Presupunem ca avem a moli de CH3Cl si b moli CH2Cl2 in amestec.
1 mol CH3Cl..... 50,5 g CH3Cl..... 35,5 g Cl
a moli CH3Cl..... (50,5*a) g CH3Cl..... (35,5*a) g Cl
1 mol CH2Cl2..... 85 g CH2Cl2..... 71 g Cl
b moli CH2Cl2..... (85*b) g CH2Cl2..... (71*b) g Cl
Amestecul are masa (50,5a+85b)g, din care (35,5a+71b)g sunt reprezentate de clor.
Deci 76,344/100(50,5a+85b)=35,5a+71b
3855,372a+6489,24b=3550a+7100b
305,372a=610,76b
Rezulta ca a/b=610,76/305,372 ≈1/2
