Question

Solid silver oxide, \(\mathrm{Ag}_{2} \mathrm{O}(\mathrm{s}),\) decomposes at temperatures in excess of \(300^{\circ} \mathrm{C},\) yielding metallic silver and oxygen gas. A 3.13 g sample of impure silver oxide yields \(0.187 \mathrm{g} \mathrm{O}_{2}(\mathrm{g}) .\) What is the mass percent \(\mathrm{Ag}_{2} \mathrm{O}\) in the sample? Assume that \(\mathrm{Ag}_{2} \mathrm{O}(\mathrm{s})\) is the only source of \(\mathrm{O}_{2}(\mathrm{g}) .\) [Hint: Write a balanced equation for the reaction.]

Short answer

The mass percent of \(\mathrm{Ag}_{2} \mathrm{O}\) in the sample is approximately 86.9%.

Step-by-step solution

Writing the Balanced Chemical Reaction

From the problem, we know that solid silver oxide, \(\mathrm{Ag}_{2} \mathrm{O}\), decomposes into silver (\(\mathrm{Ag}\)) and oxygen (\(\mathrm{O}_{2}\)). This allows us to write the following unbalanced reaction: \(\mathrm{Ag}_{2} \mathrm{O}(\mathrm{s}) \rightarrow \mathrm{Ag}(\mathrm{s}) + \mathrm{O}_{2}(\mathrm{g})\). To balance the reaction, we update it to: \(2\mathrm{Ag}_{2} \mathrm{O}(\mathrm{s}) \rightarrow 4\mathrm{Ag}(\mathrm{s}) + \mathrm{O}_{2}(\mathrm{g})\).

Applying Stoichiometry to Find Moles

From balanced equation, we know that 2 moles of \(\mathrm{Ag}_{2} \mathrm{O}\) will give 1 mole of \(\mathrm{O}_{2}\). But we need to convert grams to moles. The molecular mass of oxygen is 32 g/mol. Hence, 0.187 g of \(\mathrm{O}_{2}\) is \(0.187/32=0.00584\) moles. As, 2 moles of \(\mathrm{Ag}_{2} \mathrm{O}\) give 1 mole of \(\mathrm{O}_{2}\), then \(0.00584\) moles of \(\mathrm{O}_{2}\) has been given by \(2*0.00584=0.0117\) moles of \(\mathrm{Ag}_{2} \mathrm{O}\).

Finding the Mass of \(\mathrm{Ag}_{2} \mathrm{O}\) in the Sample

Knowing the moles of \(\mathrm{Ag}_{2} \mathrm{O}\), we can calculate the mass. The molecular weight of \(\mathrm{Ag}_{2} \mathrm{O}\) is 231.74 g/mol. So, the mass of \(\mathrm{Ag}_{2} \mathrm{O}\) in the sample is \(0.0117*231.74 = 2.72\) g.

Finding the Mass Percent

The mass percent of \(\mathrm{Ag}_{2} \mathrm{O}\) in the sample is calculated using the mass of \(\mathrm{Ag}_{2} \mathrm{O}\) and the total mass of the sample. The mass percent is given by \((\(\mathrm{Ag}_{2} \mathrm{O}\) Mass / Total mass) * 100\), which equals to \((2.72/3.13) * 100 = 86.9\)%.

Key concepts

Stoichiometry

Stoichiometry is a crucial concept in chemistry that involves the quantitative relationship between reactants and products in a chemical reaction. It is based on the law of conservation of mass, which states that matter cannot be created or destroyed in a chemical reaction. Instead, the mass of the reactants must equal the mass of the products.
In our example, silver oxide \(\mathrm{Ag}_2 \mathrm{O}\) decomposes into metallic silver and oxygen gas. To fully understand how much of each product is formed, it is necessary to write a **balanced chemical equation** first. The balanced equation tells us the **proportions** of reactants that combine to produce a certain amount of products.
The concept of moles is crucial here. It is a unit in chemistry which allows us to convert between the mass of a substance and its quantity or volume. This information can be used to predict how much of one substance will react with a given amount of another. Stoichiometry uses molecular weights and balanced reactions to solve real-world problems, like in our exercise. By calculating moles and applying stoichiometry, one determines how much of the reactants reacts and how much of each product is formed.

Mass Percent Calculation

Mass percent provides a way of expressing the concentration of an element in a compound or a component in a mixture. It's calculated by dividing the mass of the component by the total mass of the mixture, and multiplying by 100 to get a percentage.
In the given problem, our task is to find the mass percent of silver oxide \(\mathrm{Ag}_{2} \mathrm{O}\) in an impure sample. After calculating the mass of \(\mathrm{Ag}_2 \mathrm{O}\) in the sample using stoichiometry, we found it to be 2.72 g.

• First, we determine the mass of \(\mathrm{Ag}_2 \mathrm{O}\) formed based on the moles of \(\mathrm{O}_2\) produced.
• Using this mass, we calculate the mass percent by the formula:
  \[((\text{Mass of } \mathrm{Ag}_{2} \mathrm{O}) / (\text{Total sample mass})) \times 100\]
• For the problem, it's: \((2.72 / 3.13) \times 100\ = 86.9%\).

Mass percent is a useful parameter in chemical analysis to determine the purity of a compound in a sample.

Balanced Chemical Equation

Creating a balanced chemical equation involves ensuring that the number of atoms of each element is the same on both sides of the reaction. It reflects the conservation of mass during a chemical reaction.
In the reaction of silver oxide decomposition, the preliminary step is writing the unbalanced equation:
\[\mathrm{Ag}_2 \mathrm{O} \rightarrow \mathrm{Ag} + \mathrm{O}_2\]
However, this needs to be balanced since the number of atoms in the reactants should equal the number of atoms in the products.

• For silver (Ag), place a coefficient of 4 in front of \(\mathrm{Ag}\) as each molecule of \(\mathrm{O}_2\) requires four silver atoms (from two \(\mathrm{Ag}_2\) units).
• When balanced, the equation becomes:
  \[2\mathrm{Ag}_2 \mathrm{O}(s) \rightarrow 4\mathrm{Ag}(s) + \mathrm{O}_2(g)\]

Assessing the stoichiometry of the equation ensures accurate stoichiometric calculations later. A properly balanced equation is the foundation that allows us to correctly determine reactant and product masses or volumes.