I assume O2 is an abbreviation for oxygen and you DON'T mean O2 atoms.
22 g CO2 = 22/molar mass = 22/44 = 0.50 mole CO2. Since 1 mole CO2 contains 6.022E23 molecules of CO2, 0.50 mole will contain just half that number of CO2 molecules(3.011E23) and that number of atoms(6.022E23).
So 1 mole CO should contain the same number of oxygen atoms.
1 mole CO (28 grams) contains 6.02E23 molecules and 6.022E23 atoms oxygen.
calculate the weight of carbon momoxide having same number of O2 atoms as are present in 22g of carbon dioxide
9 answers
28g
CO2
W(gram)=22g
Molecular weight= 44u
N= no. of molecules
NA= Avogadro no.
N/NA=w(gram)/molecular weight
N/6.022×10^23=22/44
N=22×6.022×10^23/44
N=3.011×10^23 molecules
No. of oxygen atoms in CO2=3.011×10^23×2
=6.022×10^23 atoms
No. of oxygen atoms in CO2=no. of CO molecules=
6.022 ×10^23 atoms
CO
W(gram)=?
Molecular weight=28u
No. of molecules=6.022×10^23
NA =Avogadro no.=6.022×10^23
W(gram)/molecular weight=no. of molecules/ NA
W(gram) /28=6.022×10^23/6.022×10^23
W(gram)=1×28
=28 grams
W(gram)=22g
Molecular weight= 44u
N= no. of molecules
NA= Avogadro no.
N/NA=w(gram)/molecular weight
N/6.022×10^23=22/44
N=22×6.022×10^23/44
N=3.011×10^23 molecules
No. of oxygen atoms in CO2=3.011×10^23×2
=6.022×10^23 atoms
No. of oxygen atoms in CO2=no. of CO molecules=
6.022 ×10^23 atoms
CO
W(gram)=?
Molecular weight=28u
No. of molecules=6.022×10^23
NA =Avogadro no.=6.022×10^23
W(gram)/molecular weight=no. of molecules/ NA
W(gram) /28=6.022×10^23/6.022×10^23
W(gram)=1×28
=28 grams
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