Problem 148 - General Chemistry

Question

A metal, \(\mathrm{M}\), was converted to the chloride, \(\mathrm{MCl}_{2} .\) Then a solution of the chloride was treated with silver nitrate to give silver chloride crystals, which were filtered from the solution. \(\mathrm{MCl}_{2}(a q)+2 \mathrm{AgNO}_{3}(a q) \longrightarrow \mathrm{M}\left(\mathrm{NO}_{3}\right)_{2}(a q)+2 \mathrm{AgCl}(s)\) If \(2.434 \mathrm{~g}\) of the metal gave \(7.964 \mathrm{~g}\) of silver chloride, what is the atomic weight of the metal? What is the metal?

Step-by-Step Solution

Answer

The atomic weight of the metal, M, is 16.67 g/mol, which corresponds to sulfur, S.

Step 1: Identify Known Values

We are given the mass of metal (\( M \)) as \( 2.434 \, \text{g} \) and the mass of silver chloride (\( \text{AgCl} \)) produced as \( 7.964 \, \text{g} \). We will use these values to find the molar mass of \( M \).

Step 2: Calculate Moles of AgCl

First, find the molar mass of \( \text{AgCl} \). Silver (\( \text{Ag} \)) has a molar mass of \( 107.87 \, \text{g/mol} \) and chlorine (\( \text{Cl} \)) has a molar mass of \( 35.45 \, \text{g/mol} \). Thus, \[\text{Molar mass of AgCl} = 107.87 \, \text{g/mol} + 35.45 \, \text{g/mol} = 143.32 \, \text{g/mol}.\] Now, use the molar mass to calculate the moles of \( \text{AgCl} \): \[\text{Moles of AgCl} = \frac{7.964 \, \text{g}}{143.32 \, \text{g/mol}} = 0.05557 \, \text{mol}.\]

Step 3: Use Stoichiometry to Find Moles of MCl2

According to the reaction, one mole of \( \text{MCl}_2 \) produces two moles of \( \text{AgCl} \). Therefore, \[\text{Moles of MCl}_2 = \frac{0.05557 \, \text{mol}}{2} = 0.027785 \, \text{mol}.\]

Step 4: Calculate Molar Mass of MCl2

Since we know the moles of \( \text{MCl}_2 \) and the initial mass of the metal (\( M \)), we first need the molar mass of \( \text{MCl}_2 \) to find the molar mass of \( M \). The initial mass of \( \text{MCl}_2 \) is the same as the mass of \( M \), so: \[\text{Molar mass of MCl}_2 = \frac{2.434 \, \text{g}}{0.027785 \, \text{mol}} = 87.57 \, \text{g/mol}.\]

Step 5: Determine Atomic Weight of M

\( \text{MCl}_2 \) consists of one metal atom and two chlorine atoms. Therefore, the atomic weight of \( M \) is calculated by:\[\text{Atomic weight of } M = 87.57 \, \text{g/mol} - 2 \times 35.45 \, \text{g/mol} = 16.67 \, \text{g/mol}.\]

Key Concepts

Moles Calculation

To understand chemical reactions, we often need to determine the number of moles involved. In this exercise, calculating the moles of silver chloride, helps us find the amount of the unknown metal. The concept of moles is essential because it provides a way to relate the mass of a substance to the number of particles, such as atoms, molecules, or ions.

We start by using the formula: \[ \text{Moles} = \frac{\text{Mass (g)}}{\text{Molar Mass (g/mol)}} \] This equation allows us to convert from grams of a substance to moles by using its molar mass.

For the exercise, we calculate the moles of silver chloride (AgCl). First, find its molar mass, which is the sum of the atomic masses of silver and chlorine. Then, divide the given mass of AgCl by this molar mass to find the number of moles. This critical step bridges the mass of substances with their chemical behavior.

Molar Mass

Molar mass acts as a conversion factor between grams and moles, making it a vital concept in chemistry. Each element’s molar mass corresponds to its atomic weight in grams per mole, which can be found on the periodic table.

In the solution, to find the molar mass of compound sequences like AgCl, we combine the atomic weights of each component—Ag and Cl, which adds up to 143.32 g/mol for AgCl.

Once we have the molar mass, we are equipped to perform moles calculation. Knowing the molar mass is essential when balancing equations and predicting the outcome of reactions, as it influences the stoichiometric coefficients in the balanced chemical equation. Always ensure you use the most accurate molar masses available for precise calculations.

Stoichiometry

Stoichiometry is a fundamental concept in chemistry that deals with the quantitative relationships between reactants and products in a chemical equation. It ensures that atoms are conserved during a reaction, adhering to the law of conservation of mass.

Using the stoichiometry from this exercise, we deduced the moles of \(\text{MCl}_2\) by knowing that one mole of \(\text{MCl}_2\) produces two moles of \(\text{AgCl}\).

This one-to-two ratio is derived from the coefficients of the compounds in the balanced chemical equation. It's crucial to grasp how stoichiometry links reactants and products, ensuring the equation is properly balanced to reflect the actual chemical interaction. Consistently applying stoichiometry is key to accurately predicting the amounts of substances consumed or produced in a given reaction.

Chemical Equation

Chemical equations describe the reactants and products involved in a reaction, using chemical formulas to denote the substances and coefficients to indicate their in respective ratios. A correctly balanced equation is essential to reflect the conservation of mass and predict the reaction outcomes precisely.

In this case, the equation \( \text{MCl}_{2}(aq) + 2 \text{AgNO}_{3}(aq) \rightarrow \text{M}(\text{NO}_{3})_{2}(aq) + 2 \text{AgCl}(s) \) indicates how the chloride compound of a metal reacts with silver nitrate to form silver chloride and another compound.

The coefficients are crucial as they indicate how many moles of each substance are involved, thus, they influence calculations involving moles and molar mass. Balancing chemical equations is a skill honed with practice and is foundational for any further work involving chemical reactions and stoichiometry.