Gas Collection Lab 4u3z9

Collection of Gas Over Water Advanced Chemistry 2017-2018 Mahidol University International Demonstration School (MUIDS) November, 9 2017

: Mind Daranporn Bee Pichayathida Aom Arada

Mui Pimchanok Lilly Nitinan




Collection of carbon dioxide gas resulted from the chemical reaction between acetic acid and sodium bicarbonate was done in order to demonstrate and implement the knowledge about gas laws and collection of gas over water learned in class. The main technique used within the lab is collection of gas over water, which was explained within the lab report. Moreover, the overall idea of gas laws and calculations used within the lab was explained and the chemicals used and the reaction itself was identified and described. At the end of the experiment, the relationship between volume and mole of gas was found to be directly proportional which follows Avogadro’s law of gas. Also, it could be seen that when the value of volume is higher, the value of the pressure of the dry gas tends to be lower.

Objective

Abstract

The purpose of this experiment is to collect carbon dioxide gas from mixture of baking soda and acetic acid and also apply this experiment to the lesson about vapor pressure and gas laws, mainly, ideal gas law, Dalton’s law, and Avogadro’s law.

Introduction 


Sodium bicarbonate (NaHCO3) is a powdered substance commonly used in cleansing solutions due to its alkalinity. Also, it is used as a pH buffering agent, alkalizer, and electrolyte replenisher. Physically, it is odorless, white, crystalline, powdery and has a bitter taste[6] . Acetic acid (CH3 COOH) is one of the carboxylic acids that is commonly found in daily lives. It is considered as a weak acid. Physically, it is a clear, colorless solution with a pungent odor. Acetic acid is commonly used in the production of plastic bottles, photographic film, a component of wood glue, and production of synthetic fibers and fabrics. It also could be used as cleaning agent in households when diluted. When highly concentrated, acetic acid could be very corrosive [5]. In many cases when the amount of water insoluble gas produced by a specific reaction is being the point of interest, the technique of collecting gas over water would be used. By using this technique the amount of gas would be determined through the displacement of water in a container or the change of water level within a container. This technique involves the usage of a container in which the reaction would take place and another container, filled with water and inverted in a reservoir of water, that the gas would be collected. The gas from the reaction would be collected by using hose or tubes. One end of the tube would be attached to the reaction container, in which the container would be closed elsewhere in order to prevent gas from escaping into the atmosphere, and another end of the tube would be inserted up into the inverted gas collection container. As the gas was created from the reaction, the water from the gas collection contained would be displaced by the gas created which travels through the connected tube. The amount of gas created, then, would be able to be determined by the amount of water it displaced [1].

Co!ecting Gas over Water Technique

Ideal gas law: PV = nRT (P=absolute pressure, V=volume, n=number of moles, R= Boltzmann constant, and T=absolute temperature) is used to explain the relationship between absolute pressure, volume, and absolute temperature deduced from the kinetic theory. It also could be viewed that it has arisen from the kinetic pressure of the gas molecules colliding with the wall of the container in accordance with Newton’s law. However, there is a statistical element in determining the kinetic energy of the gas molecules. Hence, it invokes the idea of kinetic temperature since the temperature is taken as proportional to the average kinetic energy [4].

Currently, there are many other gas laws that are used to explain the relationship of different characteristics of the gas. Some of the basic gas laws are : Boyle’s law, Charles’ law, Gay-Lussac’s law, Avogadro’s law, and Combined gas law. Boyle’s law, invented by Robert Boyle (1627-1691), states that the the volume of a fixed amount of gas held at constant temperature would have an inversely proportional relationship with its pressure. Hence, this relationship could be represented by the mathematical notation: V ∝ __1P . Also, it could be interpreted that when the volume and the pressure of the gas with constant amount and temperature was multiplied, the product would be constant (PV= C). From the relationship stated by Boyle’s law, the equation: P 1V1 =P2 V2 . Charles’ law, on the other hand, was invented by Jacques Charles (1746-1823). Charles’ law explained the relationship between the temperature in Kelvin of the gas and the pressure. Charles’ law stated that the pressure and the temperature of the gas in Kelvin is directly proportional when the volume and the amount is held constant (P ∝ T). This relationship gave rise to P P P another two mathematical notation that could be used to explain the relationship: __T=C and __T11 = __T22 . Joseph Gay-Lussac (1778-1850) invented the Gay-lussac’s law in order to discusses the relationship between the volume of the gas and the temperature in Kelvin while the pressure and the number of moles of the gas is held constant. Charles’ law stated that the relationship between the volume of the gas and the temperature of the gas in Kelvins at the stated condition would be directly proportional. Hence, these three mathematical V1 V V 2 __ __ __ notation could be derived from the relationship: V ∝ T, T =C , and T1 = T2 . Avogadro’s law, also known as the volume amount law, invented by Amedeo Avogadro, relates the amount of moles of gas and the volume of that exact gas together. The law gives the relationship of the amount of gas and volume under the condition that the pressure and the temperature, in Kelvin, of the gas have to be constant. Since volume could be change, it is assumed that the container could expand or contracted somehow. The relationship between the number of moles and the volume is directly proportional and could be explained by the V V mathematical notations: V ∝ n, __Vn =C , and __n11= __n 22 . Lastly, combined gas law is the combination of all of the P1 V 1 P V [3] stated gas laws (except Avogadro’s law), creating one, unified equation: ______ = ______ . T1 T 


1

1

1

Dalton’s law or the law of partial pressure is another essential gas laws to be known in order to finish this lab. The law stated that the total pressure exerted by the mixture of gas is the sum of the partial pressure of each of the individual gases composing the mixture. This law is based on the kinetic theory of gases, which states that the gas will diffuse within a container in order to fill up the space and doesn’t have any attractive forces between each of the gas molecules; hence, they do not react and are somewhat far apart. Since the pressure of ideal gas is determined by the gas’ collisions with the container, the partial pressure of a gas would not affect the partial pressure of another gas within the same container. So, according to the ideal gas law, it could be concluded that the pressure of a gas is based on the moles, volume and temperature of itself and since the gases are in the same container, the volume and temperature of the gases are the same. Hence, the gas exerts its own pressure on the system, which could be sum up in order to find the total pressure and the partial pressure that it exerts would be proportional to the amount of mole of that exact gas over the total amount of moles of gas. Dalton’s law could be concluded as the equation: Ptotal =Pa +Pb +... and the partial pressure of an individual gas could be found using its mole ratio: number of moles of the gas (X i = total number ) by multiplying the mole ratio with the total pressure: PA =(XA)(Ptotal ) [2]. of moles of gas within the mixture

Materials vinegar

Baking Soda

r

la u t a p s

lask

goggles

r

sink

ape p g

in t h g ei

gloves

w

beaker ance

dropper

weighting bal

tubing

ric f

plastic g r a d u a t e d cylinde

distilled water

ed cylinder glass graduat

suction flask

volu met

clay

glass funnel




Procedure

1. Prepare all of the apparatus that is needed for this experiment.

5. Slowly add the baking soda until it reach the correct amount according to the calculated concentration.

2. Fold the weighting paper diagonally two times and place it on the weighting balance.

3. Press the re-zero button. 4. Use the spatula to slowly add the baking soda onto the weighting paper.

6. Pour the baking soda powder into the volumetric flask and do not forget to label the concentration. 7. Repeat step 2-6 again for two times until you get 3 concentration of baking soda. 8. Carefully pour distilled water into each of the volumetric flask until it reach the calibration mark and slowly swirl the flask. If the distilled water reach over the calibration mark then, start from obtaining the baking soda powder over again.




Procedure

9. Obtain 1M acetic acid from the lab assistant.

10.Close the strainer of the sink by using the clay and fill the sink with water until it reach half of the volume of the sink.

11. Fill the water in the plastic graduated cylinder until full.

12. Place the plastic graduated cylinder that is already filled with water into the sink. Invert the plastic graduated cylinder upside down and insert the tubing inside the plastic graduated cylinder.

13.Connect the suction flask with the tubing. Use the clay to seal any hole between the tubing and suction flask in order to prevent the gas from leaking out. 14.Place the clay on the top of the suction flask and leave some small hole to pour the chemical into the flask.

Procedure 15.Pour baking soda solution into the glass graduated cylinder until it reach 20 ml. (Only one concentration of baking soda: one by one)

17. Pour 20 ml of acetic acid into the suction flask.

18.Quickly close the lid by using the clay.

16. Pour 20 ml of baking soda solution into the suction flask.

19. Swirl the suction flask in order to stimulate the reaction between baking soda and acetic acid.




Procedure

20.Observe the differences of water level inside the plastic graduated cylinder during before and after the reaction.

21.Record the change in volume of the water level. 22. Repeat step 12-21 again for the other two concentrations of baking soda.

Set Up Below is the sketch of the set up during the experiment:




Results Results Concentration of Baking soda solution (M) 0.25 (2.1 g baking soda +100ml DI) 0.50 (4.2 g baking soda +100ml DI) 1 (8.4 g baking soda +100ml DI)

Volume of Carbon dioxide (ml) 0 70 230

Calculations Calculations PV = nRT

* We did the experiment under the condition of pressure = 1 atm, temperature = 25 celsius.

Experimental 1M -3

1 atm x 230 x 10

L = n x 0.0821_______ L.atm x (273 + 25) K -3 -3 K.mol n = ______________ 230 x 10 = 9.40 x 10 mol of CO 2 0.0821 x 298

How many grams of gas have been produced? 1 mole of CO 2 = 44 grams -3

-3

9.40 x 10 mole of CO2 = 44 x 9.40 x 10 grams = 0.4136 grams of CO 2 ________________ 1




0.5M 1 atm x 70 x 10

-3

L = n x 0.0821 _______ L.atm x (273 + 25) K -3 -3 K.mol n = ______________ 70 x 10 = 2.86 x 10 mol of CO 2 0.0821 x 298

How many grams of gas have been produced? 1 mole of CO 2 = 44 grams -3

-3

2.86 x 10 mole of CO2 = ________________ 44 x 2.86 x 10 grams = 0.1259 grams of CO2 1

0.25M -3

1 atm x 0 x 10

L = n x 0.0821 _______ L.atm x (273 + 25) K K.mol n = ______________ 0 = 0 mol of CO 2 0.0821 x 298

How many grams of gas have been produced? 1 mole of CO 2 = 44 grams 0 mole of CO 2 = 44 x 0 grams = 0 grams of CO 2 _______________ 1

Theoretical From the chemical equation of the reaction: CH 3COOH + NaHCO 3 → CH3COONa + CO2 + H 2O It could be seen that the ratio of the mole of acetic acid, baking soda, and carbon dioxide is the same, which is one. This means that the amount of moles of carbon dioxide would be equal to the amount of moles of the limiting agent.




1M

1 mole baking soda = 0.1 mol baking soda 8.4 g baking soda = 8.4 g baking soda x ________________ 84 g baking soda

There is 0.1 mol of baking soda in 100 ml of solution. Hence, in 20 ml of solution, there should mol baking soda x 20 ml = 0.02 mol baking soda ________________ be: 0.1 100 ml

mol acetic acid x 20 ml = 0.02 mol acetic acid 1 M acetic acid = 1________________ 1000 ml solution

In this case, both of the reactants are added at the same amount, which is 0.02 mol 44 g CO2 = 0.88 g CO Hence, 0.02 mol or 0.02 mol CO2 x __________ 2 1 mol CO2 actual yield grams of CO %yield = ________________ x 100% = 0.4136 __________________2 x 100% = 47% theoretical yield 0.88 grams of CO2

0.5M mole baking soda = 0.05 mol baking soda 4.2 g baking soda = 4.2 g baking soda x 1_______________ 84 g baking soda

There is 0.1 mol of baking soda in 100 ml of solution. Hence, in 20 ml of solution, there should mol baking soda x 20 ml = 0.01 mol baking soda be: 0.05 _________________ 100 ml

1 mol acetic acid 1000 ml solution

1 M acetic acid =_________________ x 20 ml = 0.02 mol acetic acid In this case, baking soda is the limiting agent; hence the amount of carbon dioxide produced in moles would be equal to the moles of the limiting agent, which is the baking soda 44 g CO2 = 0.44 g CO Hence, 0.01 mol or 0.01 mol CO 2 x __________ 2 1 mol CO2

actual yield 0.1259 grams of CO2 x 100% = 26.6% %yield = _______________ x 100% = ___________________ theoretical yield

0.44 grams of CO 2




0.5M 1 mole baking soda 84 g baking soda

2.1 g baking soda = 2.1 g baking soda x ________________ = 0.025 mol baking soda There is 0.1 mol of baking soda in 100 ml of solution. Hence, in 20 ml of solution, there should mol baking soda x 20 ml = 0.005 mol baking soda ___________________ be:0.025 100 ml

1 mol acetic acid x 20 ml = 0.02 mol acetic acid 1 M acetic acid = _________________ 1000 ml solution

In this case, baking soda is the limiting agent; hence the amount of carbon dioxide produced in moles would be equal to the moles of the limiting agent, which is the baking soda 44 g CO2 Hence, 0.005 mol or 0.005 mol CO2 x__________ = 0.22 g CO 2 actual yield theoretical yield

1 mol CO2

0 gram of CO2 %yield = ________________ x 100% = _________________ x 100% = 0% 0.22 grams of CO2

Analysed Data

The equation of the reaction happened in this experiment is CH 3COOH + NaHCO 3 → CH 3COONa + CO2 + H 2O CH 3COOH is acetic acid NaHCO3 is baking soda (sodium bicarbonate) We can read the volume from the air inside the plastic graduated cylinder because the gas produced from the reaction displaced the water inside the cylinder; hence, the volume of the gas simply could be read from the scale of the plastic graduated cylinder by eyes. According to the graph and data, the more concentrated the baking soda solution is, the more Carbon dioxide gas was produced. The result is proving the Avogadro’s gas law which stated that the mole of gas is directly proportional to the volume of gas. The increasing of concentration of baking soda (the number of mole were increased) causes more volume of carbon dioxide to be produced in the reaction. Our result shows that more concentration of reactant we put in the reaction, the more product will be produced from the reaction. We found that when we use 1 molar of baking soda solution, we can collect the most carbon dioxide gas by volume, which is 230 mL. When we use 0.5 molar of baking soda solution, we can collect less carbon dioxide gas than when 1 molar of the solution was used within the reaction with the volume of gas being 70mL. Lastly, the gas collected from the reaction using 0.25 molar of baking soda solution has the least volume which is 0 mL.

CH
 3 COOH

+

NaHCO3

→

CH3COONa +

CO2 +

H 2O

Limiting Reagent

1M 100 mL = 8.4 g of NaHCO3 20 mL = 1.68 g of NaHCO3 = 0.02 mole of NaHCO 3 1 mol NaHCO3

= 1 mol of CO 2

From the experiment

0.02 mol NaHCO 3

= 0.02 mol of CO2

V = 230 mL

PV = nRT P (230 x 10-3 L) = (0.02 mole)(0.082)(273+25K) P = 2.125 atm

0.5M 100 mL = 4.2 g of NaHCO3 20 mL = 0.84 g of NaHCO3 = 0.01 mole of NaHCO 3 0.01 mole of CO 2 was produced. PV = nRT -3 P (70 x 10 L) = (0.01 mole)(0.082)(273+25K) P = 3.49 atm

0.25M 100 mL = 2.1 g of NaHCO3 20 mL = 0.42 g of NaHCO3 = 0.005 mole of NaHCO 3 0.005 mole of NaHCO3 was produced. PV = nRT P (0 L) = (0.005 mole)(0.082)(273+25K) P = ERROR

Total pressure of the reaction Total pressure of the reaction = pressure of the dry gas + pressure of the water vapor @ 25 Celsius Where: pressure of the vapor @ 25 Celsius = 23.8 torr = approximately 0.031 atm

1M

Total pressure of the reaction = pressure of the dry gas + pressure of the water vapor @ 25 Celsius = 2.125 + 0.031 = 2.156 atm

0.5M

Total pressure of the reaction = pressure of the dry gas + pressure of the water vapor @ 25 Celsius = 3.49 + 0.031 = 3.521 atm

0.25M

Total pressure of the reaction = couldn’t be found due to the error of the experiment

Discussions When baking soda and acetic acid was mixed together and reacted, bubbles of Carbon dioxide gas was produced along with with sodium acetate and water. This reaction could be represented with the chemical equation: CH 3COOH + NaHCO 3 → CH 3COONa + CO2+ H 2O. The pressure of the dry gas resulting from the reaction could be found by using the ideal gas law: PV = nRT. As seen within the calculation section of this report, it could be seen that the pressure of the dry gas when the concentration of the baking soda solution was 1 M is 2.125 atm while the pressure of the dry gas when the concentration of the baking soda solution was 0.5 M is 3.49 atm. However, due to the error occurred during the experiment, the pressure of the dry gas while 0.26 M of baking soda solution was used couldn’t be calculated. From the observation of the pressure of the gas and the concentration of the baking soda solution, it could be seen that the pressure of the dry gas tends to be inversely proportional to the concentration of the baking soda solution or the number of moles of the gas produced. Moreover, if Dalton’s law was applied in order to find the total pressure of the reaction, the relationship between pressure of the reaction and the concentration of the baking soda or the number of moles of the gas also tends to be inversely proportional. In the first trial, where the concentration of the baking soda solution is 1 M, 0.02 mol of CO2 has been produced. On the other hand, on the second trial, where the concentration of the baking soda is halved to be 0.5 M, only 0.01 mol of CO2was produced. Lastly, from the last trial whereas the concentration has been changed into 0.25 M, there is no gas bubble present; hence, 0 M of CO2 was produced from the reaction. In the first trial, where the concentration of the baking soda solution is 1 M, there are 0.02 mol of CO that was produced, when converted to grams, the amount of of the gas produced is 0.88 grams of CO2 . In the next trial, where the concentration of the baking soda solution is halved to be 0.5 M, when converted to grams, the amount of CO 2 produced is 0.44 grams of CO2. Lastly, since there is no gas produced during the third trial, in which the concentration of the baking soda is 0.25 M, the amount of CO2 created from the reaction in grams is 0 gram. Last but not least, there are some errors during the experiment which are trapping the air inside the plastic graduated cylinder which made the amount of carbon dioxide measured inaccurate. The way to reduce the air is to increase water level inside the sink so that when flip the cylinder, the air cannot gets inside the cylinder. Secondly, the amount of carbon dioxide measured might not be the amount of carbon dioxide chemical produced. Carbon dioxide produced may leak or trap in the tube or inside the volumetric flask. This problem can be reduced by decrease the length of the tube. Lastly, the tilted base of the sink results inaccurate amount of carbon dioxide when read. This can solve by change the place to where the base is straight.

Conclusion From this experiment, our group collected the gas that was produced by the reaction between acetic acid and baking soda by using the collecting gas over water 
 technique. In order to obtain the correct amount of carbon dioxide produced, we need to use the ideal gas law (PV = nRT) to solve for the amount of moles of carbon dioxide first and in identifying the pressure of dry gas Dalton’s Law is also involved. The concentration of baking soda that can produce the highest amount of carbon dioxide is 1M with the amount of carbon dioxide of 0.4136 grams.

IIddee aall G Gaass LLaaw w

PV = nR T

References [1]
 Collection of Gas Over Water. (n.d). Retrieved November 11, 2017 from http://chemed.chem.purdue.edu/genchem/lab/techniques/gascollect.htm [2] Dalton’s Law (Law of Partial Pressure). (August, 2015) Retrieved November 11, 2017 from https://chem.libretexts.org/Core/ Physical_and_Theoretical_Chemistry/Physical_Properties_of_Matter/ States_of_Matter/Properties_of_Gases/Gas_Laws/ Dalton’s_Law_(Law_of_Partial_Pressures) [3] Gas Laws. (n.d). Retrieved November 11, from http://chemistry.bd.psu.edu/ jircitano/gases.html [4] Ideal Gas Law. (n.d.) Retrieved November 11, 2017 from http://hyperphysics.phy-astr.gsu.edu/hbase/Kinetic/idegas.html [5] National Center for Biotechnology Information. (n.d). Acetic Acid. Retrieved November 11, 2017 from https://pubchem.ncbi.nlm.nih.gov/compound/ acetic_acid#section=Top [6] National Center for Biotechnology Information. (n.d). Sodium Bicarbonate. Retrieved November 10, 2017 from https://pubchem.ncbi.nlm.nih.gov/ compound/sodium_bicar bonate #section=Top