Question

A \(0.4230-\mathrm{g}\) sample of impure sodium nitrate was heated, converting all the sodium nitrate to \(0.2864 \mathrm{~g}\) of sodium nitrite and oxygen gas. Determine the percent of sodium nitrate in the oriainal sample.

Short answer

The percentage of sodium nitrate in the original sample is 83.4%.

Step-by-step solution

Write down the balanced chemical equation for the decomposition of sodium nitrate.

The decomposition of sodium nitrate (NaNO3) produces sodium nitrite (NaNO2) and oxygen gas (O2). The balanced chemical equation is: \[2 \mathrm{NaNO}_3 \rightarrow 2 \mathrm{NaNO}_2 + \mathrm{O}_2\]

Calculate the moles of sodium nitrite formed.

We are given that 0.2864 g of sodium nitrite was formed after heating. To find the moles of sodium nitrite, we need to use its molar mass, which is 69.0 g/mol. Moles of NaNO2 formed = mass of NaNO2 / molar mass of NaNO2 Moles of NaNO2 formed = \(0.2864 \mathrm{~g} / 69.0 \mathrm{~g/mol}\) Moles of NaNO2 formed = \(4.15 \times 10^{-3} \mathrm{~mol}\)

Use stoichiometry to find the moles of sodium nitrate in the original sample.

From the balanced chemical equation, we can see that 2 moles of sodium nitrate decompose to produce 2 moles of sodium nitrite. Therefore, the moles of sodium nitrate in the original sample should be equal to the moles of sodium nitrite formed. Moles of NaNO3 = Moles of NaNO2 = \(4.15 \times 10^{-3} \mathrm{~mol}\)

Convert the moles of sodium nitrate to grams.

To convert the moles of sodium nitrate to grams, we need its molar mass, which is 85.0 g/mol. Mass of NaNO3 = moles of NaNO3 × molar mass of NaNO3 Mass of NaNO3 = \(4.15 \times 10^{-3} \mathrm{~mol} \times 85.0 \mathrm{~g/mol}\) Mass of NaNO3 = 0.3528 g

Calculate the percentage of sodium nitrate in the original sample.

Now that we have the mass of sodium nitrate in the original sample, we can calculate its percentage by dividing it by the total mass of the impure sample and multiplying by 100. Percentage of NaNO3 = (mass of NaNO3 / mass of impure sample) × 100 Percentage of NaNO3 = \((0.3528 \mathrm{~g} / 0.4230 \mathrm{~g}) \times 100\) Percentage of NaNO3 = 83.4% The percentage of sodium nitrate in the original sample is 83.4%.

Key concepts

Balanced Chemical Equation

In chemistry, a balanced chemical equation is key to understanding reactions. It tells us the ratio in which reactants turn into products. For a reaction to be balanced, the number of atoms for each element should be equal on both sides of the equation. In our exercise, we deal with the decomposition of sodium nitrate (NaNO₃). This transformation produces sodium nitrite (NaNO₂) and oxygen gas (O₂). The balanced chemical equation is:

• 2 NaNO₃ → 2 NaNO₂ + O₂

This equation reflects that two molecules of sodium nitrate yield two molecules of sodium nitrite and one molecule of oxygen gas. Ensuring the equation is balanced is crucial for stoichiometry calculations, letting us correctly predict how much of each substance forms during the reaction.

Sodium Nitrate

Sodium nitrate is a chemical compound with the formula NaNO₃. It is often used in fertilizers, glass making, and food preservation. In its breakdown process, sodium nitrate undergoes chemical decomposition, a common type of chemical reaction. Here, sodium nitrate decomposes into sodium nitrite and oxygen gas. This process can be represented by the chemical equation:

• 2 NaNO₃ → 2 NaNO₂ + O₂

Understanding sodium nitrate's behavior in a reaction helps predict the products and their quantities. Its ability to decompose makes it an interesting subject for stoichiometric calculations, as it involves calculating amounts of reactants and products involved.

Molar Mass

In stoichiometry, molar mass is a critical concept. It allows the conversion between moles and grams. The molar mass of a compound is the sum of the atomic masses of its constituent atoms. For example, sodium nitrite (NaNO₂) has a molar mass of 69.0 g/mol, calculated as follows:

• Sodium (Na): approximately 23 g/mol
• Nitrogen (N): approximately 14 g/mol
• Oxygen (O): approximately 16 g/mol (2 atoms = 32 g/mol)

Adding these gives 23 + 14 + 32 = 69 g/mol. Similarly, sodium nitrate (NaNO₃) has a molar mass of 85.0 g/mol. Molar mass allows us to convert between mass and moles to perform stoichiometric calculations, essential for finding the percentage of a compound in a mixture.

Chemical Decomposition

Chemical decomposition is a reaction where one compound breaks down into two or more simpler substances. In our scenario, sodium nitrate decomposes into sodium nitrite and oxygen gas upon heating. This kind of reaction is vital in many industrial and natural processes. It is represented by:

• 2 NaNO₃ → 2 NaNO₂ + O₂

Decomposition can occur due to heat, light, or other stimuli and often requires energy input. Understanding decomposition helps scientists and engineers develop methods for creating materials or extracting resources. In the given exercise, recognizing and applying decomposition allows us to predict the resulting products and carry out stoichiometric calculations effectively.