Question

Oblicz zawartość procentową (% masowy ) pierwiastków w solach
a) AlCl3
b) Na2So4
c) Pb(NO3)2

Answer 1

M(AlCl₃)=27u+3*35,5u=133,5u
m(Al)=27u
m(cl₃)=106,5u
133,5u ---- 100%
27u ---- %m(Al)
%m(Al)=27u*100% / 133,5u
%m(Al) = 20,22%
%m(cl₃)= 100% - 20,22% =79,88%


m(Na₂SO₄)=2*23u+32u+4*16u=143u
m(Na₂)=46u
m(s)=32u
m(O₄) = 64u
143u ---- 100%
46u -----%m(Na₂)
%m(Na₂)= 46u*100% / 143u
%m(Na₂) = 32,17%
143u ---- 100%
32u ----- %m(S)
%m(S)=32u*100 / 143u
%m(S)= 22,38%
%m(O4) = 100% - (32,17%+22,38%)
%m(O4) = 45,45%


m(Pb(NO₃)₂)= 207u+2*14u+6*16u=331u
m(Pb)=207u
m(N₂)= 28u
m(O₆)=96u
331u ---- 100%
207u ----- %m(Pb)
%m(Pb)=207 *100 / 331
%m(Pb) = 62,54%
331u ---- 100%
28u ----- %m(N₂)
%m(N₂) = 28*100/331
%m(N₂) = 8,46%
%m(O₆) = 100% - (8,46% + 62,52%)
%m(O₆) = 28,99%

to chyba wszystko ;)