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Universiti Tunku Abdul Rahman ( Kampar Campus ) Bachelor of Science (Hons) Biotechnology Year 1 Semester 1 Group 1 Laboratory 1A ( UESB 1112 ) (II) Atomic Structures and Periodicity Lecturer: Ms. Chong Foon Yee Group Member: 1. Ang Pei Wen ( 08AEB02216 ) 2. Chan Pei Sin ( 08AEB03544 ) 3. Cheah Hong Leong ( 08AEB03788 ) Experiment No. 6 Title: Determination of the Valency of Magnesium Date: 9 July 2008

Experiment 6

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Page 1: Experiment 6

Universiti Tunku Abdul Rahman ( Kampar Campus )

Bachelor of Science (Hons) Biotechnology

Year 1 Semester 1

Group 1

Laboratory 1A ( UESB 1112 )

(II) Atomic Structures and Periodicity

Lecturer: Ms. Chong Foon Yee

Group Member:

1. Ang Pei Wen ( 08AEB02216 )

2. Chan Pei Sin ( 08AEB03544 )

3. Cheah Hong Leong ( 08AEB03788 )

Experiment No. 6

Title: Determination of the Valency of Magnesium

Date: 9 July 2008

Page 2: Experiment 6

Title: Determination of the Valency of Magnesium

Objectives:- Study the quantitative relations between amounts of reactants and products of a

reaction.- Know the starting mass of magnesium and the measured collection of hydrogen gas to

determine the reaction stoichiometry.

Theory and Background:!@#$

Experiment Procedures:

Apparatus and Materials- Magnesium ribbon HCl (0.5M) Burette (50cm3) Pipette (25cm3) Retort stand Sand paper Watch glass Beaker (1000cm3) Gauze Funnel Glass rod Thermometer Analytical balance Beaker (100cm3) Distilled water Pipette filler Hand glove

Procedures-1. The burette was used upside down to collect the hydrogen. The volume of the

unmarked space in a clean, dry 50cm3 burette was determined by pipetting 25.00cm3

of water into the vertically clamped burette (right way up).The burette reading was noted, drained and repeated. The water in the burette was left for 10 min. and checked for leaks.

2. A piece of magnesium was cleaned with steel wool. A piece was cut with scissors within the length shown. The ribbon was curled up. A watch glass was tare on the four decimal balance and the magnesium ribbon was ACCURATELY weighed between 0.0300 and 0.0360g on the watch glass, which then was placed inside a 600cm3

beaker.3. A small filter funnel with a short stem (1.0-1.5cm long) was taken and covered with

gauze. The watch glass was invert and placed over the magnesium.4. The beaker was carefully filled with (tap) water until the level was approximately 0.5-

1.0cm above the end of the funnel stem. The burette was completely filled with 0.5M

Page 3: Experiment 6

HCl. The beaker was inverted and placed in the water in the beaker, and the end of the burette was placed over the stem of the funnel, no air enters was ensured and clamp it into position.

5. The excess of water was removed with a pipette until the level was just above the stem of the funnel.

6. About 100cm3 of 0.5M HCl was added to the beaker, a glass stirring rod was used to ensure complete mixing such that the HCl reached the magnesium. This was helped by the tapping of watch glass gently with a glass rod.

7. The solution was stirred to initiate the reaction and was not stir further to left the reaction proceeds unaided. At the completion of the reaction the watch glass was tapped gently to dislodge and gas bubbles.

Experiment Data:

Burette reading upside down after pipetting 25.00cm3 of water (cm3) 34.9

Burette reading upside down after the completion of the reaction (cm3) 43.3

Weight of magnesium ribbon (g) 0.0163

Analysis and Calculation:

Volume of mark region of burette = 50.00cm3 - Burette reading upside down after pipetting 25.00cm3 of water = 50.00cm3 - 34.90cm3

= 15.10cm3

Volume of unmark region of burette = 25.00cm3 - volume of mark region of burette = 25.00cm3 – 15.10cm3

= 9.9cm3

Volume of hydrogen = 50.00cm3 - (Burette reading upside down after the completion of the gas, H2 reaction) - (Volume of unmark region of burette) = 50.00cm3 - 43.2cm3 - 9.9cm3

= 16.7cm3

n(Mg) = Weight of magnesium ribbon Relative Atomic mass for Mg

= 0.0163g 24.3

= 6.7078 x 10-4mol

Page 4: Experiment 6

pV = nRT

p = 1atm R = 0.08206 L atm K-1 mol-1

T = (27 + 273)K = 300K V = volume for H2

n = number of moles for H2

pV = nRT

p = 1atmR = 0.08206 L atm K-1 mol-1

T = 300KV = 16.6 x 10-3 m3

n = n(H2)

n = pV RT

n(H2) = 1atm x (16.6 x 10 -3 m 3 ) . (0.08206 L atm K-1 mol-1) x 300K

= 6.7480 x 10-4 mol

Mg + xHCl MgClx + X/2 H2

1 (6.7078 x 10-4mol) = X/2 (6.7430 x 10-4 mol)

X = 6.7078 x 10 -4 mol 2 6.7430 x 10-4 mol

X = 1.9896

Therefore X ≈ 2

Discussion and Precaution:

Conclusion:

Reference:Lim You Sie, Yip Kim Hong (2006). Pre-U Text STPM Chemistry, Longman, Pearson Malaysia Sdn.Bhd.

Questions:

Page 5: Experiment 6

1. Note that the temperature is not taken for at least 20 min. after adding the HCl. Why is this?The temperature is taken after 20 minutes after adding the HCl is because of the heat release by the formation of MgCl (heat of summation) will affect the calculation. Based on the equation pV= nRT, temperature is inverse to the number of moles. Increase in the temperature will decrease the number of moles while decrease in temperature will increase the number of moles. The temperature has to be stable before we determine the number of moles of H2 released in the reaction.

2. Calculate the moles of hydrogen present using the given calculation method.pV=nRTp = 1 atm, R=0.08206, T=(27+273) Kelvin V = 16.7cm3 = 16.7ml/1000

=0.0167L

n=PV /RT = 1 x 0.0167 / 0.08206 x 300 = 6.784 x 10-4 mole

3. Give the Ideal Gas equation and specify what each variable is. Show one mole of gas at S.T.P. occupies 22.4 L.

Ideal gas equation is the combination of Boyle’s law, Charles’ law and Avogadro’s law.

Ideal gas equation: pV = nRT

Where p= pressure (must be in atm), V= volume (measured in dm3 or L), T = temperature in Kelvin, R = gas constant: 0.08206dm3atmK-1 and n= number of moles.

V = nRT p

= (1mol) x (0.08206 L atm K -1 mol -1 ) x (273K) 1atm

= 22.4024 L

≈ 22.4 L

Page 6: Experiment 6

4. What will be the result if hydrogen gas, H2, leaks through the stopcock of the inverted burette?If hydrogen gas, H2 leaks through the stockpile of the inverted burette, the calculation of the numbers of moles of H2 will be interrupted. Based on the ideal gas equation, the volume of gas is proportional to the number of moles. The increase of volume of H2

will increase the number of moles while the decrease of the volume will decrease the number of moles H2 produced.