Q10.79CP - Chemistry: Molecular Nature Of Matter And Change

Question

An experiment requires 50.0 mL of 0.040 M NaOH for the titration of 1.00 mmol of acid. Mass analysis of the acid shows 2.24% hydrogen, 26.7% carbon, and 71.1% oxygen. Draw the Lewis structure of the acid.

Step-by-Step Solution

Answer

The acid is oxalic acid and the Lewis structure is given below.

Step 1: The Empirical formula.

A chemical formula showing the elements in a compound rather than the total number of atoms of a molecule.

Atom	Mass present in the compound	Moles present in the compound	Ratio
Hydrogen	2.24	$\frac{\text{2.24}}{\text{1}}\text{=2.24}$	$\frac{\text{2.24}}{\text{2.23}}\text{=1}$
Carbon	26.7	$\frac{\text{26.7}}{\text{12}}\text{=2.23}$	$\frac{\text{2.23}}{\text{2.23}}\text{=1}$
Oxygen	71.1	$\frac{\text{71.1}}{\text{16}}\text{=4.44}$	$\frac{\text{4.44}}{\text{2.23}}\text{=2}$

The simplest ratio of C, O, H is the empirical formula of the compound, i.e., ${\mathbf{C}}_1{\mathbf{H}}_1{\mathbf{O}}_2$ .

Step 2: The formula of the compound.

The number of moles of NaOH required to neutralise one mole of the acid is 

 

 $\begin{array}{c}=\frac{\text{0.040moles/L}\times \text{50mL}\times \text{1moles}\times \text{1L}}{\text{0.001moles}\times \text{1000mL}}\\ \text{= 2 moles of NaOH}\end{array}$

 

This implies that the acid is diprotic and the number of hydrogens in it is 2. 

Now, multiplying the empirical formula by 2, the actual formula comes as  ${\mathbf{C}}_2{\mathbf{H}}_2{\mathbf{O}}_4$  and the acid is Oxalic Acid.

Step 3: The structure of compound.

In oxalic acid, there are two hydrogen atoms, four oxygen atoms, and 2 carbon atoms. The number of total valence electrons is 34. Two oxygen atoms are double-bonded with each carbon atom and two -OH groups are single-bonded with each carbon atom.