Pentru H20:
18 g H20.......2 g H.......16 g O
100...............x .............y
x= (2*100)/18= 11,11 % H
y= (11,11*16)/2= 88,88 % O
Pentru NH3:
Masa moleculara a NH3= 14 + 3=17
17 g NH3...... 14 g N..... 3 g H
100..............x...................y
x=(100*14)/17= 82,35 % N
y=(82,35*3)/14= 17,64 % H
Pentru CH4:
16 g CH4...... 12 g C....4 g H
100................x.............y
x=(100*12)/16= 75 % C
y= (75*4)/12= 25 %H
Pentru H2SO4:
Masa moleculara a H2SO4= 2+32+16*4=98
98 g H2SO4......2 g H....32 g S....64 g O
100....................x...........y...........z
x=(100*2)/98= 2,04 % H
y= ( 2,04*32)/2=32, 64 % S
z= (32,64*64)/32= 65, 28 % O
