YMnO3 was synthesized by MnO2 and Y2O3.I want to know how to calculate theoretical density.
I got a formula below:
Theoretical density = (Molecular weight x No. of molecules per unit cell) / (Volume of unit cell x Avogardo's number).
But the mate in my lab told me I can calculate like this:
Theoretical density(YMnO3)=Theoretical density(MnO2)+1/2Theoretical density(Y2O3)=5.03+2.505=7.535.
Is that correct?And how much is the real theoretical density?
Thank you so much!
Mr. Ingenmey
Thank you so much for your explicit explanation!
I understood most of them except for the equation below,
ρ(YMnO3) = ρ(MnO2) + 1/2 ρ(Y2O3) - 1/4 ρ(O2)
It seems like 'Law of conservation of mass',but I don't know why it's valid and where it
comes from.(As I majored in Mechanical in my bachelor)May be it's a stupid basic question,but I 'm still eager for your answer.
Best wishes,
DU
Thank you so much,Johannes!
I agree with you mostly.There is no doubt we can obtain accurate ideal density by the first formula,while differs from the consequence a lot yielded by the second one.
Then, as to the formula you gave,V = x(A) * V(A) + x(B) * V(B)
It looks similar to partial molar quantity calculation in a mixed liquid.Whereas,in our situation,there's a reaction.There's no Y2O3 and MnO2 inside YMnO3 or O2. So can we still put them together,use the formula ρ(4 YMnO2 + O2) = x(MnO2) * ρ(MnO2) + x(Y2O3) * ρ(Y2O3) = 2/3 * ρ(MnO2) + 1/3 * ρ(Y2O3)?(Sorry for my poor expression)
Cause the huge difference exists,and I haven't found any foundation to support this calculation,letting me doubt the accuracy of this formula.Could you please explain it in detail?I will be so thankful.
Best wishes
Du
Points raised/suggested by Johannes are right. The question asked is not straight forward. Add for what purpose and under what conditions. Excess properties suggested will depend on the solvent used etc( on interactions).
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