Question

Zinc phosphate is used as a dental cement. A $\mathbf{50}\mathbf{.}\mathbf{00}\mathbf{mg}$ sample is broken down into its constituent elements and gives $\mathbf{16}\mathbf{.}\mathbf{58}\mathbf{mg}$oxygen, $\mathbf{8}\mathbf{.}\mathbf{02}\mathbf{mg}$phosphorus, and $\mathbf{25}.\mathbf{40}\mathbf{mg}$zinc. Determine the empirical formula of zinc phosphate.

Short answer

The empirical formula of zinc phosphate is${\text{Zn}}_3{\text{P}}_2{\text{O}}_8$

Step-by-step solution

Step 1:Calculation of mass percentage of each elements

Mass percentage can be calculated by dividing the mass of each elements by total mass of the compound.

Given that the mass of oxygen, ${\text{m}}_{\text{O}}=16.58\text{mg}$

Mass of phosphorus, ${\text{m}}_{\text{P}}=8.02\text{mg}$

Mass of zinc, ${\text{m}}_{\text{Zn}}=25.40\text{mg}$

Total mass of zinc phosphate,

$\begin{array}{rcl}{\text{m}}_{\text{tot}}& =& 16.58+8.02+25.40\text{mg}\\ & & \text{= 50.00}\text{mg}\end{array}$

Mass percentage of oxygen,

localid="1663421668726" $\begin{array}{rcl}{\text{m}}_{\text{O}}\%& =& \frac{{\text{m}}_{\text{O}}}{{\text{M}}_{\text{tot}}}\times 100\\ & =& \frac{\text{16.58}}{\text{50.00}}\times 100\\ & =& 33.16\%\end{array}$

Mass percentage of phosphorus,

$\begin{array}{rcl}{\text{m}}_{\text{P}}\%& =& \frac{{\text{m}}_{\text{P}}}{{\text{M}}_{\text{tot}}}\times 100\\ & =& \frac{\text{8.02}}{\text{50.00}}\times 100\\ & =& 16.04\%\end{array}$

Mass percentage of zinc,

$\begin{array}{rcl}{\text{m}}_{\text{Zn}}\%& =& \frac{{\text{m}}_{\text{Zn}}}{{\text{M}}_{\text{tot}}}\times 100\\ & =& \frac{\text{25.40}}{\text{50.00}}\times 100\\ & =& 50.80\%\end{array}$

Step 2: Divide mass of each elements by their atomic masses

Atomic mass of oxygen = 16

$\text{O =}\frac{33.16}{16}=2.0725$

Atomic mass of phosphorus = 31

$\text{P =}\frac{16.04}{31}=0.517$

Atomic mass of zinc = 65

$\text{Zn =}\frac{50.80}{65}=0.7815$

Division of smallest value

The smallest value is of phosphorus, i.e., 0.5174

$\text{O =}\frac{2.0725}{0.5174}=4.0056$

$\text{P =}\frac{0.5174}{0.5174}=1$

$\text{Zn =}\frac{0.7815}{0.5174}=1.5104$

Now, we need to multiply with 2, forming O=8, P=2, and Zn=3

Thus, the empirical formula is ${\text{Zn}}_3{\text{P}}_2{\text{O}}_8$.