Дано
W(C)=85.71% = 0.8571
W(H)=14.29% = 0.1429
D(H2)=42
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CxHy-?
M(H2)= 2g/mol
M(CxHy)=D(H2)*M(H2)=42*2g/mol=84
m(CxHy)=n(CxHy)* M(CxHy) 1 mol*84
m(C)=m(CxHy)*W(C) / 100% =84*
85.71% / 100% =72g
m(H)= m(CxHy) * W(H) / 100% =84*
14.29% /100% 12g
n(C) = m(C) /M(C) = 12g/mol
n(C) =72 / 12 = 6 mol
n(H) = m(H) / M(H) , M(H) =g/mol
n(C) : n(H) = 6 : 12
C6H12
Ответ : C6H12