Zad5. Oblicz stosunek masowy węgla, wodoru i tlenu w butanolu.
Zad6. Oblicz procent masowy pierwiatków chemicznych w propanolu.
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zad5. C4H9OH
mC= 12u*4=48u
mO=16u*1=16u
mH = 10u*1u=10u
C:O:H= 48u:16u:10u = 24:8:5
zad6: mC= 12u*3=36u
mH=8*1u=8u
mO=16u
100%=60u
C - 36u*100%/60u = 60%
H - 8u*100%/60u = ok. 13,3 %
O - 16u*100%/60u = ok. 26,7%