[1800chisaki]

The lab:
Dissolve 0,1256g powder (it contains chloramine) in 100 mL water.
Add 1 g of potassium iodide R and 5 mL of dilute sulfuric acid R and leave it for 3 minutes.
Titrate with 0.1 M sodium thiosulfate using 1 mL starch solution R as an indicator.
8,44 mL was titrated
f = 1,02123

Stoichiometry:

ClO- + 3 I- + 2 H+ -> Cl- + I3- + H2O

I3- + 2 S2O3^2- -> 3 I- + S4O6^2-

Calculate mass percent (%m/m) of anhydrous chloramine

mass chloramine = [(M x f x V)sodium thiosulfate x Mr anhydrous chloramine]/2
                         = (0,1mM x 1,02123 x 8,41mL x 227,64mg/mmol)/2
                         = 195,5096 mg
mass percent = ( 195,5096 mg / 125.6 mg ) x 100 % = 156%

My Question:
Where did i go wrong? the mass percent should be above 50% and below 100%

[chenbeier]

ClO- is not chlorinamin. It is hypochorite.

[1800chisaki]

ClO- is not chlorinamin. It is hypochorite.

yes, chloramine splits into natriumhypochlorite and p-toluenesulfonamide when in water but i still dont understand my mistake /:

[chenbeier]

You forget to devide by 2.

[1800chisaki]

You forget to devide by 2.

omg i just noticed, i guess my brain is completely absent today