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 original sample.

Short answer

The percent of sodium nitrate in the original sample is 83.40%, calculated by finding the moles of sodium nitrite formed, using the 1:1 stoichiometric ratio to determine the moles of sodium nitrate, calculating the mass of sodium nitrate, and dividing by the mass of the original sample.

Step-by-step solution

1. Calculate the molecular weight of sodium nitrite and sodium nitrate

The molecular weights for the sodium nitrite (NaNO2) and sodium nitrate (NaNO3) are calculated as follows: NaNO2 = 1(22.99 g/mol Na) + 1(14.01 g/mol N) + 2(16.00 g/mol O) = 69.00 g/mol NaNO3 = 1(22.99 g/mol Na) + 1(14.01 g/mol N) + 3(16.00 g/mol O) = 85.00 g/mol

2. Calculate the moles of sodium nitrite formed in the reaction

Given the mass of sodium nitrite formed is 0.2864 g, we can now calculate the moles of sodium nitrite using the molecular weight: Moles of NaNO2 = mass / molecular weight Moles of NaNO2 = 0.2864 g / 69.00 g/mol Moles of NaNO2 = 0.00415 mol

3. Calculate the moles of sodium nitrate in the original sample

Since the balanced chemical equation indicates a 1:1 ratio between sodium nitrite and sodium nitrate, the moles of sodium nitrate that reacted will be the same as the moles of sodium nitrite formed: Moles of NaNO3 = 0.00415 mol

4. Calculate the mass of sodium nitrate in the original sample

Now, we can calculate the mass of sodium nitrate in the original sample using its molecular weight: Mass of NaNO3 = moles * molecular weight Mass of NaNO3 = 0.00415 mol * 85.00 g/mol Mass of NaNO3 = 0.35275 g

5. Calculate the percentage of sodium nitrate in the original sample

Finally, we can determine the percentage of sodium nitrate in the original sample using the mass of the original sample (0.4230 g) and the mass of sodium nitrate (0.35275 g) we just calculated: Percentage of NaNO3 = (mass of NaNO3 / mass of original sample) * 100 Percentage of NaNO3 = (0.35275 g / 0.4230 g) * 100 Percentage of NaNO3 = 83.40% The percent of sodium nitrate in the original sample is 83.40%.

Key concepts

Molecular Weight Calculation

One important step in stoichiometry is calculating the molecular weight of compounds involved in a chemical reaction. In this exercise, we needed to find the molecular weights of sodium nitrite (NaNO2) and sodium nitrate (NaNO3). This process involves summing up the atomic weights of all atoms in a molecule. Atomic weights can often be found on the periodic table.

For sodium nitrite (NaNO2), the molecular weight is determined by:

• Adding the atomic weight of sodium (Na): 22.99 g/mol
• The atomic weight of nitrogen (N): 14.01 g/mol
• Twice the atomic weight of oxygen (O): 2 \( \times \) 16.00 g/mol

The sum of these values gives us 69.00 g/mol for NaNO2. Similarly, for sodium nitrate (NaNO3), there is one more oxygen atom, making its molecular weight 85.00 g/mol.

Understanding how to calculate these weights is crucial for converting between grams and moles, which is required to solve stoichiometric problems. It's like knowing the exchange rates between currencies to convert money accurately. Once you have the molecular weights, you can calculate the moles of each compound involved in the reaction.

Chemical Reactions

Chemical reactions involve the transformation of reactants to products. In this particular exercise, the reaction converted sodium nitrate (NaNO3) into sodium nitrite (NaNO2) and oxygen gas.

This process illustrates a decomposition reaction, where a single compound breaks down into two or more simpler substances. The reaction adheres to the law of conservation of mass, meaning the mass of the reactants equals the mass of the products. To resolve stoichiometry problems, understanding this concept is essential, as it allows us to determine how much of each reactant is needed and what amounts of products are produced.

The balanced chemical equation helps us understand the mole ratio between reactants and products. For sodium nitrate converting to sodium nitrite, this ratio is 1:1. This ratio implies that one mole of sodium nitrate will produce one mole of sodium nitrite. Recognizing this helps in calculating the substances' masses or volumes required or produced in the reaction.

Sodium Nitrate Analysis

Analyzing compounds like sodium nitrate helps determine the purity of a sample, which is crucial in both educational and industrial settings. In the given exercise, a sample of sodium nitrate is analyzed to see what percent of it is pure sodium nitrate.

The analysis begins by calculating the moles of sodium nitrite formed because these directly relate to the moles of sodium nitrate originally present. Using the molecular weight found earlier, we found the moles of sodium nitrite and used the 1:1 mole ratio given by the balanced equation to find the moles of sodium nitrate.

Next, by multiplying the moles of sodium nitrate by its molecular weight, we determined the mass of pure sodium nitrate. Finally, calculating the percentage comes from comparing the mass of sodium nitrate to the total sample mass.

This process of sodium nitrate analysis exemplifies how stoichiometry can highlight the composition and purity within chemical mixtures. It is a vital skill in chemistry that allows scientists and students alike to identify the substances present and their exact ratios in mixtures rendered during experiments or industrial production.