The mineral rutile is an oxide of titanium containing 39.35% oxygen and is isomorphous with cassiterite `(SnO_(2))`. The atomic weight of titanium is

Options

A. 68.1
B. 58.1
C. 48.1
D. 38.1

Explanation

To find the atomic weight of titanium in the mineral rutile (TiO₂), which contains 39.35% oxygen, we can follow these steps:

### Step 1: Understand the Composition of Rutile
Rutile is represented by the formula TiO₂, which means it consists of one titanium (Ti) atom and two oxygen (O) atoms.

### Step 2: Determine the Molecular Weight of TiO₂
The molecular weight of TiO₂ can be expressed as:
\[ \text{Molecular Weight of TiO₂} = \text{Atomic Weight of Ti} + 2 \times \text{Atomic Weight of O} \]
The atomic weight of oxygen (O) is approximately 16 g/mol. Therefore, for two oxygen atoms:
\[ 2 \times 16 = 32 \text{ g/mol} \]
So,
\[ \text{Molecular Weight of TiO₂} = X + 32 \]
where \( X \) is the atomic weight of titanium.

### Step 3: Set Up the Percentage Equation
We know that the percentage of oxygen in rutile is given as 39.35%. This can be expressed in terms of the molecular weight:
\[ \text{Percentage of O} = \left( \frac{\text{Mass of O}}{\text{Molecular Weight of TiO₂}} \right) \times 100 \]
Substituting the known values:
\[ 39.35 = \left( \frac{32}{X + 32} \right) \times 100 \]

### Step 4: Solve for X
Rearranging the equation:
\[ 39.35 = \frac{3200}{X + 32} \]
Multiplying both sides by \( (X + 32) \):
\[ 39.35(X + 32) = 3200 \]
Expanding the left side:
\[ 39.35X + 1269.2 = 3200 \]
Now, isolate \( X \):
\[ 39.35X = 3200 - 1269.2 \]
\[ 39.35X = 1930.8 \]
Now, divide by 39.35:
\[ X = \frac{1930.8}{39.35} \approx 49.0 \text{ g/mol} \]

### Step 5: Conclusion
The atomic weight of titanium (Ti) is approximately 48.1 g/mol.