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Chapter 3: Problem 163

When aluminum metal is heated with an element from Group \(6 \mathrm{~A}\) of the periodic table, an ionic compound forms. When the experiment is performed with an unknown Group \(6 \mathrm{~A}\) element, the product is \(18.56 \%\) Al by mass. What is the formula of the compound?

Short Answer

Expert verified
The formula of the compound is Al2S3.

Step by step solution

01

Determine the mass percent of the unknown element

Since the product contains 18.56% Al by mass, the mass percent of the unknown element can be found by subtracting the mass percent of Al from 100%: Mass percent of unknown element = 100% - 18.56% = 81.44%
02

Calculate the moles of aluminum and the unknown element

Assuming 100 grams of the compound, we can find the mass of Al and the unknown element in the compound. Then, we can use their molar masses to calculate moles. Mass of Al: 18.56 g (18.56% of 100 g) Moles of Al: \(\frac{18.56\,\text{g}}{26.98\,\text{g/mol}} = 0.688\,\text{mol}\) (molar mass of Al is 26.98 g/mol) Mass of unknown element: 81.44 g (81.44% of 100 g) We can denote the Group 6A element as X. In order to find the moles of X, we need more information about its molar mass. We can look at the possible Group 6A elements (O, S, Se, Te, Po) and determine which one would potentially form an ionic compound with Al and satisfy the required mass percentage.
03

Determine the molar ratio of aluminum and the unknown element

In order to obtain the empirical formula, we need to calculate the ratio of the moles of aluminum Al to the moles of the unknown element X. Let's consider Sulfur (S) as an example. Moles of S: \(\frac{81.44\,\text{g}}{32.07\,\text{g/mol}} = 2.537\,\text{mol}\) (molar mass of S is 32.07 g/mol) Now, we can find the ratio between the moles of Al and the moles of S: Molar ratio = \(\frac{0.688\,\text{mol Al}}{2.537\,\text{mol S}} = 0.271\)
04

Determine the empirical formula of the compound

An ionic compound consists of simple whole number ratios of atoms of each element. From our calculations, the molar ratio between Al and S (0.271) is approximately 1:3. Thus, the empirical formula of the compound is Al2S3. The formula of the compound is Al2S3.

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Most popular questions from this chapter

An element \(\mathrm{X}\) forms both a dichloride \(\left(\mathrm{XCl}_{2}\right)\) and a tetrachloride \(\left(\mathrm{XCl}_{4}\right) .\) Treatment of \(10.00 \mathrm{~g} \mathrm{XCl}_{2}\) with excess chlorine forms \(12.55 \mathrm{~g} \mathrm{XCl}_{4}\). Calculate the atomic mass of \(\mathrm{X}\), and identify \(\underline{X}\)

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