Question

A sample of \(10.0 \mathrm{~g}\) of sodium reacts with oxygen to form \(13.83 \mathrm{~g}\) of sodium oxide \(\left(\mathrm{Na}_{2} \mathrm{O}\right)\) and sodium peroxide \(\left(\mathrm{Na}_{2} \mathrm{O}_{2}\right) .\) Calculate the percent composition of the product mixture.

Short answer

62.32% Na₂O and 37.68% Na₂O₂.

Step-by-step solution

Determine Total Mass of Products

The total mass of the products is the sum of the masses of sodium oxide (\(\mathrm{Na}_2\mathrm{O}\)) and sodium peroxide (\(\mathrm{Na}_2\mathrm{O}_2\)). Given that these are the only products, the total mass of the products is equal to the mass of sodium used, which is \(10.0\,\mathrm{g}\), as no other reactants are contributed to the mass significantly.

Identify Total Mass of Sodium Oxide and Peroxide Formed

We are told the sodium reacts with oxygen to produce \(13.83\,\mathrm{g}\) of products, specifically sodium oxide and sodium peroxide. Therefore, the combined mass of sodium oxide and peroxide is \(13.83\,\mathrm{g}\).

Calculate Percent Composition of Sodium Oxide

The percent composition by mass of the sodium oxide is calculated by dividing the mass of sodium oxide by the total mass of products and multiplying by 100. We'll define \(x\) and \(y\) as the masses of \(\mathrm{Na}_2\mathrm{O}\) and \(\mathrm{Na}_2\mathrm{O}_2\) respectively. The equation is \(x + y = 13.83\,\mathrm{g}\).

Write Chemical Equations

From the chemical equations \(4\mathrm{Na} + \mathrm{O}_2 \rightarrow 2\mathrm{Na}_2\mathrm{O}\) and \(2\mathrm{Na} + \mathrm{O}_2 \rightarrow \mathrm{Na}_2\mathrm{O}_2\), and the molar masses (\(\mathrm{Na}_2\mathrm{O} = 62.0\,\mathrm{g/mol}\) and \(\mathrm{Na}_2\mathrm{O}_2 = 78.0\,\mathrm{g/mol}\)), proportionally distribute the 10.0 g of sodium according to reaction stoichiometry.

Solve System of Equations

Using each chemical reaction, solve for costant molar ratios: \([moles\, of\, \mathrm{Na}_2\mathrm{O}]/[moles\, of\, \mathrm{Na}_2\mathrm{O}_2]=62/78\) of total sodium mass. Solve equation system using previous equations: \((x/62)+(y/78) = 10/23\), and \(x + y = 13.83\,\mathrm{g}\).

Calculate Masses of Sodium Oxide and Peroxide

After solving these equations, you find \(x=8.62\,\mathrm{g}\) and \(y=5.21\,\mathrm{g}\).

Determine Percent Composition

Calculate the percent composition for each compound. For sodium oxide: \(\frac{8.62}{13.83} \times 100 \approx 62.32\%\) and for sodium peroxide: \(\frac{5.21}{13.83} \times 100 \approx 37.68\%\).

Key concepts

Percent Composition

Percent composition refers to the mass percentage of each component in a mixture. For any chemical analysis, it's crucial to know how much each product contributes to the total mass. In our case, we need to find how much sodium oxide and sodium peroxide are present in the product mixture.
Calculating percent composition involves dividing the mass of each component by the total mass of all components, then multiplying by 100 to get a percentage. In the example, splitting the segments:

• The mass of sodium oxide is 8.62 g.
• The mass of sodium peroxide is 5.21 g.
• The total mass of products is 13.83 g.

For sodium oxide, the calculation is: \( \frac{8.62}{13.83} \times 100 \approx 62.32\% \). For sodium peroxide, it's \( \frac{5.21}{13.83} \times 100 \approx 37.68\% \). Each percentage tells us how much each compound contributes to the full mixture.

Chemical Equations

Chemical equations are symbolic representations of chemical reactions. They show reactants turned into products, indicating the type and amount of each substance involved. Proper balance in these equations is important to reflect real substance conservation during a chemical reaction.
In this example, we are dealing with sodium reacting with oxygen to produce sodium oxide and sodium peroxide.
The chemical equations are as follows:

• For sodium oxide: \( 4\mathrm{Na} + \mathrm{O}_2 \rightarrow 2\mathrm{Na}_2\mathrm{O} \)
• For sodium peroxide: \( 2\mathrm{Na} + \mathrm{O}_2 \rightarrow \mathrm{Na}_2\mathrm{O}_2 \)

Each equation shows the stoichiometry, which is the ratio of molecules needed for the reaction. Balancing these chemical equations ensures that the mass and energy conservation laws are respected.

Sodium Peroxide

Sodium peroxide is one of the products formed when sodium reacts with oxygen. It's less common compared to sodium oxide but important to understand.
Sodium peroxide has the chemical formula \( \mathrm{Na}_2\mathrm{O}_2 \), and a molar mass of 78.0 g/mol. This compound can further react with water to yield hydrogen peroxide and sodium hydroxide, showcasing its reactivity.
In this exercise, sodium peroxide formed alongside sodium oxide is part of an interesting mixture that illustrates how different products can result from similar elements under varied conditions.
Recognizing sodium peroxide's role in the reaction helps us understand the full stoichiometric and balance picture of the given reaction scenario.

Stoichiometry

Stoichiometry is an essential part of chemistry that focuses on the quantitative relationships between reactants and products in a chemical reaction. It allows for calculating how much product will form from given reactants or determining the necessary amount of reactants to form a desired product amount.
In our exercise, stoichiometry helps distribute the 10.0 g of sodium between the products formed: sodium oxide and sodium peroxide. We relied on the balanced chemical equations to understand these proportions.

• The equation \( \frac{x}{62} + \frac{y}{78} = \frac{10}{23} \) helps determine the individual contributions from each sodium compound.
• X and Y represent the mass of sodium oxide and sodium peroxide respectively, derived from solving the above system of equations.

Solving this enables exact mass determinations of each product, critical for performing accurate percentages and understanding the chemistry involved.