This is my Chem 1 lab report I gotta write \" E error analysis\" instructor said

Question

This is my Chem 1 lab report I gotta write

"Eerror analysis"

instructor said (exp- accepted/ accepted)* 100

Could you help me find "Error Analysis" Please?

Thank you so much!

Introduction Different types of solutions of iron react with many numbers of organic chemical Diferent types of solutions of iron react with many numbers of organic chemical compounds called phenols which produce intensity colored products. This react between iron(III) and a hydroxyl group (-OH) bonded to a benzene ring. The reaction between iron(III) and salicylic acid, for example, is used as a qualitative and quantitative test for the purity of aspirin. When treated with iron(II), pure aspirin should not produce a color. Aspirin is synthesized from salicylic acid, a phenol. Aspirin itself is not a phenol, However, if unreacted salicylic acid starting material is present then the aspirin sample will produce a purple color when treated with iron COOH OH ocoCH OcoCH3 Phenol Salicylic Acid acetylsalicylic acid (aspirin) The goal of this experiment is to determine the stoichiometry of the reaction between iron(lIl) and the phenol known as 5-sulfoalicylic acid. This phenol is chosen because it may soluble in water than phenol itself or salicylic acid. OH HO3S 5-Sulfosalicylic scid Apparatus/ Reagents required: 1.7x103 M ironl) nitrate (in 0.1 M HNO3)

Explanation / Answer

The given value for Iron = 7.87 g/mL (density) = accepted value or true value.

Now you have to calculate the experimental value for Iron to determine the error.

Moles of Fe = 8.5 x 10-7 mol

Molecular weight of Fe = 55.85 gm/mol

Weight of Fe = molecular weight x moles of Fe

                      = 8.5 x 10-7 *55.85

                       = 4.75 x 10-5 gm

Volume = 0.5 mL

Therefore density of Fe = weight of fe/ volume

                                      = (4.75 x 10-5 /0.5 ) g/mL

                                     = 9.5 x 10-5 g/mL = experimental value

Percent Error = true value – experimental value/true value

                       = (7.87 – 9.5 x 10-5/7.87)*100

                      = 99.99 %

Error analysis = (exp- accepted/ accepted)* 100

                      = (9.5 x 10-5- 7.87/7.87)*100

                       = - 99.99

From error analysis we can say that the experimental value is too far from accepted or true value.

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