In the year 2000 , the Guinness Book of World Records called ethyl mercaptan, \(\mathrm{C}_{2} \mathrm{H}_{6} \mathrm{S}\), the smelliest substance known. The average person can detect its presence in air at levels as low as \(9 \times 10^{-4} \mu \mathrm{mol} / \mathrm{m}^{3} .\) Express the limit of detectability of ethyl mercaptan in parts per billion (ppb). (Note: 1 ppb \(\mathrm{C}_{2} \mathrm{H}_{6} \mathrm{S}\) means there is \(1 \mathrm{g}\) \(\mathrm{C}_{2} \mathrm{H}_{6} \mathrm{S}\) per billion grams of air.) The density of air is approximately \(1.2 \mathrm{g} / \mathrm{L}\) at room temperature.

Short Answer

Expert verified
The limit of detectability of ethyl mercaptan in air is approximately 67.4 ppb

Step by step solution

01

Determine Molar Mass of Ethyl Mercaptan

Ethyl mercaptan is composed of Carbon (C), Hydrogen (H) and Sulfur (S). The molar masses of these elements are about 12 g/mol, 1g/mol and 32g/mol respectively. Therefore, the molar mass of ethyl mercaptan, \(C_{2} H_{6} S\), is calculated as \(2(12g/mol) + 6(1g/mol) + 32g/mol = 62 g/mol \)
02

Convert Micro-moles to Grams

To convert from moles to grams, we multiply the quantity in moles by molar mass. Given that ethyl mercaptan can be detected at \(9 × 10^{-4} µmol/m^{3}\), subtract 6 from the exponent of 10 to convert µmol to mol, and multiply by molar mass to convert to grams. Hence the detectability limit of ethyl mercaptan in air is \( (9 × 10^{-4} mol/m^{3}) × (62 g/mol) = 0.0558 g/m^{3} \)
03

Convert Cubic Meters to Liters

To convert \(m^{3}\) to liters (L), we use the equivalence \(1 m^{3} = 1000 L\). Therefore, \(0.0558 g/m^{3} = 0.0558 g/1000 L = 5.58 × 10^{-5} g/L \)
04

Convert Grams per Liter to Parts per Billion

Understanding that 1 ppb means 1g of substance per billion grams of air, to convert \(g/L\) (grams of ethyl mercaptan per Liter of air) into ppb, we multiply \(5.58 × 10^{-5} g/L\) by the density of air \(1.2 g/L\). Therefore our answer is \(5.58 × 10^{-5} g/L * 1 billion L/billion g = 67.4 ppb\)

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