Question

A compound is 24.7% Calcium, 1.2% Hydrogen, 14.8% Carbon, and
59.3% Oxygen. Write the empirical formula
please answerrrrrrr

Answer 1

Given: A compound has 24.7% Calcium, 1.2% Hydrogen, 14.8% Carbon, and 59.3% Oxygen.

To find: We have to find its emperical formula.

Solution:

Let the mass of each element in 100 g.

Number of moles of each element is a ratio of mass to molecular weight.

Number of moles will be-

Ca: $\dfrac{24.7 g}{40.078 g/mol} = 0.616 mol$

H: $\dfrac{1.2 g}{1.008 g/mol} = 1.2 mol$

C: $\dfrac{14.8 g}{12.011 g/mol} = 1.23 mol$

O: $\dfrac{59.3 g}{15.999 g/mol} = 3.71 mol$

Dividing the moles of each element by the least number of moles we get-

Ca: $\dfrac{0.616 mol}{0.616 mol} = 1$

H: $\dfrac{1.2 mol}{0.616 mol} = 1.9 = ~2$

C: $\dfrac{1.23 mol}{0.616 mol} = 2$

O: $\dfrac{3.71 mol}{0.616 mol} = 6$

So, there 1 Ca,2 H, 2C and 6O will present.

Empirical formula will be-

CaH2C2O6, which is probably Ca(HCO₃)₂.