Limone (MW 136g/mol) is a pleasant smelling liquid found in lemon and orange peels. The boiling point of limonene is 175 ˚C, but it co-distills with water at 97.5 ˚C. If the vapour pressure of water at 97.5˚C is 690 mmHg, what percent (by mass) of the steam-distillate is limonene? (Assuming that the distillation is occurring at 760 mmHg)

Limone = L

water = W
pL = partial pressure limone
pW = partial pressure W
ML = molar mass Limone
MW = molar mass Water
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(mass L/mass W) = (pL*ML/pW x MW)
pTotal = pL + pW and pW = 690 from the problem and pTotal = 760.
Solve for the mass ratio and convert to percent L. Post your work (I suggest a new post at the top but include this information--just repost this if you need further assistance.)

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To find the percent by mass of limonene in the steam-distillate, we need to compare the vapor pressure of limonene with the vapor pressure of water at the distillation temperature.

First, we can calculate the vapor pressure of water at 97.5 ˚C using the given value of 690 mmHg. This will give us the relative proportion of water in the steam-distillate.

Next, we can calculate the vapor pressure of limonene at 97.5 ˚C by using the boiling point of limonene and the atmospheric pressure of 760 mmHg.

Finally, we can calculate the percent by mass of limonene in the steam-distillate using the ratio of the vapor pressures.

Let's go step-by-step:

Step 1: Calculate the vapor pressure of water at 97.5 ˚C:

The Clausius-Clapeyron equation can be used to calculate the vapor pressure of water at a given temperature:

ln(P1/P2) = (-ΔHvap/R) * (1/T1 - 1/T2)

Where:
P1 = vapor pressure of water at T1
P2 = vapor pressure of water at T2
ΔHvap = enthalpy of vaporization of water
R = ideal gas constant (8.314 J/(mol·K))
T1 = initial temperature (boiling point of limonene, 175 ˚C + 273.15 K)
T2 = final temperature (97.5 ˚C + 273.15 K)

Since we are given the vapor pressure at 97.5 ˚C (P2 = 690 mmHg) and need to solve for P1, we rearrange the equation:

ln(P1/690) = (-ΔHvap/R) * (1/(175 + 273.15) - 1/(97.5 + 273.15))

Let's calculate the right side of the equation:

ΔHvap = Latent heat of vaporization of water = 40.7 kJ/mol = 40,700 J/mol

ln(P1/690) = (-40700 J/mol / 8.314 J/(mol·K)) * (1/448.15 K - 1/370.65 K)

Calculating the right side of the equation gives:

ln(P1/690) = -213.435

To solve for P1, we take the exponential of both sides:

P1/690 = e^(-213.435)

P1 = 690 * e^(-213.435)
P1 ≈ 7.75 x 10^-93 mmHg

Therefore, the vapor pressure of water at 97.5 ˚C is approximately 7.75 x 10^-93 mmHg.

Step 2: Calculate the vapor pressure of limonene at 97.5 ˚C:

Since limonene co-distills with water at 97.5 ˚C, we can assume that the vapor pressure is the same as the vapor pressure of water at that temperature. Therefore, the vapor pressure of limonene at 97.5 ˚C is approximately 7.75 x 10^-93 mmHg.

Step 3: Calculate the percent by mass of limonene in the steam-distillate:

To find the percent by mass of limonene, we can use the ratio of the vapor pressures:

Percent by mass of limonene = (vapor pressure of limonene / total vapor pressure) * 100

The total vapor pressure is the sum of water and limonene vapor pressures.

Total vapor pressure = vapor pressure of water + vapor pressure of limonene ≈ 7.75 x 10^-93 mmHg + 7.75 x 10^-93 mmHg ≈ 1.55 x 10^-92 mmHg

Percent by mass of limonene = (7.75 x 10^-93 mmHg / 1.55 x 10^-92 mmHg) * 100

Percent by mass of limonene ≈ 5%

Therefore, the percent by mass of limonene in the steam-distillate is approximately 5%.

To determine the percent (by mass) of limonene in the steam-distillate, we can use the Raoult's law, which states that the vapor pressure of a component in an ideal mixture is equal to the mole fraction of that component multiplied by its vapor pressure in its pure state.

First, let's calculate the mole fraction of limonene in the steam-distillate. Given that the total pressure is 760 mmHg and the vapor pressure of water at 97.5 ˚C is 690 mmHg, the pressure due to limonene can be calculated as follows:

Pressure due to limonene = Total pressure - Pressure due to water
= 760 mmHg - 690 mmHg
= 70 mmHg

Now, let's calculate the mole fraction of limonene using its vapor pressure and the total pressure:

Mole fraction of limonene = Pressure due to limonene / Total pressure
= 70 mmHg / 760 mmHg
= 0.0921

Next, we need to convert the mole fraction of limonene into mass fraction. To do this, we need to know the molecular weight of limonene, which is given as 136 g/mol.

Mass fraction of limonene = (Mole fraction of limonene) * (Molecular weight of limonene) / ((Mole fraction of limonene) * (Molecular weight of limonene) + (Mole fraction of water) * (Molecular weight of water))

The mole fraction of water can be calculated as 1 - mole fraction of limonene:

Mole fraction of water = 1 - Mole fraction of limonene

Now, let's substitute the values into the formula:

Mass fraction of limonene = (0.0921) * (136 g/mol) / ((0.0921) * (136 g/mol) + (1 - 0.0921) * (18 g/mol))

Simplifying the equation:

Mass fraction of limonene = 12.5796 g / (12.5796 g + 97.4204 g)
= 0.1145

Finally, to convert the mass fraction into a percentage, we multiply by 100:

Percent of limonene = (Mass fraction of limonene) * 100
= 0.1145 * 100
= 11.45%

Therefore, approximately 11.45% (by mass) of the steam-distillate is limonene.