Problem 122 - General Chemistry

Question

A solution of scandium chloride was treated with silver nitrate. The chlorine in the scandium compound was converted to silver chloride, \(\mathrm{AgCl}\). A \(58.9\) -mg sample of scandium chloride gave \(167.4 \mathrm{mg}\) of silver chloride. What are the mass percentages of \(\mathrm{Sc}\) and \(\mathrm{Cl}\) in scandium chloride? What is its empirical formula?

Step-by-Step Solution

Answer

Mass % of Sc = 29.7%, Cl = 70.3%. Empirical formula is ScCl₃.

Step 1: Understand the Problem

We need to find the mass percentages of scandium (Sc) and chlorine (Cl) in a sample of scandium chloride. The sample reacts with silver nitrate to form silver chloride (AgCl). By finding the moles of AgCl produced, we can determine the moles of Cl in scandium chloride. We also need to determine the empirical formula of scandium chloride.

Step 2: Calculate Moles of AgCl

The first step involves converting the mass of AgCl into moles using its molar mass. The molar mass of AgCl is approximately 143.32 g/mol.\[\text{Moles of AgCl} = \frac{167.4\, \text{mg}}{143.32\, \text{g/mol}} \times \frac{1\, \text{g}}{1000\, \text{mg}} = 0.001169\, \text{mol}\]

Step 3: Determine Moles of Cl

From the chemical reaction, each mole of AgCl corresponds directly to one mole of Cl. Therefore, the moles of Cl in the scandium chloride are also 0.001169 mol.

Step 4: Calculate Mass of Cl in Sample

Calculate the mass of Cl in scandium chloride using the moles of Cl and the molar mass of Cl (35.45 g/mol).\[\text{Mass of Cl} = 0.001169\, \text{mol} \times 35.45\, \text{g/mol} = 0.0414\, \text{g} = 41.4\, \text{mg}\]

Step 5: Determine Mass and Mass Percentage of Sc

We subtract the mass of Cl from the total mass of scandium chloride to find the mass of Sc, then find the mass percentage of Sc.\[\text{Mass of Sc} = 58.9\, \text{mg} - 41.4\, \text{mg} = 17.5\, \text{mg}\]\[\text{Mass \, % \, of \, Sc} = \left(\frac{17.5\, \text{mg}}{58.9\, \text{mg}}\right) \times 100 = 29.7\%\]

Step 6: Calculate Mass Percentage of Cl

Having calculated the mass of Cl, we can now find its mass percentage in the scandium chloride.\[\text{Mass \, % \, of \, Cl} = \left(\frac{41.4\, \text{mg}}{58.9\, \text{mg}}\right) \times 100 = 70.3\%\]

Step 7: Determine Empirical Formula

Using the mass percentages and atomic masses, we calculate the simplest mole ratio of Sc to Cl.For Sc: \[\frac{17.5\, \text{mg}}{44.96\, \text{g/mol}} = 0.389\, \text{mmol}\] For Cl: \[\frac{41.4\, \text{mg}}{35.45\, \text{g/mol}} = 1.17\, \text{mmol}\]The mole ratio of Sc to Cl is approximately 1:3 (0.389:1.17 simplifies roughly to 1:3). Therefore, the empirical formula is \(\text{ScCl}_3\).

Key Concepts

Empirical Formula

The empirical formula of a compound represents the simplest whole-number ratio of atoms of each element in the compound. It provides essential information on the composition of the compound, but not the exact number of atoms like a molecular formula would. To find the empirical formula, first, determine the moles of each element present in the compound. For instance, in the case of scandium chloride, we start by identifying the masses of scandium (Sc) and chlorine (Cl), as determined by experimental measurements.

Using these masses, convert them to moles by dividing by their respective atomic masses: scandium has an atomic mass of about 44.96 g/mol, and chlorine has an atomic mass of 35.45 g/mol. The mole calculation reveals the ratio of Sc to Cl, which is simplified to give the empirical formula. In this exercise, the ratio comes out as approximately 1:3, leading to the empirical formula \(\text{ScCl}_3\). This indicates that each scandium atom is accompanied by three chlorine atoms.

Molar Mass

Molar mass is a fundamental concept in chemistry, representing the mass of one mole of a substance, expressed in grams per mole (g/mol). It is derived from the atomic masses of the constituent elements in a compound. Understanding molar mass is essential for converting between mass and moles, a frequent task in chemical calculations.

• The molar mass of a single element is found on the periodic table, such as 35.45 g/mol for chlorine.
• For compounds like silver chloride (AgCl), the molar mass is the sum of the atomic masses of silver and chlorine, resulting in approximately 143.32 g/mol.

In calculations, molar mass acts as a conversion factor from moles to grams and vice versa. Knowing the molar mass allows you to calculate the amount of substance in a sample or the products formed in a reaction, as demonstrated in this exercise with the conversion of the mass of AgCl to moles.

Chemical Reaction

Chemical reactions involve the transformation of substances into new compounds. They follow specific reactant-product relationships governed by stoichiometry. In this exercise, scandium chloride reacts with silver nitrate to form silver chloride (AgCl) and other products. Such reactions are often represented by balanced chemical equations, which depict the conservation of mass and define the molar ratios of reactants to products.

Understanding the relationship between reactants and products is crucial. In our example, the formation of AgCl directly relates moles of chloride ions (Cl\(^-\)) released from scandium chloride through interaction with silver ions (Ag\(^+\)). The knowledge of this relationship allows for the determination of the resultant substances and their amounts. Chemical reactions thus help in unveiling the composition and properties of the initial materials and their transformation into new forms.

Moles of AgCl

The moles of AgCl are a vital part of solving this exercise. This value links directly to the chlorine content in the sample of scandium chloride. Calculating the moles of a substance from its mass involves dividing the mass by its molar mass.

With AgCl having a molar mass of 143.32 g/mol, you can use this to find out how much chlorine contributes to the compound that formed. By knowing the moles of AgCl produced in the reaction, we also know the moles of chlorine involved, since every mole of AgCl contains exactly one mole of chlorine due to its formula. Mastering this concept is essential to understand interconnections in chemical reactions and stoichiometry, where each mole measures the quantity of atoms or molecules participating in the reaction.