-calculez moli AgBr,care are masa molara M= 188g/mol(VERIFICA !)
n= m/M----> n= 3,76g/188g/mol= 0,02molAgBr------> 0,02molBr
-formula chimica generala a compusului este CxH2xBr2 si are masa molara M=12X+2X+2X80
-deduc X
in 2,16g compus.....sunt....0,02molBr
14x+160 g..........................2mol
2,16x2= 0,02(14x+160)......>x=4
compusul este C4H8Br2
-calculez compozitia procentuala
216gcompus........48gC.......8gH........160gBr
100g.........................x..............y............z--------------.> CALCULEAZA !!!!