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5.05a Methyl Salicylate is prepared

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化学

肯塔基大学

4.7（669 個評分） | 67K 名學生已註冊

This course is designed to cover subjects in advanced high school chemistry courses, correlating to the standard topics as established by the American Chemical Society. This course is a precursor to the Advanced Chemistry Coursera course. Areas that are covered include atomic structure, periodic trends, compounds, reactions and stoichiometry, bonding, and thermochemistry.

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4.7（669 個評分）

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Jun 21, 2017

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Dec 30, 2016

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從本節課中

Week 5

This unit will delve into the quantitative relationships we can determine from a balanced chemical equation to determine the relative amounts of substances needed to react or the amount of products formed. We will also explore limiting reagents and percent yield to address practical aspects of chemical reactions because there may be leftover reactants and/or incomplete formation of products.

5.01 Stoichiometry of Chemical Reactions: Mole to Mole 8:19

5.01a How many Moles of water are produced1:11

5.01b How many Moles of CO2 are produced 1:47

5.02 Stoichiometry of Chemical Reactions: Mass Relationships11:02

5.02a How many grams of magnesium are required 2:27

5.02b How many grams of magnesium do we need 4:34

5.03 Limiting and Excess Reagents11:18

5.04 Limiting and Excess Reagents: Calculations 7:57

5.04a How many grams of H2S can be formed 4:13

5.04b What mass of excess reactant will remain 6:31

5.05 Theoretical & Percent Yields 9:56

5.05a Methyl Salicylate is prepared5:48

5.05b A student needs 625 g of zinc sulfide5:32

5.06 Solution Concentration 15:16

5.06a Find the mass percentage for a solution prepared 1:56

5.06b Find the volume percentage for a solution prepared 1:51

5.06c Find the mass/volume percentage for a solution prepared 2:39

5.06d What is the molarity of a solution made 2:23

5.06e How many grams of LiCl do we need 2:44

5.07 Dilution of Solutions7:16

5.07a What is the concentration of the solution 1:45

5.08 Solution Stoichiometry 9:06

5.08a What volume of HBr is required 5:03

教學方

Dr. Allison Soult
Lecturer
Dr. Kim Woodrum
Sr. Lecturer

腳本

Methyl salicylate, which is the same as oil of wintergreen,

is prepared by heating salicylic acid with methanol.

And we're given the formula so

we can identify those species in the chemical reaction.

In an experiment,

1.5 grams of salicylic acid is reacted with 11.2 grams of methanol.

The yield of methyl salicylate is 1.31 grams.

What is the percent yield?

So we've got a lot of information given.

And we also have several substances mentioned that we're going to

need the molar masses for.

So we're actually going to go to the next page and

summarize all of this information so that we can then do our calculation.

So here we see our balance chemical equation.

We see the molar masses for each of our substances.

As well as the amounts of each substance, that was given in the problem.

Along with the actual yield.

Now, we're trying to find percent yield,

which means we need to know the actual yield and the theoretical yield.

And that's what we're going to have to actually calculate first.

Remember that we're dealing with the amounts of two different reactants, and

we don't know which one we're going to run out of first, so

we have to solve this as a limiting reagent problem.

Which means we're going to have to do two calculations.

First we're going to assume that C7H6O3 is limiting.

And that it's going to determine the maximum amount of product that can

be made, so we're going to say 1.50 grams divided by our molar mass,

138.12 grams per mole.

And again, I'm putting the grams on the bottom so

that the units will cancel out correctly.

I know that for every 1 mole of

C7H6O3 I'm going to form 1 mole of C8H8O3.

And I can go to the mass of that product by using the molar

mass of 152.15 grams per mole of C8H8O3.

So this is going to give me the maximal amount of product I can make if

this first reactant, our salicylic acid is the limiting reagent.

Now I can go thru and cancel out all my units.

So moles of each substance.

Moles.

Moles and I'm left with grams of my methyl salicylate, my product.

And when I do the calculation, what I find is that it can produce

1.65 grams of the C8H8O3.

Now I need to do a similar calculation.

This time though, I'm going to assume that our methanol,

CH3OH, is the limiting reagent.

And I'm going to see how much of that product I can make if the methanol is

limiting the amount of product, or that maximum yield of product.

So again I'm going to start with the amount of methanol given,

which is 11.2 grams.

I'm going to use the molar mass and again I'm going to put the grams on the bottom,

so that my units cancel out correctly.

Then I'm going to use the coefficients from my balanced chemical equation.

I see for every 1 mole of methanol I'm going to get 1 mole of my product C8H8O3.

And then I'm going to have to convert that back to grams, so

152.15 grams per mole of C8H8O3.

Now I can go through and cancel out my units, so grams of methanol, moles

of methanol, moles of methyl salicylate, and I get grams of methyl salicylate.

And when I do the calculation, I see that I get

53.3 grams of C8H8O3.

So that's actually not my limiting reagent.

And note this, because it produces the greater amount of product.

I can not simply look at the amount of reactants.

I have to see which one is producing the least amount of product.

And that substance is my limiting reagent.

Now, I know my limiting reagent is going to be my salicylic acid.

So this is my limiting reagent.

And that also tells me that this is the maximum amount I can produce.

So now I know my theoretical or my maximum yield and I can use that along with my

actual yield to figure out what the percent yield is for this reaction.

So now lets look at what we've got.

So we've got percent yield equals actual,

which was given in the original problem 1.31 grams,

over our theoretical yield, which we can calculate at 1.65 grams, times 100.

And we always want to use the smaller amount because once I run out of

my salicylic acid I can't make anymore of

the product regardless of how much methanol I have.

So I always have to look for the one that produces the least amount of product.

And now I can do the calculation and I see that my percent yield is

79.4% yield.

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