the website for the reaction is http://www.chem.uiuc.edu/chem233smith/smithsylla

Question

the website for the reaction is http://www.chem.uiuc.edu/chem233smith/smithsyllabus/PhCOOMeNitragte.gif

1.Find the actual yield

2.percent yield

3.atom economy calculation for the following reaction using the given date in detailed steps, please.

Cmpd

Molecular

Weight

Density

(g/mL)

g or ml used

Nitric acid

63.01 g/mol

1.51 g/mL

2 ml

Sulfuric acid

98.07 g/mol

1.83 g/mL

5.8 ml

Methyl benzoate

136.14 g/mol

1.08 g/mL

2.5 ml

Cmpd

Molecular

Weight

Density

(g/mL)

g or ml used

Nitric acid

63.01 g/mol

1.51 g/mL

2 ml

Sulfuric acid

98.07 g/mol

1.83 g/mL

5.8 ml

Methyl benzoate

136.14 g/mol

1.08 g/mL

2.5 ml

Explanation / Answer

First of all we will find weight of all components.

Mass of Nitric acid = 2 * 1.51 = 3.02 gram

Mass of Sulfuric acid = 5.8 * 1.83 = 10.614 gram

Mass of methyl benzoate = 2.5 * 1.08 =2.7 gram

Now we find moles of each component

Moles of Nitric acid = 3.02 / 63.01 = 0.0479 mol

Moles of methyl benzoate = 2.7 / 136.14 = 0.0198 mol

Moles of sulfuric acid = 10.614 / 98.07 = 0.108 mol

Here the moles of methyl benzoate is less as compared to others so it is limiting component.

so moles methyl benzoate = 0.0198 moles
nitration with nitric acid in the presence of sulfuric acid gives methyl 3-nitrobenzoate. 1 mole gives 1 mole
molar mass .. add NO2 group [46] minus H removed, = 181g

so theoretical yield will be no of moles x molar mass = 0.0198 x 181 = 3.5838 g

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