labreportofFlametest/QualAnalysisforthepresenceofion希望能给我一个外国的实验报告模式,尽量标准,反应过程描述尽量标准,现象尽量标准,谢谢

Question

lab report of Flame test/Qual Analysis for the presence of ion
希望能给我一个外国的实验报告模式,尽量标准,反应过程描述尽量标准,现象尽量标准,谢谢

Answer 1

can't find my flame test lab,but the general layout is here,hopefully it will be useful ^^
Acetic Acid Content Analysis of Consumer Vinegar
Purpose:The purpose of this experiment is to determine the percent acetic acid in supermarket vinegar.
Data:
Mass Measurement
Mass,acid Mass,base
2.24 g 1.21 g
Measurement of Burette Reading
Trial 1 Trial 2 Trial 3
Initial Burette Reading (mL) 0 0 0
Final Burette Reading (mL) 18.5 20.5 20
Volume of base (mL) 18.5 20.5 20
Trial 1 Trial 2
Initial Burette Reading (mL) 0 14
Final Burette Reading (mL) 14 29
Volume of base (mL) 14 15
Conclusions:The average molarity of NaOH was 0.112 M.The average percent of acetic acid in vinegar was 4.84%.One of the sources of error being made was that the whole experiment was discarded because NaOH was poured out before the experiment was done; therefore,the whole experiment was required to start all over again.The same concentration of NaOH was required to use throughout the experiment,but since the NaOH was poured out,all the data collected was useless.Other sources of error could be the inaccuracy of the scale or a misreading of the measuring of the potassium hydrogen phthalate.The burette might not have been thoroughly rinsed with the base so that there could have been other substances in there.Also,the overshooting or the undershooting of the endpoint could have caused an error in the experiment.
The experiment did not exactly meet the 5% requirement of acetic acid in vinegar,but the second trial had 5% acetic acid in vinegar.The average percent of acetic acid in vinegar being 4.84% was very close to 5%.
Calculations:
The molarity of NaOH:
Trial 1:
(2.24 g HC8H4O4K / 100 mL) x (1 mol HC8H4O4K / 204 g HC8H4O4K) x (1 mol NaOH / 1 mol HC8H4O4K) x (20.0 mL acid / 1 x 1 / 18.5 g NaOH) = 0.119 M
Trial 2:
(2.24 g HC8H4O4K / 100 mL) x (1 mol HC8H4O4K / 204 g HC8H4O4K) x (1 mol NaOH / 1 mol HC8H4O4K) x (20.0 mL acid / 1 x 1 / 20.5 g NaOH) = 0.107 M
Trial 3:
(2.24 g HC8H4O4K / 100 mL) x (1 mol HC8H4O4K / 204 g HC8H4O4K) x (1 mol NaOH / 1 mol HC8H4O4K) x (20.0 mL acid / 1 x 1 / 20.0 g NaOH) = 0.110 M
Average molarity of NaOH:
(0.107 + 0.110 + 0.119) / 3 = 0.112 M
The Percent of HC2H3O2 in vinegar:
Trail 1:
{(0.112 mol NaOH / 1 L) x (14.0 mL NaOH / 1) x (1 L / 1000 mL) x (1 mol HC2H3O2 / 1 mol NaOH) x (1 / 20.0 mL vinegar) x (60.0 g HC2H3O2 / 1 mol HC2H3O2) x (1 mL / 1.007 g) x 10}x 100 = 4.67% HC2H3O2
Trail 2:
{(0.112 mol NaOH / 1 L) x (15.0 mL NaOH / 1) x (1 L / 1000 mL) x (1 mol HC2H3O2 / 1 mol NaOH) x (1 / 20.0 mL vinegar) x (60.0 g HC2H3O2 / 1 mol HC2H3O2) x (1 mL / 1.007 g) x 10}x 100 = 5.00% HC2H3O2
The Average Percent of HC2H3O2 in vinegar:
(4.67 + 5.00) / 2 = 4.84% HC2H3O2