Essay on Identification And Characterization Of An Unknown Organic Acid

Submitted By gracegrace121
Words: 1185
Pages: 5

Identification and Characterization of an Unknown Organic Acid
Sado, Ehizojie
8:00am Friday
8558020-HF14

Appendix֯֯
Melting point
Known substance
Trial 1 °C
Trial 2 °C
161-166
164-167
Plateau
Plateau
153
153
The second trial is the exact melting point of the known substance.
Trial 1 °C
Trial 2 °C
183-185
183-185
Plateau
Plateau
178
178
Both of the trials were exactly the same.

NaOH molarity
The molarity of NaOH was used in the titration of the weight of the unknown acid, using phenolphthalein as an indicator. By using the concentration this NaOH solution, the weight of the unknown acid that was titrated, and the volume of NaOH that was used to titrate it, we can determine the weight of the unknown acid.

Trial 1
Trial 2
Trial 3
Final- initial volume
21.7-1.33= 20.37
24.7-4.05= 20.65
45.06-23.09 = 21.97

HCl calculations

Trial 1
Trial 2
Trial 3

22.37/10= 2.237
20.65/10=2.065
21.27/10=2.197

Average
2.137+2.065+2.127 = 2.143
3
Deviation from Average
1.) 2.143-2.240= 0.08
2.) 2.290-2.170= 0.12
3.) 2.290-2.230= 0.06
Average Deviation
0.08+0.12+0.06 = 0.0967
3
Percent Deviation
0.0967 * (100) = 4.22
2.29

Percent Deviation KHP
Trials
1
2
3 grams .3018
.3018
.3018
Final-Initial
23.7-1= 22.70
23.5-1= 22.50
23.45-1= 22.45
10-3
.0223
.0225
.02245

Molarity NaOH
1.) .3018÷204.24g/mol KHP = 0.06509M NaOH
.02270 (L)
2.) .3018÷204.24g/mol KHP = 0.06567M NaOH
.02250 (L)
3.) 3018÷204.24g/mol KHP = 0.06582M NaOH
.02245(L)
Average
0.663+0.6567+0.06582= 0.06552
3
Deviation from Average
1.) 0.0659-0.06552= 0.0004
2.) 0.0657-0.06552= 0.0002
3.) 0.06582-0.06552= 0.00029
Average Deviation
0.0004+0.0002+0.00032 = .00029
3
Percent Deviation
.00029 *(100) = .452% 0.06552
Molarity of HCl
2.143mL*0.06552=0.140

Equivalent Weight
Trials
1
2
3 grams .3094
.3097
.3094
Final-Initial
20.2-1= 19.20
20.56-1= 19.55
20.3-1=19.50
10-3 0.0192
0.01955
0.0195

1.) 0.3094 = 245.949 (0.06552)(0.0192)
2.) 0.3097 = 241.545
(0.06552)(0.01955)
3.) 0.3094 = 242.165
(0.06552)(0.01950)
Average
245.949+241.545+242.165 = 243.219
3
Deviation from Average
243.219-245.949= 2.72
243.219-241.545 = 1.67
243.219-242.165= 1.054
Average Deviation
1.054+2.72+1.67= 1.8146 3
Percent Deviation
1.8146*(100) = .746% 243.219

My calculations for the equivalent weight are incorrect.

Sodium Fusion test

Nitrogen Test
Positive, it was a blue solution
Halogen Test
Positive,

An organic acid is an organic compound with acidic properties. From the Melting point experiment to the Sodium fusion experiment, each lab completed was directed towards the identifying of unknown organic acid. The purpose of the labs was to gather enough information on the properties of the acid. The unknown organic acid that I was given is 8558020-HF14. Based on the information gathered, the identity of my unknown organic acid is 4-chloro-3-nitrobenzoic acid (ClC6H3 (NO2) COOH). .
The melting point of a compound is a purification process used in the identifying of an organic acid. It is always written in a range. A melting range of a substance can vary from about 1-4° and will still be considered a pure substance. In this experiment the melting point of the unknown acid was determined using a Bibby Sterilin device. In order to test the accuracy of the machine a known substance was placed into the Bibby Sterilin and the melting point was measured. The melting point for the first trial on the known substance was (161-166). For the second trial, the range was (164-167).This was the exact melting point of the known substance. The…