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Kjeldahl question for Analytical Chemistry?
Can you please double check my answer I got for this question..
The kjeldahl procedure for nitrogen analysis was used on a 0.9325g sample of wheat flour. The NH3 produced was collected in an excess of 4.0% boric acid. The resulting ammonium borate complex required 33.15 ml of 0.1124 M HCl for titration to the end point. What is the weight percent of nitrogen in wheat flour? Use the appropriate factor to find the %protien in the flour. The answer I found is 5.59% can someone double check for me please? The answer I got is 5.59%. Thanks!
1 Answer
- BobbyLv 79 years agoFavorite Answer
NH3+ H3BO3 = NH4H2BO3 ........1
HCl + NH4H2BO3 = NH4Cl + H3BO3 ......2
moles of HCl = .1124 x .03315 = 0.003726 moles
moles of NH4H2BO3 = 0.003726 ... equ 2
moles of NH3 = 0.003726 equ 1
so moles of N = 0.003726 moles
Mass of N = 0.0037260 x 14 = 0.05216 g
% N = 0.05216 / .9325 = 5.59% .. we agree ..
we don't have the factor to convert to protein ..
