Decomposition of Hydrogen Peroxide Lab Report PDF

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Lab Report for Decomposition of Hydrogen Peroxide...

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Decomposition of Hydrogen Peroxide Lab Report Abstract: The purpose of this experiment was to determine the mass percent of hydrogen peroxide in a solution. This was accomplished by measuring the volume of oxygen gas released during decomposition. In three trials, the mass percent of hydrogen peroxide in the solution was found to be 5.83%, 5.89%, and 5.24%, respectively.

Introduction: The purpose of this experiment was to determine the mass percent of hydrogen peroxide in a topical first aid solution. This was done by collecting oxygen gas released during decomposition. In nature, hydrogen peroxide decomposes rather slowly. However, this process can be sped up by introducing certain compounds or ions. For this experiment, potassium iodide (KI) was used as a catalyst to increase the rate of hydrogen peroxide (H2O2) decomposition and oxygen gas was collected in graduated cylinder during the reaction: 2 H2O2 (aq)

 2 H2O (l) + O2 (g)

According to Dalton’s Law, the total pressure of a mixture of gases is the sum of the partial pressures of each individual gas. Since the total atmospheric pressure was known and the vapor pressure of the water was determined by the temperature of the solution, the pressure of the oxygen gas was able to be determined through subtraction: PO2 = Ptotal – PH2O The pressure of oxygen gas was used along with its known volume and temperature to calculate the number of moles using the ideal gas law equation, PV = nRT. Then, the moles of H2O2 were determined by stochiometric ratios. Finally, the mass percent of H2O2 and percent error were calculated.

Data: Table 1 – Volume (mL), temperature (oC), and mass (g) of H2O2 solution at a constant pressure of 757.6 mmHg and the vapor pressure (torr) for each of the three experimental trials. Trial Initial graduated Final graduated Temperature Vapor Pressure Mass H2O2 (oC) cylinder volume cylinder Solution (torr) (mL) volume (mL) (g) 1 3.5 93.7 22.20 19.8 4.08 2 4.3 98.4 21.00 18.6 4.28 3 0.0 88.5 21.25 18.6 4.41 Results: Table 2 – Mass percent and the percent error of H2O2 for each experimental trial. Trial Mass H2O2 (%) Error (%) 1 5.83 5.67 2 5.89 3.67 3 5.24 25.3 Calculations: M H2O2 = 34.015 g/mol PV n= RT Ptotal = 757.6 mmHg

atm ( 7601mmHg )

= 0.997 atm

T (in K) = oC + 273.15 Trial 1:

Trial 2:

1 atm )) = 0.0261 atm 760 torr PO2 = (0.997 atm – 0.0261 atm) = 0.971 atm VO2 = (93.7 mL – 3.5 mL – 3.0 mL) = (87.2 mL * .001 L) = .0872 L (0.971 atm)(0.0872 L) nO2 = = 0.0035 L∗atm (295.35 K ) 0.08206 mol∗K mol 2 mol H 2 O 2 ) = 0.007 nH2O2 = 0.0035 mol O2 ( 1 mol O2 mol m = 0.007 mol (34.015 g/mol) = 0.238 g 0.238 g m% H2O2 = ( ) * 100 = 5.83% 4.08 g

1 atm )) = 0.0245 atm 760 torr PO2 = (0.997 atm – 0.0245 atm) = 0.973 atm VO2 = (98.4 mL – 4.3 mL – 3.0 mL) = (91.1 mL * .001 L) = .0911 L (0.973 atm)(0.0911 L) nO2 = = 0.0037 L∗atm (294.15 K ) 0.08206 mol∗ K mol 2 mol H 2 O2 )= nH2O2 = 0.0037 mol O2 ( 1 mol O2 0.0074 mol m = 0.0074 mol (34.015 g/mol) = 0.252 g 0.252 g m% H2O2 = ( ) * 100 = 5.89% 4.28 g

PH2O = (19.8 torr (

(

)

PH2O = (18.6 torr (

(

)

%error = 100 –

(|

| )=

5.83 %−3.00 % ∗100 3.00 %

5.67% Trial 3:

%error = 100 –

(|

| )=

5.89 %−3.00 % ∗100 3.00 %

3.67%

1 atm )) = 0.0245 atm 760 torr PO2 = (0.997 atm – 0.0245 atm) = 0.973 atm VO2 = (88.5 mL – 0 mL – 3.0 mL) = (85.5 mL * .001 L) = .0855 L (0.973 atm)(0.0855 L) nO2 = = 0.0034 L∗atm (294.4 K ) 0.08206 mol∗K mol 2 mol H 2 O 2 )= nH2O2 = 0.0035 mol O2 ( 1 mol O2 0.0068 mol m = 0.0068 mol (34.015 g/mol) = 0.231 g 0.231 g m% H2O2 = ( ) * 100 = 5.24% 4.41 g PH2O = (18.6 torr (

(

%error = 100 –

)

(|

| )=

5.24 %−3.00 % ∗100 3.00 %

25.3% Discussion: The purpose of this experiment was to determine the mass percent of hydrogen peroxide in a topical first aid solution. This was done by catalyzing decomposition, collecting the oxygen gas released during the reaction, and using the ideal gas law equation (PV = nRT) to determine the moles of oxygen gas. The stoichiometric ratio of oxygen gas (O2) to hydrogen peroxide (H2O2) was deduced from the reaction equation, 2 H2O2 (aq)  2 H2O (l) + O2 (g), and used to calculated the moles of H2O2. The mass percent of hydrogen peroxide in the solution from each trial was significantly higher than the accepted value of 3% obtained from the label of the topical first aid solution. It was assumed that the atmospheric pressure was equal to the total pressure of the sample. Additionally, the vapor pressure was determined by rounding the temperature of the solution to the nearest whole degree. If either of these methods resulted in a

high value for the pressure of oxygen gas, calculations for the mass percent of hydrogen peroxide would also be higher. Furthermore, if there was air present in the graduated cylinder and it was not recorded, the volume of oxygen gas would appear much higher. Due to stoichiometric ratios, this would also cause the mass percent of hydrogen peroxide to be higher than the accepted value....