Question

In the spring of 1984, concern arose over the presence f ethylene dibromide, or EDB, in grains and cereals. EDB has the molecular formula ${\mathrm{C}}_2{\mathrm{H}}_4{\mathrm{Br}}_{2_{}}$and until 1984 was commonly used as a plant fumigant. The federal limit for EDB in finished cereal products is 30.0 parts per billion (ppb), where $1.0\mathrm{ppb}=1.0\times {10}^{-9}\mathrm{g}$ of EDB for every 1.0 g of sample. How many molecules of EDB are in 1.0 lb of flour if 30.0 ppb of EDB is present?

Short answer

The EDB molecules present are$4.36\times {10}^{16}\mathrm{molecules}$

Step-by-step solution

Definition of EDB

Ethylene dibromide is apotent mutagen that is used as an

• Industrial solvent,
• Gasoline lead scavenger,
• Fumigant.

Calculation of EDB mass

In 1 lb of flour,

$$\begin{gathered} \mathrm{m}=1\mathrm{lb}\times \frac{453.59\mathrm{g}}{1\mathrm{lb}} \\ \mathrm{m}=453.59\mathrm{g} \\ \mathrm{In}30\mathrm{ppb} \\ 30\mathrm{ppb}=\frac{30\times 453.59}{1\times {10}^9} \\ 30\mathrm{ppb}=1.36\times {10}^5\mathrm{gEDB} \end{gathered}$$

Hence, the EDB mass is$1.36\times {10}^5\mathrm{gEDB}.$

Calculation of molar mass

$$\begin{gathered} =2\times 12.011+4\times 1.008+2\times 79.9 \\ =187.854g/mol \end{gathered}$$

In EDB as

$$\begin{gathered} \mathrm{n}=1.35\times {10}^5\mathrm{g}\times \frac{1\mathrm{mol}}{187.854\mathrm{g}} \\ \mathrm{n}=7.24\times {10}^8\mathrm{mol} \end{gathered}$$

In EDB molecules

$$\begin{gathered} \mathrm{N}=7.24\times {10}^8\mathrm{g}\times \frac{6.022\times {10}^{23}\mathrm{molecules}}{1\mathrm{mol}} \\ \mathrm{N}=4.36\times {10}^{16}\mathrm{molecules} \end{gathered}$$

Hence, the EDB molar mass is$4.36\times {10}^{16}\mathrm{molecules}$.