Class Notes (835,893)
United States (324,282)
Chemistry (450)
CHEM 1951 (40)
Dr.Acio (9)
Lecture

AP Chem Lab 5 Masses of Equal Volumes of Gases.docx

5 Pages
111 Views
Unlock Document

Department
Chemistry
Course
CHEM 1951
Professor
Dr.Acio
Semester
Fall

Description
Curt Kim, Jonathan Lee, Keshav Mantha, Bobbie Sheng AP Chemistry, Period 1 1/22/13 Title: Masses of Equal Volumes of Gases Goal: The goal of this lab was to measure the masses of equal volumes of two different gases and calculate the ratio between the molar mass of the gases. Procedure: First, a 560 mL Erlenmeyer flask filled with air was massed along with its connected clamp and rubber tube. The flask was then filled with methane through the rubber tube, and the flask assembly was massed again with the tube sealed shut. After the new mass was recorded, the flask was emptied and another trial of methane gas was conducted. Next, the same process was repeated for another two trials with carbon dioxide gas instead of methane. Finally, the current room temperature and pressure readings were noted. Data: Data Table for Methane Volume of Room temp. Room Mass of flask Mass of flask flask (mL) (ºC) pressure (in. assembly assembly Hg) with air (g) with CH4 (g) Trial 1 560 22.8 30.43 257.752 257.481 Trial 2 “ “ “ “ 257.481 Trial 3 “ “ “ 265.665 265.398 Trial 4 “ “ “ “ 265.413 Data Table for Carbon Dioxide Volume of Room temp. Room Mass of flask Mass of flask flask (mL) (ºC) pressure (in. assembly assembly Hg) with air (g) with CO2 (g) Trial 1 560 21.3 30.06 257.851 258.191 Trial 2 “ “ “ “ 258.183 Trial 3 “ “ “ 256.777 257.088 Trial 4 “ “ “ “ 257.090 Analysis: 1. Calculate the mass of the air in the empty flask. In all four trials for the two gases, the volume of the flasks were 560mL. The density of air was given to be about 0.001208g/cm^3 for the conditions in which the data for methane was taken and about 0.001201g/cm^3 for the conditions in which the data for carbon dioxide was taken. Mass = volume*density, so the mass of air for the methane data is 0.67648g, and the mass of the air for the carbon dioxide data is 0.67256g. 2. Calculate the actual mass of the empty flask. Actual mass of empty flask = mass of flask with air - mass of air Trials 1&2 for Methane: 257.752g - 0.67648g = 257.07552 → 257.076g Trials 1&2 for Carbon Dioxide : 257.851g - 0.67256g = 257.17844 → 257.178g Trials 3&4 for Methane: 265.665g - 0.67648g = 264.98852 → 264.989g Trials 3&4 for Carbon Dioxide: 256.777g - 0.67256g = 256.10444 → 256.104g 3. Calculate the mass of CO2 gas contained in the flask. Mass of CO2 gas = mass of flask with CO2 - mass of empty flask Trial 1: 258.191g - 257.17844g = 1.01256g → 1.013g Trial 2: 258.183g - 257.17844g = 1.00456g → 1.005g Trial 3: 257.088g - 256.10444g = 0.98356g → 0.984g Trial 4: 257.090g - 256.10444g = 0.98556g → 0.986g 4. Calculate the mass of CH4 gas contained in the flask. Mass of CO2 gas = mass of flask with CH4 - mass of empty flask Trial 1: 257.481g - 257.07552g = 0.40548g → 0.405g Trial 2: 257.481g - 257.07552g = 0.40548g → 0.405g Trial 3: 265.398g - 264.98852g = 0.40948g → 0.409g Trial 4: 265.413g - 264.98852g = 0.42448g → 0.424g 5. Calculate the ratio of the mass of CO2 contained in the flask to the mass of CH4 gas contained in the flask. The point of this lab was to prove that Avogadro’s law holds true by finding an equal ratio between the masses of equal volumes of the two gases and the molar mass of the two gases. Unfortunately, since the temperature and pressures at which the data for the two gases were taken were different, the ratios will not be equal. The ideal or non-ideal gas law needs to be applied in order to find the number of moles for both gases. Since mass is proportional to the number of moles, a proportion can be used to find what the masses of one gas would have been if the flask contained the same number of moles as the other gas. These new values can then be used in testing Avogadro’s law. PV = nRT → n = (PV)/(RT) … V and R are constant, so the ratio between the two n values should be equal to the ratio between the two values for P/T. Methane: (772.922 torr) / (295.8 K) = 2.612988506 → 2.613 torr/K Carbon Dioxide: (763.524 torr) / (294.3K) = 2.594373089 → 2.594 torr/K Ratio between the P/T of methane’s data and the P/T of carbon dioxide’s data = (2.612988506 torr/K)/(2.594373089 torr/K) = 1.007175305 → 1.007 To find what the masses of carbon dioxide would be if they were taken at the same conditions at which the masses of methane were taken, they should be multiplied by the above ratio T
More Less

Related notes for CHEM 1951

Log In


OR

Join OneClass

Access over 10 million pages of study
documents for 1.3 million courses.

Sign up

Join to view


OR

By registering, I agree to the Terms and Privacy Policies
Already have an account?
Just a few more details

So we can recommend you notes for your school.

Reset Password

Please enter below the email address you registered with and we will send you a link to reset your password.

Add your courses

Get notes from the top students in your class.


Submit