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ASHA 777 [7]
2 years ago
7

Many esters have characteristic odors. Methyl salicylate exhibits a pleasant wintergreen odor. This ester can be prepared from s

alicylic acid and methanol via Fischer esterification. Which one of these reagents would you use in excess in order to drive the reaction to the product? Explain how you would isolate the ester product from the reaction mixture for full credit. (10 pts)
Chemistry
1 answer:
hodyreva [135]2 years ago
5 0

Answer:

Methanol would be used as a reagent in excess, since it is a very low-cost solvent. For product isolation, the first thing to do is remove the methanol through a distillation process. The residue produced can be dissolved in diethyl ether. Using a NaHCO₃ solution, extraction is performed. When it separates into two phases, the product will be in the ether and the reagent in the aqueous phase. The ether can also be removed by distillation, and at the end of this process you will have the product you want.

Explanation:

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Answer:

you have to divide it

Explanation:

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Which type of bond will most likely be found in HBr?
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At 25C the density of water is 0.997044 g/mL. Use this value to determine the percent error for the two density measurements
Gnom [1K]

Given that:

  • At 25C the density of water is 0.997044 g/mL.

From the information attached below, we have the following parameters.

The density of water calculation using a bottle.

     Initial volume of    Final volume of    Mass of water   Density (g/mL)

     burette (mL)        burette   (mL)       dispensed (g)

 

Sample 1      2.33                     7.34                   5.000               -----

Sample 2      7.34                    12.37                 5.025                -----

Sample 3      12.37                   18.50                6.112                  -----

Sample 4      18.50                  24.57               6.064                 -----

Sample 5     24.57                  31.31                6.720                  -----

The first thing we need to do is to determine the change in the volume of the burette in each sample from the above information.

  • The change in the volume of the burette = (final volume - the initial volume) mL

Sample 1:

= (7.34 - 2.33) mL

= 5.01 mL

Sample 2:

= (12.37 - 7.34) mL

= 5.03 mL

Sample 3:

= (18.50 - 12.37) mL

= 6.03 mL

Sample 4:

= (24.57 - 18.50) mL

= 6.07 mL

Sample 5:

= (31.31 - 24.57) mL

= 6.74 mL

The mass of the water dispersed in sample 1 is given as = 5.000 g

Using the relation for calculating the density of each, we have:

Sample 1

\mathbf{density = \dfrac{mass}{volume}}

\mathbf{density = \dfrac{5.01 g}{5.000 ml}}

density = 0.998004 g/ml

Sample 2:

\mathbf{density = \dfrac{5.025 g}{5.03ml}}

density = 0.999006 g/ml

Sample 3:

\mathbf{density = \dfrac{6.112 g}{6.13ml}}

density = 0.997064 g/ml

Sample 4:

\mathbf{density = \dfrac{6.064 \ g}{6.07 \ ml}}

density = 0.999012 g/ml

Sample 5:

\mathbf{density = \dfrac{6.720 \ g}{6.74 \ ml}}

density = 0.997033 g/ml

Thus, the average density for all the samples is:

\mathbf{= \dfrac{( 0.998004 + 0.999006 + 0.997064 +   0.999012  + 0.997033  )}{5}}

= 0.998024

∴

The percentage error for the two densities measurement is:

=\dfrac{ (experimental \  value -theoretical  \ value)\times 100 }{theoretical  \ value}

Given that the theoretical value = 0.997044 g/ml

Then;

\mathbf{= \dfrac{(0.998024 - 0.997044)100}{0.997044}}

= 0.0983%

Therefore, we can conclude that the percent error for the two density measurements is 0.0983%

Learn more about density here:

brainly.com/question/24386693?referrer=searchResults

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