1.D за H2(CxHy) = 35
W(C) = 85.7%
W(H) = 14.3%
CxHy - ?
M(CxHy) = 35*2 = 70 g / mol
Ar(C) = 0.857 * 70 = 60.
n(C) = 60 / 12 = 5.
Ar(H) = 70-60 = 10
n(H) = 10 / 1 = 10.
C5H10
2. W(C) = 40%
W(H) = 6.66%
W(O) = 53.34%
CxHyOz - ?
x:y:z = 40 / 12 : 6.66 / 1 : 53.34 / 16 = 3.33 : 6.66 : 3.33 | / 3.33 = 1:2:1
CH2O
3.D за H2 (CxHy) = 42.
W(C) = 85.71%
W(H) = 14.29%
M(CxHy) = 42*2 = 84 g / mol
Ar(C) = 0.8571 * 84 = 72.
n(C) = 72/12 = 6.
Ar(H) = 84 - 72 = 12.
n(H) = 12 / 1 = 12.
C6H12
1.D за H2(CxHy) = 35
W(C) = 85.7%
W(H) = 14.3%
CxHy - ?
M(CxHy) = 35*2 = 70 g / mol
Ar(C) = 0.857 * 70 = 60.
n(C) = 60 / 12 = 5.
Ar(H) = 70-60 = 10
n(H) = 10 / 1 = 10.
C5H10
2. W(C) = 40%
W(H) = 6.66%
W(O) = 53.34%
CxHyOz - ?
x:y:z = 40 / 12 : 6.66 / 1 : 53.34 / 16 = 3.33 : 6.66 : 3.33 | / 3.33 = 1:2:1
CH2O
3.D за H2 (CxHy) = 42.
W(C) = 85.71%
W(H) = 14.29%
M(CxHy) = 42*2 = 84 g / mol
Ar(C) = 0.8571 * 84 = 72.
n(C) = 72/12 = 6.
Ar(H) = 84 - 72 = 12.
n(H) = 12 / 1 = 12.
C6H12