# Question a465b

Mar 6, 2016

"% yield" = 18%

#### Explanation:

You're dealing with the oxidation of benzyl alcohol, which is the common name given to phnehymethanol, $\text{C"_6"H"_5"CH"_2"OH}$, to benzoic acid, $\text{C"_6"H"_5"COOH}$, in the presence of potassium permanganate, ${\text{KMnO}}_{4}$.

The permanganate anion, ${\text{MnO}}_{4}^{-}$, will oxidize the benzyl alcohol to benzoic acid and be reduced to manganese dioxide, ${\text{MnO}}_{2}$, in the process.

Now, the reaction will also produce water and aqueous potassium hydroxide, but these products are not important here. All you really need to know is that benzyl alcohol and benzoic acid have a $1 : 1$ mole ratio in this reaction.

The reaction can be written as

${\text{C"_6"H"_5"CH"_2"OH"_text((aq]) stackrel(color(red)("KMnO"_4)color(white)(aaa))(->) "C"_6"H"_5"COOH}}_{\textrm{\left(a q\right]}}$

So, every mole of benzyl alcohol that takes part in the reaction will theoretically produce one mole of benzoic acid.

Use the molar mass of benzyl alcohol to determine how many moles you have in that $\text{1.05-g}$ sample

1.05color(red)(cancel(color(black)("g"))) * ("1 mole benzyl alc.")/(108.14color(red)(cancel(color(black)("g")))) = "0.00971 moles benzyl alc."

Theoretically, the reaction should produce $0.00971$ moles of benzoic acid. Use its molar mass to figure out how many grams that would amount to

0.00971 color(red)(cancel(color(black)("moles benzoic acd."))) * "122.12 g"/(1color(red)(cancel(color(black)("mole benzoic acd.")))) = "1.186 g benzoic acd."

This is how much benzoic acid is produced by a reaction that has a 100% yield.

However, you know that you collected $\text{0.21 g}$ of benzoic acid. This represents the actual yield of the reaction, i.e. what you actually collect after the reaction is completed.

Percent yield is defined as

$\textcolor{b l u e}{| \overline{\underline{\text{% yield" = "what you actually get"/"what you should theoretically get} \times 100}} |}$

Plug in your values to get

$\text{% yield" = (0.21 color(red)(cancel(color(black)("g"))))/(1.186color(red)(cancel(color(black)("g")))) xx 100 = "17.71%}$

You need to round this off to two sig figs, the number of sig figs you have for the mass of benzoic acid

color(green)(|bar(ul(color(white)(a/a)"% yield" = 18%color(white)(a/a)))|)#