Question

Analysis of a sample of a chemical with formula C22H30N6O4S, showed that it contained 0.0195 mol of carbon.
What mass of nitrogen was present in the sample?

Answer 1

22 — 0,0195

6 — x

x=0,0053 moll

m(N)=n*Mr(N)=0,0053*14=0,0754g