Question

A group of students in an organic chemistry course performed a lab experiment in which they synthesized a colorful compound from not-especially-colorful starting materials. This synthesis required 3mol of the starting material, with formula C8H10NBr, for every mole of product desired; and the group started with 5.0g of starting material (all other reagents in excess). At the conclusion of their reaction, they had 8.0mL of solution, intensely colored due to the presence of their product.

The concentration of colorful compounds in solution is quite simple to determine experimentally. However, the concentration must be within a range that the instrumentation can accurately measure. The group's original solution was far too concentrated to give an accurate reading, so they diluted the sample by taking 500.0 microliters (mL) of the original solution and diluting them to 100.00mL. Unfortunately, this diluted solution was still too concentrated, so the group took a 1.00mL sample and diluted it further to 50.00mL. This finally resulted in a reliable reading. If the concentration of this final sample was determined to be 3.12 x 10^-6 M, what was the percent yield of the reaction?

Answer 1

The theoretical yield is calculated:

yield = 5 g * (1 mol / 200 g) * (1 mol product / 3 mol reagent) = 8.3E-3 mol

The concentration of the 100 mL solution is calculated:

C1 = 3.12E-6 M * 50 mL / 1 mL = 1.56E-4 M

The concentration of the original solution is calculated:

C2 = 1.56E-4 M * 100 mL / 0.5 mL = 0.0312 M

The moles are calculated in the original solution:

n = M * V = 0.0312 M * 0.008 L = 2.5E-4 mol

The percentage is calculated:

% yield = 2.5E-4 * 100 / 8.3E-3 = 3%

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