• I was wondering how I'd go around calculating the equivalent weight of CaO. From my knowledge, the EW of a compound is its (given mass )/(valence factor, n). Since CaO has no charge and no change in
    the oxidation number takes place, my valence factor comes out to be 0, which I know is wrong because the compound has to have some EW, which is 28 as I know that 28grams of CaO will be formed by 8g O
  • $\begingroup$ sigh Why are you even learning this outdated stuff? EW depends on specific reaction, in typical reactions of CaO it's 2. $\endgroup$ – Mithoron May 12 '17 at 21:32
  • $\begingroup$ @Mithoron Poor kid gets homework from outdated teachers. $\endgroup$ – Pritt says Reinstate Monica May 13 '17 at 3:32

Yes, Dev, you know the correct process of calculating EW, but the problem you are facing is how to calculate the valence factor?, right?

So, in this case you have to understand that EW for any compound is reaction-specific. So, you can't calculate the EW of $\ce{CaO}$ without any reaction reference.

Still, usually $\ce{CaO}$ reacts through 2 electron exchange mechanism, in other words, $\ce{CaO}$, on dissociation, gives total cationic charge +2 and total anionic charge -2.

$$\ce{CaO = Ca^2+ + O^2-}$$

Hence, in this case of calculating EW, you have to consider $n=2$ and so the EW of $\ce{CaO}$ is $(40+16)/2 = 28$.

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