Question

Three pure compounds form when 1.00 -g samples of element \(X\) combine with,respectively, 0.472 g, \(0.630 \mathrm{~g}\), and \(0.789 \mathrm{~g}\) of element \(Z\). The first compound has the formula \(X_{2} Z_{3}\). Find the empirical formulas of the other two compounds.

Short answer

The empirical formulas of the second and third compounds are as follows: For the second compound, the formula is XZ_{2}, and for the third compound, the formula is X_{3}Z_{7}.

Step-by-step solution

Determine the Mass Ratios

Calculate the mass ratios of element Z combining with element X for each of the compounds. For the first compound, we have 0.472 g of Z combining with 1.00 g of X. For the second, we have 0.630 g of Z with 1.00 g of X. For the third, we have 0.789 g of Z with 1.00 g of X.

Calculate Molar Ratios for First Compound

Using the first compound (X_{2}Z_{3}) as a reference, calculate the molar ratio of X to Z. For every 2 moles of X, there are 3 moles of Z. Find the mass of a single mole of Z by dividing the mass of Z by 3 and multiplying it by 2 to get the mass of a mole of X.

Determine Empirical Formulas for Other Compounds

Using the obtained molar mass of X and Z from the first compound, calculate the number of moles of Z that combine with a fixed mass (1 g) of X in the second and third compounds. Then, convert these moles into the simplest whole number ratio to determine the empirical formulas for the other two compounds.

Key concepts

Mass Ratios in Chemical Compounds

Understanding mass ratios in chemical compounds is critical for revealing the quantitative relationships between elements. These ratios are determined by comparing the masses of various elements that combine to form a compound. In the exercise, we have three separate compounds formed by combining element X with different masses of element Z. Each pairing results in a unique mass ratio.

To grasp this concept, it's pivotal to recognize that these mass ratios reflect the underlying atom-to-atom ratios that govern chemical combination. When we consider a 1.00 g sample of element X combining with varying masses of element Z, the resulting ratios are unique fingerprints of the compounds' identities. The mass ratio itself, however, isn’t enough to determine the empirical formula; for that, we must venture into the realm of moles and molar ratios.

Molar Ratios Calculation

The molar ratio calculation transitions our understanding from the scale of grams to that of moles, which represent the fundamental counting unit in chemistry. By using Avogadro's number (\(6.022 \times 10^{23}\) atoms/mole), we convert grams to moles to compare the number of atoms of each element within compounds.

For instance, the exercise provides the mass of element Z that combines with a fixed mass of element X in each of the compounds. By determining the molar mass of each element (which would require knowing the elements' identities or looking them up on the periodic table—alluded to in Step 2 of the solution), students can convert these masses to moles. This molar understanding lets us see the actual atomic ratio, transporting us closer to the true empirical formula of the compounds.

Stoichiometry

Stoichiometry, at its core, is about the measurement of elements and compounds during chemical reactions. Equipped with the molar ratios of elements within a compound, stoichiometry provides the framework for understanding the larger relationship between reactants and products in a chemical reaction.

In the given exercise, stoichiometry comes into play when we use the composition of the first compound to find the empirical formulas of the others. The known formula, X₂Z₃, offers a stoichiometric map that, together with the previously calculated molar ratios, guides us to the empirical formulas of the subsequent compounds. By maintaining a fixed mass of element X and varying the mass of Z, we apply stoichiometric principles to explore the compositional landscape of these compounds.

Chemical Compound Composition

The composition of a chemical compound is essentially a blueprint of its molecular architecture. It indicates which elements are present, and in what ratios, by providing empirical formulas that signify the simplest whole-number ratio of atoms in a compound.

Our exercise hinges on discerning this composition for multiple compounds originating from the same elements. Fundamentally, the empirical formula is rooted in understanding the compound's mass and molar ratios and applying stoichiometry. When we map the mass of element Z to a fixed mass of element X, and when those masses are expressed in moles, we can crack the code of each compound's formula. The elegance of the empirical formula lies in its simplicity and how it conveys the underlying stoichiometric laws that govern chemical reactions.