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 \(\mathrm{X}^{\prime \prime}\) with "gas \(\mathrm{Y}^{\prime \prime}\) 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 \(\mathrm{Y} " \longrightarrow\) 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 element X and element Y have a ratio of approximately 4:1, meaning that the atomic mass of element X is about four times the atomic mass of element Y.

Step-by-step solution

Finding the simplest whole number ratio of atoms in the compounds

First, we will assume 100 grams of each compound. Then, we can use the mass percents to convert the grams to moles. Let's call the relative atomic masses of element X and element Y as \(m_X\) and \(m_Y\) respectively. For Compound I: \(30.43\% X \rightarrow 30.43g \rightarrow \frac{30.43}{m_X} moles\) \(69.57\% Y \rightarrow 69.57g \rightarrow \frac{69.57}{m_Y} moles\) For Compound II: \(63.64\% X \rightarrow 63.64g \rightarrow \frac{63.64}{m_X} moles\) \(36.36\% Y \rightarrow 36.36g \rightarrow \frac{36.36}{m_Y} moles\) Let's denote the ratio of moles for X and Y in Compound I as \(a_1\) and \(b_1\) and for Compound II as \(a_2\) and \(b_2\). For Compound I: \(\frac{a_1}{b_1} = \frac{\frac{30.43}{m_X}}{\frac{69.57}{m_Y}} = \frac{30.43m_Y}{69.57m_X}\) For Compound II: \(\frac{a_2}{b_2} = \frac{\frac{63.64}{m_X}}{\frac{36.36}{m_Y}} = \frac{63.64m_Y}{36.36m_X}\) We can observe that the ratio \(\frac{63.64}{36.36}\) is approximately twice the ratio \(\frac{30.43}{69.57}\), so the simplest whole number ratios for elements X and Y in the compounds are: Compound I: \(XY_2\) Compound II: \(X_2Y\)

Determining stoichiometric coefficients and formulas for reactants and products

Given the volume ratios in the chemical equations, we have: \(n_{X''}X'' + 2n_{Y''}Y'' \rightarrow 2XY_2\) \(2n_{X''}X'' + n_{Y''}Y'' \rightarrow 2X_2Y\) From these equations, we can see that the stoichiometric coefficients and formulas for reactants and products are: \(X'' = X\) \(Y'' = Y_2\)

Calculating the relative atomic masses of elements X and Y

Now we will substitute the formulas we found for the reactants and products into our mole ratios calculated in step 1: \(\frac{a_1}{b_1} = \frac{30.43m_Y}{69.57m_X} = \frac{30.43m_Y}{2 \cdot 69.57m_X/2}\) \(\frac{a_2}{b_2} = \frac{63.64m_Y}{36.36m_X} = \frac{2 \cdot 63.64m_Y/2}{36.36m_X}\) Then, we will divide the second equation by the first equation: \(\frac{\frac{63.64}{36.36}}{\frac{30.43}{69.57}} = \frac{\frac{63.64m_Y/2}{36.36m_X}}{\frac{30.43m_Y}{2 \cdot 69.57m_X/2}}\) By simplifying and solving for the ratio of atomic masses, we get: \(\frac{m_X}{m_Y} = \frac{63.64 \cdot 69.57}{36.36 \cdot 30.43}\) \(\frac{m_X}{m_Y} \approx \frac{4426}{1102} \approx 4\) Now, we can see that the ratio of the relative atomic masses is approximately 4:1, meaning that the atomic mass of element X is about four times the atomic mass of element Y.

Key concepts

Understanding Mole Ratio

The mole ratio is a crucial concept in stoichiometry that pertains to the proportional relationship between the amounts in moles of any two compounds or elements involved in a chemical reaction. Imagine each mole as a set count of items—just like a dozen eggs. In the context of chemical reactions, it reflects the number of particles, not their weight, because every mole contains the same quantity of particles (Avogadro's number, which is approximately 6.022 x 10^23 entities).

When we look at the provided exercise, the mole ratios are deduced from the mass percentages of elements X and Y in two compounds. By expressing these mass percentages as grams and dividing by the assumed relative atomic masses, we obtain the mole quantities present in 100 grams of each compound. Subsequently, through simplification, the ratios of moles reveal the simplest whole number ratios of atoms within a compound. For example, when the exercise mentions that the ratio of moles for X to Y in Compound I is approximately half that in Compound II, it guides us towards determining the formula of each compound as XY2 and X2Y respectively.

This quantification using mole ratios allows us to draw reliable conclusions about the chemical formulas of compounds from empirical data, an invaluable asset for a chemist working in any domain.

Chemical Composition Analysis

The chemical composition of a substance refers to the identity and ratio of the elements within it. Knowing the chemical composition is essential for understanding the properties and behavior of the substance during chemical reactions. To analyze chemical compositions, chemists often use mass percent information, which indicates the percentage by mass of each element in a compound. This information is foundational for determining formulas and balancing chemical equations.

In our exercise, the mass percents are given for two compounds made of elements X and Y. Compound I has 30.43% of X and 69.57% of Y, while Compound II has a higher percentage of X at 63.64%. These percentages are critical for calculating the mole ratios needed to derive the simplest formulas for the compounds. The solution to the exercise elegantly shows this analytical transition from mass percent to molar ratio, then to the simplest formula of a compound, illuminating the path from quantitative data to qualitative understanding of the substance's makeup.

Gas Volume Relationships in Reactions

Gas volume relationships, also known as the gas laws, describe the behavior of gases under different conditions and are closely related to stoichiometry in chemical reactions involving gases. In particular, the reaction provided deals with volume ratios, which, according to Avogadro's law, reflect mole ratios under constant temperature and pressure, since equal volumes of different gases contain an equal number of moles.

The exercise presents us with two reactions where gases X and Y react to form different compounds. Observing the volume ratios of reactants and products allows us to infer stoichiometric coefficients for those reactions. The first reaction shows that one volume of X and two volumes of Y yield two volumes of compound I, implying that the diatomic form of Y, Y2, is likely involved. The second reaction supports this conclusion. By using the gas volume relationships, we provide insight into the molecular composition of the reactant gases, which is then verified by the simplest formulas found from the mole ratios. Cementing this relationship between gas volumes and mole ratios ensures that the student can predict the outcome of reactions involving gases with confidence.