Ответ:
Используем относительные атомные массы: H = 1, O = 16, Na = 23, Cl = 35,5, K = 39, S = 32, N = 14, Cu = 64, Ba = 137, P = 31, Fe = 56, Ca = 40, Mn = 55. Массовая доля элемента: \(w(E)=\frac{A_r(E)\cdot n(E)}{M_r(\text{вещества})}\cdot100\%\), где \(n(E)\) — число атомов элемента в формуле.
NaClO: \(M_r=23+35.5+16=74.5\). \(w(\mathrm{Na})=23/74.5\cdot100\%=30.87\%\); \(w(\mathrm{Cl})=35.5/74.5\cdot100\%=47.65\%\); \(w(\mathrm{O})=16/74.5\cdot100\%=21.48\%\).
K₂SO₄: \(M_r=2\cdot39+32+4\cdot16=174\). \(w(\mathrm{K})=78/174\cdot100\%=44.83\%\); \(w(\mathrm{S})=32/174\cdot100\%=18.39\%\); \(w(\mathrm{O})=64/174\cdot100\%=36.78\%\).
HNO₃: \(M_r=1+14+3\cdot16=63\). \(w(\mathrm{H})=1/63\cdot100\%=1.59\%\); \(w(\mathrm{N})=14/63\cdot100\%=22.22\%\); \(w(\mathrm{O})=48/63\cdot100\%=76.19\%\).
Cu(OH)₂: \(M_r=64+2\cdot(16+1)=98\). \(w(\mathrm{Cu})=64/98\cdot100\%=65.31\%\); \(w(\mathrm{O})=32/98\cdot100\%=32.65\%\); \(w(\mathrm{H})=2/98\cdot100\%=2.04\%\).
Ba₃(PO₄)₂: \(M_r=3\cdot137+2\cdot(31+4\cdot16)=601\). \(w(\mathrm{Ba})=411/601\cdot100\%=68.39\%\); \(w(\mathrm{P})=62/601\cdot100\%=10.32\%\); \(w(\mathrm{O})=128/601\cdot100\%=21.30\%\).
FeSO₄: \(M_r=56+32+4\cdot16=152\). \(w(\mathrm{Fe})=56/152\cdot100\%=36.84\%\); \(w(\mathrm{S})=32/152\cdot100\%=21.05\%\); \(w(\mathrm{O})=64/152\cdot100\%=42.11\%\).
CaCl₂: \(M_r=40+2\cdot35.5=111\). \(w(\mathrm{Ca})=40/111\cdot100\%=36.04\%\); \(w(\mathrm{Cl})=71/111\cdot100\%=63.96\%\).
KMnO₄: \(M_r=39+55+4\cdot16=158\). \(w(\mathrm{K})=39/158\cdot100\%=24.68\%\); \(w(\mathrm{Mn})=55/158\cdot100\%=34.81\%\); \(w(\mathrm{O})=64/158\cdot100\%=40.51\%\).
