Вычислить массовые доли с6h12o6 (глюкоза)

Mr(C6H12O6) = Mr = 6Ar(C) + 12Ar(H) + 6Ar(O) = 6 * 12 + 12 * 1 + 6 * 16 = 180

w(C) = 6Ar(C) / Mr = 72 / 180 = 0.4 = 40%

w(H) = 12Ar(H) / Mr = 12 / 180 = 0.0667 = 6.67%

w(O) = 6Ar(O) / Mr = 96 / 180 = 0.5333 = 53.33%