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CHEM 1009Molar Volume of O2 and CO2
Required readings:
1. Please refer to Chapter 12 and Appendix IV of the textbook.
2. Read through the lab and answer the pre-lab questions before coming to the lab.
Required Items
Lab coat, safety goggles, shoes (no open toes, flip-flops or sleepers), one laptop or tablet,
textbook, laminated periodic table, calculator, pencil and eraser.
Purpose:
In this experiment students will:
1) Demonstrate the principles of molar gas volume.
2) Confirm the value of the molar volume of two gases (oxygen and carbon dioxide) by
generating each gas in a confined collection apparatus, measuring the volume of gas
and calculating the molar volume corrected for barometric pressure and ambient
temperature.
3) Calculate the measurement error for the experimental values by comparing each
experimental value with the theoretical value of the molar volume for any gas
(22.4 litres/mole at standard temperature and pressure (STP)).
Introduction:
Avogadro’s hypothesis states that equal volumes of all gases contain equal numbers of
molecules under the same conditions of temperature and pressure. It follows from this
hypothesis that all gas samples containing the same number of molecules will occupy the
same volume under the same conditions of temperature and pressure. A special name is
given to the volume occupied by 1 mole samples of gases at STP. This volume is called
molar volume. It has been shown that under standard conditions, (0°C and 100 kPa), the
molar volume of any gas is 22.4L. (Note: these conditions are referred to as standard
temperature and pressure or STP).
This experiment is based on the following reactions: hydrogen peroxide will decompose to
generate oxygen gas (equation 1) and sodium bicarbonate reacts with hydrochloric acid to
generate carbon dioxide gas (equation 2).
(Note, the coefficients are shown in color to help you relate the calculations below to these
equations.)
(1) 2 H2O2 (aq) à 2 H2O (l) + 1 O2 (g)
(2) 1 NaHCO3 (aq) + HCl (aq) à Na+ (aq) + Cl- (aq) + H2O (l) + 1 CO2 (g)
In the experiment a quantity of each solution will be measured accurately. The
decomposition of hydrogen peroxide will proceed with the help of a yeast-based catalyst,
while the displacement reaction will proceed once the hydrochloric acid is added to the
sodium bicarbonate solution. The gasses generated will be collected and measure. The dry
volume of each gas collected will be divided by the number of moles of product gas
generated and compared to the standard molar volume (22.4 liters/mole).
1
CHEM1009: Molar Volume
, Moles of oxygen generated (decomposed):
From equation (1) above it can be seen that for every 2 moles of H2O2 reacted there is 1
mole of O2 produced. Therefore, if 3.0 mL of H2O2 at a 3% concentration is used in the
experiment, a calculation can be done to determine how many moles of O 2 should be
produced. Remember that a 3 % solution of H2O2 means 3.00 g/per 100 mL.
So, the mass of H2O2 = 3.0 mL x (3.00g/100mL) = 0.09 g H2O2.
The molar mass of H2O2 is 34 g/mole.
So, moles of H2O2 used = 0.09 g H2O2 ÷ 34g/mol = 0.00265 mol H2O2
Then, based on the ratios in the equation:
0.00265 mol H2O2 x (1 mol O2 /2 mol H2O2) = 0.00132 mol O2
At STP, this amount of O2 gas would have a volume of:
0.00132 mol O2 x (22.4 L/mol) = 0.02965 L or 29.65 mL O2 gas at STP
In this experiment the weight of the hydrogen peroxide used to generate oxygen will be
determined using a balance, accurate to 2 decimal places. A non-toxic catalyst, an enzyme
found in Saccharomyces cerevisiae (yeast), will be added to cause the decomposition of
hydrogen peroxide into oxygen gas and water at a controlled rate. This allows time for
capping the reaction tube without losing gas once the catalyst has been added to the tube
containing the reactant solution.
Moles of carbon dioxide generated:
From equation (2) above it can be seen that for 1 mole of NaHCO3, there is generate 1
mole of CO2. Therefore, if 3.00 g of NaHCO3 at 0.4 M (moles/liter) is used in the experiment
a calculation can be done to determine how many moles of CO 2 should be produced.
Note: In this situation 3.00 g = 3.00 mL and remember that Concentration (C) is in
moles/litre and so moles = C x V, where V is volume in liters (L). Also, remember, to
convert mL into Litres divide by 1000mL/L.
So, the moles of NaHCO3 used = 0.4 moles/litre x [3mL / (1000 mL/L)] = 0.0012 mol
NaHCO3
Then, based on the ratios in the equation:
0.0012 mol NaHCO3 x (1 mol CO2 /1 mol NaHCO3) = 0.0012 mol CO2
At STP, this amount of CO2 gas would have a volume of:
0.0012 mol CO2 x (22.4 L/mol) = 0.02688 L or 26.88 mL CO2 gas at STP
2
CHEM1009: Molar Volume
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