Butyl butyrate is an ester that is a naturally occurring oil used in the flavor industry for its fruity scent. If the steam distillation of butyl butyrate with water has partial pressures of 50 mm Hg and 710 mm Hg respectively, how much of the distillate is water

The correct answer is 62.5 %.

Explanation:

Based on the given information, the partial pressure of butyl butyrate is 50 mmHg and the partial pressure of water is 710 mmHg.  

Hence, the total pressure is 710+50 = 760 mmHg

According to Dalton's law of partial pressure,  

Partial pressure = mole fraction * total pressure

Mole fraction of water is,  

Partial pressure of water/Total pressure = 710/760 = 0.93

Similarly, the mole fraction of butyl butyrate is,  

Partial pressure of butyl-butyrate/Total pressure = 50/760 = 0.07

Therefore, mole% of water is 0.93 * 100 = 93 %

For calculating mass%,  

Mass of H2O = 0.93 * 18 = 16.8 grams (The molecular mass of water is 18 grams per mole)

The molecular mass of butyl-butyrate is 144 gram per mole

The mass of butyl-butyrate = 144 * 0.07 = 10.08 grams

The mass percent of water will be,  

Mass % of water/Total mass % * 100 = 16.8 / 10.08 + 16.8 * 100 = 62.5%.  

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