Question

You have two distinct gaseous compounds made from element X and element Y. The mass percents are as follows: Compound I: \(30.43 \% \mathrm{X}, 69.57 \% \mathrm{Y}\) Compound \(\mathrm{II} : 63.64 \% \mathrm{X}, 36.36 \% \mathrm{Y}\) In their natural standard states, element X and element Y exist as gases. (Monatomic? Diatomic? Triatomic? That is for you to determine.) When you react “gas X” with “gas Y” to make the products, you get the following data (all at the same pressure and temperature): 1 volume "gas \(\mathrm{X}^{\prime \prime}+2\) volumes "gas \(\mathrm{Y}^{\prime \prime} \longrightarrow\) 2 volumes compound I 2 volumes \(^{4}\) gas \(\mathrm{X}^{\prime \prime}+1\) volume "gas \(\mathrm{Y}^{\prime \prime} \longrightarrow\) 2 volumes compound II Assume the simplest possible formulas for reactants and products in the chemical equations above. Then, determine the relative atomic masses of element X and element Y.

Short answer

The relative atomic masses of elements X and Y are \(M_\text{x} \approx 46\) and \(M_\text{y} \approx 80\), respectively.

Step-by-step solution

Study the compound formation by analyzing the data

Look at the data given: For Compound I: 1 volume gas X'' + 2 volumes gas Y'' → 2 volumes compound I For Compound II: 2 volumes gas X'' + 1 volume gas Y'' → 2 volumes compound II Assume that X'' and Y'' represent moles of X and Y, respectively, in the simplest possible formulas for reactants.

Set up a proportion based on mass percent

Let's use the mass percent information to set up a proportion. For Compound I, we have 30.43% X and 69.57% Y. Let the mass of one mole of element X be Mx and the mass of one mole of element Y be My. We can then set up a proportion: \( \frac{M_\text{x}}{M_\text{x} + 2M_\text{y}} = \frac{30.43}{100} \) For Compound II, we have 63.64% X and 36.36% Y. We can set up another proportion: \( \frac{2M_\text{x}}{2M_\text{x} + M_\text{y}} = \frac{63.64}{100} \)

Solve the equations to determine atomic masses of X and Y

We have a system of two equations for Mx and My: \( \begin{cases}\frac{M_\text{x}}{M_\text{x} + 2M_\text{y}} = \frac{30.43}{100}\\ \\ \frac{2M_\text{x}}{2M_\text{x} + M_\text{y}} = \frac{63.64}{100}\end{cases} \) Now, multiply the first equation by \(2M_\text{x} + 2M_\text{y}\) and the second equation by \((M_\text{x} + 2M_\text{y})(2M_\text{x} + M_\text{y})\) to remove the denominators: \( \begin{cases}100M_\text{x}(2M_\text{x}+2M_\text{y})=30.43M_\text{x}(M_\text{x}+2M_\text{y})\\ \\ 100M_\text{x}(M_\text{x}+2M_\text{y})(2M_\text{x}+M_\text{y})=63.64(M_\text{x}+2M_\text{y})(2M_\text{x}+M_\text{y})\end{cases} \) Solve this system of two equations with two variables (Mx and My) to determine the relative atomic masses of elements X and Y.

Key concepts

Mass Percent Composition

Understanding the concept of mass percent composition is crucial when analyzing the components of a compound. Mass percent composition tells us how much of each element is present in a compound by mass. It is calculated as the mass of a single element divided by the total mass of the compound, multiplied by 100 to convert it into a percentage.

• This helps chemists determine the proportion of elements in compounds, allowing them to analyze and predict reactions.
• For example, in Compound I of this exercise, we have 30.43% of element X and 69.57% of element Y.

This percentage representation is vital in understanding the exact formulation and breakdown of chemical compounds. By using the mass percent composition, we can set up equations to determine the unknowns like the relative atomic masses of the participating elements.

Chemical Reactions

Chemical reactions involve the transformation of reactant substances into product substances. In this scenario, understanding the stoichiometry of reactions is key, particularly for gaseous reactions. Stoichiometry refers to the relative quantities of reactants and products in chemical reactions. Such knowledge helps to predict the volumes or masses of reactants needed to produce a desired amount of product.

• The exercise above involves fixed volume ratios of gaseous reactants (gas X'' and gas Y'') converting to gaseous products, Compound I and Compound II.
• Since gases are involved, we can use their volume ratios directly as molar ratios, thanks to Avogadro's law, which states that equal volumes of gases, at the same temperature and pressure, have the same number of molecules.

Having the correct stoichiometric coefficients ensures that the chemical equations reflect the true changes and conservations in the amounts of each element between the reactants and the products.

Atomic Mass Determination

Determining atomic masses from the given compounds requires careful analysis of mass percent compositions alongside reaction stoichiometry. By setting up equations based on the percentage mass data for both compounds, you can solve for the unknown atomic masses of the elements.

• For Compound I, the equation is formulated based on the mass proportions of element X versus their total mass alongside element Y.
• In Compound II, we follow a similar approach but with a different proportion as indicated by the given percentages.

Once these equations are set up, solving them simultaneously allows you to find consistent values for the atomic masses of X and Y that satisfy both compounds simultaneously. This showcases the interplay of stoichiometry, mass percent composition, and algebra in deducing unknown atomic masses.

Gaseous Compounds

In the realm of chemistry, gaseous compounds behave according to specific principles that greatly simplify problems like the one given here. When dealing with gaseous reactions, key concepts come into play, such as Avogadro's law, the ideal gas law, and the assumption that gases behave ideally under standard conditions.

• For this exercise, "gas X''" and "gas Y''" refer to the reactants in their simplest gaseous forms.
• Analyzing compounds by their given ratios in volumes helps deduce the formulas of the gaseous reactants and products in the simplest form possible, indicating how the molar relationships translate into volume ratios.

Understanding these principles helps chemists predict outcomes in chemical reactions, particularly in reactions involving gases, and allows for further exploration into the properties and behaviors of gaseous compounds.