A student carried out the experiment and recovered 0.345g of benzoic acid and 0.157 g of...
A student carried out the experiment and recovered 0.345g of benzoic acid and 0.157 g of isoamyl alcohol. Calculate the % yield recovery of both compounds. show work. The mixture that you used in class has a total composition of 12g of Benzoic acid dissolved in 60.0ml of isoamyl alcohol and 60.0ml of of MTBE.
Homework Answers
Given starting materials:
Benzoic acid: 12g
60.0mL isoamyl alcohol
Mass of isoamyl alcohol = 0.809g/mL * 60.0mL = 48.54g
% recovery of benzoic acid = (recovered mass /
starting mass)*100 = (0.345g/12g)*100 =
2.88%
% recovery of isoamyl alcohol = (recovered mass /
starting mass)*100 = (0.157g/48.54g)*100 =
0.32%
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A student carried out the experiment and recovered 0.345g of
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