Răspuns :
% O = mO / M ×100 ⇒ M = mO·100 / (%O) =( 4×16×100) /20,25 = 316
M =12·(n+2) + 1·(2n+5) + 4·16 +32 + 23 = 14n + 148 = 316 ⇒ 14n = 168 ⇒ n = 12 ⇒ formula detergentului este: CH3-(CH2)12-(CH2)-OSO3Na ⇒ ⇒nr. atomi de C din molecula = 14
%O = masa oxigen/masa detergent * 100
masa detergent = 64/20,25 * 100 = 316 g
CH3-(CH2)n-CH2-OSO3Na => masa detergent = 15+14n+14+16+32+3*16+23=148+14n
14n+148=316=>14n=168=>n=12=>CH3 - (CH2)12 - CH2 - OSO3Na
Nr total atomi de C = 1+12+1 = 14 atomi
masa detergent = 64/20,25 * 100 = 316 g
CH3-(CH2)n-CH2-OSO3Na => masa detergent = 15+14n+14+16+32+3*16+23=148+14n
14n+148=316=>14n=168=>n=12=>CH3 - (CH2)12 - CH2 - OSO3Na
Nr total atomi de C = 1+12+1 = 14 atomi