Question

Вычислите массовые доли элементов в алкане, плотность паров по кислороду равна 2,25.

Answer 1

$D= \frac{M(C _{n} H _{2n+2} )}{32} \\M=2.25*32=72\\12n+2n+2=72\\14n=70\\n=5\\\\\\ C _{5} H _{12}$

$w(C)= \frac{12*5}{12*5+12} = \frac{60}{72} =83\\\\w(H)= \frac{12}{72} =17$

Answer 2

Mr(CnH2n+2)=2.25*32=72

12x+2x+2=72
14x=72-2
14x=70
x=70/14
x=5

C5H12

w(C)=60*100%/72=83.3%
w(H)=12*100%/72=16.7%