Calculating the Percentage Composition of a Commercial Hydrogen Peroxide Solution
The purpose of this lab was to calculate the percentage composition of a commercial hydrogen peroxide solution. Since commercial hydrogen peroxide consists of hydrogen peroxide and water, we separated the two substances with yeast, a catalyst, and measured the mass of the displaced water. We converted this non-STP mass to molar mass using the combined gas law and, thus, determined the weight percentage of the solution.
1. Fill a trough and 250 milliliter (mL) Erlenmeyer flask with water. Place and hold a glass square tightly on the top of the flask. Invert the flask and place it under the trough’s water surface. As the flask is the eudiometer, keep its mouth submerged at all time and trap as little air as possible inside it. Remove the glass square and center the flask’s mouth over the groove on the trough’s bottom.
2. Using a graduated cylinder, pour 10 mL of hydrogen peroxide into a 125 mL reaction flask. Dilute the hydrogen peroxide solution with 15 mL of water. Attach the reaction flask to a ring stand with a testube or buret clamp.
3. Obtain .40 grams of yeast.
4. Record the barometric pressure.
5. Slip the mouth of a rubber stopper’s tube into the trough under the Erlenmeyer flask’s mouth. Keep the tube in position throughout the experiment. Add .40 grams (g) of yeast to the reaction flask and immediately plug the stopper on the reaction flask. Oxygen will bubble from the reaction flask to the eudiometer for 5-10 minutes. Agitate the reaction flask to help the contact of hydrogen peroxide and yeast.
6. When the reaction is complete and oxygen has displaced a portion of the water in the eudiometer, remove the rubber tubing from the water.
7. Replace the glass square on the Erlenmeyer flask’s mouth. Remove the flask from the trough water without losing any of its contents. Place the flask in an upright position and remove its contents.
8. Record the temperature of the water in the flask. Dry the outside of the partially filled flask, then weigh and record its mass.
9. Completely fill the flask with water again; weigh its mass.
After yeast was added to docile, clear, commercial hydrogen peroxide, oxygen bubbled from the reaction flask to the Erlenmeyer flask through the reaction flask’s rubber tubing. The water level inside the eudiometer decreased. Once the reaction ended, a significant amount of oxygen resided on top of the Erlenmeyer flask; the hydrogen peroxide’s volume was notably smaller and it now had a sickly yellow color.
IV. Equations and Chemical Formulas
Decomposition of Hydrogen Peroxide: 2H2O2 —-yeast—-> 2H2O + O2
Conversion Cycle: Volume of Wet O2 -> Volume of Dry O2 @ STP -> Moles O2 -> Moles H2O2 -> Grams H2O2 -> Weight Percent H2O2
Mass Difference Between Totally and Partially Filled Flask: Mass of Totally Filled Flask – Mass of Partially Filled Flask
Density of Water: 1 g/mL
Mass Difference Between Flasks (g) = Volume of Water Displaced (mL)
Water Vapor Pressure: See p899 of “Modern Chemistry”
Oxygen Pressure at Experimental Conditions (torr): Barometric Pressure – Water Vapor Pressure – (Height Difference of Eudiometer Compared to Trough ÷ 13.6)
Combined Gas Equation: V1P1T2=V2P2T1
Used for Oxygen Volume at STP
V=Volume, P=Pressure, T=Temperature (Kelvin)
°K (Kelvin): °C + 273
STP: 273°K, 760 torr
Mol: 22,400 mL of Gas @ STP
Stoichiometry (Coefficients Derived From Balanced Equation)
(Mass of Known Substance) = Moles of Known
(Molar Mass of Known)
(Moles of Known)(Coefficient of Unknown) = Moles of Unknown
(Coefficient of Known)
Moles of Unknown * Molar Mass of Unknown = Mass of Unknown Substance
Used for Moles of Peroxide in Original Sample, Grams of Peroxide in Original Sample
Weight Percent of Peroxide in Original Sample: (Mass of Peroxide ÷ Mass of Solution) * 100
Average Percent of Peroxide in Original Sample: (Weight % From Trial 1)(Weight % From Trial 2) ÷ 2
V. Data and Calculations
Category Trial 1 Trial 2
Volume of Peroxide Sample (mL) 10 10
Barometric Pressure (torr) 743 742
Water Temperature (°C) 24 23
Mass of Flask Filled With Water (g) 363.9 364.1
Mass of Flask Partially Filled with Water (g) 250.65 253.2
Height Difference of Eudiometer Compared to Trough (mm) +85 +85
Category Trial 1 Trial 2
Mass difference between totally and partially filled flask (g) 363.9 – 250.65 = 113.25 364.1 – 252.2 = 110.9
Volume of water displaced (mL) 113.25 110.9
Water vapor pressure (torr) 22.4 21.1
Oxygen pressure at experimental conditions (torr) 743 – 22.4 – 85/13.6 = 714.35 742 – 21.1 – 85/13.6 = 714.65
Oxygen volume at standard conditions (mL) (113.25)(714.35)(273) = 97.85
(297)(760) (110.9)(714.65)(273) = 96.18
Moles of oxygen from original sample 97.85/22,400 = 0.00437 96.18/22,400 = 0.00429
Moles of peroxide in original sample 0.00437*2=0.00874 0.00429*2=0.00859
Grams of peroxide in original sample 0.00874 mol(34g)=0.297g
mol 0.00859 mol (34g) = 0.292g
Percent of peroxide in original sample 0.297g * 100 = 2.97%
10.0g 0.292g * 100 = 2.92%
Average percent of peroxide in original sample (2.97%)(2.92%) = 2.9%
The purpose of the experiment was to measure the percent composition of commercial hydrogen peroxide. The solution, billed as 3.0% hydrogen peroxide, was actually closer to 2.9%. Our experiments were accurate, so I can only assume that the company was padding its statistics. I would be upset about this, but I don’t think it makes a bit of difference.
Overall, this experiment was a great success. Ben and I performed a more accurate experiment than we did last semester and our rookie, Adam, performed well. The data and calculations were not difficult; they were simply hard to communicate well on Microsoft Word. Overall, this experiment was a victory for the cause of indoctrinating young children with the knowledge of the ancients.
1. What volume of O2 can be collection by the decomposition of 17.7 grams of Potassium Chlorate at a temperature of 39°C and a pressure of 687 torr?
KClO3 -> 2KCl + 3O2 17.7g KClO3 (mol)(3)(312K)(760 torr)(22.4L) = 6.1L O2
2. If you obtained a 91% yield of O2 in lab, what mass and volume of O2 did you obtain? (Same data as in question #1)
.91(3.8L) = 5.6L
3. What would the theoretical yield of Potassium chloride be from problem #1?
17.7g(mol)(2)(74.6g) = 21.5g KCl
4. If a catalyst were used in problem #1, would this change your results? Explain.
No. The reaction would have been the same regardless of the existence of a catalyst; it simply would have happened faster.
1. What effect would the addition of 1.0g of yeast instead of 0.4g have on your results?
More water might have diffused from the eudiometer, decreasing the measured % composition of the hydrogen peroxide.
2. If the reaction flask was not sealed tightly and had a slow leak, what effect would this have on the calculated % H2O2 in the sample?
The calculated % H2O2 would have increased. Less water would diffuse and the calculated amount of peroxide in the reaction flask would have been greater.
3. If the barometric pressure was incorrectly recorded as 735 torr instead of 753 torr, what effect would this have on the calculated % H2O2 in the sample?
The calculated % H2O2 would have decreased because the calculated volume of dry oxygen at STP would have decreased. This would decrease the measurement of moles and mass of peroxide, therefore decreasing the weight percent of peroxide in the sample.