/*! 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 128 Nitrogen forms several gaseous o... [FREE SOLUTION] | 91Ó°ÊÓ

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Nitrogen forms several gaseous oxides. One of them has a density of \(1.33 \mathrm{~g} / \mathrm{L}\) measured at \(764 \mathrm{mmHg}\) and \(150^{\circ} \mathrm{C}\). Write the formula of the compound.

Short Answer

Expert verified
The formula of the nitrogen oxide is NO.

Step by step solution

01

Convert Temperature to Kelvin

The given temperature is in Celsius, and we need to convert it to Kelvin to use it in the ideal gas law. The conversion formula is: \[T(K) = T(°C) + 273.15\]Plugging in the values:\[T(K) = 150 + 273.15 = 423.15 \, K\]
02

Convert Pressure to Atmospheres

Convert the given pressure from mmHg to atmospheres since 1 atm = 760 mmHg. Use the conversion formula:\[P(atm) = \frac{P(mmHg)}{760}\]Plugging in the values:\[P(atm) = \frac{764}{760} = 1.0053 \, atm\]
03

Use Ideal Gas Law to Find Molar Mass

Apply the ideal gas law, \(PV = nRT\), where \(n = \frac{m}{M}\) and \(m\) is the mass and \(M\) is the molar mass. Rewriting gives \(M = \frac{dRT}{P}\), where \(d\) is the density.\[M = \frac{1.33 \, g/L \times 0.08206 \, L \, atm \, K^{-1} \, mol^{-1} \times 423.15 \, K}{1.0053 \, atm}\]Calculate \(M\):\[M \approx 30.02 \, g/mol\]
04

Identify Possible Molecular Formulas

Now that we have the molar mass of the compound, we check which known nitrogen oxides have a similar molar mass. Known oxides are NO (30.01 g/mol), N2O (44.01 g/mol), NO2 (46.01 g/mol), and others. The molar mass calculated matches closest to NO, which has a molar mass of 30.01 g/mol.

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Key Concepts

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

Density Calculation
When dealing with gases, understanding density is crucial for determining various properties like molar mass. Density is essentially a measure of how much mass is contained in a given volume. For gases, it's often expressed in units of grams per liter (g/L).

In the context of the given exercise, the density provided is 1.33 g/L. This figure signifies that each liter of the gas contains 1.33 grams.

Calculating density can help us apply other equations, such as the ideal gas law, to find unknowns, like the molar mass in our example. Knowing the density and using the relationship provided by the ideal gas law assists in further calculations which can unravel more about the gas's characteristics.

Therefore, mastering the basic concept of density and how it connects to gas laws is essential for fully understanding the behavior of gaseous substances.
Molar Mass Determination
The molar mass is a key characteristic in identifying any compound, especially for gases. It tells you the weight of one mole of a substance. This becomes important when analyzing gaseous compounds, which often have known molar masses that can be used for identification.

In our example, the molar mass was determined using the ideal gas law in the form rearranged for molar mass: \[ M = \frac{dRT}{P} \]This equation relates the molar mass \(M\) to the density \(d\) of the gas, the ideal gas constant \(R\), the temperature \(T\) in Kelvin, and the pressure \(P\) in atmospheres. By plugging in the values of these parameters, we found the molar mass to be approximately 30.02 g/mol.

Once you have the molar mass, you can compare it with the known molar masses of different nitrogen oxides to deduce which one is most likely present. Thus, understanding how to determine molar mass through calculation helps students make educated identifications of compounds.
Conversion of Units
Conversion of units is an often-overlooked step that can dramatically impact the calculations and results of an experiment. In gas calculations, it is essential to use consistent units.

For temperature, the conversion from Celsius to Kelvin is done by adding 273.15 to the Celsius value. This conversion is necessary because the ideal gas law requires temperature in Kelvin to account for absolute temperature scale.

Similarly, pressure must often be converted from mmHg to atmospheres in gas calculations since the ideal gas constant \(R\) is typically expressed in terms of liters, atmospheres, and Kelvin. This conversion is calculated as:

  • Using the formula \(P(atm) = \frac{P(mmHg)}{760}\), the pressure 764 mmHg becomes approximately 1.0053 atm.
By standardizing these units, errors are minimized, and calculations align with the universally accepted gas laws, leading to accurate results.
Nitrogen Oxides
Nitrogen oxides are a group of gases important in chemistry and the environment. They consist of nitrogen and oxygen in varying ratios, resulting in differing properties and molar masses.

Common nitrogen oxides include:
  • NO (Nitric oxide) with a molar mass of 30.01 g/mol
  • N2O (Nitrous oxide) with a molar mass of 44.01 g/mol
  • NO2 (Nitrogen dioxide) with a molar mass of 46.01 g/mol

Each of these gases has unique chemical properties and effects. For instance, NO is often a colorless gas, while NO2 is reddish-brown and more toxic. Nitrous oxide, known as "laughing gas," is used as an anesthetic.

In the given problem, our calculated molar mass of 30.02 g/mol closely matches that of NO, indicating that the gas in question is likely nitric oxide. Therefore, understanding these oxides' properties aids in accurately identifying them based on molar mass calculations.

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Most popular questions from this chapter

An unknown gas evolved from the fermentation of glucose is found to effuse through a porous barrier in 15.0 min. Under the same conditions of temperature and pressure, it takes an equal volume of \(\mathrm{N}_{2} 12.0 \mathrm{~min}\) to effuse through the same barrier. Calculate the molar mass of the unknown gas, and suggest what the gas might be.

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