A natural product (MW=150) distills with steam at a boiling temperature of 99 degrees Celsius at atmospheric pressure. The vapor pressure of water at 99 degrees Celsius is 733 mm Hg.
A) Calculate the weight of the natural product that codistills with each gram of water at 99 degrees Celsius.
B) How much water must be removed by steam distillation to recover this natural product from 0.5 g of a spice that contains 10% of the desired substance?
I'll get you started.
P(H2O)/P(org) = n(H2O)/n(org). Let's simplify some by calling H2O=w and organic = o
Pw/Po = nw/no
The total pressure is 1 atm or 760 mm Hg.
Since the partial pressure of water at this temperature from the problem is 733 mm, then the partial pressure of the organic material must be 760-733 = 27 mm.
We also know that n = grams/molar mass.
Substituting we get
733/27 = (grams w/18)/(grams o/150).
The problem wants to know grams o for 1 g water so substitute 1 for grams w.
733/27 = (1/18)/grams o/150).
Solve for grams o and you will have the answer to part (A).
I solved for o and got 0.306957708 grams. Correct?
For B, I did this: 0.5 X 0.10 = 0.05 g
0.306957708 - 0.05 = 0.256957708 g of water, which is my answer. Correct?
A) To calculate the weight of the natural product that codistills with each gram of water, we need to determine the mole fractions of both the natural product and water in the vapor phase.
The mole fraction of water can be calculated using Raoult's law, which states that the vapor pressure of a component in a mixture is equal to the product of its mole fraction and the vapor pressure of the pure component.
The mole fraction of water (X_water) can be calculated as follows:
X_water = (partial pressure of water) / (total pressure)
Partial pressure of water = 733 mm Hg (given)
Total pressure = Atmospheric pressure = 760 mm Hg (at sea level and room temperature)
X_water = 733 mm Hg / 760 mm Hg ~= 0.9658
Since the system contains only water and the natural product, the mole fraction of the natural product (X_natural product) can be calculated as:
X_natural product = 1 - X_water
X_natural product = 1 - 0.9658 ~= 0.0342
Now, we need to determine the weight of the natural product (MW = 150 g/mol) that codistills with each gram of water.
Let's assume we have 1 g of water at 99 degrees Celsius.
The weight of the natural product that codistills with each gram of water can be calculated by multiplying the mole fraction of the natural product by the total weight of the natural product:
Weight of natural product = X_natural product * (1 g) * MW
Weight of natural product = 0.0342 * (1 g) * 150 g/mol
Weight of natural product = 5.13 g
Therefore, the weight of the natural product that codistills with each gram of water at 99 degrees Celsius is approximately 5.13 grams.
B) To calculate how much water must be removed by steam distillation to recover the natural product, we need to consider the percentage of the desired substance in the spice and the total weight of the spice.
Let's assume we have 0.5 g of the spice, and it contains 10% of the desired substance.
The weight of the desired substance can be calculated as follows:
Weight of desired substance = (percentage of desired substance) * (total weight of the spice)
Weight of desired substance = 0.10 * 0.5 g
Weight of desired substance = 0.05 g
Since the weight of the desired substance is the same as the weight of the natural product (as mentioned in part A), we can use the same weight, which is 5.13 g.
Now, we can calculate the amount of water that needs to be removed by steam distillation to recover the natural product.
Weight of water removed = Weight of spice - Weight of desired substance
Weight of water removed = 0.5 g - 0.05 g
Weight of water removed = 0.45 g
Therefore, to recover the natural product from 0.5 g of the spice that contains 10% of the desired substance, approximately 0.45 grams of water must be removed by steam distillation.