Question

Determina compozitia procentuala a următoarelor substanțe: a)carbonat de sodiu b)tetraclorura de carbon c)clorura de bariu d)acid carbonic e)hidroxid de aluminiu f)sulfat de amoniu

Answer 1

a) carbonat de sodiu
CaCO3
M=40+12+3*16=100
C(Ca)=40*100/100=40%
C(C)=12*100/100=12%
C(O)= 48*100/100=48%

b) tetraclorura de carbon
CCl4
M=12+35,5*4=154
C(C)=12*100/154=7,79%
C(Cl)=142*100/154=92,21%

c) clorura de bariu
BaCl2
M=137+2*35,5=208
C(Ba)=137*100/208=65,87%
C(Cl)=71*100/208=34,13%

d) acid carbonic
H2CO3
M=1*2+12+3*16=2+12+48=62
C(H)=2*100/62=3,23%
C(C)=12*100/62=19,35%
C(O)=48*100/62=77,42%

e) hidroxid de aluminiu
Al(OH)3 
M=27+3*16+3*1=27+48+3=78
C(Al)=27*100/78=34,61%
C(O)=48*100/78=61,54%
C(H)=3*100/78=3,85%

f) sulfat de amoniu
(NH4)2SO4
M=14*2+1*8+32*1+16*4=28+8+32+64=132
C(N)=28*100/132=21,21%
C(H)=8*100/132=6,06%
C(S)=32*100/132=24,24%
C(O)=64*100/132=48,48%