Gas and Bulb Essay

MOLECULAR WEIGHT OF A VOLATILE LIQUID
Prelaboratory assignment

Examples of questions related to the Dumas method:

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How many grams of ethanol would occupy 750 mL at 790.5 torr pressure at 100oC? (molar mass of ethanol is 46.0 g mol-1)

Relavent equations:

PV = nRT
1 atm = 101.3 kPa

R = 0.8206 L atm mol-1 K-1
P = (750 torr)(1atm/760 torr) = 0.987 atm
V = 0.750 L
T = 373 K

You can calculate the moles of the gas if you know can measure the volume of gas: (0.987 atm)(0.750 L) = n (0.8206 L atm mol-1 K-1)(373 K)
n = 0.0242 mol

And you can calculate the grams of gas if you know the molar mass: (0.0242 mol)(1 mol / 46.0 g) = mass
mass = 1.11 g

Dumas method:

Calculate the molar mass of a compound in the Dumas method for which a volume of the expermental container was 452 mL and the pressure was 745.1 torr. The difference in mass between the empty container and the final measurement was 1.129 g. P = (745.1 torr)/(760 torr atm-1) = 0.9804 atm

V = 0.452 L
T = 373 K
(0.9804 atm)(0.452 L) = n (0.8206 L atm mol-1 K-1)(373 K)
n = 0.01448 mol

Molar mass = M, m = mass, n = moles
M = m/n
M = (1.129 g)/(0.01448 mol)
M = 78.0 g /mol

Possible Unknowns:
Hexane Molar mass — 86.18 g/mol
1-hexene 84.1608 g/mol
Ethanol 46.07 g/mol
Propanol 60.1 g/mol
n-pentane 72.0 g/mol

Pre-lab Questions: Complete these in your class notebook, not your lab notebook. Sample data:
Mass of test tube and foil cover (g) 62.240 grams ± 0.001 g Temperature of water bath (ºC) 90.0 ºC ± 0.5 ºC
Mass of test tube and foil and gas sample (g) 62.247 g ±0.001 g Barometric pressure (kPa) 103.1 ± 0.1 kPa
Mass of test tube and foil and water (g) 87.240 g ± 0.001 g

Answer these questions for the sample data above. Report uncertainty with each calculated value. 1. Determine the mass of the condensed portion of the unknown that you placed in the test tube. Include uncertainty.

2. Use the mass of the water in the test tube and its density to calculate the volume of the test tube. Include uncertainty. Assume there is no uncertainty in the density of water (1 g/mL).

3. Use the calculations from Questions 1 and 2 above, along with the temperature of the boiling water bath and the barometric pressure of the room, to calculate the molar mass of your unknown compound. Use PV = nRT and solve for n (moles). Remember which units to use—you will need to
convert some units. Include uncertainty. 4. Identify the unknown liquid substance that you tested.

5. For the data given, what is limiting the precision of your experiment the most?

EXPERIMENT 8: MOLECULAR WEIGHT OF A VOLATILE LIQUID
INTRODUCTION: Determination of the Molecular Weight of a Volatile Liquid by the Dumas Method The Dumas method is one of the simplest ways to measure the molecular weight of a substance. It uses the ideal gas law. In this method we have to confine a sample of gas in a container of known volume, making sure when we do this that the sample is exactly at atmospheric pressure. The temperature and the mass of the gas sample have to be measured, along with the atmospheric pressure. Because liquids are much easier to handle than gases, a volatile liquid is usually used as the source of the gas. The liquid must have a boiling point substantially above room temperature and below the boiling point of water in order for this method to work well. The paragraphs below describe the experimental method generally used by scientists and the modifications to that method that we will make in our experiment. As you read, see if you can identify the possible sources of error that our simplified procedure will introduce. In the classic experiment, a round thin walled glass bulb with a long, thin bent neck is made by a glassblower and weighed. A small amount of volatile liquid is introduced into the bulb through a tiny opening at the end of the bulb’s neck. The bulb is heated in a boiling water bath to vaporize the liquid. The volume of vapor formed at 100 ºC (the approximate temperature of the boiling water – the exact temperature of the boiling water must be measured experimentally) and atmospheric pressure is greater than the volume of the bulb. The vapor first pushes all of the air out of the bulb and then begins to rush out of the opening until the pressure inside the bulb equals the pressure outside (atmospheric). With careful observation, the vapor can be seen exiting the bulb. It looks like a jet engine exhaust or a swirling cloud. When all of the liquid is vaporized and no more vapor is seen leaving the bulb, the bulb contains a sample of vapor at atmospheric pressure and 100 ºC with a volume exactly equal to the volume of the bulb. The opening in
the neck of the bulb is quickly sealed with a flame. It is important that the bulb be sealed at exactly the moment that vapor stops escaping from the bulb. (Sealing before vapor stops escaping will result too much vapor remains in the bulb. The molecular mass calculated will be too high. Waiting too long to seal the opening will result in a molecular weight that is too low. Can you explain why?) The sealed bulb is cooled, dried and carefully reweighed. The volume of the bulb must now be determined. The vapor inside has condensed, and the space above the liquid in the bulb is a near vacuum. The sealed tip of the bulb is then cut off with a file under water in a large container . The water rushes into the bulb, filling it completely. The bulb is dried on the outside and weighed once again. The volume of the bulb can be determined from the mass of water it contains. Barometric pressure and room temperature are carefully measured and recorded. The density of water at room temperature is either measured or obtained from the chemical literature. The volume (V) of the bulb is calculated from the mass of the water needed to completely fill the bulb. The mass (g) of vapor is obtained by subtraction. The temperature of the boiling water bath (T) and atmospheric pressure (P) were measured directly during the experiment. With this information, the molecular mass of the gas can be calculated using the ideal gas law. We will make a few simplifications in our experiment. First, we will use an ordinary Erlenmeyer flask covered with aluminum foil rather than a glass bulb. A tiny hole will be made in the foil to allow vapor to escape from the flask. We will not seal the flask when vapor stops escaping from the hole. We will simply remove it from the hot water bath as quickly as possible. The vapor will condense rapidly. We will measure the volume of the flask by filling it with water and then measuring the volume of water using a graduated cylinder. The information needed to determine the molecular mass of the unknown is the same as in the classic Dumas method: pressure (P), volume ,(V) the mass of the vapor,(g), and the temperature (T). Using this procedure, we should be able to determine the molecular mass of a volatile liquid to within 10% error.

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