Stoichiometry And Stoichiometric Calculations

Step 1: Analyze the CCP structure

In a cubic close-packed structure, the oxide ions (O²⁻) form the close-packed arrangement, which leaves octahedral voids. The number of octahedral voids is equal to the number of oxide ions in the structure.

Step 2: Consider the Aluminium ions

We are told that aluminium ions (Al³⁺) occupy 1/4th of the octahedral voids in the CCP structure. This means that for every 4 oxide ions, there is 1 aluminium ion occupying an octahedral void.

Step 3: Determine the ratio of ions

- The ratio of Al to O is 1:4 because Al occupies 1/4th of the octahedral voids.
- The beryllium ions (Be²⁺) are likely in tetrahedral voids in the structure.

Step 4: Construct the formula

Given that there are 4 oxygen ions (O²⁻) for every aluminium ion (Al³⁺), the number of Be ions will balance the structure. The most likely formula for the compound would be BeAl₂O₄.

Final Answer: The formula of the compound is BeAl₂O₄.

Given:
Mass of Carbon (C) = 2.4 g
Number of hydrogen atoms = atoms
Moles of Oxygen (O) = 0.2 moles

Step 1: Calculate the number of moles of each element.

Carbon (C):
Molar mass of carbon = 12 g/mol.
Number of moles of carbon =
Hydrogen (H):
The number of hydrogen atoms is .
Molar mass of hydrogen = 1 g/mol.
Number of moles of hydrogen =
Oxygen (O):
Number of moles of oxygen = 0.2 moles.

Step 2: Find the ratio of moles of elements.
The ratio of moles of C : H : O = 1 : 1 : 1.

Step 3: The empirical formula is CHO.

The correct answer is (D) : CHO.