behavior of gases: molar mass of a vapor?

Question

what are the major sources of error in your determine of the molar gases?

Answer 1

You must be taking Chemistry. I assume you are talking about an experiment where you heat up and vaporize a liquid in an Erlenmeyer flask whose top is covered with some aluminum foil with a pin-hole in it to allow the excess to escape. The Erlenmeyer is then cooled and the mass of the liquid determined. This is a classic experiment that usually works pretty good. You have the mass of the liquid (that was a gas at the higher temperature) you have the volume of the container (usually measured with a graduated cylinder after filling the flask with water) you should have your atmospheric pressure when you did your measurements (given to you) you know the temperature the flask was heated to (remember to change to K) and you know the value of the gas constant R.

Use the equation PV = nRT but rearrange to n = PV/RT and solve for n, the number of moles. Take the mass of the compound in g and divide by the number of moles and you have the molar mass in g/mole.

Error - inaccurate volume measurement
Error - inaccurate temperature measurement
Error - a gas close to it's condensation temperature is not an ideal gas and the ideal gas only applies to a limited extent.

Answer 2

Well, I would think that using ideal gas law to find the molar mass would introduce a little error, although it is usually "good enough"

Answer 3

For the best answers, search on this site https://shorturl.im/aviD2

1. We need to figure out how many moles of methane burned. We use the ideal gas law PV=nRT P=1 atm (STP) T=298K (STP) V=0.50L R=.0820 L atm/mol K Plug and chug. (1)(0.50)=n(.082)(298) n= .0205 moles CH4 Looking at the reaction equation, we see that 2 moles of O2 are needed to combust 1 mole of CH4, so for that number of moles we need .0416 moles O2. Now we can do the ideal gas law again to find the volume of O2 that corresponds to that. P=1 atm (STP) T=298 K (STP) R=.082 L atm/mole K n=.0416 moles (1)V=(.0416)(.082)(298) V= 0.984 L of O2 needed For each mole of methane combusted, we get 1 mole of CO2, so we'll get .0208 moles of CO2. Since all the other conditions are the same, the volume will turn out to be the same as the volume of CH4 or 0.50 L (work it out if you don't believe me) 2) You again must figure out how many moles of gas are there. P=700 mmHg V=.200 L T=308 K R=62.36 mmHg L/mol K ( you have to use this value of R or convert the pressure to atm to use the other one) (700)(.2)=n(62.36)(308) n= 7.29 x 10^-3 moles gas So at STP, that number of moles would have these values P=760 mmHg T=298K n=7.29 x 10^-3 moles R=62.36 mmHg L/mole K So solve for V (760)V=(7.29x10^-3)(62.36)(298) V= 5.61L 3) Since PV=nRT, and R, T, and P are constant, rearrange it this way to help you understand: V/n=RT/P That RT/P value is constant, so if n gets bigger, V must also get bigger so that their ratio stays the same. If n gets smaller, V just also get smaller. Plug some random values in to see for yourself. 4) Density is g/L, so we need to get the ideal gas equation rearranged so that we can get those units. We'll do this by first saying that molar mass is grams/mole, so grams/molar mass=moles We can plug in grams/molar mass for the n in the equation PV=(g/MM)RT Now we want to rearrange it so we have g/V on one side of the equation, since that is our value for density. PV(MM)=gRT PV(MM)/RT=g P(MM)/RT=g/V So our density is D=P(MM)/RT We have values for all of those P=.850 atm MM=32 g/mol because its O2 R=.082 Latm/moleK T=298 K So D=(.850)(32)/(.082)(298) D=1.11 g/L