Question

The percentage composition of sodium phosphate as determined by analysis is $42.1\%$sodium, $18.9\%$phosphorus, and $39\%$oxygen. What will be the empirical formula of the compound?

• A. ${\mathrm{NaPO}}_4$
• B. ${\mathrm{Na}}_2{\mathrm{PO}}_4$
• C. ${\mathrm{Na}}_3{\mathrm{PO}}_4$
• D. ${\mathrm{Na}}_4{\mathrm{PO}}_4$

Answer 1

The correct option is C

${\mathrm{Na}}_3{\mathrm{PO}}_4$

Explanation for correct option

(C) ${\mathrm{Na}}_3{\mathrm{PO}}_4$

Empirical formula

• The empirical formula of a compound is defined as the simplest whole-number ratio of each element in a compound.

Steps to find the empirical formula of the compound

Step 1: The first step is to change the percentage to grams.

• $$\begin{gathered} 42.1\%\to 42.1\mathrm{gram}\mathrm{of}\mathrm{Na}\left(\mathrm{sodium}\right) \\ 18.9\%\to 18.9\mathrm{gram}\mathrm{of}\mathrm{P}\left(\mathrm{phosphorus}\right) \\ 39\%\to 39\mathrm{gram}\mathrm{of}\mathrm{O}\left(\mathrm{oxygen}\right) \end{gathered}$$

Step 2: Here, divide all the given masses with the atomic mass to find the mole ratio.

• We know that, the atomic mass of $$\begin{gathered} \mathrm{Na}=23 \\ \mathrm{P}=31 \\ \mathrm{O}=16 \end{gathered}$$
• $\mathrm{Mole}\mathrm{ratio}=\frac{\mathrm{Given}\mathrm{mass}}{\mathrm{Atomic}\mathrm{mass}}$
• Then, the mole ratio of $$\begin{gathered} \mathrm{Na}=\frac{42.1}{23} \\ =1.83 \\ \mathrm{P}=\frac{18.9}{31} \\ =0.61 \\ \mathrm{O}=\frac{39.0}{16} \\ =2.44 \end{gathered}$$

Step 3:Then, pick the smallest moles from the previous step and divide all of the answers by that.

• $\frac{1.83}{0.61}\approx 3of\mathrm{sodium}$
• $\frac{0.61}{0.61}=1of\mathrm{phosphorus}$
• $\frac{2.44}{0.61}\approx 4\mathrm{of}\mathrm{oxygen}$

Step 4: In the last step, the calculated coefficients in the previous step will be written as the subscript in the chemical formula.

• $$\begin{gathered} \mathrm{Na}=3 \\ \mathrm{P}=1 \\ \mathrm{O}=4 \end{gathered}$$
• The chemical formula will be ${\mathrm{Na}}_3{\mathrm{PO}}_4$

Explanation for incorrect options

Since, the correct empirical formula is ${\mathrm{Na}}_3{\mathrm{PO}}_4$. THEREFORE, OPTIONS (A), (B) AND (D) STAND INCORRECT.

Thus the empirical formula of the compound is ${\mathbf{Na}}_{\mathbf{3}}{\mathbf{PO}}_{\mathbf{4}}$