Question #15944

How many atoms are in 1.8675g of a recently synthesized organic compound:
1.which has a molar mass between 350 and 400 grams/mole
2.which consists of only iron, carbon and hydrogen
3.when analyzed using combustion analysis on a 0.9437 g sample, produced 2.233 g of CO2 and 0.457 g of H2O?

Expert's answer

We have compound with& formula (C)x(H)y(Fe)o

Let's calculate x,y and o using combustion analysis

2.233 & m=0.6090 g n=0.6090/12= 0.0508

CO2 --------& C

& 44 12

0.457 m=0.0508 g n=0.0508/1=0.0508 mol

H2O------------2H

18 2*1

m of Fe =0.9437-0.609-0.0508=0.2839 n=0.2839/56= 0.00510

n of C : n of H : n of Fe =10:10:1,& it is something like this (C10H10Fe)z& together it have molar mass& (120+10+56)z= 186z

z=400/186=2 (it's closer to 2) , so it is C20H20Fe2

& 1.8672 & N

C20H20Fe2------------- 42 atoms

& 372 42*6.02 x 10^23

N=1.8672* 42*6.02 x 10^23 /& 372 =3.02 x 10^21 atoms

Let's calculate x,y and o using combustion analysis

2.233 & m=0.6090 g n=0.6090/12= 0.0508

CO2 --------& C

& 44 12

0.457 m=0.0508 g n=0.0508/1=0.0508 mol

H2O------------2H

18 2*1

m of Fe =0.9437-0.609-0.0508=0.2839 n=0.2839/56= 0.00510

n of C : n of H : n of Fe =10:10:1,& it is something like this (C10H10Fe)z& together it have molar mass& (120+10+56)z= 186z

z=400/186=2 (it's closer to 2) , so it is C20H20Fe2

& 1.8672 & N

C20H20Fe2------------- 42 atoms

& 372 42*6.02 x 10^23

N=1.8672* 42*6.02 x 10^23 /& 372 =3.02 x 10^21 atoms

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