Odpowiedź :
a) mC6H12O6=6*12u+12*1u+6*16u=180 u
72u---x%
180u---100%
x=40%C
12u---x%
180u---100%
x=6,67%H
100%-(40%+6,67%)=53,33%O
b) mFe(OH)2=56u+2*16u+2*1u=90u
56u---x%
90u---100%
x=62,22%Fe
32u---x%
90u---100%
x=35,56%O
100%-(62,22%+35,56%)=%H
%H=2,22%
c) mHCOOH=12u+2*1u+2*16u=46u
12u---x%
46u---100%
x=26,09%C
2u---x%
46u---100%
x=4,35%H
100%-(26,09%+4,35%)=69,56%O