Question

How many atoms of Oxygen (O) are in 2Ge
3
​
(PO
4
​
)
4
​
?

Answer 1

To determine how many atoms of oxygen (O) are in \(2 \text{Ge}_3(\text{PO}_4)_4\), we first need to break down the components of the formula.

1. **Molecular Structure**: The formula \( \text{Ge}_3(\text{PO}_4)_4 \) indicates that there are 3 germanium (Ge) atoms, and 4 phosphate (PO\(_4\)) groups.

2. **Phosphate Group**: Each phosphate group \( \text{PO}_4 \) contains 4 oxygen atoms.

3. **Total Oxygen Atoms in One Unit of \( \text{Ge}_3(\text{PO}_4)_4 \)**:
\[
\text{Total O in one } \text{Ge}_3(\text{PO}_4)_4 = 4 \text{ O in one } \text{PO}_4 \times 4 \text{ PO}_4 = 16 \text{ O}
\]

4. **Total Oxygen Atoms in 2 Units**:
\[
\text{Total O in } 2 \text{Ge}_3(\text{PO}_4)_4 = 2 \times 16 = 32 \text{ O}
\]

Therefore, there are **32 atoms of oxygen** in \(2 \text{Ge}_3(\text{PO}_4)_4\).