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Chapter 10: Problem 48

Rank the following gases from least dense to most dense at 1.00 atm and \(298 \mathrm{K} : \mathrm{SO}_{2}, \mathrm{HBr}, \mathrm{CO}_{2} .\)

Short Answer

Expert verified
The gases can be ranked in terms of density at 1.00 atm and 298 K as CO2 < SO2 < HBr, with CO2 being the least dense and HBr being the most dense. Their densities are: CO2: \(1.86 \frac{\text{g}}{\text{L}}\), SO2: \(2.63 \frac{\text{g}}{\text{L}}\), and HBr: \(3.31 \frac{\text{g}}{\text{L}}\).

Step by step solution

01

Determine the molar mass of each gas

Firstly, we need to find the molar mass of the given gases. The molar mass can be found by summing the atomic mass of each element in the gas molecule. For SO2: Molar mass = S + 2O = (1)(32.06 g/mol) + (2)(16.00 g/mol) = 64.06 g/mol For HBr: Molar mass = H + Br = (1)(1.01 g/mol) + (1)(79.90 g/mol) = 80.91 g/mol For CO2: Molar mass = C + 2O = (1)(12.01 g/mol) + (2)(16.00 g/mol) = 44.01 g/mol
02

Calculate the density of each gas using the ideal gas law

We will use the ideal gas law in the form of density formula: Density = \(\frac{PM}{RT}\) Where: P = pressure = 1.00 atm M = molar mass R = ideal gas constant = \(0.0821 \frac{\text{L atm}}{\text{mol K}}\) T = temperature = 298 K For SO2: Density = \(\frac{(1.00 \text{ atm})(64.06 \text{ g/mol})}{(0.0821 \frac{\text{L atm}}{\text{mol K}})(298\text{ K})} = 2.63 \frac{\text{g}}{\text{L}}\) For HBr: Density = \(\frac{(1.00\text{ atm})(80.91 \text{ g/mol})}{(0.0821 \frac{\text{L atm}}{\text{mol K}})(298\text{ K})} = 3.31 \frac{\text{g}}{\text{L}}\) For CO2: Density = \(\frac{(1.00\text{ atm})(44.01 \text{ g/mol})}{(0.0821 \frac{\text{L atm}}{\text{mol K}})(298\text{ K})} = 1.86 \frac{\text{g}}{\text{L}}\)
03

Rank the gases based on their densities

Now rank the gases as per their densities: - CO2: \(1.86 \frac{\text{g}}{\text{L}}\) - SO2: \(2.63 \frac{\text{g}}{\text{L}}\) - HBr: \(3.31 \frac{\text{g}}{\text{L}}\) So, in terms of densities at 1.00 atm and 298 K, the gases can be ranked as CO2 < SO2 < HBr, with CO2 being the least dense, and HBr being the most dense.

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Key Concepts

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

Understanding Molar Mass
Molar mass is a fundamental concept in chemistry and crucial for calculating gas densities. It represents the mass of one mole of a substance, typically expressed in grams per mole (g/mol). One mole of any substance contains exactly Avogadro's number of entities, which is approximately 6.022 x 1023 entities.

To calculate the molar mass of a compound, like the gases mentioned in the exercise, you'll need to sum the atomic masses of each element in a molecule, based on their molar proportions. For example, carbon dioxide (CO2) consists of one carbon atom and two oxygen atoms. If you know the atomic mass of carbon (12.01 g/mol) and oxygen (16.00 g/mol), you can easily calculate the molar mass of CO2 using simple arithmetic: Molar mass of CO2 = 1(12.01 g/mol) + 2(16.00 g/mol), which equals 44.01 g/mol.

  • Remember that the molar mass of an element is numerically equal to its atomic mass in grams per mole.
  • When compounds are involved, each atom's contribution depends on its atomic mass and the amount present in the molecular formula.
The Ideal Gas Law and Density Calculations
The ideal gas law is a critical equation in chemistry and physics, encapsulating the relationship among pressure (P), volume (V), number of moles (n), and temperature (T) for an ideal gas. Mathematically, it is given by PV = nRT, where R denotes the ideal gas constant.

To determine gas density using the ideal gas law, a rearranged form of the equation is used: Density = \(PM/RT\), wherein M represents the molar mass of the gas. It follows that the density is directly proportional to molar mass and pressure but inversely proportional to temperature. This direct relationship is why in the exercise, as molar mass increases, so does the density of the gas.

  • Always check the units when using the ideal gas law; units of pressure, volume, and temperature must be compatible with the gas constant's units being used.
  • The ideal gas law assumes that particles are point masses with no volume and that there are no intermolecular forces, which is an approximation but useful for a wide range of conditions.
Atomic Mass and Its Role in Molar Mass
Atomic mass, sometimes referred to as atomic weight, is the weighted average mass of an atom of an element based on the abundance of each of its isotopes. It is measured in atomic mass units (amu), where 1 amu is defined as 1/12th of the mass of a carbon-12 atom.

This concept is intimately tied to the calculation of molar mass. Since molar mass depends on the sum of the atomic masses of the atoms within a molecule, understanding both concepts is key for performing accurate calculations in chemistry. For instance, the atomic mass of sulfur (S) is 32.06 amu, and oxygen (O) is 16.00 amu; thus the molar mass of sulfur dioxide (SO2) is (1)(32.06 g/mol) + (2)(16.00 g/mol), or 64.06 g/mol.

  • Atomic mass allows chemists to predict the masses of different moles of substances and to balance chemical equations.
  • The use of atomic masses is particularly significant when determining the ratios of elements within a compound.

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Most popular questions from this chapter

Carbon dioxide, which is recognized as the major contributor to global warming as a "greenhouse gas," is formed when fossil fuels are combusted, as in electrical power plants fueled by coal, oil, or natural gas. One potential way to reduce the amount of \(\mathrm{CO}_{2}\) added to the atmosphere is to store it as a compressed gas in underground formations.Consider a 1000 -megawatt coal-fired power plant that produces about \(6 \times 10^{6}\) tons of \(\mathrm{CO}_{2}\) per year. (a) Assuming ideal-gas behavior, 1.00 atm, and \(27^{\circ} \mathrm{C},\) calculate the volume of \(\mathrm{CO}_{2}\) produced by this power plant. (b) If the \(\mathrm{CO}_{2}\) is stored underground as a liquid at \(10^{\circ} \mathrm{C}\) and 120 \(\mathrm{atm}\) and a density of \(1.2 \mathrm{g} / \mathrm{cm}^{3},\) what volume does it possess?(c) If it is stored underground as a gas at \(30^{\circ} \mathrm{C}\) and \(70 \mathrm{atm},\) what volume does it occupy?

Many gases are shipped in high-pressure containers. Consider a steel tank whose volume is 55.0 gallons that contains \(\mathrm{O}_{2}\) gas at a pressure of \(16,500 \mathrm{kPa}\) at \(23^{\circ} \mathrm{C}\) . (a) What mass of \(\mathrm{O}_{2}\) does the tank contain? (b) What volume would the gas occupy at STP? (c) At what temperature would the pressure in the tank equal 150.0 atm? (d) What would be the pressure of the gas, in kPa, if it were transferred to a container at \(24^{\circ} \mathrm{C}\) whose volume is 55.0 \(\mathrm{L}\) ?

A set of bookshelves rests on a hard floor surface on four legs, each having a cross-sectional dimension of \(3.0 \times 4.1 \mathrm{cm}\) in contact with the floor. The total mass of the shelves plus the books stacked on them is 262 kg. Calculate the pressure in pascals exerted by the shelf footings on the surface.

Calculate the pressure that \(\mathrm{CCl}_{4}\) will exert at \(80^{\circ} \mathrm{C}\) if 1.00 mol occupies \(33.3 \mathrm{L},\) assuming that (a) \(\mathrm{CCl}_{4}\) obeys the ideal-gas equation; (b) \(\mathrm{CCl}_{4}\) obeys the van der Waals equation. (Values for the van der Waals constants are given in Table \(10.3 .\) ) (c) Which would you expect to deviate more from ideal behavior under these conditions, \(\mathrm{Cl}_{2}\) or \(\mathrm{CCl}_{4}\) ? Explain.

A 15.0 -L tank is filled with helium gas at a pressure of \(1.00 \times 10^{2}\) atm. How many balloons (each 2.00 L) can be inflated to a pressure of 1.00 atm, assuming that the temperature remains constant and that the tank cannot be emptied below 1.00 atm?
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