Counting Atoms, Ions, Electrons, Protons and Neutrons
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How to count atoms, ions, electrons, protons and neutrons in any sample for JEE: the three-factor counting rule (moles, particles per formula unit, Avogadro constant), electrons in ions and the sign of the charge, neutrons from mass number minus atomic number, water versus heavy water, the stay-in-moles trick, and two solved problems on methane and hydroxide ions.
Chapters
- 0:00Introduction
- 0:45The counting rule: three factors
- 2:32Electrons and protons: add the atomic numbers
- 3:45Neutrons: mass number minus atomic number
- 4:54Topper's trick: stay in moles till the end
- 5:23Example 1: electrons in 1.6 g of methane
- 6:33Example 2: electrons in 3.4 g of OH- ions
- 7:58Quick revision: counting electrons, protons and neutrons
Transcript
0:00Introduction
Hello everyone! Okay, one small puzzle before we start. Take one point six grams of methane gas. That is just one tenth of a mole of methane molecules, okay? But this small sample holds one full mole of electrons: six point zero two two times ten to the power twenty three electrons! One tenth of a mole of gas, and one whole mole of electrons? Wait, how can that be? Today we will learn to count every tiny particle in a sample: atoms, ions, electrons, protons and neutrons. And trust me, JEE loves these counting questions, because one small slip changes the whole answer. So, let's begin.
0:45The counting rule: three factors
So, let's start with what you already know. If you know the moles of a substance, how do you get the number of molecules? Yes, exactly, multiply by N A, the Avogadro constant, six point zero two two times ten to the power twenty three. But JEE asks for atoms, ions, electrons and neutrons too. For those, we need one more factor. Here is the counting rule. The number of any particle equals n, the moles of the substance, times that particle per formula unit, times N A, okay? Three factors: moles, then how many in one unit, then Avogadro. Write this down. Clear? The middle factor is where the chemistry lives. One H two S O four has four oxygen atoms. So when the question asks for oxygen atoms, the middle factor is four. Now you tell me. One formula unit of calcium chloride, C a C l two, gives how many ions? Three, that's right! One calcium ion with charge plus two, and two chloride ions with charge minus one. One plus two, three ions. So for ions, the middle factor is three. Alright, now the exam question: find the oxygen atoms in some H two S O four. Many students find the moles, multiply by N A, and write the answer. Correct, right? No, no, careful! This is the first JEE trap. n times N A alone counts only the formula units, the molecules. For oxygen atoms, you must also multiply by four. So before you calculate anything, ask one question: which particle does the question want? Got it?
2:32Electrons and protons: add the atomic numbers
Now, electrons and protons. The atomic number of an element is the number of protons in one of its atoms, okay? And a neutral atom has exactly as many electrons as protons. So in a neutral molecule, the electrons are just the sum of the atomic numbers of all its atoms. Add them up, atom by atom. That's it! And the protons? The same sum. But what about an ion? An ion carries a charge. So think: what changes, the electrons or the protons? Yes, only the electrons! The charge changes the electrons, not the protons. So for an ion, the electrons equal the sum of atomic numbers, minus the charge. The protons stay exactly as before. Now careful. Suppose the charge is minus one. We subtract minus one. What does that give? Plus one! Absolutely right. So a charge of minus one means one extra electron, and a charge of plus two means two electrons fewer. Many students go the wrong way here, so remember: negative charge, more electrons. With me?
3:45Neutrons: mass number minus atomic number
Next, neutrons. The mass number counts protons plus neutrons, and the atomic number counts only the protons. So, neutrons per atom equal mass number minus atomic number. Simple subtraction. Take oxygen sixteen, with the little sixteen written on top. Mass number sixteen, atomic number eight. Sixteen minus eight: eight neutrons. Now ordinary hydrogen, hydrogen one. How many neutrons? One neutron, right? No, no, careful! This is the second JEE trap. Mass number one, atomic number one. One minus one is zero. Ordinary hydrogen has no neutron at all. But deuterium, written D, is an isotope of hydrogen with mass number two. So two minus one: one neutron. Now put them together. One ordinary water molecule has zero plus zero plus eight: eight neutrons. One heavy water molecule, D two O, has one plus one plus eight: ten neutrons. They look alike on paper, but the neutrons are different. Remember this pair, it comes again and again.
4:54Topper's trick: stay in moles till the end
Now let me share a trick that toppers use. Do not jump to big numbers too early. First find the moles of the particle you want: n, times that particle per formula unit. Often, this comes out as a neat number. Then multiply by N A only once, in the very last step. Why does this help? Yes, fewer big numbers means fewer silly mistakes with powers of ten. Nice and clean. You will see it working in both examples.
5:23Example 1: electrons in 1.6 g of methane
Now, remember our puzzle from the start? Let's solve it now. How many electrons are present in one point six grams of methane, C H four? Pause the video, and try it first. Done? Okay, let's check. Step one, moles. The molar mass of methane is sixteen grams per mole. So n equals one point six over sixteen: zero point one mole. Now the middle factor. How many electrons in one methane molecule? See, carbon has atomic number six, and each of the four hydrogens has atomic number one. So, six plus four: ten electrons in every methane molecule. It is neutral, so there is no charge to worry about. Now multiply, staying in moles. Zero point one times ten is exactly one mole of electrons. So the electrons are exactly N A, six point zero two two times ten to the power twenty three. And there you go, mystery solved! One tenth of a mole of molecules, but every molecule brings ten electrons. Together, one full mole of electrons.
6:33Example 2: electrons in 3.4 g of OH- ions
Now a slightly harder one, JEE level. How many electrons are there in three point four grams of O H minus ions? Take oxygen as sixteen, and hydrogen as one. Pause and try. Hint: look carefully at the charge. Done? Okay, let's check. Molar mass, capital M: sixteen plus one, seventeen. So n is three point four over seventeen: zero point two mole of ions. Now, electrons in one ion. Oxygen has atomic number eight, and hydrogen one. So eight plus one, nine electrons, right? No, no, careful! We forgot the charge. The charge is minus one, and we subtract the charge. Subtracting minus one adds one. So eight plus one plus one: ten electrons per ion. Interesting, isn't it? The same ten as in methane. Now the topper's trick: stay in moles. Zero point two times ten: two moles of electrons. Nice and neat. Only now, multiply by the Avogadro constant. Two times six point zero two two times ten to the power twenty three: one point two zero times ten to the power twenty four electrons. If you had taken nine, the whole answer would be wrong. So for ions, always check the charge first!
7:58Quick revision: counting electrons, protons and neutrons
So, quick revision. The number of particles equals moles, times that particle per formula unit, times the Avogadro constant. Electrons: add the atomic numbers, and for an ion, subtract the charge. Neutrons per atom: mass number minus atomic number. Now go and try the practice problems on this page. You will find them easy, I promise. All the best!
More videos in Unit 1 · Some Basic Concepts in Chemistry
- Matter and its Classification
- Physical Quantities, SI Units and Unit Conversion
- Scientific Notation and Significant Figures
- Laws of Chemical Combination
- Gay-Lussac's Law, Avogadro's Law and Dalton's Atomic Theory
- Atomic Mass and Isotopes
- Molecular Mass and Formula Mass
- The Mole and the Avogadro Constant
- Gas Volumes in the Mole Map
- Percentage Composition
- Minimum Molar Mass
- Empirical and Molecular Formulae
- Formulae from Combustion Data
- Hydrates, Metal Compounds and Non-stoichiometric Oxides
- Carbon and Hydrogen: Liebig's Method
- Nitrogen: Dumas' Method
- Nitrogen: Kjeldahl's Method
- Halogens: Carius' Method
- Sulphur and Phosphorus
- Oxygen by Difference and Complete Analysis
- Reading a Balanced Equation
- The Limiting Reagent
- Yield and Purity
- Sequential and Parallel Reactions
- The Principle of Atom Conservation and Mixtures
- Moles of a Gas from P, V and T
- Vapour Density and Average Molar Mass
- Volume Relations in Reactions
- Dalton's Law of Partial Pressures
- Eudiometry
- Mass Percent, Volume Percent and ppm
- Mole Fraction
- Molarity and Molality
- Converting Between Concentration Terms
- Dilution and Mixing
- Ions in Solution and Reactions in Solution
- Equivalent Mass and the n-Factor
- Normality and the Law of Equivalence
- Acid–Base Titrations and Back Titration
- Double Titration
- Volume Strength of Hydrogen Peroxide
- Hardness of Water
- Overview
- Mixed Problems: Set 1 (Mole Concept, Composition and Formulae)
- Mixed Problems: Set 2 (Stoichiometry and Gases)
- Mixed Problems: Set 3 (Concentration, Equivalents and Titrations)
- Overview
- Part A: JEE Main Level
- Part C: Concept Checks