A mineral of iron contains an oxide containing 72.36% iron by mass and has a density of 5.2 g/cc. its unit cell is cubic with edge length of 839 pm. What is the total number of atoms (ions) present in each unit cell ? (Fe -56, O-16)

Explanation

To solve the problem, we need to determine the total number of atoms (ions) present in the unit cell of the iron oxide mineral. Here's a step-by-step breakdown of the solution:

### Step 1: Determine the Molar Mass of the Oxide
Given that the mineral contains 72.36% iron by mass, we can calculate the molar mass of the oxide.

Let the molar mass of the oxide be \( M \). The mass fraction of iron is given by:

\[ \frac{\text{mass of Fe}}{\text{mass of oxide}} = 0.7236 \]

Let \( x \) be the mass of iron in the oxide, then the mass of the oxide is \( M \). Thus:

\[ \frac{56}{M} = 0.7236 \]

Where 56 g/mol is the molar mass of iron (Fe). Rearranging gives:

\[ M = \frac{56}{0.7236} \approx 77.5 \text{ g/mol} \]

### Step 2: Calculate the Mass of the Unit Cell
Using the density of the oxide (5.2 g/cc), we can find the mass of the unit cell.

The volume of the cubic unit cell can be calculated using the edge length:

\[ \text{Volume} = a^3 = (839 \text{ pm})^3 = (839 \times 10^{-10} \text{ cm})^3 = 5.9 \times 10^{-23} \text{ cm}^3 \]

Now, using the density (\( \rho \)) to find the mass (\( m \)) of the unit cell:

\[ m = \rho \times \text{Volume} = 5.2 \text{ g/cm}^3 \times 5.9 \times 10^{-23} \text{ cm}^3 \approx 3.07 \times 10^{-22} \text{ g} \]

### Step 3: Calculate the Number of Moles in the Unit Cell
To find the number of moles of the oxide in the unit cell, we use the molar mass calculated earlier:

\[ \text{Number of moles} = \frac{\text{mass of unit cell}}{\text{molar mass}} = \frac{3.07 \times 10^{-22} \text{ g}}{77.5 \text{ g/mol}} \approx 3.96 \times 10^{-24} \text{ mol} \]

### Step 4: Calculate the Number of Formula Units in the Unit Cell
Using Avogadro's number (\( N_A = 6.022 \times 10^{23} \text{ mol}^{-1} \)), we can find the number of formula units in the unit cell:

\[ \text{Number of formula units} = \text{Number of moles} \times N_A \approx 3.96 \times 10^{-24} \text{ mol} \times 6.022 \times 10^{23} \text{ mol}^{-1} \approx 2.38 \text{ formula units} \]

### Step 5: Determine the Total Number of Atoms in the Unit Cell
Assuming the oxide is of the form \( \text{Fe}_x \text{O}_y \), we can deduce the ratio of iron to oxygen. Given the oxide contains iron and oxygen, let’s assume it is \( \text{Fe}_2\text{O}_3 \) (common iron oxide).

Each formula unit \( \text{Fe}_2\text{O}_3 \) contains 2 Fe atoms and 3 O atoms, giving a total of:

\[ \text{Total atoms per formula unit} = 2 + 3 = 5 \text{ atoms} \]

Thus, the total number of atoms in the unit cell is:

\[ \text{Total atoms in unit cell} = 2.38 \text{ formula units} \times 5 \text{ atoms/formula unit} \approx 11.9 \text{ atoms} \]

Since the number of atoms must be a whole number, we round it to 12 atoms.

### Final Answer
The total number of atoms (ions) present in each unit cell is approximately **12**.

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