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Chapter 5: Problem 93

At elevated temperatures, sodium chlorate decomposes to produce sodium chloride and oxygen gas. A \(0.8765\) -g sample of impure sodium chlorate was heated until the production of oxygen gas ceased. The oxygen gas collected over water occupied \(57.2 \mathrm{~mL}\) at a temperature of \(22^{\circ} \mathrm{C}\) and a pressure of 734 torr. Calculate the mass percent of \(\mathrm{NaClO}_{3}\) in the original sample. (At \(22^{\circ} \mathrm{C}\) the vapor pressure of water is \(19.8\) torr.)

Short Answer

Expert verified
The mass percent of sodium chlorate (\(\mathrm{NaClO}_{3}\)) in the original sample is \(18.45\%\).

Step by step solution

01

Calculate the moles of Oxygen gas produced

To determine the moles of oxygen gas (O₂) collected, we will use the ideal gas law formula: PV=nRT. First, we have to correct the pressure for the partial pressure of the vapor in the collected gas. Total Pressure = Pressure of Oxygen gas + Pressure of Water vapor Pressure of Oxygen gas = Total Pressure - Pressure of Water vapor \(P_{O_2} = 734\,\text{torr} - 19.8\,\text{torr} = 714.2\,\text{torr}\) Now, we will convert the pressure from torr to atm, the volume from mL to L, and the temperature from Celsius to Kelvin, to apply the ideal gas law correctly: \(P_{O_2} = \frac{714.2\,\text{torr}}{760\,\text{torr/atm}} = 0.9397\,\text{atm}\) \(V_{O_2} = 57.2\,\text{mL} \times \frac{1\,\text{L}}{1000\,\text{mL}} = 0.0572\,\text{L}\) \(T = 22+273.15 = 295.15\,\text{K}\) Using the ideal gas law formula (PV=nRT), we can calculate the moles of oxygen gas: \(n_{O_2} = \frac{P_{O_2}V_{O_2}}{RT} = \frac{0.9397\,\text{atm} \times 0.0572\,\text{L}}{0.08206\,\text{L atm/mol K} \times 295.15\,\text{K}} = 0.00228\,\text{mol}\)
02

Calculate the moles of Sodium Chlorate decomposed

Using the stoichiometry of the balanced chemical equation, we will determine the moles of sodium chlorate (NaClO₃) that decomposed: The balanced chemical equation is: \(2\,\text{NaClO}_3 \rightarrow 2\,\text{NaCl} + 3\,\text{O}_{2}\) From the equation, we can see that 2 moles of NaClO₃ yield 3 moles of O₂. So, we can set up the proportion: \(\frac{\text{moles of NaClO}_3}{\text{moles of O}_{2}} = \frac{2}{3}\) Now, we will use the moles of O₂ determined in step 1 to find the moles of NaClO₃: \(\text{moles of NaClO}_3 = \frac{2}{3} \times 0.00228\,\text{mol} = 0.00152\,\text{mol}\)
03

Calculate the mass percent of Sodium Chlorate in the sample

Now, we will calculate the mass of NaClO₃ decomposed using the molar mass of NaClO₃: Molar Mass of NaClO₃ = \(22.99 \,\text{g/mol (Na)} + 35.45\, \text{g/mol (Cl)} + 48.00\,\text{g/mol (3O)} = 106.44\,\text{g/mol}\) Mass of NaClO₃ decomposed = 0.00152 mol × 106.44 g/mol = 0.1618 g Finally, we will determine the mass percent of NaClO₃ in the original sample: Mass Percent of NaClO₃ = \(\frac{\text{Mass of NaClO}_3}{\text{Mass of the original sample}} \times 100\) Mass Percent of NaClO₃ = \(\frac{0.1618\,\text{g}}{0.8765\,\text{g}} \times 100 = 18.45\%\)

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Key Concepts

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

Ideal Gas Law
The ideal gas law is a fundamental equation in chemistry that relates the pressure, volume, temperature, and number of moles of a gas. It is represented by the formula:
\( PV = nRT \),
where
\( P \) is the pressure of the gas, \( V \) is the volume, \( n \) is the number of moles, \( R \) is the gas constant, and \( T \) is the temperature in Kelvin.

In the context of the exercise, the ideal gas law is used to calculate the moles of oxygen gas produced during the decomposition of sodium chlorate. By knowing the volume of gas collected and the temperature and pressure of the system, one can calculate the number of moles of gas using the corrected pressure for the vapor. This calculation is the first critical step in finding the mass percent composition of the substance in question.

The real-life application of the ideal gas law extends to fields like respiratory medicine, automotive industry, and environmental engineering, where understanding gas behavior under various conditions is crucial.
Molar Mass
Molar mass is the mass of one mole of a substance and it is typically expressed in grams per mole (g/mol). It is a bridge between the mass of a substance and the number of particles or moles depicted in a chemical equation.

To calculate the molar mass, one must sum the atomic masses of all the atoms in the molecule. For example, sodium chlorate (\( \text{NaClO}_3 \)) has a molar mass calculated by adding the atomic masses of sodium (Na), chlorine (Cl), and three oxygen atoms (O). In the given exercise, the molar mass of sodium chlorate is used to convert the number of moles into grams, which is crucial for calculating the mass percent composition of the sample.

Understanding molar mass allows chemists to count molecules by weighing them, a task that would be impossible by sheer enumeration due to their microscopic size and vast quantity.
Chemical Reaction
A chemical reaction involves the transformation of one set of chemical substances to another. During a chemical reaction, the bonds in the reactants break, and new bonds form in the products.

The given exercise involves the decomposition of sodium chlorate into sodium chloride and oxygen gas. The stoichiometry, or the quantitative relationship between reactants and products in a chemical reaction, is essential for calculations involving chemical reactions. In this case, the balanced chemical equation shows that two moles of sodium chlorate produce three moles of oxygen gas, which allows chemists to predict the amount of products formed from a given quantity of reactant. This balanced equation serves as a recipe for the chemical change that occurs, showing the proportional relationship between reactants and products.
Mass Percent Composition
Mass percent composition is a measurement that expresses the concentration of a particular element or compound within a mixture. It is defined as the mass of the component divided by the total mass of the mixture, multiplied by 100 to convert to a percentage.

In the task given, the mass percent composition is used to determine the purity of the sodium chlorate sample. By finding the mass of the compound that reacted (which is calculated using molar mass and stoichiometry from the number of moles), and dividing it by the mass of the original sample, students can calculate what percentage of the sample was sodium chlorate. The remainder could be impurities or other substances mixed with the targeted compound.

This concept is widely used in the pharmaceutical industry to assess drug purity, in food science for nutrient content, and any application where the composition of a mixture is of importance.

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Most popular questions from this chapter

An organic compound containing only C, H, and N yields the following data. i. Complete combustion of \(35.0 \mathrm{mg}\) of the compound produced \(33.5 \mathrm{mg} \mathrm{CO}_{2}\) and \(41.1 \mathrm{mg} \mathrm{H}_{2} \mathrm{O}\). ii. A \(65.2\) -mg sample of the compound was analyzed for nitrogen by the Dumas method (see Exercise 129), giving \(35.6 \mathrm{~mL}\) of dry \(\mathrm{N}_{2}\) at 740 . torr and \(25^{\circ} \mathrm{C}\). iii. The effusion rate of the compound as a gas was measured and found to be \(24.6 \mathrm{~mL} / \mathrm{min}\). The effusion rate of argon gas, under identical conditions, is \(26.4 \mathrm{~mL} / \mathrm{min}\). What is the molecular formula of the compound?

Natural gas is a mixture of hydrocarbons, primarily methane \(\left(\mathrm{CH}_{4}\right)\) and ethane \(\left(\mathrm{C}_{2} \mathrm{H}_{6}\right)\). A typical mixture might have \(\chi_{\text {methane }}=\) \(0.915\) and \(\chi_{\text {ethane }}=0.085\). What are the partial pressures of the two gases in a \(15.00\) - \(\mathrm{L}\) container of natural gas at \(20 .{ }^{\circ} \mathrm{C}\) and \(1.44\) atm? Assuming complete combustion of both gases in the natural gas sample, what is the total mass of water formed?

Methane \(\left(\mathrm{CH}_{4}\right)\) gas flows into a combustion chamber at a rate of 200. L/min at \(1.50 \mathrm{~atm}\) and ambient temperature. Air is added to the chamber at \(1.00 \mathrm{~atm}\) and the same temperature, and the gases are ignited. a. To ensure complete combustion of \(\mathrm{CH}_{4}\) to \(\mathrm{CO}_{2}(\mathrm{~g})\) and \(\mathrm{H}_{2} \mathrm{O}(g)\), three times as much oxygen as is necessary is reacted. Assuming air is 21 mole percent \(\mathrm{O}_{2}\) and \(79 \mathrm{~mole}\) percent \(\mathrm{N}_{2}\), calculate the flow rate of air necessary to deliver the required amount of oxygen. b. Under the conditions in part a, combustion of methane was not complete as a mixture of \(\mathrm{CO}_{2}(g)\) and \(\mathrm{CO}(g)\) was produced. It was determined that \(95.0 \%\) of the carbon in the exhaust gas was present in \(\mathrm{CO}_{2}\). The remainder was present as carbon in \(\mathrm{CO}\). Calculate the composition of the exhaust gas in terms of mole fraction of \(\mathrm{CO}, \mathrm{CO}_{2}, \mathrm{O}_{2}, \mathrm{~N}_{2}\), and \(\mathrm{H}_{2} \mathrm{O} .\) Assume \(\mathrm{CH}_{4}\) is completely reacted and \(\mathrm{N}_{2}\) is unreacted.

We state that the ideal gas law tends to hold best at low pressures and high temperatures. Show how the van der Waals equation simplifies to the ideal gas law under these conditions.

Consider the following reaction: $$ 4 \mathrm{Al}(s)+3 \mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{Al}_{2} \mathrm{O}_{3}(s) $$ It takes \(2.00 \mathrm{~L}\) of pure oxygen gas at STP to react completely with a certain sample of aluminum. What is the mass of aluminum reacted?
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