"\\% O= 100 \\% - (48.63 \\% + 8.18\\%)=43.19 \\%"
"\\dfrac{74g\/mol*48.63\\%}{12g\/mol*100\\%}=2.998(~3)atoms(C)"
"\\dfrac {74g\/mol*8.18\\%}{1g\/mol*100\\%}=6.05 atoms(H)"
"\\dfrac{74g\/mol*43.19 \\%}{16g\/mol*100\\%}" ~ 3 atoms(O)
molecular formula of lactic acid: C3H6O3
Comments
Leave a comment