Question

An oxide of phosphorus contains 56.4% phosphorus and 43.6% oxygen. It's relative molecular mass is 220. Find both the empirical and the molecular formula of the oxide'

Answer 1

Moles of P = 56,4g/30,974g/mole = 1,82 moles P
moles of O = 43,6/15,999 = 2,73 moles of O

converting to the simplest ratio:
For P : 1,82/1,82 = 1
For O : 2,73/1,82 = 1,5

1 P and 2 oxygens.
PO2 -> the empirical formula

hope this help

Answer 2

Answer: The empirical and molecular formula for the given compound is $P_2O_3$ and $P_4O_6$ respectively.

Explanation:

We are given:

Percentage of P = 56.4 %

Percentage of O = 43.6 %

Let the mass of compound be 100 g. So, percentages given are taken as mass.

Mass of P = 56.4 g

Mass of O = 43.6 g

To formulate the empirical formula, we need to follow some steps:

• Step 1: Converting the given masses into moles.

Moles of Phosphorus =$\frac{\text{Given mass of Phosphorus}}{\text{Molar mass of Phosphorus}}=\frac{56.4g}{31g/mole}=1.82moles$

Moles of Oxygen = $\frac{\text{Given mass of oxygen}}{\text{Molar mass of oxygen}}=\frac{43.6g}{16g/mole}=2.725moles$

• Step 2: Calculating the mole ratio of the given elements.

For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is 1.82 moles.

For Phosphorus = $\frac{1.82}{1.82}=1$

For Oxygen = $\frac{2.725}{1.82}=1.5$

Converting the calculated moles into lowest whole number, we multiply the number of moles by '2'

Moles of Phosphorus = (1 × 2) = 2

Moles of oxygen = (1.5 × 2) = 3

• Step 3: Taking the mole ratio as their subscripts.

The ratio of P : O = 2 : 3

Hence, the empirical formula for the given compound is $P_2O_3$

• For determining the molecular formula, we need to determine the valency which is multiplied by each element to get the molecular formula.

The equation used to calculate the valency is :

$n=\frac{\text{Molecular mass}}{\text{Empirical mass}}$

We are given:

Mass of molecular formula = 220 g/mol

Mass of empirical formula = 110 g/mol

Putting values in above equation, we get:

$n=\frac{220g/mol}{110g/mol}=2$

Multiplying this valency by the subscript of every element of empirical formula, we get:

$P_{(2\times 2)}O_{(3\times 2)}=P_4O_6$

Hence, the empirical and molecular formula for the given compound is $P_2O_3$ and $P_4O_6$ respectively.