The empirical formula of this compound is 
<h3>Empirical formula </h3>
To calculate the empirical formula of a compound, the value of moles of each element is needed.
As we have the information of the mass value, we will use the molar mass expression, which corresponds to:





As the value of the empirical formula must be an integer, simply multiply the two values by a common factor:


So, the empirical formula of this compound is
.
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Answer and Explanation:
The balanced chemical equations are as follows:
The chemical formula of oxalic is 
In the case when oxalic acts reacted with the water so here the oxalic acid eliminates one proton that leads to the development of mono acids
After that, the second step derives that when oxalic acid is in aqueous solution eliminates other proton so it represent the polyprotic acid
Now the chemical equations are as follows:
Elimination of one proton

Now the elimination of other proton

Substitution Reactions are those reactions in which one nucleophile replaces another nucleophile present on a substrate. These reactions can take place via two different mechanism i.e SN¹ or SN². In SN¹ substitution reactions the leaving group leaves first forming a carbocation and nucleophile attacks carbocation in the second step. While in SN² reactions the addition of Nucleophile and leaving of leaving group take place simultaneously.
Example:
OH⁻ + CH₃-Br → CH₃-OH + Br⁻
In above reaction,
OH⁻ = Incoming Nucleophile
CH₃-Br = Substrate
CH₃-OH = Product
Br⁻ = Leaving group
Organic reactions are typically slower than ionic reactions because in organic compounds the covalent bonds are first broken, this breaking of bonds is a slower step, while, in ionic compounds no bond breakage is required as it consists of ions, so only bond formation takes place which is a quicker and fast step.