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You add 0.59 mL of n-propanol (0.803 g/mL, 60.1 g/mol) to an excess of acetic acid to create the pear ester propyl acetate (102.1 g/mol). If you obtain 0.54 grams of the product, what was your percent yield

Respuesta :

The percentage yield of the reaction between n-propanol and excess of acetic acid is 67.5%

How to determine the mass of n-propanol

  • Volume = 0.59 mL
  • Density = 0.803 g/mL
  • Mass of n-propanol =?

Mass = Density × Volume

Mass of n-propanol = 0.803 × 0.59

Mass of n-propanol = 0.47 g

Balanced equation

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Molar mass of n-propanol = 60.1 g/mol

Mass of n-propanol from the balanced = 1 × 60.1 = 60.1 g

Molar mass of propyl acetate = 102.1 g/mol

Mass of proply acetate from the balanced equation = 1 × 102.1 = 102.1 g

SUMMARY

From the balanced equation,

60.1 g of n-propanol reacted to produce 102.1 g of propyl acetate

How to determine the theoretical yield

From the balanced equation,

60.1 g of n-propanol reacted to produce 102.1 g of propyl acetate

Therefore,

0.47 g of n-propanol will react to produce = (0.47 × 102.1) / 60.1 = 0.8 g of propyl acetate

How to determine the percentage yield

  • Actual yield of propyl acetate = 0.54 g
  • Theoretical yield of propyl acetate = 0.8 g
  • Percentage yield =.?

Percentage yield = (Actual / Theoretical) × 100

Percentage yield = (0.54 / 0.8) × 100

Percentage yield = 67.5%

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