Question

a compound is 24.7% calcium, 1.2% hydrogen, 14.8% carbon, and 59.3% oxygen. write the empirical formula and name the compound.

Answer 1

For simplicity, assume that the total mass of the compound is 100 g.

Therefore, by mass
Ca = 24.7 g, H = 1.2 g, C = 14.8 g, and O = 59.3 g

Converting these to moles

Mol = mass*1/atomic mass
Ca = 24.7*1/40.078 =0.6163 mol Ca
H = 1.2*1/1.01 = 1.1881 mol H
C = 14.8*1/12.01 = 1.2323 mol C
O = 59.3/16 = 3.7062 mol O

Next, divide all the mols by the smallest value obtained.

Ca: 0.6163/0.6163 = 1 mol Ca
H: 1.1881/0.6163 = 2 mol H
C: 1.2323/0.6163 = 2mol C
O: 3.7062/0.6163 = 6 mol O

Therefore, empirical formula of the compound is

CaH2C2O6

This compound is referred to as Calcium Bicarbonate