Oblicz zawartość procentową węgla,wodoru,tlenu: -penten -metanol

PENTEN mC5H10=60u+10u=70u %C = (60/70)*100% = 85,7% %H = (10/70)*100% = 14,3%   METANOL mCH3OH=12u+4u+16u=32u %C = (12/32)*100% = 37,5% %H = (4/32)*100% = 12,5% %O = (16/32)*100% = 50%

penten: C5H10 masa: (12g x 5)+(1g x 10)=70g C: 60/70 x 100%=85,7% H: 10/70 x 100%=14,3%   metanol: CH3OH masa: 12g+16g+(1g x 4)=32g C: 12/32 x 100%=37,5% O: 16/32 x 100%=50% H: 4/32 x 100%=12,5%