Problem 122 - Chemistry An Atoms First Approach

Question

Trace organic compounds in the atmosphere are first concentrated and then measured by gas chromatography. In the concentration step, several liters of air are pumped through a tube containing a porous substance that traps organic compounds. The tube is then connected to a gas chromatograph and heated to release the trapped compounds. The organic compounds are separated in the column and the amounts are measured. In an analysis for benzene and toluene in air, a \(3.00-\mathrm{L}\) sample of air at \(748\) torr and \(23^{\circ} \mathrm{C}\) was passed through the trap. The gas chromatography analysis showed that this air sample contained \(89.6\) ng benzene \(\left(\mathrm{C}_{6} \mathrm{H}_{6}\right)\) and \(153 \mathrm{ng}\) toluene \(\left(\mathrm{C}_{7} \mathrm{H}_{8}\right) .\) Calculate the mixing ratio (see Exercise 121 ) and number of molecules per cubic centimeter for both benzene and toluene.

Step-by-Step Solution

Answer

The mixing ratio for benzene is \(9.59 \cdot 10^{-9}\), and the mixing ratio for toluene is \(1.38 \cdot 10^{-8}\). The number of molecules per cubic centimeter of benzene is \(2.30 \cdot 10^{11} \, molecules/cm^3\), and the number of molecules per cubic centimeter for toluene is \(3.32 \cdot 10^{11} \, molecules/cm^3\).

Step 1: Calculate moles of air

First, we need to find the moles of air. We are given the volume V, the pressure P, and the temperature T. We can use the ideal gas law equation to find the moles of air: \(PV = nRT\) where: P = pressure, in atm V = volume, in L n = moles of air R = ideal gas constant, equal to 0.0821 \(L\cdot atm\cdot K^{-1} mol^{-1}\) T = temperature, in K First, convert pressure from torr to atm: \(748 \, torr * \frac{1 \, atm}{760 \, torr} = 0.9842 \, atm\) Next, convert temperature from Celsius to Kelvin: \(23^{\circ} C + 273.15 = 296.15 \, K\) Now, we can use the ideal gas law to calculate n: n = (P * V) / (R * T)

Step 2: Convert benzene and toluene amounts to moles

Given the mass of benzene and toluene, we can now find the moles of each compound using their respective molar masses: For benzene (\(C_6H_6\)), the molar mass is \(78.11 \, g/mol\). For toluene (\(C_7H_8\)), the molar mass is \(92.14 \, g/mol\). Convert the mass of each compound to moles: For benzene: \(n_{C_6H_6} = \frac{m_{C_6H_6}}{M_{C_6H_6}} = \frac{89.6 \cdot 10^{-9} g}{78.11 \, g/mol} = 1.15 \cdot 10^{-9} \, mol\) For toluene: \(n_{C_7H_8} = \frac{m_{C_7H_8}}{M_{C_7H_8}} = \frac{153 \cdot 10^{-9} g}{92.14 \, g/mol} = 1.66 \cdot 10^{-9} mol\)

Step 3: Calculate the mixing ratios

Now we can calculate the mixing ratios for benzene and toluene: For benzene: \(Mixing \, ratio_{C_6H_6} = \frac{n_{C_6H_6}}{n_{Air}} = \frac{1.15 \cdot 10^{-9} mol}{n_{Air}}\) For toluene: \(Mixing \, ratio_{C_7H_8} = \frac{n_{C_7H_8}}{n_{Air}} = \frac{1.66 \cdot 10^{-9} mol}{n_{Air}}\) To find the actual values, we need to calculate n using the ideal gas law: n = (P * V) / (R * T) = (0.9842 \, atm * 3.00 \, L) / (0.0821 * 296.15 \, K) = 0.1199 mol Now we can find the mixing ratios: For benzene: \(Mixing \, ratio_{C_6H_6} = \frac{1.15 \cdot 10^{-9} mol}{0.1199 \, mol} = 9.59 \cdot 10^{-9}\) For toluene: \(Mixing \, ratio_{C_7H_8} = \frac{1.66 \cdot 10^{-9} mol}{0.1199 \, mol} = 1.38 \cdot 10^{-8}\)

Step 4: Calculate the number of molecules per cubic centimeter

Finally, we need to find the number of molecules per cubic centimeter for benzene and toluene. We can use the following relation: \(Number \, of \, molecules = \frac{number \, of \, moles \cdot Avogadro's \, number}{volume \, in \, cm^3}\) For benzene: \(Number \, of \, molecules_{C_6H_6} = \frac{1.15 \cdot 10^{-9} mol \cdot 6.022 \cdot 10^{23} molecules/mol}{3000 \, cm^3} = 2.30 \cdot 10^{11} \, molecules/cm^3\) For toluene: \(Number \, of \, molecules_{C_7H_8} = \frac{1.66 \cdot 10^{-9} mol \cdot 6.022 \cdot 10^{23} molecules/mol}{3000 \, cm^3} = 3.32 \cdot 10^{11} \, molecules/cm^3\) The mixing ratio for benzene is \(9.59 \cdot 10^{-9}\), and the mixing ratio for toluene is \(1.38 \cdot 10^{-8}\). The number of molecules per cubic centimeter of benzene is \(2.30 \cdot 10^{11} \, molecules/cm^3\), and the number of molecules per cubic centimeter for toluene is \(3.32 \cdot 10^{11} \, molecules/cm^3\).

Key Concepts

Understanding the Ideal Gas Law

The ideal gas law, expressed as the equation PV = nRT, is a pivotal concept in chemistry that describes the behavior of an ideal gas under varying conditions of pressure (P), volume (V), temperature (T), and amount in moles (n). It hinges on the notion that these four properties are interrelated. For instance, as demonstrated in the exercise solution, the ideal gas law allows us to calculate the number of moles of air in a given volume at a known temperature and pressure. This is essential in atmospheric analysis, such as when assessing the concentration of pollutants using gas chromatography.
By converting known units of pressure to atmospheres (atm) and temperature to Kelvin (K), the ideal gas constant (R), which is 0.0821 L⋅atm⋅K⁻¹⋅mol⁻¹, facilitates the computation of moles of gas present. This step is fundamental for accurate mixing ratio calculations, ensuring precise measurements when analyzing atmospheric samples.

Calculating Mixing Ratios

The concept of a mixing ratio is integral in environmental science, particularly when quantifying the concentration of a substance within a mixture. The mixing ratio is defined as the amount of a particular compound relative to the total amount of air in a sample. This can be expressed using the mole ratio, which is the number of moles of the compound divided by the number of moles of air.

Applying the Mixing Ratio in Practice

In the context of atmospheric analysis through gas chromatography, precise mixing ratio calculations can reveal the level of pollutants, like benzene and toluene, within an air sample. After calculating the moles of air with the ideal gas law, as detailed in the original solution, the subsequent step is to determine the moles of the pollutants. The ratio of these respective moles then yields the mixing ratios, pivotal for ascertaining atmospheric pollution levels. Understanding this calculation fosters greater insight into environmental monitoring and the effectiveness of pollutant mitigation strategies.

Deciphering Avogadro's Number

Avogadro's number (6.022 × 10²³) is a fundamental constant that represents the number of atoms or molecules in one mole of a substance. This immense number plays a critical role in converting between the macroscopic scale of substances we can handle and measure and the microscopic scale of atoms and molecules.

Converting Moles to Molecules

When conducting atmospheric analyses with tools like gas chromatography, Avogadro's number allows us to transition from moles of a substance, which is a rather abstract concept, to something more tangible like the number of molecules present in a given volume. By multiplying the number of moles of a substance, for instance benzene or toluene as shown in the exercise, by Avogadro's number, we can estimate the number of individual molecules in any sample size, providing a microscopic view of chemical quantities.