Question

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)

Answer 1

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%

 

Answer 2

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%