Problem 39 - Advanced Problems in Physical Chemistry for Competitive Examinations

Question

If \(0.250 \mathrm{~g}\) of an element, \(\mathrm{M}\), reacts with excess fluorine to produce \(0.547 \mathrm{~g}\) of the hexafluoride, \(\mathrm{MF}_{6}\), the element should be \((\mathrm{Cr}=52, \mathrm{Mo}=95.94, \mathrm{~S}=32, \mathrm{Te}=127.6\) \(\mathrm{F}=19\) ) (a) \(\mathrm{Cr}\) (b) Mo (c) \(S\) (d) Te

Step-by-Step Solution

Answer

The element is Mo (Molybdenum), as its molar mass (95.94 g/mol) is closest to the calculated molar mass of 95.97 g/mol.

Step 1: Determine the mass of fluorine in the compound

Subtract the mass of the element M from the mass of the compound to find the mass of fluorine in MF6. Mass of fluorine = Mass of the compound - Mass of the element = 0.547 g - 0.250 g = 0.297 g.

Step 2: Calculate the number of moles of fluorine

Use the molar mass of fluorine (19 g/mol) to calculate the moles of fluorine in the compound. Number of moles = Mass of fluorine / Molar mass of fluorine = 0.297 g / 19 g/mol = 0.01563 moles.

Step 3: Calculate the moles of the element M in the compound

Since there are 6 fluorine atoms for every atom of M in MF6, divide the number of moles of fluorine by 6 to find the moles of M. Moles of M = Moles of fluorine / 6 = 0.01563 moles / 6 = 0.002605 moles.

Step 4: Calculate the molar mass of the element M

Use the mass of the element M and the moles of M to calculate the molar mass. Molar mass of M = Mass of M / Moles of M = 0.250 g / 0.002605 moles = 95.97 g/mol.

Step 5: Identify the element M

Compare the calculated molar mass of M with the given molar masses of the potential elements to identify M. The molar mass of Mo is closest to the calculated molar mass (95.97 g/mol), so the element should be Mo.

Key Concepts

Stoichiometry

Stoichiometry is like a mathematical playground within chemistry where we can calculate relationships between the reactants and products in a chemical reaction. It allows us to precisely determine how much of each substance is involved in a reaction, and with this understanding, we can predict the outcomes of chemical processes.

Take for example the reaction of an element with fluorine to produce a hexafluoride; stoichiometry guides us to balance the equation and figure out the exact amounts of reactants used and products formed. It deals with concepts such as moles, mass, and molar ratios, which bridge the gap between the microscopic world of atoms and the macroscopic world we can measure and observe.

Molar Mass Calculation

The molar mass of a substance is the weight of one mole of that substance, typically expressed in grams per mole (g/mol). It's critical for quantifying substances in chemistry and is found by summing the atomic masses of all atoms in a molecule.

Consider a substance ‘M’ that forms a compound by reacting with fluorine. To find the molar mass of ‘M’, we first need to identify how much of ‘M’ is in the compound. This involves observing mass changes before and after the reaction, and the concept of molar mass allows us to convert these mass measurements into moles, providing a quantitative measure of atoms present. This conversion is pivotal in identifying unknown substances or verifying purity and composition in chemical syntheses.

Chemical Reaction Quantification

Quantifying chemical reactions is a fundamental part of chemistry, allowing scientists and engineers to understand exactly how much product is created from given amounts of reactants. This process involves calculations to determine the number of moles, masses, volumes, and concentrations of each substance in a reaction.

For instance, when an unknown element 'M' reacts to form its hexafluoride, MF6, understanding the stoichiometry of the reaction enables us to calculate not only the mass of F involved but also the stoichiometric coefficients which tell us the ratio in which reactants combine. This ratio is critical in practical applications ranging from cooking recipes to manufacturing pharmaceuticals, ensuring that reactions proceed as desired without excess waste or unreacted materials.