a/ %mC = \(\frac{12}{12+16.2}.100\%=27,27\%\)
b/ %mAl = \(\frac{27.2}{27.2+16.3}.100\%=52,94\%\)
=> %mO = 100% - 52,94% = 47,06%
a/ %mC = \(\dfrac{12}{12+16.2}\).100%=27,27%
b/ %mAl = \(\dfrac{27.2}{27.2+16.3}\).100%=52,94%
%mO = 100% - 52,94% = 47,06%