
The haemoglobin from red corpuscles of most mammals contains approximately 0.33% of iron by weight. The molecular weight of haemoglobin is 67200. The number of iron atoms in each molecule of haemoglobin is:
A. 1
B. 2
C. 3
D. 4
Answer
225k+ views
Hint: The question deals with the concentration of iron atoms in terms of its mass in haemoglobin. We have to find the mole of iron atoms in one mole of haemoglobin, and then the number of iron atoms in each molecule of haemoglobin will be obtained.
Complete step by step answer:
Let us first write the meaning of the term 0.33% of iron by weight; it means that ‘0.33 grams of iron is present in 100 grams of haemoglobin’. Find the mass of iron present in 1 gram of haemoglobin. It will be obtained when it will be divided by 100 as$\dfrac{0.33}{100}$. Now convert the grams term in moles to reach the answer. Mole is defined as the ratio of given weight to molecular weight of that compound. Mathematically it is represented as
$\text{moles = }\dfrac{\text{given mass}}{\text{molar mass}}$. Molar mass of haemoglobin is 67200 grams which is given. If we multiply 1 gram of haemoglobin with 67200; then 67200 will be equal to 1 mole of haemoglobin as$\left( \dfrac{0.33}{100}\times 67200 \right)$. So, 1 mole of haemoglobin has 221.8 grams of iron. The atomic weight of iron is 56 grams. So, the moles of iron $\dfrac{221.8}{56}$ equals to 3.96 which is approx. 4. So, one mole of haemoglobin has 4 moles of iron. Similarly, one molecule of haemoglobin has 4 molecules of iron present in it, which matches option ‘d’.
Note: The ratio of moles of any two compounds will be the same as the ratio of the number of molecules present in those compounds. This is because moles are converted to molecules by multiplying moles with Avogadro's number. So,$\dfrac{\text{mole}{{\text{s}}_{1}}}{\text{mole}{{\text{s}}_{2}}}=\dfrac{\text{moles}\times {{\text{N}}_{\text{A}}}}{\text{moles}\times {{\text{N}}_{\text{A}}}}=\dfrac{\text{molecule}{{\text{s}}_{1}}}{\text{molecule}{{\text{s}}_{2}}}$.
Complete step by step answer:
Let us first write the meaning of the term 0.33% of iron by weight; it means that ‘0.33 grams of iron is present in 100 grams of haemoglobin’. Find the mass of iron present in 1 gram of haemoglobin. It will be obtained when it will be divided by 100 as$\dfrac{0.33}{100}$. Now convert the grams term in moles to reach the answer. Mole is defined as the ratio of given weight to molecular weight of that compound. Mathematically it is represented as
$\text{moles = }\dfrac{\text{given mass}}{\text{molar mass}}$. Molar mass of haemoglobin is 67200 grams which is given. If we multiply 1 gram of haemoglobin with 67200; then 67200 will be equal to 1 mole of haemoglobin as$\left( \dfrac{0.33}{100}\times 67200 \right)$. So, 1 mole of haemoglobin has 221.8 grams of iron. The atomic weight of iron is 56 grams. So, the moles of iron $\dfrac{221.8}{56}$ equals to 3.96 which is approx. 4. So, one mole of haemoglobin has 4 moles of iron. Similarly, one molecule of haemoglobin has 4 molecules of iron present in it, which matches option ‘d’.
Note: The ratio of moles of any two compounds will be the same as the ratio of the number of molecules present in those compounds. This is because moles are converted to molecules by multiplying moles with Avogadro's number. So,$\dfrac{\text{mole}{{\text{s}}_{1}}}{\text{mole}{{\text{s}}_{2}}}=\dfrac{\text{moles}\times {{\text{N}}_{\text{A}}}}{\text{moles}\times {{\text{N}}_{\text{A}}}}=\dfrac{\text{molecule}{{\text{s}}_{1}}}{\text{molecule}{{\text{s}}_{2}}}$.
Recently Updated Pages
JEE Main 2026 Session 1 Correction Window Started: Check Dates, Edit Link & Fees

JEE Isolation, Preparation and Properties of Non-metals Important Concepts and Tips for Exam Preparation

Isoelectronic Definition in Chemistry: Meaning, Examples & Trends

Ionisation Energy and Ionisation Potential Explained

Iodoform Reactions - Important Concepts and Tips for JEE

Introduction to Dimensions: Understanding the Basics

Trending doubts
JEE Main 2026: City Intimation Slip and Exam Dates Released, Application Form Closed, Syllabus & Eligibility

JEE Main 2026 Application Login: Direct Link, Registration, Form Fill, and Steps

Understanding the Angle of Deviation in a Prism

Hybridisation in Chemistry – Concept, Types & Applications

How to Convert a Galvanometer into an Ammeter or Voltmeter

Ideal and Non-Ideal Solutions Explained for Class 12 Chemistry

Other Pages
JEE Advanced Marks vs Ranks 2025: Understanding Category-wise Qualifying Marks and Previous Year Cut-offs

Thermodynamics Class 11 Chemistry Chapter 5 CBSE Notes - 2025-26

Organic Chemistry Some Basic Principles And Techniques Class 11 Chemistry Chapter 8 CBSE Notes - 2025-26

Hydrocarbons Class 11 Chemistry Chapter 9 CBSE Notes - 2025-26

JEE Advanced 2026 - Exam Date (Released), Syllabus, Registration, Eligibility, Preparation, and More

JEE Advanced 2026 - Exam Date (Released), Syllabus, Registration, Eligibility, Preparation, and More

