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M) * MM(CH3COOH)
(0.5967M) * (65.05 g/mol) = 35.85 g/L
4) Percent Acid ((g/L)/(given density)) * 100
((35.85 g/L)/(1005 g/L)) * 100 = 3.57%
UNKNOWN ACID 1) Moles of NaOH ((Vf - Vi)/1000) * M
Run #1: ((31.11 – 0.51)/1000) * 0.2116m = .006475
Run #2: ((31.35 – 0.38)/1000) * 0.2116m = .006553
2) Calc Moles of Acid because 1:1 ratio moles acid equals moles NaOH moles acid = Run 1: .006475 and Run #2: .006553
3) Molar Mass (grams used/moles acid)
Run #1: (1.3160g/.006475) = 203.24 g/mol
Run #2: (1.3276g/.006553) = 202.59 g/mol 4) Average Molar Mass (203.24 + 202.59)/2 = 202.92 …show more content…
When solving for the percent of acetic acid present in vinegar the end result of 3.57% was not far off from the standard 4%. The small error could be a result of error in experiment or an actual failure of the acid to meet the federal standard. An example of error in experimentation is while the solution is titrated completely, the flask being swirled is removed from underneath the buret and a drop or two of NaOH can sneak out before the valve is closed. Those few missed drops as well as excess NaOH in titration can affect the accuracy of the results. Human error is almost always a factor for example when measuring any values without digital assistance leaves room for the naked eye to make mistakes. As far as the second part of the experiment was, the 0.0025% deviation indicates that the titration was accurate. The ending molar mass (202.59 g/mol) of the mystery acid points to two possible suspects. The first being Cesium Diuranate, and the second Sodium Diuranate. Without the information it wasn’t possible to determine which one it would be. There is also a chance that the results had small deviation (accurate) but were not precise. For example if both titrations had error due to the mass of the samples being incorrect. There was also a potential error during the second titration. While adding the indicator to the acid solution the tip came in
(0.5967M) * (65.05 g/mol) = 35.85 g/L
4) Percent Acid ((g/L)/(given density)) * 100
((35.85 g/L)/(1005 g/L)) * 100 = 3.57%
UNKNOWN ACID 1) Moles of NaOH ((Vf - Vi)/1000) * M
Run #1: ((31.11 – 0.51)/1000) * 0.2116m = .006475
Run #2: ((31.35 – 0.38)/1000) * 0.2116m = .006553
2) Calc Moles of Acid because 1:1 ratio moles acid equals moles NaOH moles acid = Run 1: .006475 and Run #2: .006553
3) Molar Mass (grams used/moles acid)
Run #1: (1.3160g/.006475) = 203.24 g/mol
Run #2: (1.3276g/.006553) = 202.59 g/mol 4) Average Molar Mass (203.24 + 202.59)/2 = 202.92 …show more content…
When solving for the percent of acetic acid present in vinegar the end result of 3.57% was not far off from the standard 4%. The small error could be a result of error in experiment or an actual failure of the acid to meet the federal standard. An example of error in experimentation is while the solution is titrated completely, the flask being swirled is removed from underneath the buret and a drop or two of NaOH can sneak out before the valve is closed. Those few missed drops as well as excess NaOH in titration can affect the accuracy of the results. Human error is almost always a factor for example when measuring any values without digital assistance leaves room for the naked eye to make mistakes. As far as the second part of the experiment was, the 0.0025% deviation indicates that the titration was accurate. The ending molar mass (202.59 g/mol) of the mystery acid points to two possible suspects. The first being Cesium Diuranate, and the second Sodium Diuranate. Without the information it wasn’t possible to determine which one it would be. There is also a chance that the results had small deviation (accurate) but were not precise. For example if both titrations had error due to the mass of the samples being incorrect. There was also a potential error during the second titration. While adding the indicator to the acid solution the tip came in