1. oblicz zawartość procentową węgla w octanie metylu.
M C3H6O2 =74 g/mol (lub "u")
%C= (mC*100%)/M C3H6O2
%C= (36*100% )/74
%C=48,65%
mCH₃COOCH₃=12u*3+6*1u+16u*2=74u
74u-100%
36u-x
x=48,65%
Odp zawartość % węgla w octanie metylu wynosi 48,65%
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M C3H6O2 =74 g/mol (lub "u")
%C= (mC*100%)/M C3H6O2
%C= (36*100% )/74
%C=48,65%
mCH₃COOCH₃=12u*3+6*1u+16u*2=74u
74u-100%
36u-x
x=48,65%
Odp zawartość % węgla w octanie metylu wynosi 48,65%