19.51 x 10^-3 L*0.1060M=2.07 x 10^-3 moles of HCl was used, so 2.07 moles of H2BO3- reacted with HCl., which means that 2.07 x 10^-3 moles of NH4^+ was collected. Since we know that 2.07 x 10^-3 moles of N is present, we need to find out how much C2H7N5 is present in the sample.
2.07 x 10^-3 moles of N*(1mole of C2H7N5/5moles of N)=4.14 x 10^-4 moles of C2H7N5
4.14 x 10^-4 moles of C2H7N5*(101.1g/mole)=4.18 x 10^-2 g of C2H7N5
4.18 x 10^-2 g of C2H7N5/0.055 g sample of impure biguanide)*100=76%
A Kjeldahl analysis was preformed upon a 0.055 g sample of impure biguanide (C2H7N5, fw=101.1). The liberated ammonia, collected in 40 mL of 4% boric acid was titrated with 19.51 mL of 0.1060 M HCl. Calculate the percentage of biguanide in the sample.
(Result: 76%)
This is how i did the problem, but i had to make a mistake somewhere, because my result is 7.6%, and i also don't know what to do with the volume of boric acid, where should i include it in the calculations? I would really appreciate if someone would help me. At least to tell me where i did the mistake and i will try to do the problem again.
(NH4)2SO4 + 2NaOH -> Na2SO4 + 2H2O + 2NH3
NH3 + H3BO3 -> NH4+ + H2BO3-
H2BO3- + HCl -> H3BO3 + Cl-
n(HCl)=n(H2BO3-)=n(NH3)= 0.01951L*0.1060M=0.0020681 mol
n((NH4)2SO4)=n(NH3)/2=0.0020681/2=0.00103403 mol
n(Compound)=0.00103403*0.04=0.000041361 mol
m(Compound)=0.000041361*101.1=0.004181617 g
%=0.004181617/0.0550 *100%=7.6%
2 answers
thank you so much, again :)
2 answers