Parts per million of vapor above a liquor in a sealed jar

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Discussion Overview

The discussion revolves around calculating the concentration of vapor above a liquid in a sealed jar, specifically focusing on a 10% solution of Citral in diethyl phthalate. Participants explore methods to express this concentration in parts per million (ppm) and discuss the implications of vapor pressure and solution behavior.

Discussion Character

  • Technical explanation
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • Some participants propose using the ideal gas law (pV = nRT) to calculate concentration from vapor pressure.
  • Others clarify that ppm is defined as milligrams of solute per kilogram of solution, not grams per kilogram.
  • There is a request for clarification on whether the 10% solution is by mass or volume, and whether it can be treated as an ideal solution for calculations.
  • Participants discuss the need to determine both mass and molar concentrations, as well as the mole fraction of Citral in the solution.
  • Some participants provide specific calculations for the number of moles and mole fractions based on given densities and molecular weights.
  • There is a suggestion to focus on the concentration of Citral in the vapor phase rather than the total concentration of air above the liquid.

Areas of Agreement / Disagreement

Participants express differing views on the assumptions regarding the ideality of the solution and the definitions of concentration. There is no consensus on the best approach to calculate the vapor concentration in ppm, as various methods and interpretations are presented.

Contextual Notes

Participants note potential limitations in their calculations, including assumptions about ideal behavior, the specific conditions of the solution, and the initial state of the air in the vial.

Who May Find This Useful

This discussion may be useful for individuals interested in physical chemistry, particularly those exploring vapor-liquid equilibria and concentration calculations in solutions.

costanzolab
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How would you calculate the concentration of a vapor above a liquid in a sealed jar in parts per million? Supposing that the liquid is a 10% solution of X with known vapor pressure.

We've gotten to this point:
1ppm = 1mgX/ g solution
1ppm = 1mgX/1mL solution (assuming it's water and water has 1g/mL)
1ppm = 1000mg / liter = 1 gm/liter

A tougher issue is how can you calculate the concentration of a vapor above a solid in a sealed jar.

Any help is appreciated!
 
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ppm are parts per million, so not gram/kilogram, but milligram per kilogram.
If you know the vapour pressure, you can use pV = nRT to calculate the concentration n/V.
 
costanzolab said:
How would you calculate the concentration of a vapor above a liquid in a sealed jar in parts per million? Supposing that the liquid is a 10% solution of X with known vapor pressure.

We've gotten to this point:
1ppm = 1mgX/ g solution
1ppm = 1mgX/1mL solution (assuming it's water and water has 1g/mL)
1ppm = 1000mg / liter = 1 gm/liter

A tougher issue is how can you calculate the concentration of a vapor above a solid in a sealed jar.

Any help is appreciated!
Can you please provide an example of a specific solution that you might be interested in.
 
Chestermiller said:
Can you please provide an example of a specific solution that you might be interested in.
Sure.

Given 10mL of 10% Citral solution (dissolved in diethyl pthalate) ( vapor pressure 0.2 mm of Hg (@ 20°C)) in a closed 20mL vial, what is the concentration of the air above the liquid in parts per million.
 
BvU said:
ppm are parts per million, so not gram/kilogram, but milligram per kilogram.
If you know the vapour pressure, you can use pV = nRT to calculate the concentration n/V.

After getting to moles per liter how do you convert to ppm?
 
Do you want the mass concentration or the molar concentration? Does the solution of Citral in diethyl pthalate represent something close to an "ideal solution" (so that Raoult's law can be applied), or, in your judgment, is it non-ideal? Is that 10 % solution by mass or is it molar? Was the air in the vial initially at 1 atm?
 
Chestermiller said:
Do you want the mass concentration or the molar concentration? Does the solution of Citral in diethyl pthalate represent something close to an "ideal solution" (so that Raoult's law can be applied), or, in your judgment, is it non-ideal? Is that 10 % solution by mass or is it molar? Was the air in the vial initially at 1 atm?

We would like the mass concentration, and then to eventually convert this to ppm.

The solution can be considered ideal.

10% is by mass (1mL pure Citral in 10 total mL of Diethyl phthalate)

The air in the vial was initially at 1atm
 
costanzolab said:
We would like the mass concentration, and then to eventually convert this to ppm.

The solution can be considered ideal.

10% is by mass (1mL pure Citral in 10 total mL of Diethyl phthalate)

The air in the vial was initially at 1atm
So, if you had 10 grams of solution, how many grams would you have each of Citral and Diethyl phthalate?
How many moles would you have each of Citral and Diethyl phthalate?
What would the total number of moles be for 10 gm of solution?
What would be the mole fraction of Citral?

Is this a homework problem?
 
Chestermiller said:
So, if you had 10 grams of solution, how many grams would you have each of Citral and Diethyl phthalate?
How many moles would you have each of Citral and Diethyl phthalate?
What would the total number of moles be for 10 gm of solution?
What would be the mole fraction of Citral?

Is this a homework problem?

This is not a homework problem. General curiosity and tangentially related to some work we are doing with citral.

Working through your questions now.
 
  • #10
Chestermiller said:
So, if you had 10 grams of solution, how many grams would you have each of Citral and Diethyl phthalate?
How many moles would you have each of Citral and Diethyl phthalate?
What would the total number of moles be for 10 gm of solution?
What would be the mole fraction of Citral?

Is this a homework problem?

Using 893kg/m3 as the density for citral, and 1.12g/cm3 as the density for diethyl phthalate, I have:

5.866*10^-3 moles citral and 4.536*10^-2 moles diethyl phthalate in 10mL of 10% solution

The mole fraction for citral is 0.1145

What do I need the moles of citral in 10grams of solution for?
 
  • #11
We are using 10 gm as a "basis" for the calculation. You don't need to know the densities. If you have 10 gm of solution, then you have 1 gm of Citril and 9 gm of diethyl phthalate. The molecular weight of diethyl phthalate is 229, and the molecular weight of Citral is 152. So the number of moles are
Citral: 0.00659 moles
Diethyl phthakate: 0.0393
Total moles = 0.0459
mole fraction citral = 0.144
From Raolt's law, if the mole fraction Citral is 0.144 and the equilibrium vapor pressure of pure Citral is 0.2 mm Hg, what is the partial pressure of Citral in the head space? What is the total pressure of the gas in the head space?
 
  • #12
A 10% solution usually refers to the mass percent, not the volume percent.
 
  • #13
costanzolab said:
10% is by mass (1mL pure Citral in 10 total mL of Diethyl phthalate)
Somewhat contradictory ! (So Chet goes that way)

So your citral has a vapour pressure of 0.22 mm Hg at 20 C. And the diethylphtalate 0.002 at 25 C. Ignore the latter and work out how many moles there are in the gas volume (20 ml minus the 11 ml liquid volume). Idem air. And I don't assume you want
costanzolab said:
the concentration of the air above the liquid in parts per million
but instead the ppm citral, so 106 times mass of citral / mass of air
 
Last edited:
  • #14
And (lazy me) you don't have to work it out completely: with m = nM and pV = nRT (both for air and for citral) V, R and T divide out and things become simple.

(note I ignored the much lower partial pressure from the solvent)
 

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