Gwhaltsbestimmung?Chemie?Hilfe?

You first dissolve the paraben (a 4-hydroxybenzoic acid ester) in alkali and simmer it, it probably being hydrolyzed to free 4-hydroxybenzoic acid. Then you add a quantity of bromide and a precisely known quantity of bromate and acidify. Bromide and bromate are combined in the acid to form bromine:

BrO3 ̄ + 5 Br ̄ + 6 H3O+ ⟶ 3 Br2 + 9 H2O

To know how much bromine is produced in your soup, we just need to know how much KBrO3 has been added. I interpret your term “0.1 N with titer 0.98” as c= 0.0163 mol/l, then we have n=cV=817 μmol of bromate from which 2.45 mmol of Br2 are obtained three times as much. The amount of AI required for this is less than 0.5 g, so your 3 g is definitely sufficient, so it looks reasonable.

The bromine is now attacked by the 4-hydroxybenzoic acid and thus, i.e., part of the bromine is eliminated, and what is left is converted by addition of iodide to iodine I2 (dissolved under these conditions as I3 ̄) and then titrated with thiosulfate:

Br2 + 3 I ̄ ⟶ I3 ̄ + 2 Br ̄
2 S2O32 ̄ + I3 ̄ ⟶ S4O62 ̄ + 3 I ̄

You consume V=20.9 ml of a 0.098 mol/l Na2 S2 O3 solution, which corresponds to n=cV=2.05 mmol Na2 S2 O3, which indicates half as much Br2 (because thiosulphate reacts with half the amount of iodine, but which was formed 1:1 from the bromine). Consequently, after the bromination of the 4-hydroxybenzoic acid, 1.02 mmol of Br2 are still left, so the 4-hydroxybenzoic acid has swallowed 1.43 mmol of bromine. I suspect that the reaction leads to 2,4,6-tribromophenol (as in the case of 2-hydroxybenzoic acid), i.e. the 4-hydroxybenzoic acid and the bromine in a ratio of 1:3, i.e. n=0.475 mmol of 4-hydroxybenzoic acid or just as many parabens in the sample.

In order to get to the desired mass fraction, we first need the mass which is obtained from the quantity of material with m=nM. I use the molar mass of ethyl paraben M=166.18 g/mol, because this is in the task, and that you speak in your description of methyl paraben; then we get m=79 mg, the weight was m0=97 mg, so the mass fraction w=m/m0=81.4%

The whole calculation compact:

(1/60*.98 * 50 *3 - 20.9*0.1*.98/2) /3 * 166.18 / 97
0.81428

You have made a total of four series of measurements, which result in 81.4%, 82.4%, 82.7% and even 82.7%. The average value of this is 82.3%, but since you only know weights and titers exactly on 2 positions, you can only specify 2 significant grades: 82%.

There are some inconveniences in it that may need correction:

1. Concentration data. I didn’t have too much effort to straighten your historical N-data.
2. Reaction equation. I am quite sure that the 4-hydroxybenzoic acid reacts with bromine in a ratio of 1:3 to 2,4,6-tribromophenol (with elimination of the COOH mixture), but I do not really know exactly.
3. For the molar mass, I took ethyl paraben (4-hydroxybenzoic acid ethyl ester, M=166.18 g/mol), although you are speaking in your text of methyl paraben (M= 152.15 g/mol). Theoretically, the sodium salts could also be meant; for ethyl paraben, this would then be M=174.13 g/mol. repeat the invoice with the other molar mass if necessary; for ethyl paraben Na salt it would then be, for example, 86.2%
4. I haven’t expected such a thing for many, many decades, so caveat emptor.

By the way, to specify a titre exactly on only 2 places is bullshit. But this is still quadraticly exaggerated by the schnappssis to write a titre for the potassium bromate solution into the data, that is, a urtite substance and can therefore be weighed exactly (if the laboratorynt was not previously titrated with ethanol until the envelope was made).