Question

What are the empirical formulas for tetraphosphorus decaoxide $(P_4O_{10})$ and ethane $(C_2H_6)$ ?

Answer 1

Hint: Empirical formula of a compound is a simple positive integer ratio which shows the bonding ratio of the elements in the compound. To find the empirical formula from the molecular formula, we must take the ratio of the subscripts of each element and reduce it down to the lowest whole number ratio possible. We take the obtained ratio as the subscripts of the elements in the same ratio, which gives us the empirical formula.

Complete step-by-step answer:
We can define empirical formulas as the simple whole number ratio of the elements in the compound, which only shows the bonding ratio (Bonding ratio is defined as how many atoms of one element is bonded with how many atoms of other elements).

Empirical formulas only show us the bonding ratio, they cannot provide any information about the arrangement or number of atoms of a particular element actually present in a compound.

i) Let us consider the compound tetraphosphorus decaoxide $(P_4O_{10})$
Here, we can see from the formula that phosphorus and oxygen are present in the compound in the ratio $4:10\;$ meaning four phosphorous atoms are bonded with ten oxygen atoms.
- Now, let us divide the ratio by two, by which we get the ratio as $2:5\;$ . We cannot reduce the ratio further, as they do not have a common factor.
- So let us substitute the ratio with the elements in the same manner it was noted. Hence, we get the formula as $P_2O_5$ which is the empirical formula for tetraphosphorus decaoxide.

ii) Let us consider the compound ethane $(C_2H_6)$
Here, we can see from the formula that carbon and hydrogen are present in the compound in the ratio $2:6\ $ meaning two carbon atoms are bonded with six hydrogen atoms.
- Now, let us divide the ratio by two, by which we get the ratio as $1:3\ $ . We cannot reduce the ratio further, as they do not have a common factor.
- So let us substitute the ratio with the elements in the same manner it was noted. Hence, we get the formula as $C{{H}_{3}}$ which is the empirical formula for ethane.

Note: While taking the ratio of the subscripts of elements, we must take the ratio in the same manner as they are mentioned in the molecular formula. And after reducing the ratio, we must check if the ratio hasn’t inverted or changed, as it might completely change the molecular formula. Also, empirical formulas can only be used to find the bonding ratio between elements.