ_{2}\), we use the given information to set up equations that account for the weight contribution of each atom in the molecule. As evidenced in the solution steps, we assume variables for the unknown atomic weights and use the mole to mass relationship to form equations that, once solved, reveal these atomic weights. This approach underscores the interconnectivity of atomic weight to stoichiometric calculations, allowing us to express chemical relationships in a precise and quantitative manner.

Delving into the mole concept reveals an understanding of the counting unit used by chemists to express amounts of a chemical substance, analogous to the term 'dozen' in everyday life. One mole represents \(6.022 \times 10^{23}\) entities, be it atoms, ions, or other particles, and is tied to the quantity known as Avogadro's number. In our exercise, the mole plays a critical role as it bridges the gap between the microscopic world of atoms and the macroscopic world we can measure.

The use of the mole concept allows us to convert between mass and number of particles through the atomic weight. For instance, given the mass and the substance's molar mass, we can calculate the number of moles present. This calculation is vital in setting up appropriate stoichiometric equations to solve for unknown variables, such as atomic weights. By mastering the mole concept, students can better understand chemical quantities and the calculations that stem from them, deepening their comprehension of the very foundation of chemical reactions and stoichiometry.

The system of linear equations is a powerful mathematical tool that is employed in various fields, including chemistry, to find solutions to problems with multiple variables. When addressing our chemical conundrum, the application of linear equations allows us to handle multiple variables – in this case, the atomic weights of \(P\) and \(Q\).

These equations are constructed from the relationships dictated by the molecular formulas and stoichiometry of the compounds. By methodically organizing the information from the problem, we can set up equations that represent the mass of the molecules in terms of the atomic weights we are looking to find. Solving this system through methods like substitution or elimination gives us the exact values of the variables, providing clarity and accuracy in the otherwise abstract concept of atomic weights. This process illustrates how linear equations are at the heart of finding quantitative solutions in chemistry and enhance the student's problem-solving skill set.