/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 156 A \(0.4230-\) g sample of impure... [FREE SOLUTION] | 91Ó°ÊÓ

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

Short Answer

Expert verified
The percentage of sodium nitrate in the original sample is approximately \(83.36\%\).

Step by step solution

01

Write a balanced chemical equation

First, we need to write a balanced chemical equation for the decomposition of sodium nitrate (NaNO3) into sodium nitrite (NaNO2) and oxygen gas (O2). 2NaNO3 ⟶ 2NaNO2 + O2
02

Determine the molar mass of sodium nitrate and sodium nitrite

Next, we need to determine the molar mass of sodium nitrate (NaNO3) and sodium nitrite (NaNO2). NaNO3: 1(22.99) + 1(14.01) + 3(16.00) = 85.00 g/mol NaNO2: 1(22.99) + 1(14.01) + 2(16.00) = 69.00 g/mol
03

Convert the mass of sodium nitrite produced to moles

Now, let's convert the mass of sodium nitrite (0.2864 g) produced into moles using the molar mass of sodium nitrite (69.00 g/mol). moles of NaNO2 = mass of NaNO2 / molar mass of NaNO2 moles of NaNO2 = 0.2864 g / 69.00 g/mol = 0.00415 mol
04

Calculate the mass of sodium nitrate used

Using the balanced chemical equation, we can see that 1 mole of NaNO2 is produced from 1 mole of NaNO3. Therefore, the moles of NaNO3 used are equal to the moles of NaNO2 produced. moles of NaNO3 = 0.00415 mol Next, convert the moles of sodium nitrate to mass using the molar mass of sodium nitrate (85.00 g/mol). mass of NaNO3 = moles of NaNO3 * molar mass of NaNO3 mass of NaNO3 = 0.00415 mol * 85.00 g/mol = 0.35275 g
05

Determine the percentage of sodium nitrate in the original sample

Finally, we will determine the percentage of sodium nitrate in the original impure sample (0.4230 g). percentage = (mass of NaNO3 / mass of impure sample) * 100 percentage = (0.35275 g / 0.4230 g) * 100 = 83.36% Therefore, the percentage of sodium nitrate in the original sample is approximately \(83.36\%\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Chemical Reactions
A chemical reaction involves the transformation of reactants into products. In our example, sodium nitrate (\(\text{NaNO}_3\)) breaks down into sodium nitrite (\(\text{NaNO}_2\)) and oxygen gas (\(\text{O}_2\)). This type of reaction is known as decomposition. It involves a compound being reduced into smaller chemical constituents.
In chemical reactions, it's essential to know what products are formed and in what ratio they appear. Sometimes reactions even release gases or change in temperature. Understanding these aspects helps in predicting the outcome of a reaction, which is essential in experiments and industrial applications.

When learning about chemical reactions, remember these key points:
  • Identify the reactants and products.
  • Investigate if the reaction is endothermic or exothermic.
  • Consider if energy, light, or a gas is released.
Molar Mass Calculation
In stoichiometry, molar mass is critical for converting between the mass of a substance and the amount in moles. For sodium nitrate (\(\text{NaNO}_3\)), the molar mass is calculated by adding the molar masses of all the atoms in the compound. We have:
  • Sodium (\(\text{Na}\)) = 22.99 g/mol
  • Nitrogen (\(\text{N}\)) = 14.01 g/mol
  • Oxygen (\(\text{O}\)) = 16.00 g/mol (and three oxygen atoms, hence 48.00 g/mol for NaNO3)
  • Total = 85.00 g/mol for \text{NaNO}_3
For sodium nitrite (\(\text{NaNO}_2\)), you do a similar calculation:
  • Two oxygen atoms contribute 32.00 g/mol
  • Total = 69.00 g/mol for \(\text{NaNO}_2\)
Knowing these values enables us to convert from grams to moles and vice versa, which is fundamental in stoichiometry.

Remember to always check atomic weights from the periodic table, as they are precise to two decimal places.
Percent Composition
Percent composition refers to the percentage, by mass, of each element in a compound. But when working with mixtures or impure samples, you often need to find the percentage of a specific compound in your sample. In our exercise, we determined that sodium nitrate (\(\text{NaNO}_3\)) made up approximately 83.36% of the original sample.

To find percent composition, use the formula:\[\text{{Percentage}} = \left(\frac{{\text{{mass of desired compound}}}}{{\text{{total mass of sample}}}}\right) \times 100\]This provides insight into the proportion of a specific component in a mixture and is valuable in assessing purity or making quantitative decisions in chemical manufacturing.
Key steps in calculating percent composition:
  • Determine the mass of the compound of interest.
  • Divide by the total mass of the sample or mixture.
  • Multiply the result by 100 to get a percentage.
Balanced Chemical Equation
A balanced chemical equation accurately represents the conservation of mass, stating that matter is neither created nor destroyed. In our decomposition reaction:\[2\text{NaNO}_3 \rightarrow 2\text{NaNO}_2 + \text{O}_2\]The number of atoms for each element is the same on both sides. This ensures that calculations based on the reaction are correct.

Here's why balancing equations is important:
  • It reflects the law of conservation of mass.
  • It helps in stoichiometric calculations, determining reactant and product quantities needed or produced.
  • It predicts the amount of energy exchange.
The steps to balance a chemical equation include:
  • Write the unbalanced equation.
  • Count atoms for each element on both sides.
  • Add coefficients to balance each element.
  • Check your work, ensuring atoms and charges are balanced.
Keeping equations balanced allows for accurate predictions and is foundational in chemical problem-solving.

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Most popular questions from this chapter

Hydrogen peroxide is used as a cleansing agent in the treatment of cuts and abrasions for several reasons. It is an oxidizing agent that can directly kill many microorganisms; it decomposes on contact with blood, releasing elemental oxygen gas (which inhibits the growth of anaerobic microorganisms); and it foams on contact with blood, which provides a cleansing action. In the laboratory, small quantities of hydrogen peroxide can be prepared by the action of an acid on an alkaline earth metal peroxide, such as barium peroxide: $$ \mathrm{BaO}_{2}(s)+2 \mathrm{HCl}(a q) \longrightarrow \mathrm{H}_{2} \mathrm{O}_{2}(a q)+\mathrm{BaCl}_{2}(a q) $$ What mass of hydrogen peroxide should result when 1.50 \(\mathrm{g}\) barium peroxide is treated with 88.0 \(\mathrm{mL}\) hydrochloric acid solution containing 0.0272 \(\mathrm{g} \mathrm{HCl}\) per mL? What mass of which reagent is left unreacted?

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