Pewien alkohol zawiera 37,5% węgla (% masowy). Oblicz jego mase cząsteczkowa, napisz wzór sumaryczny i nazwe.
zakladamy ze 37,5%=37,5g
CnH2n-1OHM=12n+2n+1+17
M=14n+18C=12n*100/14n+1837,5=12n*100/14n+1837,5(14n+18)=1200n525n+675=1200n
n=1CH3OHM=32u
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zakladamy ze 37,5%=37,5g
CnH2n-1OH
M=12n+2n+1+17
M=14n+18
C=12n*100/14n+18
37,5=12n*100/14n+18
37,5(14n+18)=1200n
525n+675=1200n
n=1
CH3OH
M=32u