h3c-ch2-ch=hc-ch2-ch3 + 4[O] = 2 ch3-ch2-cooh
m= 50,4 g
masa molara= 6ori12 + 12=84 g/mol
masa molara acid = 2 ori 74 g= 148g /mol
deci vom avea 148 ori 50,4/84=88,1 g acid la randament 100%
iar la 75% => 75ori 88,1/100= 66,075g acid
md= 66,075
ms=433,4+66,075=499,47g
deci c= md/ms ori 100= 66,075/499,47 ori 100=13,22%