In: Chemistry
A sample of grain (0.5348) was digested on an electrical heating mantle with 50 ml of concentrated H2SO4 and a concentrated solution of NaOH was then added. The NH3 formed was collected by bubbling through 30.00ml of 0.045 M HCl and the excess HCl titrated to equivalence with 5.35 ml 0.0400 M NaOH. Calculate the percent nitrogen in the grain sample.
moles of NaOH = excess of HCl
= 5.35 x 0.0400/1000
= 2.14 x 10^-4
actual moles of HCl = 30 x 0.045 / 1000
= 1.35 x 10^-3
moles of HCl reacted with NH3 = 1.35 x 10^-3 - 2.14 x 10^-4
= 1.136 x 10^-3 mol
NH3 + HCl ---------------> NH4Cl
moles of NH3 = moels of HCl = 1.136 x 10^-3 mol
moles of N = moles of NH3 = 1.136 x 10^-3 mol
mass of N = 0.0159 g
percent nitrogen = (0.0159 / 0.5348 ) x 100
= 2.97 %