Question

You have two distinct gaseous compounds made from element \(\mathrm{X}\) and element \(\mathrm{Y}\). The mass percents are as follows: Compound I: \(30.43 \% \mathrm{X}, 69.57 \% \mathrm{Y}\) Compound II: \(63.64 \% \mathrm{X}, 36.36 \% \mathrm{Y}\) In their natural standard states, element \(\mathrm{X}\) and element \(\mathrm{Y}\) 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 "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 \(\mathrm{X}\) and element \(\mathrm{Y}\).

Short answer

The relative atomic masses of elements X and Y are 64 and 32, respectively.

Step-by-step solution

Analyzing data of the gaseous compounds

From the information given, we have the following reaction equations: 1 volume of gas X + 2 volumes of gas Y -> 2 volumes of compound I 2 volumes of gas X + 1 volume of gas Y -> 2 volumes of compound II

Assume simplest formula for reactants and products

Let's assume the simplest possible formulas for the reactants to be X (monatomic) and Y (diatomic). So, in the chemical equations above, the reactants and products would be written as follows: X + 2(Y2) -> 2(XY) 2X + Y2 -> 2(X2Y) As we can see, the formulas make sense if Y is diatomic and X is monatomic.

Analyze mass percentages in each compound

Consider the mass percentages of X and Y in compound I and II. Let's assume there are n moles of X and m moles of Y in each compound. For compound I: \(0.3043 = \frac{n \times M_x}{n \times M_x + m \times M_y}\) \(0.6957 = \frac{m \times M_y}{n \times M_x + m \times M_y}\) For compound II: \(0.6364 = \frac{n \times M_x}{n \times M_x + m \times M_y}\) \(0.3636 = \frac{m \times M_y}{n \times M_x + m \times M_y}\)

Establish a relation between the masses in two compounds

From the chemical equations of the reaction, we know the ratio of masses in compound I and II is proportional to the ratio between X and Y in their reaction. For compound I: \(M_x = 1 \times M_x\), \(M_y = 2 \times M_y\) For compound II: \(M_x = 2 \times M_x\), \(M_y = 1 \times M_y\) Now we can write an equation as: \(\frac{{{M_x}_I}}{{{M_y}_I}} = \frac{{{M_x}_2}}{{{M_y}_2}}\)

Calculate relative atomic masses of elements X and Y

Combining steps 3 and 4, we can solve for M_x and M_y numerically: \(\frac{0.3043}{0.6957} = \frac{0.6364}{0.3636}\) \(M_x = 2.01 M_y\) Now, using the mass percentages in either compound I or compound II: \(0.3043 = \frac{2.01M_y}{2.01M_y + M_y}\), or: \(\frac{2.01M_y}{3.01M_y} = 0.3043\) Solve for M_y: \(M_y = 32\) Substitute M_y back into the equation for M_x: \(M_x = 2.01 \times 32\) \(M_x = 64\) Hence, the relative atomic masses of elements X and Y are 64 and 32, respectively.

Key concepts

Mass Percent Composition

Mass percent composition of a chemical compound is a crucial concept in chemistry that conveys the proportion of each element present in the compound relative to its total mass. Put simply, it tells you how much of each ingredient is in the recipe. For example, if we have Compound I with mass percentages of 30.43% for element X and 69.57% for element Y, it means that for every 100 grams of Compound I, there are about 30.43 grams of element X and 69.57 grams of element Y. This information becomes incredibly useful when trying to deduce the chemical formula of an unknown compound or when carrying out stoichiometric calculations.

Improvement can be achieved by considering that the mass percent composition is proportional to the molar mass and number of atoms of each element in a molecule. From the mass percent, we can create a ratio that reflects the atoms' relative quantities and their atomic masses in the compound.

Gas Volume Ratios

Gas volume ratios are rooted in the ideal gas law, which describes the behavior of gases in terms of pressure, volume, temperature, and number of moles. When examining chemical reactions involving gases at constant temperature and pressure, the volumes of reacting gases and their products can be directly related to their mole ratios. This is because, under the same conditions, one mole of any ideal gas occupies the same volume. For instance, in our exercise, 1 volume of gas X reacting with 2 volumes of gas Y to form 2 volumes of compound I mirrors a simple 1:2:2 molar ratio.

An insight into improvement would be that considering the gases may not be ideal, deviations from these simple ratios can occur due to the real behavior of gases. However, typically, gas volume ratios provide an excellent first approximation that helps to establish the stoichiometry of a gaseous reaction.

Relative Atomic Mass

Relative atomic mass, often simply termed as atomic weight, is a dimensionless quantity that expresses how heavy, on average, an element's atoms are, compared to one-twelfth of the mass of an atom of carbon-12. This measurement doesn't reflect a single atom's actual mass but is rather a comparison or a ratio of average atomic masses. In our exercise, using the stoichiometry and mass percent compositions, the relative atomic masses of elements X and Y are deduced, giving us important information about their identity and how they might behave in chemical reactions.

When students improve their understanding of relative atomic mass, they should consider how these values are averages derived from isotopic abundances, which means they reflect the mixture of an element's isotopes in nature, not just a single isotope's mass.

Chemical Compounds

Chemical compounds are substances composed of two or more different types of atoms bonded together in definite proportions. They have unique properties that are distinct from the elements that form them. For instance, water is a compound of hydrogen and oxygen. It's neither flammable like hydrogen nor does it support combustion like oxygen. The behavior and properties of chemical compounds can often be predicted and understood by knowing the types and ratios of the constituent elements' atoms, a process which involves considering the mass percent composition and stoichiometry of the compounds. In the context of our exercise, correctly assuming that Y is a diatomic molecule while X is monatomic is crucial for establishing the true nature of the compounds formed.

Stoichiometric Calculations

Stoichiometric calculations are an integral part of chemistry that deal with the quantitative relationships between reactants and products in a chemical reaction. Stoichiometry hinges on the law of conservation of mass and the concept of the mole, allowing chemists to predict how much product will result from given amounts of reactants and vice versa. Our textbook problem serves as a perfect example of using stoichiometric principles to determine unknown quantities, such as the relative atomic masses of elements from the mass percent compositions and reaction volume ratios.

One way to improve proficiency in stoichiometric calculations is through practice with various balanced chemical equations and by paying close attention to the molar relationships between compounds and elements within those equations. Additionally, understanding that coefficients in balanced chemical equations represent molar ratios can greatly simplify the process of these calculations.