Question

Isopentyl acetate (C₇H₁₄O₂) is the compound responsible for the scent of bananas. Interestingly, bees release about 1-micro-gram of this compound when this sting. The resulting scent attracts other bees to join the attack.

How many molecules of isopropyl acetate are released in typical bee sting?

Answer 1

Approximately 4.6 x 10^15 molecules of isopentyl acetate are released in a typical bee sting, which can be calculated using the molar mass of isopentyl acetate and Avogadro's number.

Step-by-step explanation:

The typical bee sting releases about 1 microgram of isopentyl acetate, which is the compound responsible for the scent of bananas. To calculate the number of molecules, we utilize Avogadro's number (6.022 x 1023 molecules/mole). First, we need to determine the molar mass of isopentyl acetate (C7H14O2), which is 130.19 g/mol. Then, we convert the mass of isopentyl acetate released from the sting from micrograms to moles:

1 microgram = 1 x 10-6 grams,
Number of moles = mass (in grams) / molar mass,
Number of moles = 1 x 10-6 g / 130.19 g/mol.

Then to find the number of molecules, we multiply the number of moles by Avogadro's number:

Number of molecules = number of moles x Avogadro's number.

By carrying out the calculations, you can find that a bee sting releases approximately 4.6 x 1015 molecules of isopentyl acetate.