Вычислить массовые доли (в процентах) элементов в соединениях HNO3; KOH; FeSO4; MgO

1) Mr (HNO₃) = Ar (H) * N (H) + Ar (N) * N (N) + Ar (O) * N (O) = 1 * 1 + 14 * 1 + 16 * 3 = 63;ω (H в HNO₃) = N (H в HNO₃) * Ar (H) * 100% / Mr (HNO₃) = 1 * 1 * 100% / 63 = 1,59%;ω (N в HNO₃) = N (N в HNO₃) * Ar (N) * 100% / Mr (HNO₃) = 1 * 14 * 100% / 63 = 22,22%;ω (O в HNO₃) = 100% - ω (H в HNO₃) - ω (N в HNO₃) = 100% - 1,59% - 22,22% = 76,19%;

2) Mr (KOH) = Ar (K) * N (K) + Ar (O) * N (O) + Ar (H) * N (H) = 39 * 1 + 16 * 1 + 1 * 1 = 56;ω (K в KOH) = N (K в KOH) * Ar (K) * 100% / Mr (KOH) = 1 * 39 * 100% / 56 = 69,64%;ω (O в KOH) = N (O в KOH) * Ar (O) * 100% / Mr (KOH) = 1 * 16 * 100% / 56 = 28,57%;ω (H в KOH) = 100% - ω (K в KOH) - ω (O в KOH) = 100% - 69,64% - 28,57% = 1,79%;

3) Mr (FeSO₄) = Ar (Fe) * N (Fe) + Ar (S) * N (S) + Ar (O) * N (O) = 56 * 1 + 32 * 1 + 16 * 4 = 152;ω (Fe в FeSO₄) = N (Fe в FeSO₄) * Ar (Fe) * 100% / Mr (FeSO₄) = 1 * 56 * 100% / 152 = 36,84%;ω (S в FeSO₄) = N (S в FeSO₄) * Ar (S) * 100% / Mr (FeSO₄) = 1 * 32 * 100% / 152 = 21,05%;ω (O в FeSO₄) = 100% - ω (Fe в FeSO₄) - ω (S в FeSO₄) = 100% - 36,84% - 21,05% = 42,11%;

4) Mr (MgO) = Ar (Mg) * N (Mg) + Ar (O) * N (O) = 24 * 1 + 16 * 1 = 40;ω (Mg в MgO) = N (Mg в MgO) * Ar (Mg) * 100% / Mr (MgO) = 1 * 24 * 100% / 40 = 60%;ω (O в MgO) = 100% - ω (Mg в MgO) = 100% - 60% = 40%.