Question

under suitable conditions, thiourea is oxidized to sulfate by bromate:
3CS(NH2)2 + 4BrO3- + 3H2O --- 3CO(NH2)2 + 3SO4 + 4Br- + 6H+
a 0.0 715g sample of material was found to consume 14.1mL of 0.00833 M KBrO3. What was the % purity of the thiourea sample?
P.S this is analytical chem..... Thanks for the answer in advance

Answer 1

$$3CS(NH_2)_2 + 4BrO_3^- + 3H_2O \rightarrow 3CO(NH_2)_2 + 3SO_4^- + 4Br^- + 6H^+$$ $$n(KBrO_3) = (14.1 * 10^{-3} * 0.00833) * 6 = 1.17 * 10^{-4} * 6 = 7.05 * 10^{-4}$$ $$m(CS(NH_2)_2) = 7.05 * 10^{-4} * 76/8 = 6.7 * 10^{-3}$$ $$\omega = \frac{m(CS(NH_2)_2)}{m_{sample}} = \frac{6.7 * 10^{-3}}{7.15 * 10^{-2}} = 0.09 = 9\%$$

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