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The nitrogen content of organic compounds can be determined by the Dumas method. The compound in question is first reacted by passage over hot \(\mathrm{CuO}(s)\) : $$ \text { Compound } \underset{\text { Cvoss }}{\stackrel{\text { Hot }}{\longrightarrow}} \mathrm{N}_{2}(g)+\mathrm{CO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(g) $$ The product gas is then passed through a concentrated solution of \(\mathrm{KOH}\) to remove the \(\mathrm{CO}_{2}\). After passage through the \(\mathrm{KOH}\) solution, the gas contains \(\mathrm{N}_{2}\) and is saturated with water vapor. In a given experiment a \(0.253-g\) sample of a compound produced \(31.8 \mathrm{~mL} \mathrm{~N}_{2}\) saturated with water vapor at \(25^{\circ} \mathrm{C}\) and 726 torr. What is the mass percent of nitrogen in the compound? (The vapor pressure of water at \(25^{\circ} \mathrm{C}\) is \(23.8\) torr.)

Short Answer

Expert verified
The mass percent of nitrogen in the compound is 14.8%.

Step by step solution

01

Calculate the partial pressure of nitrogen gas

To find the partial pressure of nitrogen gas, subtract the vapor pressure of water from the total pressure: Partial pressure of nitrogen gas (P_N2) = total pressure - vapor pressure of water = 726 torr - 23.8 torr = 702.2 torr
02

Convert pressure to atm and volume to L

For using the ideal gas law, we need the pressure in atm and the volume in L: P_N2 (atm) = P_N2 (torr) / 760 = 702.2 torr / 760 = 0.924 atm Volume of nitrogen gas (V_N2) = 31.8 mL / 1000 = 0.0318 L
03

Use the ideal gas law to find moles of nitrogen gas

Using the ideal gas law \(PV = nRT\), we can determine the moles of nitrogen gas (n_N2). Note that the gas constant R = 0.0821 L atm K⁻¹ mol⁻¹, and the temperature T = 25°C = 298 K. n_N2 = (P_N2 * V_N2) / (R * T) = (0.924 atm * 0.0318 L) / (0.0821 L atm K⁻¹ mol⁻¹ * 298 K) = 0.00134 mol
04

Calculate the mass of nitrogen

To find the mass of nitrogen in the compound, multiply the moles by the molar mass of nitrogen (N₂ = 28.02 g/mol): Mass of nitrogen = n_N2 * M_N2 = 0.00134 mol * 28.02 g/mol = 0.0375 g
05

Calculate the mass percent of nitrogen

Finally, find the mass percent of nitrogen in the compound by dividing the mass of nitrogen by the mass of the compound and multiplying by 100: Mass percent of nitrogen = (Mass of nitrogen / mass of the compound) * 100 = (0.0375 g / 0.253 g) * 100 = 14.8% The mass percent of nitrogen in the compound is 14.8%.

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Key Concepts

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

Nitrogen Content
In chemistry, determining the nitrogen content of a compound is essential for understanding its composition. This is particularly important in fields such as agriculture and pharmaceuticals. The Dumas method is a classical technique used to achieve this. By analyzing the nitrogen gas (\( \text{N}_2 \)) released during the reaction, we can estimate the amount of nitrogen in the original sample. This method involves heating the compound with copper oxide (\( \mathrm{CuO} \)), which results in the production of nitrogen gas. Through this process, the nitrogen in the compound is converted into a measurable quantity of nitrogen gas, allowing for a precise determination of its content. In the discussed exercise, the objective was to use the nitrogen gas collected to calculate the mass percent of nitrogen in the compound.
Ideal Gas Law
The ideal gas law is a powerful equation that relates the pressure, volume, temperature, and number of moles of a gas. It is expressed as \( PV = nRT \), where \( P \)is the pressure, \( V \)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, it was used to determine the amount of nitrogen gas by calculating its moles using the conditions of volume and pressure given.
  • Convert pressure from torr to atm as the constant \( R \)is in terms of atm.
  • Convert temperature from degrees Celsius to Kelvin by adding 273 to the Celsius temperature.
  • Ensure the volume is in liters to match the units of \( R \)
    • Using these conversions ensures that the ideal gas law provides an accurate calculation of the moles of nitrogen gas, pivotal for calculating the mass percent.
Partial Pressure of Gases
In mixed gases, each component exerts its own pressure known as partial pressure. This is especially crucial when collecting gases over water, as water vapor will always contribute to the total pressure. In this exercise, to find the partial pressure of nitrogen gas, you subtract the vapor pressure of water from the total gas pressure measured in the system. This adjustment provides an accurate measure of just the pressure exerted by the nitrogen gas.
  • Total pressure must account for all gases present including water vapor.
  • To isolate nitrogen, subtract the vapor pressure of water from the total pressure recorded.
    • Recognizing the partial pressures helps in precisely applying the ideal gas law to determine the moles of nitrogen produced. Without correcting for this, the calculations may significantly overestimate the actual amount of nitrogen.
Mass Percent Calculation
The mass percent of an element within a compound is a useful way to express its concentration. It is calculated by dividing the mass of the element by the total mass of the compound, and then multiplying by 100. This measure indicates what portion of the compound's mass comes from that element.
  • Calculate the mass of the nitrogen using its moles and molar mass.
  • Divide the nitrogen mass by the total mass of the sample.
  • Multiply by 100 to convert to percentage form.
    • For the exercise, this calculation reveals that nitrogen constitutes 14.8% of the compound's mass. Understanding this percentage is fundamental for chemists to ascertain the nutritional and reactive properties of the compound.

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Most popular questions from this chapter

Methane \(\left(\mathrm{CH}_{4}\right)\) gas flows into a combustion chamber at a rate of 200\. \(\mathrm{L} / \mathrm{min}\) 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 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 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.

You have a helium balloon at \(1.00 \mathrm{~atm}\) and \(25^{\circ} \mathrm{C}\). You want to make a hot-air balloon with the same volume and same lift as the helium balloon. Assume air is \(79.0 \%\) nitrogen and \(21.0 \%\) oxygen by volume. The "lift" of a balloon is given by the difference between the mass of air displaced by the balloon and the mass of gas inside the balloon. a. Will the temperature in the hot-air balloon have to be higher or lower than \(25^{\circ} \mathrm{C}\) ? Explain. b. Calculate the temperature of the air required for the hot-air balloon to provide the same lift as the helium balloon at \(1.00\) atm and \(25^{\circ} \mathrm{C}\). Assume atmospheric conditions are \(1.00 \mathrm{~atm}\) and \(25^{\circ} \mathrm{C}\).

Consider two different containers, each filled with 2 moles of \(\mathrm{Ne}(\mathrm{g}) .\) One of the containers is rigid and has constant volume. The other container is flexible (like a balloon) and is capable of changing its volume to keep the external pressure and internal pressure equal to each other. If you raise the temperature in both containers, what happens to the pressure and density of the gas inside each container? Assume a constant external pressure.

Hydrogen azide, \(\mathrm{HN}_{3}\), decomposes on heating by the following unbalanced reaction: $$ \mathrm{HN}_{3}(g) \longrightarrow \mathrm{N}_{2}(g)+\mathrm{H}_{2}(g) $$ If \(3.0 \mathrm{~atm}\) of pure \(\mathrm{HN}_{3}(\mathrm{~g})\) is decomposed initially, what is the final total pressure in the reaction container? What are the partial pressures of nitrogen and hydrogen gas? Assume the volume and temperature of the reaction container are constant.

Write reactions to show how nitric and sulfuric acids are produced in the atmosphere.

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