Count conserved atoms, translate through moles, and make every approximation visible.
JEE Main · JEE Advanced · NEET UG38 min study guide9 indexed sections
BY THE END, YOU CAN
Use conservation laws without assuming mass ratios equal atom ratios.
Translate between particles, moles and mass with the correct entity.
Infer empirical and molecular formulae, including from combustion data.
Identify limiting reagents and calculate excess, purity and yield.
Report calculated quantities with defensible significant figures.
BEFORE YOU START
Ratios, proportionality and scientific notation.
Chemical symbols and simple formulae.
Original educational lesson. Work through the examples before revealing the practice answers. Check the current official syllabus separately; historical question frequency does not establish eligibility or predict future questions.
CONNECT THE CONCEPT TO THE EXAM
Selected exam questions & worked solutions
These selected examples are separate from the original practice below. Each identifies its exam, year and wording. They do not represent every paper or every selected year.
Available in this lesson: JEE Advanced 2019. This is a limited lesson sample, not a complete exam-question bank.
1 sourced sample question available across the listed exams and years.
Matched topics: Chemical Composition and Stoichiometry.
JEE Advanced · 2019 · Paper 2 · Chemistry · Q12
Paraphrased exam question
Independently derived and checked solution. Archived key finality is unverified; this is not presented as an official final answer.
An aqueous solution contains 900 g water, has urea mole fraction 0.05, and density 1.2 g/cm³. Using molar masses 60 g/mol for urea and 18 g/mol for water, calculate urea molarity.
Molarity = (50/19)/(16.75/19)=200/67 mol/L. Use total solution volume, not water volume.
Source: JEE (Advanced) 2019, Paper 2, Q12; independently derived; key finality unverified. Solution written for this lesson.
01 / UNDERSTAND & APPLY
1. Why chemical arithmetic works
A chemical equation is an inventory statement about atoms. In an ordinary chemical reaction, nuclei retain their identities while electrons and bonds are rearranged. If every atom is accounted for, the same total mass is present before and after reaction to normal laboratory accuracy. A flask can nevertheless appear to lose mass when a gaseous product escapes. The conservation statement applies to the closed system, including any gas, not just to whatever remains on a balance pan.
The law of definite proportions says a given pure stoichiometric compound has a fixed elemental mass ratio. Different compounds of the same elements need not have the same ratio. Carbon monoxide and carbon dioxide are different substances: for a fixed amount of carbon, the oxygen amounts stand in the simple ratio 1:2. This illustrates the law of multiple proportions. Mixtures have variable composition, so a sample described merely as an iron–sulfur mixture cannot be assigned one fixed formula.
Dalton’s atomic picture explained these regularities with discrete atoms combining in whole-number ratios. Its historical claims need qualifications: atoms contain subatomic particles, isotopes of an element have different masses, and nuclei can change in nuclear reactions. The useful chemical bookkeeping survives these corrections. Also, some solids have composition ranges caused by defects; the elementary law of definite proportions is most directly applied to well-defined stoichiometric substances.
Gay-Lussac’s combining-volume observation concerns reacting gases measured at the same temperature and pressure. Avogadro’s principle supplies the bridge: equal gas volumes under those conditions contain equal numbers of molecules in the ideal limit. Thus volume ratios can reflect balanced gaseous coefficients. They do not authorize using an equal-volume rule for liquids, solids or gases measured under different conditions.
ORIGINAL WORKED EXAMPLE
Distinguish two laws
Two pure oxides contain 14.0 g nitrogen combined respectively with 16.0 g and 32.0 g oxygen. What regularity is demonstrated, and what is not established?
Fix the nitrogen mass at 14.0 g in both samples.
Compare oxygen masses: 16.0:32.0 = 1:2, a small whole-number ratio.
This comparison concerns two compounds, so it demonstrates multiple proportions; testing repeated samples of one compound would concern definite proportions.
The data give empirical compositions NO and NO₂ using N = 14.0 and O = 16.0, but cannot distinguish NO₂ from a molecular species such as N₂O₄ without molar-mass information.
Answer Multiple proportions; molecular formulae are not uniquely determined.
02 / UNDERSTAND & APPLY
2. The mole is a count with a named entity
A mole contains exactly 6.02214076 × 10²³ specified entities. The word “specified” prevents many errors: one mole of O₂ molecules is two moles of oxygen atoms, whereas one mole of NaCl formula units represents one mole of sodium ions and one mole of chloride ions in a fully dissociated ideal accounting. Crystalline sodium chloride is not made of isolated NaCl molecules. Counting formula units expresses its composition without asserting molecular structure.
Molar mass M converts amount to mass: m = nM. If M is in g mol⁻¹, use mass in grams. Relative atomic mass is dimensionless; the mass of an individual atom may be expressed in u; molar mass carries mass-per-mole units. These quantities are numerically related at normal educational precision but they are not interchangeable physical quantities. Isotopic abundance makes the tabulated atomic mass an average, not the mass number of every atom.
Always calculate the molar mass of the complete formula before converting. Parentheses multiply all atoms inside them; water of crystallization is part of a hydrate’s formula mass. In Ca(NO₃)₂, there are one calcium, two nitrogen and six oxygen atoms. If a question asks for oxygen atoms, first count formula units, then multiply by six. A multiplier for atoms is not another factor of Avogadro’s constant.
n = m/M; N = nN_A
N counts the named particles; n is in mol.
n(element atoms) = subscript × n(compound)
Use the total subscript after expanding brackets.
ORIGINAL WORKED EXAMPLE
Count atoms in a salt
Using Ca = 40.0, N = 14.0 and O = 16.0 g mol⁻¹, find the amount of oxygen atoms in 8.20 g Ca(NO₃)₂.
M = 40.0 + 2(14.0 + 3 × 16.0) = 164.0 g mol⁻¹.
n(salt) = 8.20/164.0 = 0.0500 mol.
Each formula unit contains six oxygen atoms, so n(O atoms) = 0.300 mol.
Multiply by N_A only if an actual particle count is needed: approximately 1.81 × 10²³ oxygen atoms.
Answer 0.300 mol oxygen atoms, approximately 1.81 × 10²³ atoms.
03 / UNDERSTAND & APPLY
3. Recover composition from measured masses
An empirical formula is the simplest whole-number atom ratio. A molecular formula gives the atom counts in an actual molecule and is an integer multiple of its empirical formula. Start with masses, divide each by its atomic molar mass, and compare the resulting amounts. Comparing mass percentages directly as subscripts confuses grams with atoms. A convenient 100 g basis turns percentages into grams without claiming the experimental sample actually weighed 100 g.
Divide all elemental amounts by the smallest, then inspect the ratios. Values near 1, 1.5 and 2 should become 2, 3 and 4 by multiplying all ratios by two. Do not independently round 1.5 to 2, which changes composition. Ratios near thirds or quarters may require multiplication by three or four. Real measurements have uncertainty; a proposed formula should reproduce the measured composition within that uncertainty rather than force every long decimal into enormous subscripts.
For molecular compounds, calculate the empirical formula mass E, then use k = M/E with an independently supplied molecular molar mass. The factor k must be a plausible positive integer. If a measured molar mass yields 2.03 instead of 2, uncertainty may explain the difference. If it yields 2.50 with small uncertainty, recheck the data or chemical assumptions. An ionic network generally has a formula unit rather than a molecular formula, so avoid applying molecular language indiscriminately.
Molecular formula = (empirical formula)ₖ; k = M/E
Multiply every subscript by the same integer k.
ORIGINAL WORKED EXAMPLE
From percentages to molecule
A molecular compound is 40.0% C, 6.67% H and 53.33% O by mass. Its molar mass is 180 g mol⁻¹. Use C = 12.0, H = 1.00 and O = 16.0.
On a 100 g basis, amounts are 40.0/12.0 = 3.333 mol C; 6.67/1.00 = 6.67 mol H; 53.33/16.0 = 3.333 mol O.
Divide by 3.333 to obtain approximately 1:2:1. Empirical formula: CH₂O.
Empirical formula mass is 12 + 2 + 16 = 30 g mol⁻¹.
k = 180/30 = 6. Multiply all subscripts, including the implicit ones, by six.
Answer Empirical formula CH₂O; molecular formula C₆H₁₂O₆. This formula alone does not identify the compound’s structure.
04 / UNDERSTAND & APPLY
4. Indirect composition: combustion and hydrates
Complete combustion routes carbon atoms from the sample into CO₂ and hydrogen atoms into H₂O. Consequently n(C) = n(CO₂), but n(H) = 2n(H₂O). The oxygen atoms in combustion products do not all come from the sample: the oxygen supply also contributes. For a compound known to contain only C, H and O, determine carbon and hydrogen masses from products, then obtain oxygen mass by subtracting them from the original sample mass.
This subtraction is justified only by the stated elemental composition and quantitative product collection. If nitrogen, halogen, residual solvent or ash is present, assigning the entire remainder to oxygen is incorrect. Incomplete combustion may form carbon monoxide or soot, causing the assumed CO₂ accounting to fail. The purpose is to recognize which atoms a measurement genuinely tracks, not to memorize one universal subtraction recipe.
A hydrate calculation follows a similar inventory. If controlled heating removes only water and leaves a known anhydrous salt, the mass loss gives water and the residue gives salt. Their mole ratio gives the hydration number. Overheating can decompose the salt; insufficient heating can leave water behind. Reaching constant mass supports completion under the chosen conditions but does not, by itself, prove the residue’s identity.
ORIGINAL WORKED EXAMPLE
Combustion with oxygen by difference
A 0.900 g compound containing only C, H and O gives 1.320 g CO₂ and 0.540 g H₂O on complete combustion. Find its empirical formula using masses 12, 1 and 16.
n(C) = 1.320/44 = 0.0300 mol, so carbon mass = 0.360 g.
n(H) = 2 × 0.540/18 = 0.0600 mol, so hydrogen mass = 0.0600 g.
Oxygen in the sample has mass 0.900 − 0.360 − 0.0600 = 0.480 g; n(O) = 0.0300 mol.
The atomic ratio is 1:2:1. No independent molar mass is supplied.
Answer CH₂O is the empirical formula; the molecular formula remains undetermined.
05 / UNDERSTAND & APPLY
5. A balanced equation is a mole-ratio map
Coefficients in a balanced equation compare numbers of reacting entities and therefore mole amounts. They do not generally compare masses. For 2H₂ + O₂ → 2H₂O, the molecule ratio is 2:1:2, but the mass ratio using rounded atomic masses is 4:32:36. The same equation conserves both atoms and mass when each quantity is interpreted correctly.
Balance by changing coefficients, never by changing a substance’s subscripts. Replacing H₂O by H₂O₂ to balance oxygen changes the product. For ionic equations, check electric charge as well as each element. Physical state labels matter for subsequent volume or equilibrium reasoning even though they do not change the atom count. After balancing, draw a calculation path: given mass → given moles → required moles → requested units.
The stoichiometric ratio is conditional on the specified reaction actually occurring with that product distribution. A balanced equation does not prove a reaction is fast, complete or exclusive. If a question supplies conversion, selectivity, competing reactions or an equilibrium composition, incorporate that information rather than assume every mole follows one ideal path. In a multistep synthesis, write each step’s ratio; cancellation between intermediates can reveal the overall requirement.
n(B) = n(A) × ν_B/ν_A
For the specified reaction and extent, ν denotes positive stoichiometric coefficient magnitudes.
ORIGINAL WORKED EXAMPLE
Predict a gas mass
Pure CaCO₃ decomposes completely: CaCO₃ → CaO + CO₂. How much CO₂ forms from 12.5 g CaCO₃? Use molar masses 100 and 44.0 g mol⁻¹.
Convert reactant mass: n(CaCO₃) = 12.5/100 = 0.125 mol, treating the supplied rounded masses as the intended calculation constants.
The balanced equation gives one mole CO₂ per mole CaCO₃.
m(CO₂) = 0.125 × 44.0 = 5.50 g.
The remaining solid has mass 12.5 − 5.50 = 7.00 g; their sum checks conservation in a closed accounting.
Answer 5.50 g CO₂.
06 / UNDERSTAND & APPLY
6. Find what runs out before calculating products
A recipe may demand two units of one reactant for every three of another. The smaller starting mass or mole amount is therefore not automatically limiting. Divide each available reactant amount by its coefficient. The smallest quotient sets the maximum common reaction extent. The other reactants have enough capacity to continue, but cannot do so after the limiting partner is exhausted.
For aA + bB → cC, compare n₀(A)/a with n₀(B)/b. If the first is smaller, consume all A, consume b n₀(A)/a of B, and form c n₀(A)/a of C. The leftover amount must be nonnegative. If your calculation produces a negative leftover, you selected the wrong limiting reagent or used the wrong equation. A tie means a stoichiometric mixture under the assumed single complete reaction.
An inert impurity does not participate in this comparison. First extract the amount of chemically active material from the sample purity, then compare capacities. A dissolved reactant’s amount is concentration times solution volume, with compatible units. “Excess acid” is a qualitative statement that acid is not limiting; it does not supply a numerical leftover concentration without the initial acid amount and final solution volume.
ξ_max = min[n₀(A)/a, n₀(B)/b, …]
Applies to a single reaction proceeding completely as specified; n(C) formed = cξ_max.
ORIGINAL WORKED EXAMPLE
Limiting reagent with leftover
Mix 5.40 g Al and 16.0 g O₂ and form only Al₂O₃ completely. Use Al = 27.0 and O₂ = 32.0 g mol⁻¹.
Balance: 4Al + 3O₂ → 2Al₂O₃.
Initial amounts: Al = 0.200 mol, O₂ = 0.500 mol. Capacities are 0.200/4 = 0.0500 mol and 0.500/3 = 0.1667 mol.
Al limits. Product amount = 2 × 0.0500 = 0.100 mol, giving 10.2 g Al₂O₃ using M = 102 g mol⁻¹.
Mass check: 10.2 + 11.2 = 21.4 g, equal to the combined starting mass.
Answer 10.2 g Al₂O₃ and 11.2 g unreacted O₂.
07 / UNDERSTAND & APPLY
7. Separate purity, conversion and isolated yield
Purity describes the composition of the starting sample. Conversion describes the fraction of a reactant that reacts. Isolated percentage yield compares the amount of desired product actually recovered with its theoretical amount. These are different stages of the calculation. A sample may be pure but give a low isolated yield through incomplete conversion, competing pathways or product losses during separation.
For a simple stated model, active reactant mass equals sample mass times purity fraction. Calculate theoretical product from that active amount and the limiting reagent. Multiply by the stated yield fraction only at the stage to which it applies. Two successive isolated yields multiply if the recovered intermediate from the first stage is the input to the second. Adding 80% and 75%, or taking their average, does not give the overall yield.
Mixture problems often become simultaneous atom or mole balances. If two carbonates each release one CO₂ per formula unit, total CO₂ amount equals the sum of their amounts, while sample mass supplies a differently weighted sum. Two independent measurements can then determine two unknowns. A single total mass ordinarily cannot separate two unknown composition fractions. Check that inferred fractions lie between zero and one.
ORIGINAL WORKED EXAMPLE
Purity followed by yield
A 20.0 g limestone sample is 80.0% CaCO₃; its other material is inert. Complete decomposition would give CO₂, but only 90.0% of the theoretical CO₂ is recovered. Use M(CaCO₃) = 100 and M(CO₂) = 44.0.
Active carbonate mass = 20.0 × 0.800 = 16.0 g.
Theoretical amount CO₂ = 16.0/100 = 0.160 mol; theoretical mass = 7.04 g.
Recovered CO₂ mass = 7.04 × 0.900 = 6.336 g.
Report to three significant figures, matching the supplied percentages and sample mass.
Answer 6.34 g CO₂ recovered. The 90.0% collection yield does not change the limestone’s 80.0% purity.
08 / UNDERSTAND & APPLY
8. Numerical precision is part of the answer
A measured value carries a resolution or uncertainty. Significant-figure rules are a compact classroom way of respecting that information, not a replacement for formal uncertainty propagation. Leading zeros locate a decimal point and are not significant; zeros between nonzero digits are significant. Trailing zeros after a decimal point can communicate precision. Scientific notation removes ambiguity: 1.20 × 10³ has three significant figures, whereas the bare integer 1200 may be unclear.
For multiplication or division, the result is normally reported with the number of significant figures in the least precise measured factor. For addition or subtraction, compare decimal places or absolute precision instead. Exact counts, balanced-equation coefficients and defined conversion factors do not limit precision. Keep guard digits through intermediate steps and round once at the end; repeated rounding can create a visible error in a final subtraction or ratio.
Do not confuse a precise result with an accurate one. Repeated measurements can cluster tightly while a calibration error shifts all of them from the true value. A calculation cannot repair a biased balance. Conversely, a formula-derived coefficient such as the two hydrogens per H₂O molecule is exact, so it should not force a combustion result to one significant figure.
ORIGINAL WORKED EXAMPLE
Apply the right rounding rule
Evaluate (12.46 g − 2.1 g)/(2.00 mL) with appropriate classroom precision.
The unrounded difference is 10.36 g, but subtraction leaves precision at the tenths place because 2.1 g is the limiting measurement.
Keep 10.36 internally rather than prematurely replacing it in subsequent arithmetic.
The difference is effectively 10.4 g, with three significant figures. Divide the guard-digit value by 2.00 mL to get 5.18 g mL⁻¹.
Both the numerator’s effective precision and the denominator support three significant figures.
Answer 5.18 g mL⁻¹.
09 / UNDERSTAND & APPLY
9. Original practice with complete solutions
These exercises are original learning problems, not claims about questions appearing in an examination. Unless supplied otherwise, use C = 12, H = 1, O = 16, Mg = 24, S = 32, N = 14 and Ca = 40 g mol⁻¹ as the intended rounded calculation values. Show the entity attached to every mole amount. A numerical answer without that label can conceal an atom-versus-molecule mistake.
Use three checks after each problem: the balanced equation must conserve every element; the answer must have the requested units; any leftover, purity or yield must be physically plausible. For inferred formulae, calculate composition back from the proposed formula. A backwards check often catches a lost factor of two faster than repeating the original arithmetic.
ORIGINAL PRACTICE · 1
How many moles of oxygen atoms are present in 0.250 mol H₂SO₄?
Reveal answer & reasoning
1.00 mol O atoms.
Each formula unit contains four oxygen atoms. Thus 4 × 0.250 = 1.00 mol; the coefficient four is an exact count.
ORIGINAL PRACTICE · 2
A compound contains 2.4 g Mg and 1.6 g O. Find its empirical formula.
Reveal answer & reasoning
MgO.
Amounts are 2.4/24 = 0.10 mol Mg and 1.6/16 = 0.10 mol O. Their ratio is 1:1, even though the mass ratio is 3:2.
ORIGINAL PRACTICE · 3
An empirical formula is CH and the molecular molar mass is 78 g mol⁻¹. Find the molecular formula.
Reveal answer & reasoning
C₆H₆.
The empirical mass is 13 g mol⁻¹. The multiplier is 78/13 = 6, applied to both elements. The formula does not by itself establish connectivity.
ORIGINAL PRACTICE · 4
For N₂ + 3H₂ → 2NH₃, start with 0.40 mol N₂ and 0.90 mol H₂. Find ammonia formed and nitrogen left.
Reveal answer & reasoning
0.60 mol NH₃ formed; 0.10 mol N₂ left.
Capacities are 0.40/1 = 0.40 and 0.90/3 = 0.30 mol, so H₂ limits. Consume 0.30 mol N₂ and form 2 × 0.30 = 0.60 mol NH₃.
ORIGINAL PRACTICE · 5
A 10.0 g sample containing 60.0% CaCO₃ and 40.0% inert, non-gas-producing material reacts completely with excess acid. How much CO₂ forms?
Reveal answer & reasoning
2.64 g CO₂.
Active CaCO₃ is 6.00 g, or 0.0600 mol. CaCO₃ + 2H⁺ → Ca²⁺ + CO₂ + H₂O gives CO₂ in a 1:1 ratio. Multiply 0.0600 by 44.0.
ORIGINAL PRACTICE · 6
A synthesis has two consecutive isolated yields of 80.0% and 75.0%. What fraction of the overall theoretical product is isolated?
Reveal answer & reasoning
60.0%.
The second stage receives only 0.800 of the theoretical intermediate. Recovering 0.750 of that amount gives 0.800 × 0.750 = 0.600. This assumes the stated yields refer to the sequential recovered material.
ORIGINAL PRACTICE · 7
Heating 2.46 g MgSO₄·xH₂O leaves 1.20 g MgSO₄, with only water lost. Find x.
Reveal answer & reasoning
x = 7.
Water mass is 1.26 g, so its amount is 1.26/18 = 0.0700 mol. Anhydrous salt amount is 1.20/120 = 0.0100 mol. Water:salt = 7:1. The conclusion assumes the residue is undecomposed MgSO₄.
ORIGINAL PRACTICE · 8
How should 0.004560 be described, and how should 6.20 × 1.4 be reported?
Reveal answer & reasoning
0.004560 has four significant figures; the product is 8.7.
Leading zeros are placeholders, while the final decimal zero communicates precision. The unrounded product is 8.68; multiplication is limited by the two significant figures of 1.4.
KEEP THE BIG IDEAS
Quick revision
Balance atoms and charge before using ratios; coefficients compare moles, not masses.
Name the entity being counted, especially atoms versus molecules or formula units.
Empirical formulae use mole ratios; molecular formulae additionally require a molar mass.
Compare available moles divided by stoichiometric coefficients to identify the limiting reagent.
Apply purity before reaction arithmetic and yield to the appropriate product stage.
Keep guard digits and round the final answer according to the relevant operation.
The teaching notes, worked explanations and original practice are written for this lesson. Separately labeled selected exam questions identify their exam, year, source and whether wording is paraphrased or an exact excerpt. Educational references do not certify current exam coverage.