Răspuns :
daer=28,9
dgaz=M/Mgaz =>M=dgaz*Mgaz=1,103*28,9=32g/mol
C=37,5/12=3,125
H=12,5/1=12,5
O=50/16=3,125
Dupa imparitrea nr cel mai mic la toate rezultatele => (CH4O)n=32
12n+4n+16n=32=>32n=32=>n=1 => F.m => CH3-OH - metanolul (alcoolul metilic)
dgaz=M/Mgaz =>M=dgaz*Mgaz=1,103*28,9=32g/mol
C=37,5/12=3,125
H=12,5/1=12,5
O=50/16=3,125
Dupa imparitrea nr cel mai mic la toate rezultatele => (CH4O)n=32
12n+4n+16n=32=>32n=32=>n=1 => F.m => CH3-OH - metanolul (alcoolul metilic)
d = masa molara/28.9 => masa molara = 28,9 * 1,103 = 32 g/mol
37,5/12 = 3,125
12,5/1 = 12,5
50/16 = 3,125
Imparti la 3,125 => FB: CH4O
FM: n*FB
M = 32 => 12n+4n+16n = 32 => 32n = 32 => n =1
Dc. n = 1 => FM: CH4O
37,5/12 = 3,125
12,5/1 = 12,5
50/16 = 3,125
Imparti la 3,125 => FB: CH4O
FM: n*FB
M = 32 => 12n+4n+16n = 32 => 32n = 32 => n =1
Dc. n = 1 => FM: CH4O