N Awasthi Stoichiometry Formula Sheet PDF for JEE (Free)

Here are our three free PDFs for JEE stoichiometry: a formula sheet for quick reference, a cheat sheet with key ideas and shortcuts, and a common mistakes sheet to save you marks. They’re designed to help you master chemical calculations without the usual coaching noise. All three PDFs are completely free for you to download.

What these PDFs are (and what they are not)

Let’s be clear. These aren’t scanned pages from some textbook. These are original, distilled resources created by us—the 1% Syndicate. We’ve been through the JEE grind and know what actually works. These sheets are your companion, not your textbook. They are designed to sit right next to your problem book, like N Awasthi (N Avasthi), and help you bridge the gap between theory and application.

Think of them as the notes a senior who aced this stuff would hand you. They are raw, to the point, and focused on the systems and patterns behind the problems, not just blind formula memorization. They are our own work, in our own words, and are free for your personal study. Don’t try to sell them.

Not affiliated with the author or publisher of any book mentioned.

Download the Stoichiometry PDFs

We’ve split the material into three distinct files, each with a specific job. Use them strategically.

Stoichiometry formula sheet (PDF)

This is your master list. It’s a no-frills, single-source document for every important formula in the chapter. Use it when you’re solving problems to quickly find the tool you need, or use it for rapid revision before a test. It’s about efficiency, not explanation.

Mole concept and molar masses
Empirical and molecular formula
Stoichiometric calculations
Limiting reagent and percentage yield
Concentration terms (Molarity, Molality, Normality, etc.)
Eudiometry (gas analysis)
Redox and the equivalent concept
Relationships between different concentration terms

Download the Stoichiometry formula sheet (PDF)

Stoichiometry cheat sheet (PDF)

This isn’t just formulas; it’s the thinking behind the formulas. We built this sheet to give you the key insights and shortcuts that save time in an exam. It’s for when you understand the basics but need to get faster and more accurate under pressure. This is the “how-to-think” guide.

Key ideas for stoichiometry
Exam shortcuts for calculations
A quick method for identifying limiting reagents
Handling gas reactions at constant T and P (Avogadro’s principle)
Converting between molarity and molality
Tips for balancing complex redox equations
Calculating percentage yield efficiently
A framework for combustion analysis of hydrocarbons

Download the Stoichiometry cheat sheet (PDF)

Stoichiometry mistakes and traps (PDF)

This is probably the most valuable PDF here. It lists the dumb mistakes and clever traps that cost students the most marks. We’ve seen every one of these errors trip up even smart students. Read this after you solve a set of problems to see which traps you fell for, and why.

Not balancing the chemical equation first
Ignoring the limiting reagent in reaction calculations
Calculating the wrong n-factor for normality
Incorrectly adding volumes of solutions
Simple unit mix-ups (mL vs L, g vs kg)
Confusing the definitions of molarity and molality
Common errors in eudiometry calculations
Using the molar volume of gas (22.4 L/mol) under non-STP conditions

Download the Stoichiometry mistakes and traps (PDF)

The chapter ideas in plain words

Forget the textbook jargon. Here’s what the core ideas of stoichiometry actually mean.

The Mole Concept

The mole is just a number, like a “dozen.” It’s Avogadro’s number (6.022 × 10^23) of things—atoms, molecules, ions, whatever. It’s the bridge that connects the microscopic world of atoms to the macroscopic world of grams that you can weigh in a lab.

Number of moles (n) = Given mass (g) / Molar mass (g/mol)

This holds for any pure substance. It’s the most fundamental formula in all of chemistry.

Empirical and Molecular Formula

The empirical formula is the simplest whole-number ratio of atoms in a compound (e.g., CH2O for glucose). The molecular formula is the actual number of atoms in a molecule (e.g., C6H12O6 for glucose). The molecular formula is always a whole-number multiple of the empirical formula.

Molecular Formula = (Empirical Formula) × n

Here, n = Molecular Mass / Empirical Formula Mass. This holds true for all molecular compounds.

The Limiting Reagent

In a chemical reaction, you almost never have the exact right amounts of reactants. One will run out first, stopping the reaction. That’s the limiting reagent. It dictates the maximum amount of product you can possibly make (the theoretical yield).

To find it: Calculate the moles of product formed from each reactant. The one that produces the least amount of product is the limiting reagent.

This concept applies to every reaction with more than one reactant.

Molarity (M)

This is the most common measure of concentration. It’s simply the number of moles of solute dissolved in one liter of the total solution. It’s temperature-dependent because the volume of the solution can change with temperature.

Molarity (M) = Moles of solute / Volume of solution (in Liters)

This is valid for any liquid solution.

Molality (m)

Molality is another concentration term, but it’s based on the mass of the solvent, not the volume of the solution. This makes it independent of temperature, which is why it’s used in topics like colligative properties.

Molality (m) = Moles of solute / Mass of solvent (in kg)

This is valid for any liquid solution. Don’t mix it up with molarity.

Normality (N) and n-factor

Normality is like molarity but for reactive species. It considers the number of “equivalents” per liter. The key is the n-factor, which is the number of H+ ions (for an acid), OH- ions (for a base), or electrons transferred (for a redox agent) per molecule.

Normality (N) = Molarity (M) × n-factor

This relation is a powerful shortcut, especially for titrations, but only if you calculate the n-factor correctly for the specific reaction.

Gas Stoichiometry (at constant T and P)

Avogadro’s Law gives us a huge shortcut for reactions involving gases. At the same temperature and pressure, equal volumes of all gases contain the same number of molecules (moles). This means the ratio of volumes is the same as the ratio of moles (and the stoichiometric coefficients).

V1 / n1 = V2 / n2

This holds only for gases and only when temperature and pressure are constant throughout the reaction.

Which formula, when

Here’s a quick-reference table to help you pick the right tool for the job.

Situation	Formula to use	Watch out for
Finding moles from a given mass	`n = mass / Molar mass`	Units. Mass must be in grams, molar mass in g/mol.
Finding moles from number of particles	`n = Number of particles / 6.022 × 10^23`	Using this for mass calculations without converting.
A reaction with known amounts of two reactants	Calculate moles of product from each; smaller one wins.	Identifying the limiting reagent. Don’t assume it’s the one with less mass.
Diluting a stock solution	`M1V1 = M2V2`	V1 and V2 must be in the same units (mL or L). Moles of solute are constant.
Mixing two non-reacting solutions of same solute	`M_final = (M1V1 + M2V2) / (V1 + V2)`	Assuming volumes are always additive (it’s an approximation, but safe for JEE).
A titration between an acid and a base	`N1V1 = N2V2` (for equivalents)	Calculating the n-factor correctly. H2SO4 has n=2, HCl has n=1.
A reaction involving only gases at constant T, P	`Ratio of volumes = Ratio of moles (coefficients)`	This shortcut does not apply to solids or liquids in the same reaction.
Calculating percentage yield	`% Yield = (Actual Yield / Theoretical Yield) × 100`	The “Theoretical Yield” must be calculated based on the limiting reagent.
Relating Kp and Kc for a gas-phase equilibrium	`Kp = Kc(RT)^Δng`	`Δng` is (moles of gaseous products) – (moles of gaseous reactants). R = 0.0821.

Mistakes that cost the most marks

We’ve seen these errors sink thousands of students. Burn them into your brain so you don’t repeat them.

Not balancing the equation.
This is the most fundamental error. If your equation isn’t balanced, your mole ratios are wrong, and every single calculation that follows will be wrong. Always, always balance the equation first.

Ignoring the limiting reagent.
When given amounts for more than one reactant, you must find the limiting reagent. Assuming the reactant with the smaller mass or volume is limiting is a common trap that leads to a completely incorrect answer for the amount of product formed.

Wrong n-factor for normality.
The n-factor for a compound isn’t fixed; it depends on the reaction. For KMnO4, the n-factor is 5 in acidic medium but 3 in neutral medium. Using the wrong one in N = M × n-factor or N1V1 = N2V2 is an easy way to lose marks.

Confusing molarity and molality.
Molarity uses the volume of the solution in Liters. Molality uses the mass of the solvent in kilograms. They are not the same. Check the question carefully to see which one is required, especially in problems connecting to colligative properties.

Using 22.4 L/mol for all conditions.
The molar volume of an ideal gas is 22.4 L/mol only at STP (Standard Temperature and Pressure: 0°C or 273 K and 1 atm). If the conditions are different (like at NTP or room temperature), you must use the ideal gas law PV = nRT to find the volume.

Adding volumes of reacting solutions.
When you mix 50 mL of an acid with 50 mL of a base, the final volume is not always 100 mL, and you can’t just add concentrations. You must calculate the moles of H+ and OH-, find out which is in excess after neutralization, and then calculate the final concentration based on the total volume.

Two worked examples (our own)

Let’s walk through two classic problem types. No tricks, just the system.

Example 1: Limiting Reagent Calculation

Problem: 12 g of magnesium (Mg) is burned in a vessel containing 8 g of oxygen (O2). What mass of magnesium oxide (MgO) is formed? (Atomic masses: Mg = 24 g/mol, O = 16 g/mol)

1. Given Data:

Mass of Mg = 12 g
Mass of O2 = 8 g
Molar mass of Mg = 24 g/mol
Molar mass of O2 = 2 × 16 = 32 g/mol
Molar mass of MgO = 24 + 16 = 40 g/mol

2. Balanced Equation:
First, balance the reaction.
2Mg + O2 → 2MgO

3. Find the Limiting Reagent:

Calculate moles of each reactant:
- Moles of Mg = 12 g / 24 g/mol = 0.5 mol
- Moles of O2 = 8 g / 32 g/mol = 0.25 mol
Use the mole ratio from the balanced equation to see which runs out first.
- From the equation, 2 moles of Mg react with 1 mole of O2.
- Therefore, 0.5 moles of Mg would need (1/2) × 0.5 = 0.25 moles of O2.
We have exactly 0.25 moles of O2. This is a perfect stoichiometric mixture, so neither is in excess. We can use either reactant to calculate the product.

4. Calculate Product Formed:

Using Mg: 2 moles of Mg produce 2 moles of MgO. So, 0.5 moles of Mg will produce 0.5 moles of MgO.
Mass of MgO = moles × Molar mass = 0.5 mol × 40 g/mol = 20 g.

Answer: 20 g of MgO is formed.

Example 2: Molarity of a Mixture

Problem: 400 mL of a 0.5 M HCl solution is mixed with 100 mL of a 2.0 M HCl solution. What is the molarity of the final solution?

1. Given Data:

Solution 1: V1 = 400 mL = 0.4 L, M1 = 0.5 M
Solution 2: V2 = 100 mL = 0.1 L, M2 = 2.0 M

2. Formula Choice:

HCl is being mixed with HCl. There is no chemical reaction.
We can find the total moles and divide by the total volume.
The formula is M_final = (M1V1 + M2V2) / (V1 + V2).

3. Step-by-step Calculation:

Calculate moles from Solution 1:
- Moles1 = M1 × V1 = 0.5 mol/L × 0.4 L = 0.20 mol
Calculate moles from Solution 2:
- Moles2 = M2 × V2 = 2.0 mol/L × 0.1 L = 0.20 mol
Calculate total moles:
- Total Moles = Moles1 + Moles2 = 0.20 + 0.20 = 0.40 mol
Calculate total volume:
- Total Volume = V1 + V2 = 400 mL + 100 mL = 500 mL = 0.5 L
Calculate final molarity:
- M_final = Total Moles / Total Volume = 0.40 mol / 0.5 L = 0.8 M

Answer: The molarity of the final solution is 0.8 M.

How to use these sheets with a problem book

Don’t just read the PDFs. Integrate them into your workflow. Here’s a simple system that works.

Prime Your Brain: Before you start a new set of problems, spend 5-10 minutes reading our Formula Sheet. This loads the necessary tools into your short-term memory so you’re not fumbling for formulas.
Attempt the Problems: Dive into your problem book (like N Awasthi). Struggle with the problems. Try to solve them from scratch without looking at the solutions. The struggle is where the learning happens.
Analyze Your Mistakes: After you’ve tried a block of questions, check your answers. For every single one you got wrong, open our Mistakes and Traps Sheet. See if your error is listed there. Understand the pattern of the mistake, not just the single instance.
Consolidate and Revise: Before your next study session or a chapter test, use the Cheat Sheet for a quick 15-minute review. It will reinforce the key ideas, shortcuts, and problem-solving frameworks.

A three-session plan for stoichiometry

This chapter is a foundation for all of physical chemistry. Give it the time it deserves. Break it down into three focused sessions.

Session 1: The Basics (90 minutes)
- Goal: Master the mole and its applications.
- Action: Read the theory of the mole concept, molar mass, and empirical/molecular formulas. Use our Formula Sheet as a guide. Solve 20-25 basic problems focused only on these concepts. Get comfortable converting between mass, moles, and number of particles.
Session 2: Reaction Stoichiometry (90 minutes)
- Goal: Master limiting reagent and percentage yield.
- Action: This is the core of the chapter. Focus entirely on problems involving chemical reactions. For every problem, write the balanced equation, identify the limiting reagent, and calculate the theoretical yield. Use our Mistakes Sheet to debug your errors.
Session 3: Solutions and Gases (90 minutes)
- Goal: Master concentration terms and gas analysis (eudiometry).
- Action: Practice problems on molarity, molality, and normality, including dilution and mixing. Then, tackle eudiometry problems, applying Avogadro’s law and principles of gas analysis. Use our Cheat Sheet to see the patterns in these problem types.

Quick self-test

Test yourself. No calculators for the first pass.

A compound has an empirical formula of CH2 and a molecular mass of 70 g/mol. What is its molecular formula? (C=12, H=1)
How many moles of oxygen atoms are present in 4.9 g of sulfuric acid (H2SO4)? (H=1, S=32, O=16)
10 L of nitrogen gas reacts with 30 L of hydrogen gas to form ammonia at constant temperature and pressure. What is the final volume of the gaseous mixture?
What is the molarity of a solution containing 10 g of NaOH dissolved in water to make a 250 mL solution? (Na=23, O=16, H=1)
If you use the molar volume of 22.4 L/mol to calculate the volume of 1 mole of gas at 27°C and 1 atm, will your answer be larger or smaller than the true volume?

Answers:

Empirical formula mass of CH2 = 12 + 2 = 14 g/mol. n = Molecular Mass / Empirical Mass = 70 / 14 = 5. Molecular Formula = (CH2) × 5 = C5H10.
Molar mass of H2SO4 = 2(1) + 32 + 4(16) = 98 g/mol. Moles of H2SO4 = 4.9 g / 98 g/mol = 0.05 mol. Each molecule has 4 oxygen atoms, so moles of O atoms = 0.05 mol × 4 = 0.2 mol.
N2 + 3H2 → 2NH3. The ratio is 1:3:2. 10 L of N2 needs 30 L of H2, which is exactly what we have. All reactants are consumed. 10 L of N2 produces 20 L of NH3. Final volume = 20 L.
Molar mass of NaOH = 40 g/mol. Moles of NaOH = 10 g / 40 g/mol = 0.25 mol. Volume = 250 mL = 0.25 L. Molarity = 0.25 mol / 0.25 L = 1.0 M.
Smaller. The ideal gas law is V = nRT/P. At STP (273 K), V = 22.4 L. At 27°C (300 K), the temperature is higher, so the true volume will be larger than 22.4 L. Using 22.4 L is an underestimate.

FAQ

How do I calculate moles from mass or number of particles?
It’s a two-way street. To get moles from mass, use moles = mass / molar mass. To get moles from the number of particles (atoms/molecules), use moles = number of particles / Avogadro's number (6.022 × 10^23). Master these two conversions.

What is the difference between empirical and molecular formula?
The empirical formula is the simplest ratio (e.g., CH). The molecular formula is the actual composition (e.g., C6H6). The molecular formula is always a whole number multiple of the empirical formula. You need the molecular mass to get from one to the other.

How do I perform stoichiometric calculations for gases?
If the reaction is at constant temperature and pressure, you can use Avogadro’s Law as a shortcut: the ratio of the volumes of reacting gases is equal to the ratio of their stoichiometric coefficients. If T and P are not constant, you must use the ideal gas law, PV = nRT, to relate volume to moles.

When should I use the concept of limiting reagent?
Always. As a rule, any time a problem gives you the starting amounts of more than one reactant, you must determine the limiting reagent. It’s the one that dictates how much product can be formed. Failing to do this is one of the most common mistakes.

What are the common concentration terms and their formulas?
The big three are Molarity (M), Molality (m), and Normality (N). Molarity = moles solute / L solution. Molality = moles solute / kg solvent. Normality = gram equivalents / L solution, or more simply, N = M × n-factor. Know the difference, especially between the denominators (solution vs. solvent).

Is it better to use Normality or Molarity for titrations?
For simple acid-base titrations, N1V1 = N2V2 is faster if you can spot the n-factors quickly. However, for complex redox reactions or if you’re ever unsure about the n-factor, it’s safer to use the molarity method: write the balanced equation and use mole ratios (M1V1/n1 = M2V2/n2). The molarity method is more fundamental and less prone to error.

How do I handle percentage purity problems?
Treat percentage purity as a filter. If you’re given 100 g of an 80% pure sample of reactant A, the first step is to calculate the actual mass of A that can react: 100 g × 0.80 = 80 g. Use this pure mass for all subsequent mole calculations.

Once you have a solid grip on chemical calculations, you need to apply them. The next logical step is understanding the states of matter where these reactions happen.

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