Question

1) 45.04 grame de amestec de acid ftalic si benozoic se nutralizeaza cu 150 ml de sol de NaOH 2M. Calculeaza compozitia in procente a amestecului.

2) Care este cetoacidul care are %C=47.05%?

3) Care este acidul dihidroxidicarboxilic care are %O=64%?

4) Calculeaza masa de triglicerida care:
a) prin saponificare formeaza 7.65 g de sapun cu 7.51% Na
b) prin saponificare formeaza 13.9 g de sapun cu 8.273% Na

Answer 1

n1 moli    HOOC-C6H4-COOH + 2NaOH ⇒ NaOOC-C6H4-COONa + 2H2O
n2 moli    C6H5-COOH + NaOH ⇒ C6H5-COONa + H2O
166n1 + 122n2 = 45,4    83n1 + 61n2 = 22,7 
2n1 + n2 = 0,25·2 = 0,5 moli       39n1 = 7,8   n1 = 0,2 moli  ac. ftalic ;
n2 = 0,1 moli ac. benzoic
m1 =33,2 g      % ac.ft. = 33,2·100/45,4 = 73,13%  
m2 = 12,2 g     % ac. benz. = 1220/45,4 = 26,87%
CxH(2x+1)-CO-COOH     12(x + 2)·100/(14x + 74) = 47,05 
25,5x + 51 = 14x + 74      11,5x = 23  x = 2    CH3-CH2-CO-COOH
3.     6·16·100/M = 64   ⇒⇒⇒ M = 150g/mol
  150 - 2·45 - 2·17 = 26 ⇒⇒⇒ acid = HOOC-CH(OH)CH(OH)-COOH·
4.  a) sapun : R-COONa     23·100/M = 7,516    M = 306g/mol
daca R = saturat    CxH2x+1-COONa    14x + 68 = 306   x = 17  
acid = C18H36O2  (acid stearic)  triglicerida = tristearina
triglicerida + 3NaOH ⇒3 CH3-(CH2)16)-COONa + glicerina 
 M = 890g/mol                            M = 306g/mol
n sapun = 7,65/306 = 0,025 moli = 3n glicerida  ⇒  n gl. = 0,025/3
m = 0,025/3 ·890 = 7,416g
b) 23·100/M = 8,273   M = 278g/mol  14x + 68 = 278    x = 15
acid : C16H32O2  (acid palmitic) trigicerida = tripalmitina  M = 806 g/mol
n sapun = 13,9/278 = 0,05 moli  ⇒ n glicerida = 0,05/3
m = 0,05/3 ·806 = 13,43 g