Isopentyl acetate (c7h14o2), the compound responsible for the scent of bananas, can be produced commercially. interestingly, bees release about 1μg (1 × 10-6 g) of this compound when they sting. the resulting scent attracts other bees to join the attack. (a)calculate the number of molecules of isopentyl acetate released in a typical bee sting. (b)calculate the number of carbon atoms present.

1 μg of C₇H₁₄O₂ contains 4.62 × 10¹⁵ molecules and there are 3.23 × 10¹⁶ atoms of carbon present.

First, we want to calculate the number of molecules in 1 μg (1 × 10⁻⁶ g) of C₇H₁₄O₂. We will use the following relationships.

The molar ratio of C₇H₁₄O₂ to C is 1:7. The number of atoms of C in 4.62 × 10¹⁵ molecules of C₇H₁₄O₂ is:

1 μg of C₇H₁₄O₂ contains 4.62 × 10¹⁵ molecules and there are 3.23 × 10¹⁶ atoms of carbon present.

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The answer to your question is:

a) 4.64 x 10 ¹⁵ molecules

b) 9.28 x 10 ¹⁵ atoms of O2

Explanation:

MW C7H14O2 = 84 + 14 + 32 = 130 g

a)        130 g of C7H14O2 1 mol of C7H14O2

           1 x 10 ⁻⁶ g                    x

           x = 7.7 x 10 ⁻⁹ mol

          1 mol of C7H14O2     6 .023 x 10 ²³ molecules

          7.7 x 10⁻⁹ mol              x

          x = 4.64 x 10¹⁵ molecules

b)      130 g of C7H14O2     1 mol of C7H14O2

         1 x 10⁻⁶  C7H14O2       x

         x = 7.7 x 10 ⁻⁹ mol of C7H14O2

        1 mol of C7H14O2       2 mol of O2

        7.7 x 10 ⁻⁹                   x

         x = 1.54 x 10⁻⁸ mol of O2

       1 mol of O2  6.023 x 10 ²³ atoms

       1.54 x 10 ⁻⁸     x

        x = 9.28 x 10 ¹⁵ atoms of O2

the answer is 0.42 is that is an option. this question provides no info