ω = $A_{r}$ * n : $M_{r}$ * 100%, где n − индекс
1) HBr:
$M_{r}$ (HBr) = 1+ 80 = 81
ω (H) = 1 * 1 : 81 * 100% = 1,23%
ω (Br) = 100% −1,23% = 98,77%.
2) $K_{2}CO_{3}$:
$M_{r}$ ($K_{2}CO_{3}$) = 2 * 39 + 12 + 3 * 16 = 138
ω(K) = 2 * 39 : 138 * 100% = 57%
ω(C) = 12 * 1 : 138 = 9%
ω(O) = 100% − 57% − 9% = 34%
3) $Mg(OH)_ {2}$:
$M_{r}$ ($Mg(OH)_{2}$) = 24 + 2 * 16 + 2 * 1 = 58
ω(Mg) = 24 * 1 : 58 * 100% = 41%
ω(O) = 2 * 16 : 58 * 100% = 55%
ω(H) = 100% − 41% − 55% = 4%
4) $P_{2}O_{5}$:
$M_{r}$ ($P_{2}O_{5}$) = 2 * 31 + 5 * 16 = 142
ω(P) = 2 * 31 : 142 * 100% = 44%
ω(O) = 100% − 44% = 56%
5) $FeCl_{3}$:
$M_{r}$ ($FeCl_{3}$) = 56 + 3 * 35,5 = 162,5
ω(Fe) = 56 * 1 : 162,5 * 100% = 34%
ω(Cl) = 100% − 34% = 66%
6) $Cu(NO_{3})_ {2}$:
$M_{r}$ ($Cu(NO_{3})_{2}$) = 64 + 2 * 14 + 6 * 16 = 188
ω(Cu) = 64 * 1 :188 * 100% = 34%
ω(N) = 2 * 14 : 188* 100% = 15%
ω(O) = 100% − 34% − 15% = 51%