Question

In: Chemistry

A sample of grain (0.5348) was digested on an electrical heating mantle with 50 ml of...

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.

Solutions

Expert Solution

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 %


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