consideram : ms = m oleum = 100g md = mSO3 = c·100/100 = c
n SO3 = c/ 80 ⇒ mS din oleum = 32·c/80 = 0,4c
mH2SO4 = ms - md = 100 - c nH2SO4 = (100-c)/98
m S din H2SO4 = 32·(100-c)/98 = 16(100-c)49
2c/5 + 16(100-c)/49 = 34,122
98c + 80(100-c) = 8359,89
18c = 359,89 c ≈ 20%