Chapter 10, Problem 165A

Interpretation Introduction

Interpretation:

The percent composition and empirical formula of compound which is formed when 35.07 g of Phosphorus react with oxygen to form 71.00 g of phosphorus oxide needs to be determined.

Concept introduction:

Empirical formula is the simplest formula of any organic or inorganic compound that represents the simple ratio of all atoms present in the molecule. It can be calculated with the help of elemental composition of molecule. Certain steps must be used to get the empirical formula:

• Consider mass % as mass in grams and calculate moles of element with the help of molar mass.
• Calculate the moles of each element in least whole number.
• Write the number of each atom as subscript to write the empirical formula.

Answer to Problem 165A

Empirical formula = ${\text{PO}}_{\text{2}}$.

Explanation of Solution

Given information:

Mass of phosphorus = 35.07 g

Mass of phosphorus oxide = 71.00 g

Molar Mass of phosphorus = 31.0 g/mol

Mass % of P = $\frac{\text{35}\text{.07 g}}{\text{71}\text{.00 g}}\times \text{100= 49}\text{.4\%}$

Mass % of O = $\frac{\text{71}\text{.00-35}\text{.07 g}}{\text{71}\text{.00 g}}\times \text{100= 50}\text{.6\%}$

Calculate moles of P and O :

$\begin{array}{l}\text{Moles of P = }\frac{\text{35}\text{.07g}}{\text{31}\text{.0g/mol}}\text{= 1}\text{.131 mole}\\ \text{Moles of O = }\frac{\text{71}\text{.00-35}\text{.07g}}{\text{15}\text{.99 g/mol}}\text{= 2}\text{.247 mole}\end{array}$

Divide by least moles to get the whole ratio:

$\begin{array}{l}\text{Moles of P = }\frac{\text{ 1}\text{.131 mole}}{\text{ 1}\text{.131 mole}}\text{=1}\\ \text{Moles of O = }\frac{\text{2}\text{.247 mole}}{\text{ 1}\text{.131 mole}}\text{= 1}\text{.98=2}\end{array}$

Thus, the empirical formula must be ${\text{PO}}_{\text{2}}$.

Conclusion

Thus, empirical formula = ${\text{PO}}_{\text{2}}$.