# Equivalent weight for CaO

• 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
• sigh Why are you even learning this outdated stuff? EW depends on specific reaction, in typical reactions of CaO it's 2. May 12, 2017 at 21:32
• @Mithoron Poor kid gets homework from outdated teachers. May 13, 2017 at 3:32

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 $(40+16)/2 = 28$.