Вопрос по химии:

найдите массовую долю Оксигена:а)в углекислом газе СО2;В)в метиловом спирте СH3OH:Б)в гашеной извести Ca(OH)2:г) в глюкозе

1) Mr (CO₂) = Ar (C) * N (C) + Ar (O) * N (O) = 12 * 1 + 16 * 2 = 44;2) ω (O) = N (O) * Ar (O) * 100% / Mr (CO₂) = 2 * 16 * 100% / 44 = 72,73%;3) Mr (CH₃OH) = Ar (C) * N (C) + Ar (H) * N (H) + Ar (O) * N (O) = 12 * 1 + 1 * 4 + 16 * 1 = 32;4) ω (O) = N (O) * Ar (O) * 100% / Mr (CH₃OH) = 1 * 16 * 100% / 32 = 50%;5) Mr (Ca(OH)₂) = Ar (Ca) * N (Ca) + Ar (O) * N (O) + Ar (H) * N (H) = 40 * 1 + 16 * 2 + 1 * 2 = 74;6) ω (O) = N (O) * Ar (O) * 100% / Mr (Ca(OH)₂) = 2 * 16 * 100% / 74 = 43,24%;7) Mr (C₆H₁₂O₆) = Ar (C) * N (C) + Ar (H) * N (H) + Ar (O) * N (O) = 12 * 6 + 1 * 12 + 16 * 6 = 180;8) ω (O) = N (O) * Ar (O) * 100% / Mr (C₆H₁₂O₆) = 6 * 16 * 100% / 180 = 53,33%.