A compound contains 40% Ca, 12%C and 48% O find it's empirical formula and name the compound ?

To find the empirical formula of the compound, we need to determine the ratio of each element present in the compound.

1. Convert the percentages to grams:
- 40% Ca = 40g Ca
- 12% C = 12g C
- 48% O = 48g O

2. Determine the moles of each element:
- Moles of Ca = 40g Ca / 40.08 g/mol = 0.998 mol Ca
- Moles of C = 12g C / 12.01 g/mol = 0.999 mol C
- Moles of O = 48g O / 16.00 g/mol = 3.00 mol O

3. Divide each mole value by the smallest mole value to get the simplest ratio:
- Ca: 0.998 / 0.998 = 1
- C: 0.999 / 0.998 = 1
- O: 3.00 / 0.998 = 3

Therefore, the empirical formula of the compound is CaCO3, which is calcium carbonate.