ZAD Oblicz skład procentowy ( % masowy) podanych związków chemicznych. a) tlenku magnezu b) tlenku siarki (IV) c) tlenku siarki (VI) d) tlenku fosforu (V) e) tlenku ołowiu (IV) f) siarczku bizmutu(III)

a) MgO   mcz MgO = 24u +16u = 40u   40u – 100% Mg - 24u – x %   x = 24u * 100%/ 40 = 240/4 = 60 100% - 60% = 40% Mg – 60%, O – 40% b) SO2 mczSO2  = 32u + 16u * 2 = 32u + 32u = 64u   64u – 100% S - 32u – x % (bez liczenia 50%)   x = 32u * 100%/64% = 100/2 = 50% 100% - 50% == 50% S – 50% , O2 – 50%   c) SO3 mczSO3 = 32u + 16u * 3 = 32u + 48u = 80u   80u – 100% S- 32u – x %   x = 32u * 100%/80u = 40% 100% - 40% = 60% S – 40% , O3 – 60%   d) P4O10 mczP4O10 = 31u * 4 + 16u* 10 = 124u + 160u = 284u   284u – 100% P4 – 124u -  x %   x = 124u * 100%/284u = ~ 43,65% 100% - 43,65% = ~56,35% P4 – 43,65% , O10 – 56,35% e) PbO2 mczPbO2 = 207u + 16u * 2 = 207u + 32u = 239u   239u – 100% Pb – 207u – x %   x = 207u * 100%/239u = ~86,6 % 100% - 86,6% = ~13,4% Pb – 86,6% , O2 - 13,4%   f) Bi2S3 mcz Bi2S3 = 209u * 2 + 32u * 3 = 418u + 96u = 514u   514u – 100% Bi2 - 418u – x % x = 418u * 100%/514u = ~ 81,3% 100% - 81,3% = ~ 18,7% Bi2 - 81,3%, S3 - 18,7%  

a) tlenek magnezu - MgO Masa cząsteczkowa MgO = 24u + 16u = 40u %Mg = 24u x 100% : 40u = 60% %O = 16u x 100% : 40u = 1% b) tlenek siarki (IV) - SO2 MSO2 = 32u + 2x 16u = 32u +32u = 64u %S = 32 x 100% : 64 = 50% %O = 32 x 100% : 64 = 50% c) tlenek siarki(VI) - SO3 MSO3 = 32u + 3 x 16u = 32u + 48u = 80u %S = 32 x 100% : 80 = 40% %O = 48 x 100% : 80 = 60% d) tlenek fosforu (V) - P4O10 mP4O10 = 4x 31u + 10 x 16u = 124u + 160u = 284u %P = 124 x 100% : 284 = 43,66% %O = 160 x 100% : 284 = 56,34% e) tlenek ołowiu(IV) - PbO2 MPbO2 = 207u + 2 x 16u = 207u + 32u = 239u %Pb = 207 x 100% : 239 = 86,61% %O = 32 x 100% : 239 = 13,39% f) siarczek bizmutu(III) - Bi2S3 MBi2S3 = 2 x 209u + 3 x 32u = 418u + 96u = 514u %Bi = 418 x 100% : 514 = 81,32% %S = 96 x 100% : 514 = 18,68%