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-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

01

Calculate moles of benzene and toluene

To calculate the moles of benzene and toluene, we'll first need to convert the given masses to moles using their respective molar masses. Benzene: \(\mathrm{C_{6}H_{6}}\), Molar mass = 78.11 g/mol Toluene: \(\mathrm{C_{7}H_{8}}\), Molar mass = 92.14 g/mol Convert the given mass in nanograms (ng) to grams. Benzene: \(89.6 \times 10^{-9}\) g Toluene: \(153 \times 10^{-9}\) g Now, calculate moles: Moles of benzene = \(\frac{89.6 \times 10^{-9} \ \text{g}}{78.11 \ \text{g/mol}}\) Moles of toluene = \(\frac{153 \times 10^{-9} \ \text{g}}{92.14 \ \text{g/mol}}\)

02

Calculate total moles of air and mole fraction of benzene and toluene

Use the Ideal Gas Law to find the total moles of air present in the sample, knowing volume, temperature, and pressure: \(PV = nRT\) where: P = pressure in atm (748 torr * \(1\,\text{atm}/760\,\text{torr}\)) V = volume in liters (3.00 L) n = moles of gas (unknown) R = gas constant (\(0.0821\, (\text{L.atm})/(\text{mol.K})\)) T = temperature in Kelvin (23 °C + 273.15 = 296.15 K) Rearrange the Ideal Gas Law to solve for moles (n): \(n = \frac{PV}{RT}\) Calculate the mole fraction of benzene and toluene as: Mole fraction of benzene = \(\frac{\text{moles of benzene}}{\text{total moles of air}}\) Mole fraction of toluene = \(\frac{\text{moles of toluene}}{\text{total moles of air}}\)

03

Calculate the mixing ratio for benzene and toluene

To calculate the mixing ratio, divide the mole fraction of benzene and toluene by the mole fraction of air. The mixing ratio is usually expressed in units of moles/mole or parts per million (ppm). Mixing ratio of benzene = \(\frac{\text{mole fraction of benzene}}{\text{mole fraction of air}} \% \, or\, ppm\) Mixing ratio of toluene = \(\frac{\text{mole fraction of toluene}}{\text{mole fraction of air}} \% \, or\, ppm\)

04

Calculate the number of molecules per cubic centimeter for benzene and toluene

To find the number of molecules, we'll use Avogadro's number (\(6.022 \times 10^{23}\) molecules/mol) and the mole fraction: Number of molecules of benzene per cubic centimeter = \(\text{mole fraction of benzene} \times \frac{6.022 \times 10^{23} \ \text{molecules/mol}}{1000 \ \text{cm}^{3}/\text{L}}\) Number of molecules of toluene per cubic centimeter = \(\text{mole fraction of toluene} \times \frac{6.022 \times 10^{23} \ \text{molecules/mol}}{1000 \ \text{cm}^{3}/\text{L}}\) Now, you can calculate the mixing ratio and number of molecules per cubic centimeter for both benzene and toluene using the steps outlined above.

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