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 Trial 1 Trial 2 Trial 3

Magnesium Ribbon (Mg) 0.0283 g 0.0290 g 0.0260 g

2 M Hydrochloric Acid (HCl) 75 mL

Tap water 500 mL

Distilled water 1200 mL

Eudiometer 2 eudiometers

Copper wire 50 cm

600-mL beaker 2 beakers

500-mL graduated cylinder 1 cylinder

25-mL graduated cylinder 1 cylinder

Thermometer 1 thermometer

Funnel 1 funnel

Plier with wire cutter 1 plier with wire cutter

One-hole rubber stopper 2 rubber stoppers with hole

Ring stand 2 stands

Universal clamp 2 clamps

0.0001-g precision balance 1 balance

Procedures:

1.) Add about 500 mL of distilled water to a 600-mL beaker

2.) Use a plier to cut out a piece of about 0.029 g magnesium

3.) Weigh the piece of magnesium

4.) Repeat step 2 and 3 until the piece of magnesium is within the range of 0.028 g and 0.030 g

5.) 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)

Molar Volume Determination Table

Trial 1 Trial 2 Trial 3

Mass of magnesium piece (g) 0.0283 0.0290 0.0260

Evidence of chemical reaction Bubbles were produced from the from the magnesium piece and the piece slowly gets smaller

Volume of H2 gas (mL) 27.6 28.3 25.1

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Corrected volume of H2 (mL) 28.44 29.16 25.82

Ideal volume of H2 (mL) 25.31 25.95 23.10

Barometric pressure (atm) 0.9978

Barometric pressure (torr) 758.3

Temperature of the tap water (K) 297.15 297.15 296.15

Pressure of water at the temperature (torr)

22.38 22.38 21.07

Partial pressure of H2 (torr) 735.9 735.9 737.2

Partial pressure of H2 (atm) 0.968 0.968 0.970

Theoretical moles of H2 gas (mols) 1.16 × 10-3 1.19 × 10-3 1.07 × 10-3

Molar Volume of H2 (L/mols) 21.74 21.75 21.59

Literature molar volume of a gas at STP (L/mol) (http://wwwchem.csustan.edu/chem1102/molarvol.htm )

22.41

Percent Error (%) 2.99 2.95 3.66

Calculated Density of H2 gas at STP (g/L) 9.292 × 10-2 9.287 × 10-2 9.356 × 10-2

Literature Density of H2 gas at STP (g/L) (http://en.wikipedia.org/wiki/Hydrogen)

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.

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