If 83.0 ml of 0.150 m hcl(aq) is needed to neutralize all the nh3(g) from a 2.25-g sample of organic material, calculate the mass percentage of nitrogen in the sample.

7.75% nitrogen in the sample. The balanced reaction of HCl and NH3 is HCl + NH3 = NH4Cl Which means that 1 mole of HCl will neutralize 1 mole of HN3. Since we used 83.0 ml of 0.150 m HCl, that means that we used 0.083 l * 0.150 mol/l = 0.01245 mol So we had 0.01245 mol of NH3 which means that we had 0.01245 mol of N Looking up the atomic weight of nitrogen, we get 14.0067. So 0.01245 * 14.0067 = 0.174383415 g We now know that the sample had 0.17438 g of nitrogen. Divide by the original mass of the sample, getting 0.17438 / 2.25 = 0.077502222 = 7.7502222% Rounding the result to 3 significant figures, gives us a mass percentage of 7.75% nitrogen in the original sample.

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