Question

Doi compusi clorurati ai metanului,A si B,contin 70,3 % clor si respectiv 83,53% clor. Care este raportul maselor moleculare ale celor doi compusi?

Answer 1

avem 4 compusi: CH3Cl, CH2Cl2, CHCl3 si CCl4
incepem cu masele moleculare:
MCH3Cl=50,5g/mol => %Cl=35,5*100/50,5=70,3%
MCH2Cl2=85g/mol => %Cl=71*100/85=83,53%
MCHCl3=119,5g/mol => %Cl=106,5*100/119,5=89,12%
MCCl4=154g/mol => %Cl=142*100/154=92,21%
deci A - clorometanul si B - diclorometanul
MCH3Cl/MCH2Cl2=50,5/85=0,5941