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enyata [817]
3 years ago
7

Predict the carboxylic acid product of the following reaction:CH3COOCH2CH3ethyl ethanoate+H2OH+⟶carboxylic acid+CH3CH2OHethanolE

xpress your answer using the IUPAC name.
Chemistry
1 answer:
GaryK [48]3 years ago
8 0

<u>Answer:</u> The IUPAC name of the carboxylic acid formed is ethanoic acid.

<u>Explanation:</u>

The basic rules for naming of organic compounds are :

  • First select the longest possible carbon chain.
  • The longest possible carbon chain should include the carbons of double or triple bonds.
  • The naming of alkane is done by adding the suffix -ane, alkene by adding the suffix -ene, alkyne by adding the suffix -yne and carboxylic acid by adding the suffix -oic acid.
  • The numbering is done in such a way that first carbon of double or triple bond gets the lowest number.
  • The carbon atoms of the double or triple bond get the preference over the other substituents present in the parent chain.

For the given chemical reaction, the equation follows:

CH_3COOCH_2CH_3+H_2O\rightarrow CH_3COOH+CH_3CH_2OH

When hydrolysis of ester takes place, it leads to the formation of an alcohol and a carboxylic acid.

So, when hydrolysis of ethyl ethanoate occurs, it produces ethanol and a 2- Carbon carboxylic acid named as ethanoic acid.

Hence, the IUPAC name of the carboxylic acid formed is ethanoic acid.

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Toluene (C6H5CH3 ), an organic compound often used as a solvent in paints, is mixed with a similar organic compound, benzene (C6
Sergeeva-Olga [200]

Explanation:

The given data is as follows.

 Weight of solute = 75.8 g,   Molecular weight of solute (toulene) = 92.13 g/mol,    volume = 200 ml

  • Therefore, molarity of toulene is calculated as follows.

      Molarity = \frac{\text{weight of solute}}{\text{molecular weight of solute}} \times \frac{1000}{\text{volume of solution in ml}}

                    = \frac{75.8 g}{92.13 g/mol} \times \frac{1000}{200 ml}

                    = 4.11 M

Hence, molarity of toulene is 4.11 M.

  • As molality is the number of moles of solute present in kg of solvent.

So, we will calculate the molality of toulene as follows.

   Molality = \frac{\text{given weight of solute}}{\text{given molecular weight of solute}} \times \frac{1000}{\text{weight of solvent in grams}}

             = \frac{75.8 g}{92.13 g/mol} \times \frac{1000}{95.6 g}

             = 8.6 m

Hence, molality of given toulene solution is 8.6 m.

  • Now, calculate the number of moles of toulene as follows.

       No. of moles = \frac{mass}{\text{molar mass}}

                             = \frac{75.8 g}{92.13 g/mol}

                             = 0.8227 mol

Now, no. of moles of benzene will be as follows.

     No. of moles = \frac{mass}{\text{molar mass}}

                             = \frac{95.6 g}{78.11 g/mol}

                             = 1.2239 mol

Hence, the mole fraction of toulene is as follows.

         Mole fraction = \frac{\text{moles of toulene}}{\text{total moles}}

                             = \frac{0.8227 mol}{(0.8227 + 1.2239) mol}

                             = 0.402

Hence, mole fraction of toulene is 0.402.

  • As density of given solution is 0.857 g/cm^{3} so, we will calculate the mass of solution as follows.

         Density = \frac{mass}{volume}

     0.857 g/cm^{3} = \frac{mass}{200 ml}      (As 1 cm^{3} = 1 g)

                      mass = 171.4 g

Therefore, calculate the mass percent of toulene as follows.

      Mass % = \frac{\text{mass of solute}}{\text{mass of solution}} \times 100

                   = \frac{75.8 g}{171.4 g} \times 100

                   = 44.22%

Therefore, mass percent of toulene is 44.22%.

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The step by step calculations is shown as attached below.

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gtnhenbr [62]
V ( HCl ) = 45.00 mL in liters : 45.00 / 1000 => 0.045 L

M ( HCl ) = ?

V ( NaOH ) = 25.00 / 1000 => 0.025 L 

M ( NaOH) = 0.2000 M

number of moles NaOH :

n = M x V = 0.2000 x 0.025 => 0.005 moles of NaOH

Mole ratio:

HCl + NaOH = NaCl + H2O

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moles HCl = 0.005 x 1 / 1

= 0.005 moles of HCl :

M ( HCl ) = n / V

M ( HCl ) = 0.005 / 0.045

= 0.1111 M

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4 0
3 years ago
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