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Chapter 9: Problem 94

A \(0.4230-g\) sample of impure sodium nitrate (contains sodium nitrate plus inert ingredients) was heated, converting all the sodium nitrate to \(0.2339 \mathrm{~g}\) of sodium nitrite and oxygen gas. Determine the percent of sodium nitrate in the original sample.

Short Answer

Expert verified
The percent of sodium nitrate in the original impure sample is approximately \(68.1\%\).

Step by step solution

01

Write the balanced chemical equation for the reaction

Sodium nitrate decomposes upon heating to form sodium nitrite and oxygen gas according to the following balanced equation: \[2 \mathrm{NaNO}_3 (s) \rightarrow 2 \mathrm{NaNO}_2 (s) + \mathrm{O}_2 (g)\] This equation will be used for the stoichiometry calculations.
02

Find the number of moles of sodium nitrite formed

To find the number of moles of sodium nitrite (\(\mathrm{NaNO}_2\)), we will use its molar mass: Molar mass of \(\mathrm{NaNO}_2\) = \(22.99\;\text{g/mol (Na)} + 14.01\;\text{g/mol (N)} + 16.00\;\text{g/mol (O)} \times 2 = 69.00\;\text{g/mol}\) Now, we will find the number of moles of sodium nitrite (\(n_{\mathrm{NaNO}_2}\)): \(n_{\mathrm{NaNO}_2}= \frac{m_{\mathrm{NaNO}_2}}{\text{Molar mass of}\; \mathrm{NaNO}_2} = \frac{0.2339\;\text{g}}{69.00\;\text{g/mol}}\) \(n_{\mathrm{NaNO}_2} \approx 0.00339\;\text{mol}\)
03

Calculate the moles of sodium nitrate present in the sample

According to the balanced chemical equation, 2 moles of sodium nitrate react to produce 2 moles of sodium nitrite. Therefore, the moles of sodium nitrate (\(n_{\mathrm{NaNO}_3}\)) present in the sample initially will be the same as the moles of sodium nitrite formed: \(n_{\mathrm{NaNO}_3}= n_{\mathrm{NaNO}_2} \approx 0.00339\;\text{mol}\)
04

Convert the moles of sodium nitrate into mass

First, we need to find the molar mass of sodium nitrate (\(\mathrm{NaNO}_3\)): Molar mass of \(\mathrm{NaNO}_3\) = \(22.99\;\text{g/mol (Na)} + 14.01\;\text{g/mol (N)} + 16.00\;\text{g/mol (O)} \times 3 = 85.00\;\text{g/mol}\) Now, we will convert the moles of sodium nitrate back into mass: \(m_{\mathrm{NaNO}_3} = n_{\mathrm{NaNO}_3} \cdot \text{Molar mass of}\; \mathrm{NaNO}_3 \approx 0.00339\;\text{mol} \times 85.00\;\text{g/mol}\) \(m_{\mathrm{NaNO}_3} \approx 0.288\;\text{g}\)
05

Calculate the percentage of sodium nitrate in the original sample

Finally, we will find the percent of sodium nitrate in the original impure sample: \(\text{Percent of sodium nitrate} = \frac{\text{Mass of sodium nitrate}}{\text{Mass of original sample}} \times 100\%\) \(\text{Percent of sodium nitrate} = \frac{0.288\;\mathrm{g}}{0.423\mathrm{g}} \times 100\% \approx 68.1\%\) Therefore, the percent of sodium nitrate in the original sample is approximately \(68.1\%\).

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Key Concepts

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

Chemical Reactions
Understanding chemical reactions is crucial when solving stoichiometry problems, like the decomposition of sodium nitrate. A chemical reaction involves the transformation of one or more substances into different substances. This happens when atoms rearrange, forming new bonds. Chemical reactions can be represented by balanced chemical equations.

In this particular example, we have sodium nitrate (\(\mathrm{NaNO}_3\)) decomposing into sodium nitrite (\(\mathrm{NaNO}_2\)) and oxygen gas (\(\mathrm{O}_2\)). To solve this kind of problem, first write out the balanced equation: \[2 \mathrm{NaNO}_3 (s) \rightarrow 2 \mathrm{NaNO}_2 (s) + \mathrm{O}_2 (g)\]
This equation shows that two units of sodium nitrate produce two units of sodium nitrite and one of oxygen gas. Balancing the chemical equation is essential as it ensures that mass is conserved and that the number of each type of atom is equal on both sides of the equation.

Properly understanding chemical reactions and their equations is necessary for effectively calculating the amounts of reactants and products, ensuring you get accurate stoichiometry results.
Molar Mass
Molar mass is a key concept in chemical calculations. It represents the mass of one mole of a given substance and is expressed in units of grams per mole (g/mol). Knowing the molar mass allows us to convert between the mass of a substance and the amount of substance in moles.

To determine the molar mass of a compound, you sum up the atomic masses of all the atoms present in its molecular formula. For sodium nitrite (\(\mathrm{NaNO}_2\)), its molar mass is calculated by adding the atomic masses of sodium (Na), nitrogen (N), and oxygen (O). That is \(22.99\ \text{g/mol (Na)} + 14.01\ \text{g/mol (N)} + 16.00 \ \text{g/mol (O)} \times 2 = 69.00\ \text{g/mol}\).

The understanding of molar mass allows us to move between amounts in grams and moles, which forms the basis for stoichiometric calculations in chemical reactions.
Chemical Equations
Chemical equations are symbolic representations of chemical reactions. They provide information about the reactants and products involved, as well as their relative proportions. This is evident through the coefficients, which are the numbers placed before compounds.

The task here involves the decomposition of sodium nitrate, represented by \[2 \mathrm{NaNO}_3 (s) \rightarrow 2 \mathrm{NaNO}_2 (s) + \mathrm{O}_2 (g)\].This equation is balanced, meaning both sides have equal numbers of each type of atom, maintaining the conservation of mass. The coefficients also give us a ratio of moles: 2 moles of sodium nitrate produce 2 moles of sodium nitrite and 1 mole of oxygen gas.

Balancing a chemical equation is crucial because it ensures our calculations in a stoichiometric problem are accurate. By understanding the role of chemical equations, students can more easily determine how much reactant is needed or how much product can be made.
Percent Composition
Percent composition tells us the percentage by mass of each element in a compound or the proportion of a chemical within a mixture. In the problem given, we determine the percentage of sodium nitrate in an impure sample by finding its mass contribution to the total sample.

To find percent composition, divide the mass of the component (sodium nitrate) by the total mass of the sample and multiply by 100. For example, if the mass of sodium nitrate is \(0.288\ \mathrm{g}\) and the mass of the original sample is \(0.423\ \mathrm{g}\), the percent composition is calculated as follows:\[\text{Percent of sodium nitrate} = \frac{0.288\ \mathrm{g}}{0.423\ \mathrm{g}} \times 100\% \approx 68.1\%\].

This shows that the original sample contains approximately 68.1% sodium nitrate by mass. Understanding percent composition is beneficial in fields such as chemical analysis and material characterization.

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Most popular questions from this chapter

Explain how one determines which reactant in a process is the limiting reactant. Does this depend only on the masses of the reactant present? Give an example of how to determine the limiting reactant by using a Before-Change- After (BCA) table with a balanced chemical cquation and reactant starting amounts.

One process for the commercial production of baking soda (sodium hydrogen carbonate) involves the following reaction, in which the carbon dioxide is used in its solid form ("dry ice" enough for the sodium hydrogen carbonate to precipitate: $$ \mathrm{NaCl}(a q)+\mathrm{NH}_{3}(a q)+\mathrm{H}_{2} \mathrm{O}(l)+\mathrm{CO}_{2}(s) \rightarrow \mathrm{NH}_{4} \mathrm{Cl}(a q)+\mathrm{NaHCO}_{3}(s) $$ Because they are relatively cheap, sodium chloride and water are typically present in excess. What is the expected yield of \(\mathrm{NaHCO}_{3}\) when one performs such a synthesis using \(10.0 \mathrm{~g}\) of ammonia and \(15.0 \mathrm{~g}\) of dry ice, with an excess of \(\mathrm{NaCl}\) and water?

Using the average atomic masses given inside the front cover of the text, calculate the mass in grams of cach of the following samples. a. 5.0 moles of nitric acid b. 0.000305 mole of mercury c. \(2.31 \times 10^{-5}\) mole of potassium chromate d. 10.5 moles of aluminum chloride e. \(4.9 \times 10^{4}\) moles of sulfur hexafluoride f. 125 moles of ammonia g. 0.01205 mole of sodium peroxide

When clemental copper is strongly heated with sulfur, a mixture of \(\mathrm{CuS}\) and \(\mathrm{Cu}_{2} \mathrm{~S}\) is produced, with CuS predominating. $$ \begin{array}{c} \mathrm{Cu}(s)+\mathrm{S}(s) \rightarrow \mathrm{CuS}(s) \\ 2 \mathrm{Cu}(s)+\mathrm{S}(s) \rightarrow \mathrm{Cu}_{2} \mathrm{~S}(s) \end{array} $$ What is the theoretical yield of CuS when \(31.8 \mathrm{~g}\) of \(\mathrm{Cu}(s)\) is heated with \(50.0 \mathrm{~g}\) of \(\mathrm{S} ?\) (Assume only CuS is produced in the reaction.) What is the percent yield of CuS if only 40.0 g of CuS can be isolated from the mixture?

Which of the following statements is(are) true? a. A balanced equation relates the numbers of molecules of reactants and products (or numbers of moles of reactants and products). b. To convert between moles of reactants and moles of products, we use mole ratios derived from the balanced equation. c. Often reactants are not mixed in stoichiometric quantities (they do not "run out" at the same time). In that case, we must use the limiting reactant to calculate the amounts of products formed. d. When a chemical reaction occurs, it must follow the law of conservation of mass.
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