Question

when 97.5g of zine was heated in oxygen?121.5g of zinc oxide was formed. calculate the empirical formula of oxide?

Answer 1

When 97.5 g of zinc was heated in oxygen? 121.5 g of zinc oxide was formed.

Calculate the empirical formula of oxide?

**Solution:**

We will designate a formula that is necessary to be found: Znₓ Oᵧ

Knowing the mass of zinc oxide and zinc, which can react to find the mass of oxygen in the oxide:

$$m(O) = m(Zn_xO_y) - m(Zn)$$ $$m(O) = 121,5 - 97,5 = 24\,g$$

Find a relationship between the number of atoms of the elements in zinc oxide, determine the indices X and Y. To do this, divide the mass of each element to its relative atomic mass:

$$n(Zn): n(O) = \frac{m(Zn)}{A_r(Zn)} \cdot \frac{m(O)}{A_r(O)}$$ $$n(Zn): n(O) = \frac{97,5}{65,4} \cdot \frac{24}{16} = 1,5 : 1,5 = 1 : 1$$

So the empirical formula of zinc oxide – ZnO.

**Answer:** The empirical formula of zinc oxide – ZnO.