Question

The concentration of ethyl acetate in an alcoholic solution was determined by diluting a 10.0ml sample to 100.00ml. A 20.00ml portion of the diluted solution was refluxed with 40.00ml of 0.04672M KOH: C₄H₈O₂ + OH⁻ → C2H3O₂- + C₂H₅OH. After cooling, the excess OH⁻ was back-titrated with 3.41ml of 0.05042M H₂SO₄. Calculate the amount of ethyl acetate (88.11g/mol) in the original sample in grams.

Answer 1

After diluting, reacting with KOH, and back-titrating with H2SO4, the original concentration of ethyl acetate is determined via stoichiometry and the mass is then found by multiplying the moles of ethyl acetate by its molar mass of 88.11g/mol.

Step-by-step explanation:

To calculate the amount of ethyl acetate in the original sample, we first need to understand the process and reactions involved. The solution was diluted, reacted with KOH, and then back-titrated with H2SO4. The original concentration of the ethyl acetate can be determined using stoichiometry.

Here are the general calculations you will need to follow:

1. Calculate the moles of OH- that reacted with the ethyl acetate by using the initial KOH concentration and volume.
2. Calculate the moles of excess OH- from the back-titration using the concentration and volume of H2SO4.
3. Subtract the excess OH- from the total to find the moles of OH- that reacted with ethyl acetate.
4. Use stoichiometry to find the moles of ethyl acetate that reacted.
5. Calculate the original concentration of ethyl acetate in the sample before dilution.
6. Multiply by the molar mass of ethyl acetate to find the mass.

The mass of ethyl acetate can be calculated using the following formula: mass of ethyl acetate = mol ethyl acetate × molar mass ethyl acetate. Given that the molar mass of ethyl acetate (C4H8O2) is 88.11g/mol, we can find the mass after calculating the moles as described in the steps above.