Gas Laws
In this experiment, we will use yeast to accelerate the decomposition of the hydrogen peroxide into water and O2 gas. Yeast contains the enzyme catalase, which is a catalyst for this reaction. You will add yeast activated in warm water to a known amount of hydrogen peroxide and quickly seal off the system so that the O2 gas formed is collected in a graduated cylinder. After measuring the total volume of gas produced, its temperature, and the atmospheric pressure, the ideal gas law can then be used to calculate how many moles of O2 gas is formed. We can do this by solving the ideal gas law equation for n.
PV n = RT Once the number of moles of O2 gas is calculated, the percent of H2O2 present in the solution can be determined. To do this, you first need to calculate the theoretical number of moles of O2 there would be if the solution was 100% hydrogen peroxide. This can be found by using the following equation: For this experiment: mL H2O2 used is the volume of H2O2 you actually use (approximately 5 mL). H2O2 density is 1.02 g/mL 1 mol H2O2 / 34.0 g H2O2 is the reciprocal (inverted fraction). of the molar mass of H2O2 . The molar mass of H2O2 is 34.0 g /mol, so this is equal to 1 mol H2O2 / 34.0 g H2O2. 1 mol O2 / 2 mol H2O2 is used since the decomposition produces 1 mole O2 from 2 moles of H2O2 . The units in the entire equation cancel to give moles of O2. The percent hydrogen peroxide can now be found. To do this, divide (n), the actual number of moles you calculated, by the theoretical moles of O2 there would be if the hydrogen peroxide were 100%. This number is then multiplied by 100%. This value can now be compared to the 3% hydrogen peroxide shown on the label to see if any decomposition has occurred. 22 2 2 22 22 22 22 1 mol HO 1 mol O Theoretical moles O = H O used × H O density × × 34.0 g HO 2 mol H O
