Two intermetallic compounds, \(A_{3} B\) and \(\mathrm{AB}_{3},\) exist for elements \(\mathrm{A}\) and \(\mathrm{B}\). If the compositions for \(A_{3} B\) and \(A B_{3}\) are 91.0 wt \(\%\) \(\mathrm{A}-9.0 \mathrm{wt} \% \mathrm{B}\) and \(53.0 \mathrm{wt} \% \mathrm{A}-47.0 \mathrm{wt} \% \mathrm{B}\) respectively, and element A is zirconium, identify element B.

Understanding the molar mass of elements is vital in materials science, particularly when dealing with intermetallic compounds. The molar mass, which is the weight of one mole of an element or compound, is typically measured in grams per mole (g/mol). It is a fundamental property that is necessary for comprehending the composition of materials.

To calculate the molar mass of a compound such as an intermetallic compound like \(A_3B\) or \(AB_3\), you first need to know the molar masses of the individual elements involved, obtained from the periodic table. In the context of the exercise provided, we focus on element A, zirconium, which has a known molar mass, and use the weight percentages and stoichiometric relationships to deduce the molar mass of the unknown element B.

Using simple algebra, we set up an equation for each compound as demonstrated in the steps provided. The key is to relate the molar mass of the entire compound with its component elements and their proportions. By setting up a proportion of weight percentages and using stoichiometry, we can isolate the molar mass of the unknown component, in this case, element B. This process is critical for determining the exact composition of the compounds being analyzed.

Analyzing the chemical composition of compounds is a frequent task in materials science. It involves determining what elements are present and in what proportions. In intermetallic compounds, the ratios in which different metals combine can significantly affect their properties. Therefore, understanding their composition is essential.

Chemical composition is often expressed in weight percentages, which indicate the fraction of the total mass contributed by each component. For a compound such as \(A_{3}B\), where \(A\) is zirconium, we see it is composed of 91 wt% \(A\) and 9 wt% \(B\). This tells us that in a sample of this compound weighing 100 grams, 91 grams are zirconium, and 9 grams are the unknown element B.

By using the weight percentages provided and the known molar mass of zirconium, the analysis becomes a straightforward math problem, leveraging stoichiometry—the study of quantitative relationships in chemical formulas and reactions. With carefully formed equations and known stoichiometric ratios as shown in the exercise, one can deduce not only the molar mass of unknown components but also confirm the empirical formula of the compound itself.

The periodic table is a crucial tool for anyone studying materials science. It provides a vast array of information about elements, including atomic numbers, element symbols, atomic weights, and more. Looking up data in the periodic table is often the initial step when analyzing or synthesizing chemical compounds.

When searching for an element's molar mass to complete our compound analysis, as done in the example, we directly use the periodic table. Each entry in the periodic table is standardized, making it relatively straightforward to locate the molar mass. For zirconium, the average atomic mass listed can be approximated to the molar mass in grams per mole, a value critical for further calculations.

In the final step of the exercise, once we have calculated the approximate molar mass of element B, the periodic table is again consulted. This time it helps to identify the element by comparing the calculated molar mass with those listed in the periodic table. A close match confirms the identity of the element, completing the analysis and shedding light on the full composition of the intermetallic compound in question.