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CnH2n+1OH = 102
Obliczenia
C = 12u
H = 1u
O = 16u
12n + 2n+1 + 17u = 102
14n = 102 - 18
14n = 84 /:14
n = 6
wzór sumaryczny : C6H13OH
Wzór półstrukturalny: CH3 - (CH2)4 - OH
Nazwa : heksanol