Archer G11
10 November 2011
Determining the Molar Volume of a Gas
Purpose: The purpose of this lab is to determine the molar volume of H2 at room temperature and
standard pressure. This can be done by producing H2 gas from the reaction between Mg ribbon and HCl
then calculate the pressure of H2 using Dalton’s law. The calculated pressure can then be used to find
the molar volume of H2 using the ideal gas equation. The significance of this lab is that the aeronaut can
determine the thermo-heat capacity of the balloon and calculate the expansion of volume of gas at
certain elevations.
Hypothesis: The hypothesis is that, with correct pressure and temperature, an accurate molar volume of
H2 can be calculated using the ideal gas equation. In the eudiometer, there’s high volume and low
pressure. Thus, the molecules in the eudiometer act similarly to ideal gases because the volume of each
molecule would be so tiny that it is negligible compared to the volume of the container that it takes up.
Low pressure would lower the intermolecular attraction between molecules. Thus, the molecules would
behave similarly to the ideal gas, which has no volume or intermolecular attraction.
Materials:
Materials
Magnesium Ribbon (Mg)
2 M Hydrochloric Acid (HCl)
Tap water
Distilled water
Eudiometer
Copper wire
600-mL beaker
500-mL graduated cylinder
25-mL graduated cylinder
Thermometer
Funnel
Plier with wire cutter
One-hole rubber stopper
Ring stand
Universal clamp
0.0001-g precision balance
Trial 1
0.0283 g
Trial 2
Trial 3
0.0290 g
0.0260 g
75 mL
500 mL
1200 mL
2 eudiometers
50 cm
2 beakers
1 cylinder
1 cylinder
1 thermometer
1 funnel
1 plier with wire cutter
2 rubber stoppers with hole
2 stands
2 clamps
1 balance
Procedures:
1.)
2.)
3.)
4.)
5.)
Add about 500 mL of distilled water to a 600-mL beaker
Use a plier to cut out a piece of about 0.029 g magnesium
Weigh the piece of magnesium
Repeat step 2 and 3 until the piece of magnesium is within the range of 0.028 g and 0.030 g
Cut the copper wire so that it is about 25 cm long
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6.) Strip the insulation off the copper wire
7.) Twist the copper wire around the magnesium piece tightly
8.) Put the straight end of the wire through a one-hole rubber stopper
9.) Measure 25 mL of 2 M HCl to a 25-mL graduated cylinder
10.) Slowly add the HCl into the eudiometer
11.) Gently add distilled water into the eudiometer until it’s full
12.) Plug the eudiometer with the one-hole rubber stopper so that the magnesium about 7 cm into
the eudiometer
13.) Twist the copper wire that is out through the hole around the neck of the eudiometer
14.) Cover the hole with a finger
15.) Turn the eudiometer upside down
16.) Put the eudiometer in the 600-mL beaker with water
17.) Let go of the finger
18.) Attach the eudiometer to a ring stand with a universal clamp
19.) Adjust the clamp so that the hole of the eudiometer is under water but not at the bottom
20.) Wait until magnesium completely reacts with HCl
21.) Fill a 500-mL graduated cylinder with tap water
22.) Cover the hole of the eudiometer with a finger
23.) Move the eudiometer to the 500-mL graduated cylinder with tilting it
24.) Let go of the finger
25.) Gently put the eudiometer at the bottom of the cylinder
26.) Regulate the amount of water so that the water level inside and outside the eudiometer are the
same
27.) Record the data
28.) Measure the temperature of the water in the cylinder
29.) Repeat step 1 to 28 for two more trials
Results: The magnesium ribbon was soft enough to be bend and break apart by hands. Copper was a lot
harder to bend into small loops compared to the magnesium ribbon. Some time was needed before the
Mg start reacting with HCl. This was because the surface of Mg was oxidized and coated with a thin layer
of magnesium oxide (MgO). Thus, it took a while before the HCl came in touch with the Mg. When
magnesium reacted with HCl, lots of small bubbles were produced. Occasionally, there will be some big
bubble produced from the reaction. The magnesium piece became smaller in the process of reaction.
The magnesium piece turn white when react with HCl before dissolving into colorless liquid.
Mg(s) + 2HCl(aq) → MgCl2(aq) + H2(g)
Mass of magnesium piece (g)
Evidence of chemical reaction
Volume of H2 gas (mL)
Molar Volume Determination Table
Trial 1
Trial 2
Trial 3
0.0283
0.0290
0.0260
Bubbles were produced from the from the magnesium
piece and the piece slowly gets smaller
27.6
28.3
25.1
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Corrected volume of H2 (mL)
Ideal volume of H2 (mL)
Barometric pressure (atm)
Barometric pressure (torr)
Temperature of the tap water (K)
Pressure of water at the temperature
(torr)
Partial pressure of H2 (torr)
Partial pressure of H2 (atm)
Theoretical moles of H2 gas (mols)
Molar Volume of H2 (L/mols)
Literature molar volume of a gas at STP
(L/mol)
28.44
25.31
297.15
29.16
25.95
0.9978
758.3
297.15
25.82
23.10
296.15
22.38
22.38
21.07
735.9
0.968
1.16 × 10-3
21.74
735.9
0.968
1.19 × 10-3
21.75
737.2
0.970
1.07 × 10-3
21.59
22.41
(http://wwwchem.csustan.edu/chem1102/molarvol.htm )
Percent Error (%)
Calculated Density of H2 gas at STP (g/L)
Literature Density of H2 gas at STP (g/L)
2.99
9.292 × 10-2
(http://en.wikipedia.org/wiki/Hydrogen)
2.95
9.287 × 10-2
3.66
9.356 × 10-2
8.988 × 10-2
Analysis:
Barometric pressure in torr = (Barometric pressure in atm) × 760 torr/atm
Barometric pressure in torr equals barometric pressure in atm times 760 torr/atm
0.9978 × 760 = 758.3 torr
Theoretical moles of H2 gas = [(Mass of Mg) ÷ (Molar mass of Mg)] × (Mole Ratio)
Theoretical moles of hydrogen gas equal mass of Mg divide by molar mass of Mg then times by the mole
ratio
Trial 1: (0.0283 ÷ 24.31) × (1 ÷ 1) = 1.164 × 10-3 mole H2
Trial 2: (0.0290 ÷ 24.31) × (1 ÷ 1) = 1.193 × 10-3 mole H2
Trial 3: (0.0260 ÷ 24.31) × (1 ÷ 1) = 1.070 × 10-3mole H2
Partial pressure of H2 in torr = (Barometric pressure in torr) – (Pressure of water at the recorded
temperature)
Partial pressure of H2 in torr equal the barometric pressure in torr minus the pressure of water at the
temperature
Trial 1: 758.3 – 22.38 = 735.92 torr
Trial 2: 758.3 – 22.38 = 735.92 torr
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Trial 3: 758.3 – 21.07 = 737.23 torr
Partial pressure of H2 in atm = (Partial pressure of H2 in torr) ÷ 760 torr/atm
Partial pressure of H2 in atm equals partial pressure of H2 in torr divided by 760 torr/atm
Trial 1: 735.92 ÷ 760 = 0.968 atm
Trial 2: 735.92 ÷ 760 = 0.968 atm
Trial 3: 737.23 ÷ 760 = 0.970 atm
Corrected Volume of H2 = [(Barometric pressure in torr) × (Volume of H2 gas)] ÷ (Partial pressure of H2)
Corrected volume of H2 equals barometric pressure in torr times volume of H2 gas then divided by the
partial pressure of H2
Trial 1: (758.3 × 27.6) ÷ 735.9 = 28.44 mL H2
Trial 2: (758.3 × 28.3) ÷ 735.9 = 29.16 mL H2
Trial 3: (758.3 × 25.1) ÷ 737.2 = 25.82 mL H2
Ideal Volume of H2 = [(Partial pressure of H2 in atm) × (Corrected volume of H2)× (273.15K)] ÷
[(Temperature of the tap water) × (1 atm)]
Ideal volume of H2 equals the product of partial pressure of H2 in atm, corrected volume of H2 and
273.15K divide by the product of temperature of the tap water and 1 atm
Trial 1: (0.968 × 28.44 × 273.15) ÷ (297.15 × 1) = 25.31 mL
Trial 2: (0.968 × 29.16 × 273.15) ÷ (297.15 × 1) = 25.95 mL
Trial 3: (0.970 × 25.82 × 273.15) ÷ (296.15 × 1) = 23.10 mL
Molar Volume of H2 = (Ideal volume of H2) ÷ (Theoretical moles of H2 gas)
Molar volume of H2 equals ideal volume of H2 divide by theoretical moles of H2 gas
Trial 1: 0.02531 ÷ (1.164 × 10-3) = 21.74 L/moles H2
Trial 2: 0.02595 ÷ (1.193 × 10-3) = 21.75 L/moles H2
Trial 3: 0.02310 ÷ (1.070 × 10-3) = 21.59 L/moles H2
Percent error = |(Molar volume of H2) – (Literature molar volume of a gas at STP)| ÷ (Literature molar
volume of a gas at STP
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Percent error equals to the differences between molar volume of H2 and the literature molar volume of
a gas at STP divide by literature molar volume of a gas at STP
Trial 1: |21.74 – 22.41| ÷ 22.41 = 2.99%
Trial 2: |21.75 – 22.41| ÷ 22.41 = 2.95%
Trial 3: |21.59 – 22.41| ÷ 22.41 = 3.66 %
Calculated Density of H2 gas at STP = [1 ÷ (Molar volume of H2)] × (Molar mass of H2)
Calculated density of H2 gas at STP equals the reciprocal of the molar volume of H2 times by the molar
mass of H2
Trial 1: (1 ÷ 21.74) × 2.02 = 9.292 × 10-2 g/L
Trial 2: (1 ÷ 21.75) × 2.02 = 9.287 × 10-2 g/L
Trial 3: (1 ÷ 21.59) × 2.02 = 9.356 × 10-2 g/L
The hypothesis could be confirmed by the experiment. There was only about 3.2 percent difference
between the calculated molar volume and the literature molar volume. The difference in between the
density calculated from the result of the experiment and the literature density was not even 1 g/L, so
there was not much difference between them. If a bubble of air leaded into the eudiometer tube, the
measured volume of hydrogen gas would become higher. This is because the bubble of air would be
regarded as part of hydrogen’s volume. If the measured volume of hydrogen gas is higher, the calculated
molar volume of hydrogen would also be too high. The volume of air would be included in the
calculations causing the molar volume of hydrogen to be higher than normal. If the magnesium ribbon
had been oxidized, the measured volume of hydrogen gas would be lower than it should be because the
oxidized part of the ribbon would not be involve in the reaction that produces H2. Thus, there would be
less reaction which occur and so less of the product, H2, would be formed. Lower measured volume of
hydrogen gas would also mean that the calculated molar volume of hydrogen would be too low. This is
because the too-low volume of H2 that was used to calculate the molar volume would be divided by the
normal amount of moles of hydrogen.
Conclusion: The hypothesis was verified through the experiment. However, some errors could have
occurred during the experiment. An example of the errors that could have occurred was that the surface
of Mg was oxidized. If the Mg was coated with a considerably thick layer of MgO, there would be a
significant difference in the molar volume. The volume of H2 produced would be less than what should
have been produced with the number of moles of magnesium that was used. Thus, molar volume
(L/mols) would be lower than normal. Another error that could have happened during the course of the
experiment was that the volume of H2 measured was incorrect. There were still some little tiny bubbles
stuck to the side of the eudiometer when the volume was measured. This could have caused the volume
of H2 measured to be lower than the real volume since the volumes of the tiny bubbles were not added
to the calculations. This error would have caused the molar volume of H2 to be lower than what it
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should have been. This is one likely error that might have happened in this experiment because the
results gotten from the experiment was lower than the literature value. Some ways to prevent the
errors could be to use some sand paper to rub on the magnesium piece in order to remove the MgO
coating from the Mg piece. This would allow the volume of H2 produced to be relatively similar to the
theoretical volume of H2, calculated through stoichiometry, which would have been produced with the
number of moles of Mg used. The other error could be prevent by gently knocking the bottom of the
eudiometer to combine the volume of gas in the eudiometer. This would add the volume of the tiny
bubbles with the volume of H2 at the top which will result in an accurate amount of H2 produced from
the reaction.
Archer G11