# Chemistry help: Density of moist air **URL:** \ **Category:** Factual Questions\ **Created:** [January 13, 2013, 7:08pm UTC](https://boards.straightdope.com/t/chemistry-help-density-of-moist-air/647083 "2013-01-13T19:08:53Z")\ **Posts on this page:** 8\ **Page:** 1
**Author:** ![Johnny\_L.A](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/johnny_l.a/32/1084_2.png) [@Johnny\_L.A](https://boards.straightdope.com/u/Johnny_L.A)\ **Post date:** [January 13, 2013, 7:08pm UTC](https://boards.straightdope.com/t/chemistry-help-density-of-moist-air/647083/1 "2013-01-13T19:08:53Z")
Yes, this is homework. I’m not looking for the answer, but guidance on how to come up with the answer. (I’ve emailed my lab partner and the instructor, but I’ve never received a reply.) How do I determine the density of air in a collection flask? Given: [ul][li]V = volume of the flask is 141.5769 mL.[/li][li] P = air pressure is 1.006 atm \* 101325 Pa = 101932.95 Pa[/li][li] T = temperature = 292.9K[/li][li] R = 0.0821 L_atm/K_mol[/li][li] Density of water at room temperature = 0.998227 g/mL[/ul][/li] PV = nRT ==\> n = PV/RT So n = (1.006 atm \* .1415769 L) / (0.0821 (L_atm/K_mol) \* 292.9K = 0.005922811 mol. The goal is to determine the mass of CO[sub]2[/sub] in the flask (after it has been filled with CO[sub]2[/sub]) and from that calculate the molecular weight of CO[sub]2[/sub] (which should be close to 44.01 g/mol). It goes like this: If I know the density of moist air at 292.9K and 1.006 atm, then I can use D=M/V ==\> M=DV to find the mass of the air. When I know the mass of the air, I can subtract that from the mass of the flask, air, and stopper and end up with the weight of the flask and stopper without air. Once I know that, I can determine the mass of the CO[sub]2[/sub] in the flask. ISTM that I need the mass of the moist air to figure out the density of the moist air so that I can figure out the mass of the moist air. I’m missing something somewhere. Can someone point it out to me so that I can get on with the rest of the calculations? ---
**Author:** ![njtt](https://avatars.discourse-cdn.com/v4/letter/n/ecd19e/32.png) [@njtt](https://boards.straightdope.com/u/njtt)\ **Post date:** [January 13, 2013, 8:29pm UTC](https://boards.straightdope.com/t/chemistry-help-density-of-moist-air/647083/2 "2013-01-13T20:29:05Z")
Are you sure you are not overthinking this? I am not sure I understand what the problem is that you have been set, or where the moist air is supposed to be coming from, but ISTM that if you just weigh the flask (and stopper) on a balance in a room filled with an atmosphere of moist air, then you are going to be measuring the mass of the flask (and stopper), _not_ the mass of the flask, stopper, and the air inside the flask. Consider: if the flask is open (with the stopper on the balance pan beside it), the glass that comprises the flask is all surrounded by air, but no more so than the weight on the other balance pan is. You won’t change the mass of the flask if you put the stopper in, or even if you melt the glass into a blob, so that it does not “contain” any air. The mass of the CO2 is the mass of the stoppered flask with the CO2 in it, minus the mass of the flask, same as if you were calculating the mass of water in a flask by weighing the flask filled with water and subtracting the mass of the flask. The mass of air that would otherwise be in the flask is irrelevant, because there is air all around, buoying everything up. Now, if you were doing the experiment in a vacuum, but were only allowed to weigh the flask when it was filled with either air or CO2, _then_ you would need to know the weight of the air - but that is a rather unlikely scenario. ---
**Author:** ![njtt](https://avatars.discourse-cdn.com/v4/letter/n/ecd19e/32.png) [@njtt](https://boards.straightdope.com/u/njtt)\ **Post date:** [January 13, 2013, 8:37pm UTC](https://boards.straightdope.com/t/chemistry-help-density-of-moist-air/647083/3 "2013-01-13T20:37:55Z")
Also, if you weighed an evacuated flask in a room full of air, the mass you measured would be too low (by the amount of the mass of air that would otherwise fill the flask), because the flask is being buoyed up by that amount of air, that it is displacing. ---
**Author:** ![Johnny\_L.A](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/johnny_l.a/32/1084_2.png) [@Johnny\_L.A](https://boards.straightdope.com/u/Johnny_L.A)\ **Post date:** [January 13, 2013, 8:57pm UTC](https://boards.straightdope.com/t/chemistry-help-density-of-moist-air/647083/4 "2013-01-13T20:57:09Z")
The object is to calculate by experimentation the molecular weight of carbon dioxide. To do this, we need to determine the mass of the CO2 by subtracting the weight of the stopper and flask. But the ‘empty’ flask isn’t empty. It contains air, and air has mass. From the handout: > [@](#): > > In order to find the mass in grams of the CO2 sample, you need to find the difference in weight between the flask and the flask with the CO2 sample in it. Unfortunately, the weight of your flask before you started the experiment also includes the weight of air in the flask. To find the weight of the flask without this air, you must subtract out the weight of this air. The weight of air in the flask may be calculated if you know the volume of the flask and the density of air at room temperature. This density is ~1.2 g/L. A more precise value will be determined by you now based on information you will obtain from the CRC Handbook of Chemistry and Physics, the barometer in the room and the temperature in the room. FWIW, I found online that dry air is 28.85 g. So: Mass = n \* M.W. 0.00592281 mol \* 28.85 g= 0.16999231 g Density = Mass/Volume 0.16999231 g / 0.1415769 L = 1.201 g/l Which is what the density is in the handout. But that’s for dry air, and I don’t know how to get the answer for moist air. … OK, let’s assume we have the density just so we can move forward. Density of moist air at room temperature/presure: 1.201 g/l Mass of air in the collection flask: Mass = Density \* Volume ==\> Mass = 1.201 \* 0.1415769 = 0.16999231 g Mass of collection flask & stopepr: Flask minus air = 80.7341 g - 0.16999231 g = 80.56410769 g Mass of CO[sub]2[/sub] in the collection flask: Flask with CO[sub]2[/sub] minus flask = 80.8110 g - 80.5641 g = 0.2469 g CO2 If we have 0.2469 g of CO[sub]2[/sub] and we divide it by 0.1415769 L, then we have 1.744 g/L – which is heavier than air, which is what we expect. M.W (CO[sub]2[/sub]) = (0.2469 g CO[sub]2[/sub] \* 0.0821 \* 292.9K) / (1.006 atm \* 0.1415769 L) = 41.9364 g/mol Pretty close to 44.01 g/mol of CO[sub]2[/sub]. The numbers looks reasonable, given experimental errors, but without the actual density of the moist air I’m not seeing how we could get the right answer. If I knew how to get the actual density of the moist air, rather than using the number for dry air, the final calculated value of the M.W. of CO[sub]2[/sub] would be different. AND… I suck at math. ---
**Author:** ![johnpost](https://avatars.discourse-cdn.com/v4/letter/j/f17d59/32.png) [@johnpost](https://boards.straightdope.com/u/johnpost)\ **Post date:** [January 13, 2013, 9:09pm UTC](https://boards.straightdope.com/t/chemistry-help-density-of-moist-air/647083/5 "2013-01-13T21:09:36Z")
have you looked in the CRC Handbook of Chemistry and Physics for the values of the density of moist air? ---
**Author:** ![Johnny\_L.A](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/johnny_l.a/32/1084_2.png) [@Johnny\_L.A](https://boards.straightdope.com/u/Johnny_L.A)\ **Post date:** [January 13, 2013, 9:14pm UTC](https://boards.straightdope.com/t/chemistry-help-density-of-moist-air/647083/6 "2013-01-13T21:14:25Z")
I don’t have a copy. We were provided with: Pressure = 1.006 atm Temperature = 292.9K Density of water & room temp. & pressure: 0.998227 g/L ---
**Author:** ![ZenBeam](https://avatars.discourse-cdn.com/v4/letter/z/3ab097/32.png) [@ZenBeam](https://boards.straightdope.com/u/ZenBeam)\ **Post date:** [January 13, 2013, 10:15pm UTC](https://boards.straightdope.com/t/chemistry-help-density-of-moist-air/647083/7 "2013-01-13T22:15:59Z")
> [@Johnny\_L.A](#): > > Density of water & room temp. & pressure: 0.998227 g/L This is the density of liquid water, so shouldn’t be relevant. I don’t see where you have any way of knowing the partial pressure of water vapor in moist air. You’d need that, or its equivalent, to make the calculation you want. I guess you could assume 100% of the pressure comes from water vapor, and get a _lower bound_ on the density of moist air. ---
**Author:** ![Johnny\_L.A](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/johnny_l.a/32/1084_2.png) [@Johnny\_L.A](https://boards.straightdope.com/u/Johnny_L.A)\ **Post date:** [January 13, 2013, 10:21pm UTC](https://boards.straightdope.com/t/chemistry-help-density-of-moist-air/647083/8 "2013-01-13T22:21:56Z")
Yep, we needed that for something else.