Răspuns:
INSUMAND procentele date, mai ramane,pana la 100%, procentul26,2%,care apartine,probabil, oxigenului
100g compus..45,9gC...2,73gH....17,48gS....7,65gN....26,2gO
183.........................a................b............c.................d..................e
a=84gC----> niu=m/M= 84/12mol= 7molC
b=5 gH niu= 5/1=5molH
c=32gS niu=...32/32mol= 1molS
d=14gN niu= 14/14mol= 1molN
e=48gO niu= 48/16mol= 3molO
F.M. C7H5O3SN
Explicație:
