Question #16ddb

1 Answer
Dec 14, 2015


#"3.16 g"#


Percent yield problems are all about how much product should be formed by the reaction versus how much product is actually formed.

As you know, percent yield is defined as the actual yield of the reaction, which tells you how much product is actually produced, divided by the theoretical yield of the reaction, which tells you how much product should be produced, and multiplied by #100#.

#color(blue)("% yield" = "actualy yield"/"theoretical yield" xx 100)#

Now, what does should be produced mean?

The theoretical yield of a reaction is determined by assuming that all the moles of reactants that take part in the reaction are converted to product.

Simply put, theoretical yield is what you get when the reaction has a #100%# yield.

Now, take a look at the balanced chemical equation

#"CH"_3"COOH"_text((l]) + "C"_5"H"_11"OH"_text((l]) -> "CH"_3"COOC"_5"H"_text(11(l]) + "H"_2"O"_text((l])#

Acetic acid, #"CH"_3"COOH"#, and isoamyl alcohol, #"C"_5"H"_1"OH"#, will react in a #1:1# mole ratio to produce isoamyl acetate, #"CH"_3"COOC"_5"H"_11# and water.

This tells you that the reaction will always consume equal numbers of moles of the two reactants.

Now, you need to determine if one of the two reactants will act as a limiting reagent. To do that, use their molar masses to determine how many moles you're mixing

#3.58 color(red)(cancel(color(black)("g"))) * ("1 mole CH"_3"COOH")/(60.05color(red)(cancel(color(black)("g")))) ="0.05962 moles CH"_3"COOH"#

#4.75 color(red)(cancel(color(black)("g"))) * ("1 mole C"_5"H"_11"OH")/(88.15 color(red)(cancel(color(black)("g")))) = "0.05389 moles C"_5"H"_11"OH"#

Notice that you have fewer moles of isoamyl alcohol than you have of acetic acid. Since the reaction consumes the reactants in a #1:1# mole ratio, the isoamyl alcohol will be completely consumed before the acetic acid is consumed.

In other words, isoamyl alcohol will be a limiting reagent, since it will bring the reaction to a halt before all the moles of acetic acid are consumed.

Now, isoamyl acetate is produced in a #1:1# mole ratio with the two reactants. Since isoamyl alcohol is a limiting reagent, it will determine how many moles of isoamyl acetate are produced.

Therefore, you can say that

#0.05389 color(red)(cancel(color(black)(" moles C"_5"H"_11"OH"))) * ("1 mole CH"_3"COOC"_5"H"_11)/(1color(red)(cancel(color(black)("mole C"_5"H"_11"OH")))) = "0.05389 moles CH"_3"COOC"_5"H"_11#

The reaction's theoretical yield will be determined by assuming that all these moles of product are actually formed - remember, theoretical yield is equivalent to #100%# yield!

Use isoamyl acetate's molar mass to find how many grams would contain this many moles

#0.05389 color(red)(cancel(color(black)("moles"))) * "130.19 g"/(1color(red)(cancel(color(black)("mole")))) = "7.016 g"#

However, you know that the reaction has a #45%# yield, which means that only #45%# of the number of moles of product that could be formed are actually formed.

This means that you can write

#45% = "actual yield"/"7.016 g" xx 100#

Rearrange to get

#"actual yield" = 45/100 * "7.016 g" = color (green)("3.16 g")#

The answer is rounded to three sig figs.