Answer:
D) empirical formula is: C₃P₂O₈
Explanation:
Given:
Mass % Calcium (Ca) = 38.7%
Mass % Phosphorus (P) = 19.9%
Mass % oxygen (O) = 41.2 %
This implies that for a 100 g sample of the unknown compound:
Mass Ca = 38.7 g
Mass P = 19.9 g
Mass O = 41.2 g
Step 1: Calculate the moles of Ca, P, O
Atomic mass Ca = 40.08 g/mol
Atomic mass P = 30.97 g/mol
Atomic mass O = 16.00 g/mol
[tex]Moles\ Ca = \frac{38.7g}{40.08g/mol} =0.966\ mol\\\\Moles\ P = \frac{19.9g}{30.97g/mol} =0.643\ mol\\\\Moles\ O = \frac{41.2g}{16.00g/mol} =2.58\ mol[/tex]
Step 2: Calculate the molar ratio
[tex]C = \frac{0.966}{0.643} =1.50\\\\P = \frac{0.643}{0.643} = 1.00\\\\O = \frac{2.58}{0.643} =4.00[/tex]
Step 3: Calculate the closest whole number ratio
C: P: O = 1.50 : 1.00 : 4.00
C : P : O = 3:2:8
Therefore, the empirical formula is: C₃P₂O₈
