Question

Consider an equimolar mixture (equal number of moles) of two diatomic gases \(\left(A_{2} \text { and } B_{2}\right)\) in a container fitted with a piston. The gases react to form one product (which is also a gas) with the formula \(A_{x} B_{y}\). The density of the sample after the reaction is complete (and the temperature returns to its original state) is \(1.50\) times greater than the density of the reactant mixture. a. Specify the formula of the product, and explain if more than one answer is possible based on the given data. b. Can you determine the molecular formula of the product with the information given or only the empirical formula?

Short answer

a. The possible empirical or molecular formulas for the product are \(B_3, AB_2, A_2B, A_3\). More than one answer is possible based on the given data. b. We can only determine possible empirical or molecular formulas with the given information. Further experiments or information would be needed to confirm the true molecular formula.

Step-by-step solution

Calculate the initial and final densities.

Since the reactants and products have the same temperature and are under the same pressure, the ratio of their densities is equal to the ratio of their molar masses: \(\frac{\rho_{final}}{\rho_{initial}} = \frac{M_{final}}{M_{initial}}\) Since we are given that the density of the reaction mixture after the reaction is 1.5 times the initial density, the ratio of the molar masses is also 1.5: \(\frac{M_{final}}{M_{initial}} = 1.5\)

Assess the possible formulas for the product.

Let's consider the general formula for the product: \(A_xB_y\). The molar mass of the product, M_final, will be given by: \(M_{final}=xM_A+yM_B\) where \(M_A\) and \(M_B\) are the molar masses of elements A and B, respectively. The molar masses of the initial reactants, A₂ and B₂, are given by: \(M_{initial}=\frac{1}{2}(M_A+M_B)\). Using the equation from Step 1, we obtain: \(\frac{xM_A+yM_B}{\frac{1}{2}(M_A+M_B)} = 1.5\)

Determine the possible values for x and y.

Now, solve the equation for the ratio of the molar masses for x and y: \(1.5 = \frac{xM_A+yM_B}{\frac{1}{2}(M_A+M_B)} \Rightarrow 2(1.5)=3 = \frac{xM_A+yM_B}{M_A+M_B}\) This gives us: \(3(M_A+M_B)=xM_A+yM_B\) Rearrange the above equation: \(xM_A-M_B(3-y)=0\) This equation is true if any of the following conditions are met: 1. \(x = 0\) and \(y = 3\) 2. \(x = 1\) and \(y = 2\) 3. \(x = 2\) and \(y = 1\) 4. \(x = 3\) and \(y = 0\) From the conditions, the following formulas for the product are possible: \(B_3, AB_2, A_2B, A_3\)

Step 4a: Identify the possible molecular or empirical formulas.

The results show that there are four possible product formulas (\(B_3, AB_2, A_2B, A_3\)). These represent the possible empirical or molecular formulas of the product, not necessarily the true molecular formulas. Further experiments or information would be needed to confirm which formula corresponds to the actual molecular formula of the product.

Step 4b: Determine if more than one answer is possible.

More than one answer is possible based on the given data. Without additional information, we cannot determine which formula is the true molecular formula of the product. The formulas provided are the only possibilities based on the given information. Answer: a. The possible empirical or molecular formulas for the product are \(B_3, AB_2, A_2B, A_3\). More than one answer is possible based on the given data. b. We can only determine possible empirical or molecular formulas with the given information. Further experiments or information would be needed to confirm the true molecular formula.

Key concepts

Empirical and Molecular Formulas

Understanding the distinction between empirical and molecular formulas is crucial when discussing chemical compounds. An empirical formula represents the simplest whole-number ratio of the elements in a compound, while a molecular formula shows the exact number of each type of atom in one molecule of the compound.

Empirical formulas are particularly useful because they provide an insight into the composition of the compound, irrespective of the molecule's size. However, if we want to know the exact structure of the molecule, we need the molecular formula. For example, glucose has an empirical formula of CH₂O and a molecular formula of C₆H₁₂O₆. Both convey different levels of information: the former tells us the ratios, and the latter tells us the actual molecular structure.

In the context of the exercise provided, the potential formulas for the product (B₃, AB₂, A₂B, A₃) represent empirical formulas, as they give no indication of the molecule's size or the number of atoms involved. They simply suggest a ratio between elements A and B. To determine the actual molecular formula of the product, additional information or experimental data would be needed.

Molar Mass Ratio

Molar mass ratio is a comparison of the molar masses of two compounds. It is essential for balancing chemical equations and for calculating reactants or products in a chemical reaction. Molar mass is measured in grams per mole (g/mol) and can be calculated by summing the atomic masses of all atoms in a molecule, as provided on the periodic table.

Returning to our exercise, we used a proportionality relationship between the molar mass ratio of the initial reactants and the product to deduce potential formulas for the gas product. We can state this relationship as \(\frac{M_{final}}{M_{initial}} = 1.5\), where \(M_{final}\) and \(M_{initial}\) are the molar masses of the product and reactant, respectively. It is the key to understanding the stoichiometry of the reaction, as it helps predict the product formula based on the increased density after the reaction.

The molar mass ratio provided a means to derive possible ratios between the atoms in the product molecule, guiding us to several empirical formulas that are consistent with the data. It is a fundamental concept for any stoichiometry analysis.

Chemical Reaction Density Changes

The change in density during a chemical reaction can provide significant insights into the nature of the reaction. In reactions involving gases, as in the given exercise, density can reflect changes in molar mass if we assume constant pressure and temperature conditions.

Density here serves as an indirect measure of the molecular mass of the product relative to the reactants. The data from the exercise suggests that the density of the gas after the reaction is 1.50 times that of the reactant mixture; this hints at an increase in molar mass for the product. Such a change in density indicates the formation of a heavier (or more densely packed) gas molecule from lighter reactant gases.

Thus, if the volume and temperature remained constant, the proportional increase in density would directly correlate with an increase in the molar mass of the gas. This implies the possible formation of a compound with a higher molecular weight than the reactants. Understanding these density changes allows us to predict the nature of chemical reactions and their products, reinforcing the importance of the concept in stoichiometric calculations.