CHM 420 Laboratory Report Experiment 8 PDF

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Summary

Experiment 8 : Gas LawsObjectiveThe objective of this experiment is to verify the Graham’s law by measuring the distances traveled during the same period of time by two different gases of known molecular mass and to determine the molar mass of a volatile liquid by measuring mass of vapour of the liq...

Description

Experiment 8 : Gas Laws

Objective The objective of this experiment is to verify the Graham’s law by measuring the distances traveled during the same period of time by two different gases of known molecular mass and to determine the molar mass of a volatile liquid by measuring mass of vapour of the liquid is needed to fill a flask of known value at a particular temperature and pressure.

Introduction

A. Graham’s Law Graham's law, also known as Graham's law of effusion, was formulated by Scottish physical chemist Thomas Graham.Graham’s law is an empirical relationship that states that the ratio of the rates of diffusion or effusion of two gases is the square root of the inverse ratio of their molar masses. The formula can be written as :

Rate1 = Rate2

M2 M1

Where ; Rate1 is the rate of diffusion of one gas, expressed as volume or as moles per unit time, Rate2 is the rate of diffusion of the second gas, M1 is the molar mass of gas 1 and M2 is the molar mass of gas 2.

In this experiment, the rates of diffusion of two gases, ammonia NH3 and hydrogen chloride HCl, will be investigated. Rates of diffusion yield information that can lead to calculation of the molecular weights of gases. These gases are convenient

to use for such an experiment because, when they meet and react, a ring of smoke consisting of ammonium chloride, NH4Cl will form at the location where they meet.

NH3(g) + HCl(g)→NH4Cl(s)

B. Molar Mass of Volatile Liquid Volatile liquid is Liquid that evaporates at room temperature, or vaporizes easily. Volatile liquids have low boiling point. If the substance in question is a volatile liquid, a common method to determine its molar mass is to use the ideal gas law :

PV = nRT

Where P is the pressure, V is the volume, n is the number of moles, R is the ideal gas constant (8.21×10-2 L atm mol-1 K-1), and T is the absolute temperature (in Kelvin).

Volume, pressure and temperature can be measure while the substance in the gas phase. Ideal gas law also can be used to calculate the number of moles of the substance. From the measurement of the mass of the gas sample, the molar mass of the gas sample can be calculated .

n=

g M

Where n is the number of moles, g is the number of grams of the volatile substance per mol and M is the molar mass.

In this experiment, a small amount of easily volatilized liquid will be placed in a flask. The flask will be heated in a boiling water bath and will be equilibrated with atmospheric pressure. From the volume of the flask used ( V), the temperature (T) of the boiling water bath, and the atmospheric pressure (P), the number of moles of gas (n) contained in the flask may be calculated. From the mass of liquid required to fill the flask with vapor when it is in the boiling water bath, the molar mass of the liquid may be calculated.

Chemicals and Apparatus NH3 (conc.)

Stopper

HCl (conc.)

Medicine dropper

Acetone ( to dry the glass tube)

Tweezers or crucible tongs

Retort stand

Stock watch

Glass tube (50 cm length, 10 mm internal Marker pen diameter)

Long ruler (1 meter)

Cotton

Retort stand

Boiling chips

400/500 mL beaker

Unknown volatile liquid - acetone (or 100/125 conical flask other

compounds

lecturer)

recommended

by

500 mL graduated cylinder

Thermometer Aluminium

Bunsen burner foil

(square

shape

just Pin (to make a hole)

enough to cover the mouth of a conical flask)

Procedure A. Graham’s Law

1. A 50 cm length of glass tubing (10 mm internal diameter) was obtained. The glass tube must be dried. Two cotton swabs were obtained or prepared. 2. By using dropper pipettes, about 10 to 15 drops of concentrated HCl (number of drops should be equal for conc. HCl and conc. NH3) were placed on the cotton swab and 10 to 15 drops of concentrated NH3 solution were placed on another cotton swab. 3. By using tweezers, the moistened ends of the cotton swabs were inserted immediately and simultaneously into opposite ends of the tube. The time needed for the appearance of the faint white cloud ( ring ) of ammonium chloride was noted. 4. After several minutes, a white ring was formed where the gases HCl and NH3 meet inside the tube to form the white compound NH4Cl (ammonium chloride). The point on the tube where the white ring is formed was marked. The distance traveled by each gas were measured. Data was recorded in Datasheet 8A. 5. Tweezers was used to remove the cotton swabs and it was immersed into a beaker of tap water. 6. The tubing was rinsed with water. The tubing was dried by rinsing it with acetone.

B. Molar mass of volatile liquid

1. A boiling water up was set up using a 400 mL beaker containing 250 mL of water (or enough water to immerse the flask). 2. A 125 mL conical flask was obtained and boiling chips were added in. The mouth of the flask was tightly covered with a small square of aluminium foil. A straight pin was used to make a small hole on foil cap. 3. The empty, capped flask was weighed together with boiling chips. 4. The foil cap was removed. A 2 mL sample of the liquid to be studied was placed into the flask and the foil was replaced. 5. The flask was clamped with a single burette clamp. The flask was transferred to the boiling water bath, it was immersed and heated. The flames must not get too high because volatile liquids are flammable. A boiling chips was put in the water bath too if necessary. 6. The liquid refluxing inside the flask was noted. The flask was kept slightly tilted because the liquid disappears was noticed easily. 7. The flask was heated until liquid was no longer visible and no vapor was seen emerging from pinhole. The heating was continued for 30 seconds beyond this time. 8. The flask was removed. It was set on a hot pad or tile. The clamp was removed and the flask was cooled to room temperature. 9. The flask was dried. The flask, cap and condensed vapor was weighed. 10. The contents of the flask was disposed in waste bottle or as instructed. The flask was filled with tap water (to overflowing). The water was poured into a 500mL graduated cylinder, the volume was measured and recorded. 11. The barometic pressure (assume 760 torr or equal to 1 atm) was measured and recorded. Data was recorded in Datasheet 8B.

Data

A. Graham’s Law

Observation of NH4Cl appearance : The ring of NH4Cl forms nearer to the HCl end of the tube. The ring of NH4Cl is a white ring. Trial 1

Trial 2

0.00

0.00

1619.00

1485.00

Distance traveled by NH3 (cm)

69.00

78.00

Distance traveled by HCl (cm)

29.00

22.00

Ammonia diffusion rate (cm/sec)

0.04

0.05

HCl diffustion rate (cm/sec)

0.02

0.01

Start time (s) Finish time (first visible smoke) (s)

B. Molar Mass of Volatile Liquid

Mass of flask, foil, boiling chips and condensed vapor

72.7669 g

Mass of flask, boiling chips and foil

72.5649 g

Mass of condensed vapor ( mass of vapor) Temperature of vapor Barometric pressure (pressure of vapor) Volume of flask (volume of vapor)

0.2020 g 99 °C 760 mmHg (Torr) 155 mL

Calculations

A. Graham’s Law

From the Trial 1, Rate of diffusion of NH3

Rate of diffusion of HCl

Distance traveled by NH3 = 69 cm

Distance traveled = 29 cm

Time required = 1619 s

Time required = 1619 s

Rate of diffusion =

69 cm = 0.04 cm s-1 1619 s

Rate of diffusion =

29 cm = 0.02 cm s −1 1619 s

From Trial 2, Rate of diffusion of NH3

Rate of diffusion of HCl

Distance traveled by NH3 = 78 cm

Distance traveled = 22 cm

Time required = 1485 s

Time required = 1485 s

Rate of diffusion =

78 cm = 0.05 cm s-1 1485 s

Rate of diffusion =

22 cm = 0.01 cm s −1 1485 s

From Trial 1, Ratio of the rate of diffusion of NH3 to Theoretical

ratio

of

the

rate

the rate of diffusion of HCl,

diffusion by using molecular mass,

Rate of diffusion of NH3 = 0.04 cm s-1

Molecular mass of NH3 = 17.03 g mol-1

Rate of diffusion of HCl = 0.02 cm s-1

Molecular mass of HCl = 36.46 g mol-1

=

Rate1 Rate2

=

M2 M1

=

36.46 gmol −1 17.03 gmol

−1

0.04 cm s −1 0.02 cm s = 2.0

of

−1

=

= 1. 5 Thus, the theoretical ratio of the rate of

Thus, the ratio of the rate of diffusion of diffusion is 1.5 : 1 NH3 to the rate of diffusion of HCl is 2:1

(1.5 - 2.0)  100% 1.5 = 33.33% (absolute value)

Percentage error =

B. Molar Mass of Volatile Liquid

P = 760mmHg = 1 atm V = 0.155 L

PV = nRT n=

R = 0.08206 L atm K-1

=

mol-1

PV RT

(1 atm )(0.155 L)

(0.08206 L atm K −1 mol−1 )(372.15 K)

= 5.08  10 −3 mol

T = 99 + 273.15 = 372.15 K

n =5.08×10-3 mol g =0.2020 g

g M g M= n n=

M=

0.2020 g 5.08 10−3 mol

= 39.76 g mol−1

Molecular weight of unknown liquid = 39.76 g mol-1

Questions

1. Experiment 1

a) Calculate the rate of diffusion for each gas by dividing the the distance traveled (cm) by the time required (sec) for the appearance of the white deposit.

From the Trial 1, Rate of diffusion of NH3

Rate of diffusion of HCl

Distance traveled by NH3 = 69 cm

Distance traveled = 29 cm

Time required = 1619 s

Time required = 1619 s

Rate of diffusion =

69 cm = 0.04 cm s-1 1619 s

Rate of diffusion =

29 cm 1 = 0.02 cm s − 1619 s

From Trial 2, Rate of diffusion of NH3

Rate of diffusion of HCl

Distance traveled by NH3 = 78 cm

Distance traveled = 22 cm

Time required = 1485 s

Time required = 1485 s

Rate of diffusion =

78 cm = 0.05 cm s-1 1485 s

Rate of diffusion =

22 cm 1 = 0.01 cm s − 1485 s

b) Calculate the ratio of the rate of diffusion of NH3 to the rate of diffusion of HCI. Trial 1

Trial 2

Ratio of the rate of diffusion of NH3 to Ratio of the rate of diffusion of NH3 to the rate of diffusion of HCl,

the rate of diffusion of HCl,

Rate of diffusion of NH3 = 0.04 cm s-1

Rate of diffusion of NH3 = 0.04 cm s-1

Rate of diffusion of HCl = 0.02 cm s-1

Rate of diffusion of HCl = 0.02 cm s-1

=

Rate1 Rate2

= −1

0.04 cm s −1 0.02 cm s = 2.0 =

Rate1 Rate2 −1

0.05 cm s −1 0.01 cm s = 5.0 =

Thus, the ratio of the rate of diffusion of Thus, the ratio of the rate of diffusion of NH3 to the rate of diffusion of HCl is

NH3 to the rate of diffusion of HCl is

2:1

5:1

c) Using the molecular masses of NH3 and HCI, calculate the theoretical ratio of the rates of diffusion of these gases.

Theoretical ratio of the rate of diffusion by using molecular mass, Molecular mass of NH3 = 17.03 g mol-1 Molecular mass of HCl = 36.46 g mol-1

=

M2 M1

=

36.46 gmol −1 17.03 gmol −1

= 1. 5

Thus, the theoretical ratio of the rate of diffusion is 1.5 : 1

d) Calculate the % error in your experimentally determined value for the ratio of the rates of diffusion of NH3 and HCl. Use the theoretical ratio calculated in (c) as the accepted value for the ratio. [% error = absolute value of (theoretical ratio-experimental theoretical ratio) / theoretical ratio) ] × 100%.

Trial 1

Trial 2 (1.5 - 2.0) (1.5 - 5.0)  100% Percentage error =  100% 1.5 5.0 = 33.33% (absolute value) = 70% (absolute value)

Percentage error =

2. Experiment 2

Calculate the molecular weight of the unknown liquid. Show your calculations, and include units of the different quantities in your calculations.

P = 760mmHg = 1 atm V = 0.155 L R = 0.08206 L atm K-1 mol-1 T = 99 + 273.15 = 372.15 K

PV = nRT n= =

PV RT

(1 atm )(0.155 L)

(0.08206 L atm K −1 mol−1 )(372.15 K)

= 5.08  10 −3 mol

n =5.08×10-3 mol

g M g M= n

n=

g =0.2020 g

M=

0.2020 g 5.08 10−3 mol

= 39.76 g mol−1

Molecular weight of unknown liquid = 39.76 g mol-1

Discussions

A. Graham’s Law

The reaction which involve is :

NH3(g) + HCl(g)→NH4Cl(s)

From this experiment, ammonia react with hydrogen chloride to form ammonium chloride. The formation of ammonia chloride can be seen when a ring of white smoke

was formed in the tube. During experiment, the number of drops of NH3 should be equal to the number of drops of HCl and both of the cotton swabs containing NH3 and HCl should be insert into the tube at the same time. This is important to get the accurate time taken and measurement.

Based on observation, the white ring was formed nearer to the end of the tube of HCl cotton swab. This is because NH3 diffuses faster than HCl. From the calculation, the ratio of the rate of diffusion of NH3 to the HCl is 2:1. Hence, HCl has almost twice the molecular weight of ammonia. The rate of diffusion was inversely proportional to the square root of the molecular mass of the gas.

It is very important that the tube is clean and completely dry for this experiment. If necessary, the tube can be dried by pushing a cotton wool pad soaked in acetone through the tube and leaving it for a few minutes. Both concentrated NH3 solution and concentrated HCl and the vapors that they emit are extremely caustic and corrosive. Wear gloves to avoid these liquids comes into contact with skin. This step must be done in the fume hood.

B. Molar Mass of Volatile Liquid

The temperature, 99°C of the gas was determined by measuring the temperature of the water bath surrounding the flask. The mass of the vapor, 0.2020 g is measured by weighing the condensate remaining in the flask. Mass of flask, boiling chips and foil was substracted from the mass of flask boiling chips, foil and condensed vapor to get Mass of condensed vapor. The volume of the vapor is equal to the volume of the flask which is 0.155 mL.

When weighing the flask, the boiling chips must be completely dried and does not contain any excess water to avoid affecting to the mass recorded. The flask also must be weigh several times to get accurate reading.

Volatile liquid was used because volatile liquid have low boiling point and vaporizes easily. A small hole must be made on the aluminium foil that covers mouth of the flask to allow any extra vapor to escape the flask until the volume of the flask is completely filled with the gas. A straight pin was used to make a tiny hole as possible to reduce an increased rate of effusion.

The unknown liquid that was used may be harmful to skin. Avoid contact, and wash immediately if the liquid is spilled. Use tongs or a towel to protect hands from hot glassware

Conclusion Ammonia travel a greater length of the tube and diffuse faster compared to hydrogen chloride since ammonia have a lower molecular weight compared to hydrogen chloride. The molar mass of volatile liquid by measuring mass of vapour of the liquid is 39.76 g mol-1.

References Neil D. Jespersen & Alison Hyslop. (2014). Chemistry: The Molecular Nature of Matter, 7th Edition. John Wiley and Sons, Incorporated.

Helmenstine, Anne Marie. (2019). Graham's Law Definition. Retrieved on November 8, 2019 from https://www.thoughtco.com/definition-of-grahams-law-604513

Julian Rubin. (2013). Gas Diffusion and Effusion Experiment. Retrieved on November 8, 2019 from https://www.juliantrubin.com/encyclopedia/physics/ graham_law.html Randy Sullivan. (2012). Graham's Law: Diffusion of ammonia gas and HCl(g). Retrieved

on

November

8,

2019

from

https://chemdemos.uoregon.edu/demos/Grahams-Law-Diffusion-of-ammonia-gas -and-HClg

Winona. (2019) Molar Mass of a Volatile Liquid using the Ideal Gas Law. Retrieved on November 8, 2019 from http://course1.winona.edu/wng/ C212F15-NgLabs/MW-Gas-Law/Molar%20Mass%20Volatile%20Liquid_F15-1.pdf...