In chemical scale, the relative mass of the isotopic mixture of oxygen atoms `(O^(16),O^(17),O^(18))` is assumed to be equal to

Options

A. (a)`16.002`
B. (b)`16.00`
C. (c )`17.00`
D. (d)`11.00`

Answer

To determine the relative mass of the isotopic mixture of oxygen atoms (O-16, O-17, O-18), we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Isotopes of Oxygen**:
- Oxygen has three stable isotopes:
- O-16 (with a mass of approximately 16 amu)
- O-17 (with a mass of approximately 17 amu)
- O-18 (with a mass of approximately 18 amu)

2. **Understand the Concept of Relative Atomic Mass**:
- The relative atomic mass of an element is the weighted average of the masses of its isotopes, taking into account their natural abundance.

3. **Natural Abundance of Oxygen Isotopes**:
- The natural abundance of these isotopes is approximately:
- O-16: 99.76%
- O-17: 0.04%
- O-18: 0.20%

4. **Calculate the Weighted Average**:
- To find the relative atomic mass of the isotopic mixture, we can use the formula:
\[ \text{Relative Atomic Mass} = \left(\frac{(mass \ of \ O-16 \times abundance \ of \ O-16) + (mass \ of \ O-17 \times abundance \ of \ O-17) + (mass \ of \ O-18 \times abundance \ of \ O-18)}{100}\right) \]
- Plugging in the values:
\[ \text{Relative Atomic Mass} = \left(\frac{(16 \times 99.76) + (17 \times 0.04) + (18 \times 0.20)}{100}\right) \]
- Calculating each term:
- O-16 contribution: \( 16 \times 99.76 = 1596.16 \)
- O-17 contribution: \( 17 \times 0.04 = 0.68 \)
- O-18 contribution: \( 18 \times 0.20 = 3.60 \)

5. **Sum the Contributions**:
- Total contribution = \( 1596.16 + 0.68 + 3.60 = 1600.44 \)

6. **Divide by 100**:
- Now, divide by 100 to get the average:
\[ \text{Relative Atomic Mass} = \frac{1600.44}{100} = 16.0044 \]

7. **Round to Appropriate Significant Figures**:
- The relative atomic mass of the isotopic mixture of oxygen is approximately 16.00 when rounded to two decimal places.

### Conclusion:
The relative mass of the isotopic mixture of oxygen atoms (O-16, O-17, O-18) is approximately **16.002**.