Sample of haemoglobin is found to contain `0.333%` of iron. Assuming that each molecule of haemoglobin contains four iron atoms, which of these is the correct molecular mass of haemoglobin ?

Options

A. 67267g/mole
B. 76543g/mole
C. 11218g/mole
D. 30033g/mole

Explanation

To find the correct molecular mass of hemoglobin given that it contains 0.333% iron and each molecule contains four iron atoms, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Percentage of Iron**:
- The sample of hemoglobin contains 0.333% iron. This means that in 100 grams of hemoglobin, there are 0.333 grams of iron.

2. **Calculating the Mass of Iron in One Molecule of Hemoglobin**:
- Since each molecule of hemoglobin contains 4 iron atoms, we need to find the total mass of iron in one molecule.
- The atomic mass of iron (Fe) is approximately 56 g/mol.
- Therefore, the total mass of iron in one molecule of hemoglobin is:
\[ \text{Mass of Iron in 1 molecule} = 4 \times 56 \, \text{g/mol} = 224 \, \text{g/mol} \]

3. **Setting Up the Proportion**:
- We know that 0.333 grams of iron is present in 100 grams of hemoglobin.
- We can set up a proportion to find the molecular mass of hemoglobin (M).
- If 224 grams of iron is present in M grams of hemoglobin, we can write:
\[ \frac{0.333 \, \text{g}}{100 \, \text{g}} = \frac{224 \, \text{g}}{M} \]

4. **Cross-Multiplying to Solve for M**:
- Cross-multiplying gives us:
\[ 0.333 \times M = 224 \times 100 \]
- Simplifying this:
\[ 0.333M = 22400 \]

5. **Calculating M**:
- Now, divide both sides by 0.333 to find M:
\[ M = \frac{22400}{0.333} \approx 67267.26 \, \text{g/mol} \]

6. **Final Result**:
- The calculated molecular mass of hemoglobin is approximately 67267.26 g/mol.

### Conclusion:
The correct molecular mass of hemoglobin, based on the provided information, is approximately 67267.26 g/mol.