empirijska formula

Submitted by ibrahim 123 on Mon, 11/10/2014 - 22:49
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Sagorevanjem 0,864 grama uzorka spoja koji je odgovoran za neugoda miris užeglog putera a koji sadrži C, H i O dobijeno je 1,727 grama CO2 i 0,7068 grama H2O.

Koja je empirijska formula spoja????????? 

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milicaradenkovic - Tue, 11/11/2014 - 06:54

CxHyOz+O2--------------------->CO2+H2O

n(CO2)=m/Mr=1.727/44=0.03925 mol

n(H2O)=m/Mr=0.7068/18=0.03927 mol

n(C)=n(CO2)=0.03925 mol

n(H)=n(H2O)*2=0.03927*2=0.07854 mol

m(C)=n(C)*Mr=12*0.03925=0.471 g

m(H)=n(H)*Mr=0.07854*1=0.07854 g

m(O)=m(jedinjenja)-m(C)-m(H)=0.864-0.471-0.07854=0.31446 g

n(O)=m(O)/Mr=0.31446/16=0.01965 mol

n(C):n(H):n(O)=0.03925:0.07854:0.01965/:0.01965

n(C):n(H):n(O)=2:4:1 i formula je C2H4O ili CH3CHO