86g..............100%
..xg........... 83,72%
x=86*83,72/100=72g C
y=14g H
C: 72/12=6
H: 14/1=14
f.m: C6H14
b)
NE=[2+6(4-2)+14(1-2)]/2=[2+12-14]/2=0
d)deoarece NE=0 => atomii de C sunt legati doar prin legaturi sigma, si sunt hibridizati sp³
c)
H3C-CH(CH3)-CH(CH3)-CH3
2,3-dimetil-butan