A List of Percentages Is Enough to Rebuild a Compound's Formula
Turn percentage composition or combining masses into an empirical formula, use vapour density or molecular mass to reach the molecular formula, and solve mass-to-mass, mass-to-volume and limiting-reactant problems from balanced equations with a layout that checks itself.
How can a list of percentages reveal a compound's formula?
A laboratory analyses a white powder and reports only this: ** sodium, sulphur and oxygen. No formula, no name. From those three numbers alone, you can work out that the powder is sodium sulphate, — and this part shows how.
The trick is the mole. A percentage is a mass, and dividing a mass by an atomic mass gives a number of moles of atoms. Moles of atoms are proportional to numbers of atoms, so once each percentage becomes a number of moles, the ratio of those numbers is the ratio of atoms in the formula.
That gives the empirical formula — the simplest whole-number ratio of atoms. A second piece of information, the molecular mass, turns it into the molecular formula, the actual number of each atom in one molecule.
The rest of this part runs the reasoning in the opposite direction. Given a balanced equation, the same mole ratios tell you:
- How much product a given mass of reactant will give
- How much reactant is needed for a required product
- What volume of gas is released at STP
- Which reactant runs out first when both amounts are given
This is stoichiometry — measuring out chemistry by numbers — and it is used every time anything is manufactured.
- A fertiliser factory calculates how much ammonia a given mass of nitrogen will produce
- A cement works calculates how much lime a load of limestone will give when heated
- A pharmacist checks the formula of a compound from its analysis before accepting a batch
Every calculation on this page uses the three relations from Part 2 — mass to moles, moles to particles, moles to gas volume — plus one new step: the mole ratio from a formula or an equation.
Atomic masses used throughout:** H , C , N , O , Na , Mg , S , Cl , K , Ca , Fe , Cu , Zn , Pb .
This page covers the third part of the ICSE Class 10 Chemistry chapter on mole concept and stoichiometry: empirical formula from percentages and from masses, molecular formula, and calculations based on balanced equations.
The trick is the mole. A percentage is a mass, and dividing a mass by an atomic mass gives a number of moles of atoms. Moles of atoms are proportional to numbers of atoms, so once each percentage becomes a number of moles, the ratio of those numbers is the ratio of atoms in the formula.
That gives the empirical formula — the simplest whole-number ratio of atoms. A second piece of information, the molecular mass, turns it into the molecular formula, the actual number of each atom in one molecule.
The rest of this part runs the reasoning in the opposite direction. Given a balanced equation, the same mole ratios tell you:
- How much product a given mass of reactant will give
- How much reactant is needed for a required product
- What volume of gas is released at STP
- Which reactant runs out first when both amounts are given
This is stoichiometry — measuring out chemistry by numbers — and it is used every time anything is manufactured.
- A fertiliser factory calculates how much ammonia a given mass of nitrogen will produce
- A cement works calculates how much lime a load of limestone will give when heated
- A pharmacist checks the formula of a compound from its analysis before accepting a batch
Every calculation on this page uses the three relations from Part 2 — mass to moles, moles to particles, moles to gas volume — plus one new step: the mole ratio from a formula or an equation.
Atomic masses used throughout:** H , C , N , O , Na , Mg , S , Cl , K , Ca , Fe , Cu , Zn , Pb .
This page covers the third part of the ICSE Class 10 Chemistry chapter on mole concept and stoichiometry: empirical formula from percentages and from masses, molecular formula, and calculations based on balanced equations.
How do you find an empirical formula from percentage composition?
Divide each element's percentage by its atomic mass, divide every result by the smallest, and convert the ratio to whole numbers — multiplying rather than rounding when a result ends near a half or a third.
Definitions.
- Empirical formula — the simplest whole-number ratio of atoms of each element in a compound
- Molecular formula — the actual number of atoms of each element in one molecule
- Molecular formula empirical formula, where is a whole number
The method, in four columns:
- Percentage of each element
- Divide by atomic mass — this gives relative moles of atoms
- Divide by the smallest of those values
- Convert to whole numbers
Worked example 1. A compound contains carbon, hydrogen and oxygen.
- Carbon:
- Hydrogen:
- Oxygen:
- **Divide by **:
**Empirical formula .
Worked example 2.** A compound contains sodium, sulphur and oxygen.
- Sodium:
- Sulphur:
- Oxygen:
- **Divide by **:
**Empirical formula .
Worked check — does sodium sulphate really have those percentages?** Formula mass .
All three match, and they add up to .
Worked example 3 — a ratio that must be multiplied. An oxide of iron contains iron and oxygen.
- Iron:
- Oxygen:
- **Divide by **:
- **Multiply by **:
**Empirical formula .
Worked example 4 — a percentage found by difference.** A hydrocarbon contains carbon. Find its empirical formula.
- Hydrogen
- Carbon: ; hydrogen:
- **Divide by **:
**Empirical formula .
The rule for converting to whole numbers — the boundary case of this whole topic.
- A value within a few hundredths of a whole number**, such as or , may be rounded — the difference is experimental error
- **A value near must be multiplied by
- A value near or must be multiplied by
- A value near must be multiplied by
Rounding to in worked example 3 would give — a formula for a compound that is not iron(III) oxide at all.
An everyday example.** The acid that gives vinegar its sour taste, acetic acid, analyses as carbon, hydrogen and oxygen — exactly the figures in worked example 1, so its empirical formula is also . The next section shows how its molecular mass tells it apart from glucose, which has the same percentages.
Definitions.
- Empirical formula — the simplest whole-number ratio of atoms of each element in a compound
- Molecular formula — the actual number of atoms of each element in one molecule
- Molecular formula empirical formula, where is a whole number
The method, in four columns:
- Percentage of each element
- Divide by atomic mass — this gives relative moles of atoms
- Divide by the smallest of those values
- Convert to whole numbers
Worked example 1. A compound contains carbon, hydrogen and oxygen.
- Carbon:
- Hydrogen:
- Oxygen:
- **Divide by **:
**Empirical formula .
Worked example 2.** A compound contains sodium, sulphur and oxygen.
- Sodium:
- Sulphur:
- Oxygen:
- **Divide by **:
**Empirical formula .
Worked check — does sodium sulphate really have those percentages?** Formula mass .
All three match, and they add up to .
Worked example 3 — a ratio that must be multiplied. An oxide of iron contains iron and oxygen.
- Iron:
- Oxygen:
- **Divide by **:
- **Multiply by **:
**Empirical formula .
Worked example 4 — a percentage found by difference.** A hydrocarbon contains carbon. Find its empirical formula.
- Hydrogen
- Carbon: ; hydrogen:
- **Divide by **:
**Empirical formula .
The rule for converting to whole numbers — the boundary case of this whole topic.
- A value within a few hundredths of a whole number**, such as or , may be rounded — the difference is experimental error
- **A value near must be multiplied by
- A value near or must be multiplied by
- A value near must be multiplied by
Rounding to in worked example 3 would give — a formula for a compound that is not iron(III) oxide at all.
An everyday example.** The acid that gives vinegar its sour taste, acetic acid, analyses as carbon, hydrogen and oxygen — exactly the figures in worked example 1, so its empirical formula is also . The next section shows how its molecular mass tells it apart from glucose, which has the same percentages.
How do you get an empirical formula from combining masses, and then the molecular formula?
Treat the masses of combining elements exactly like percentages to find the empirical formula, then divide the molecular mass by the empirical formula mass to find how many empirical units make one molecule.
Combining masses work just like percentages, because the method only needs the ratio of masses.
Worked example 1. of magnesium combines with of oxygen.
**Empirical formula .
Worked example 2.** of iron combines with of oxygen.
**Empirical formula .
From empirical to molecular formula.**
If the vapour density is given instead, first use .
Worked example 3. A hydrocarbon contains of carbon and of hydrogen. Its vapour density is . Find its molecular formula.
- Carbon: ; hydrogen:
- Ratio: , **multiplied by **:
- Empirical formula , empirical formula mass
- Molecular mass
-
**Molecular formula — butane, the main fuel in cooking gas cylinders.
Worked example 4.** A compound has empirical formula and molecular mass .
Worked example 5. The empirical formula from the previous section belongs to a compound with vapour density .
**Molecular formula — glucose. With vapour density instead**, and the compound is , acetic acid. One empirical formula, two quite different substances — which is why the molecular mass is essential.
Worked example 6 — water of crystallisation. of hydrated copper sulphate, , leaves of anhydrous salt on heating. Find .
- Water lost:
- Formula mass of
- Moles of :
- Moles of water:
-
**The formula is — the blue crystals from the acids, bases and salts chapter.
The boundary case. For an ionic compound such as magnesium oxide, there are no molecules, so the empirical formula is the only formula — there is nothing further to find. Molecular formulae belong to covalent substances made of discrete molecules.**
Combining masses work just like percentages, because the method only needs the ratio of masses.
Worked example 1. of magnesium combines with of oxygen.
**Empirical formula .
Worked example 2.** of iron combines with of oxygen.
**Empirical formula .
From empirical to molecular formula.**
If the vapour density is given instead, first use .
Worked example 3. A hydrocarbon contains of carbon and of hydrogen. Its vapour density is . Find its molecular formula.
- Carbon: ; hydrogen:
- Ratio: , **multiplied by **:
- Empirical formula , empirical formula mass
- Molecular mass
-
**Molecular formula — butane, the main fuel in cooking gas cylinders.
Worked example 4.** A compound has empirical formula and molecular mass .
Worked example 5. The empirical formula from the previous section belongs to a compound with vapour density .
**Molecular formula — glucose. With vapour density instead**, and the compound is , acetic acid. One empirical formula, two quite different substances — which is why the molecular mass is essential.
Worked example 6 — water of crystallisation. of hydrated copper sulphate, , leaves of anhydrous salt on heating. Find .
- Water lost:
- Formula mass of
- Moles of :
- Moles of water:
-
**The formula is — the blue crystals from the acids, bases and salts chapter.
The boundary case. For an ionic compound such as magnesium oxide, there are no molecules, so the empirical formula is the only formula — there is nothing further to find. Molecular formulae belong to covalent substances made of discrete molecules.**
How do you do mass–mass and mass–volume calculations from a balanced equation?
Convert the known quantity to moles, use the coefficients of the balanced equation as a mole ratio, and convert the moles of the required substance into grams or into litres at STP.
The method:
- Write the balanced equation
- Convert the given mass or volume to moles
- Use the coefficients to find moles of the substance asked for
- Convert to mass with its molar mass, or to volume with per mole for a gas at STP
- Check by conservation of mass where you can
Worked example 1 — mass to mass and mass to volume. of calcium carbonate is heated strongly. Find the mass of quicklime formed and the volume of carbon dioxide released at STP.
- Molar masses: , ,
- Moles of calcium carbonate:
- **Mole ratio **, so mol of each product
- Quicklime:
- Carbon dioxide: , which is
- Check: — mass conserved
Worked example 2 — volume to mass. What mass of potassium chlorate must be heated to give of oxygen at STP?
- Molar mass of
- Moles of oxygen:
- Ratio , so moles of
- Mass:
- Check: potassium chloride formed , oxygen , and
Worked example 3 — one reactant, several answers. of zinc reacts completely with dilute hydrochloric acid.
- Moles of zinc:
- Hydrogen: mol at STP
- Hydrogen chloride used: mol
- Zinc chloride formed:
- Check: and
Worked example 4. What mass of water is formed when of hydrogen burns completely, and what mass of oxygen is used?
- Moles of hydrogen:
- Water: mol ; oxygen: mol
- Check:
The boundary case in worked example 4. Hydrogen's molar mass is , not , because it is . **Dividing by gives moles and doubles every answer — the diatomic trap from Part 2 reappearing inside a stoichiometry problem.
An everyday example. A lime kiln converts limestone into quicklime for whitewash and mortar. Worked example 1 says every kilograms of pure limestone gives kilograms of quicklime**, the other kilograms leaving the chimney as carbon dioxide — a calculation that matters to anyone buying limestone by weight.
The method:
- Write the balanced equation
- Convert the given mass or volume to moles
- Use the coefficients to find moles of the substance asked for
- Convert to mass with its molar mass, or to volume with per mole for a gas at STP
- Check by conservation of mass where you can
Worked example 1 — mass to mass and mass to volume. of calcium carbonate is heated strongly. Find the mass of quicklime formed and the volume of carbon dioxide released at STP.
- Molar masses: , ,
- Moles of calcium carbonate:
- **Mole ratio **, so mol of each product
- Quicklime:
- Carbon dioxide: , which is
- Check: — mass conserved
Worked example 2 — volume to mass. What mass of potassium chlorate must be heated to give of oxygen at STP?
- Molar mass of
- Moles of oxygen:
- Ratio , so moles of
- Mass:
- Check: potassium chloride formed , oxygen , and
Worked example 3 — one reactant, several answers. of zinc reacts completely with dilute hydrochloric acid.
- Moles of zinc:
- Hydrogen: mol at STP
- Hydrogen chloride used: mol
- Zinc chloride formed:
- Check: and
Worked example 4. What mass of water is formed when of hydrogen burns completely, and what mass of oxygen is used?
- Moles of hydrogen:
- Water: mol ; oxygen: mol
- Check:
The boundary case in worked example 4. Hydrogen's molar mass is , not , because it is . **Dividing by gives moles and doubles every answer — the diatomic trap from Part 2 reappearing inside a stoichiometry problem.
An everyday example. A lime kiln converts limestone into quicklime for whitewash and mortar. Worked example 1 says every kilograms of pure limestone gives kilograms of quicklime**, the other kilograms leaving the chimney as carbon dioxide — a calculation that matters to anyone buying limestone by weight.
How do you solve problems with masses and gas volumes of both reactants and products?
Find the moles of every reactant given, identify the limiting reactant from the equation's ratio, base every product on it, and report what is left of the other reactant — giving each answer in the units asked.
Worked example 1 — all quantities from one mass. of nitrogen reacts with excess hydrogen to form ammonia. Find the mass and volume at STP of ammonia formed, and the mass and volume of hydrogen used.
- Moles of nitrogen:
- Ammonia: mol
- Hydrogen: mol
- Check:
Worked example 2 — a limiting reactant. of hydrogen and of oxygen are ignited. Find the mass of water formed and the mass of reactant left over.
- Moles: hydrogen ; oxygen
- ** mol of hydrogen would need mol of oxygen** — only mol is present, so oxygen is limiting
- Water: mol
- Hydrogen used: mol ; hydrogen left:
- Check: and
Worked example 3 — a reaction with four products and reactants. of calcium carbonate reacts completely with dilute hydrochloric acid.
- Moles of calcium carbonate:
- Hydrogen chloride needed:
- Calcium chloride:
- Water:
- Carbon dioxide: , or at STP
- Check: and
Worked example 4 — burning methane. of methane burns completely.
- Moles of methane:
- Oxygen used: mol
- Carbon dioxide: mol
- Water: mol
- Check: and
Worked example 5 — a decomposition giving two gases. of lead nitrate is heated until it decomposes completely.
- Molar mass of ; moles
- Lead oxide: mol
- Nitrogen dioxide: mol at STP
- Oxygen: mol at STP
- Total gas:
- Check:
The boundary case in worked example 2. The limiting reactant is not the one with the smaller mass — hydrogen's is far smaller than oxygen's , yet oxygen runs out. Nor is it simply the one with fewer moles. It is found only by comparing the moles present with the moles the equation requires.
An everyday example. A cook with cups of flour and eggs, using a recipe that needs cups of flour per egg, can make only enough batter for eggs — **the eggs are limiting, and cup of flour is left over.** Worked example 2 is the same reasoning with hydrogen and oxygen.
Worked example 1 — all quantities from one mass. of nitrogen reacts with excess hydrogen to form ammonia. Find the mass and volume at STP of ammonia formed, and the mass and volume of hydrogen used.
- Moles of nitrogen:
- Ammonia: mol
- Hydrogen: mol
- Check:
Worked example 2 — a limiting reactant. of hydrogen and of oxygen are ignited. Find the mass of water formed and the mass of reactant left over.
- Moles: hydrogen ; oxygen
- ** mol of hydrogen would need mol of oxygen** — only mol is present, so oxygen is limiting
- Water: mol
- Hydrogen used: mol ; hydrogen left:
- Check: and
Worked example 3 — a reaction with four products and reactants. of calcium carbonate reacts completely with dilute hydrochloric acid.
- Moles of calcium carbonate:
- Hydrogen chloride needed:
- Calcium chloride:
- Water:
- Carbon dioxide: , or at STP
- Check: and
Worked example 4 — burning methane. of methane burns completely.
- Moles of methane:
- Oxygen used: mol
- Carbon dioxide: mol
- Water: mol
- Check: and
Worked example 5 — a decomposition giving two gases. of lead nitrate is heated until it decomposes completely.
- Molar mass of ; moles
- Lead oxide: mol
- Nitrogen dioxide: mol at STP
- Oxygen: mol at STP
- Total gas:
- Check:
The boundary case in worked example 2. The limiting reactant is not the one with the smaller mass — hydrogen's is far smaller than oxygen's , yet oxygen runs out. Nor is it simply the one with fewer moles. It is found only by comparing the moles present with the moles the equation requires.
An everyday example. A cook with cups of flour and eggs, using a recipe that needs cups of flour per egg, can make only enough batter for eggs — **the eggs are limiting, and cup of flour is left over.** Worked example 2 is the same reasoning with hydrogen and oxygen.
Exam tip
What layout earns full marks in a stoichiometry calculation?
Write the balanced equation, list molar masses, convert everything to moles, show the mole ratio, and finish with a conservation-of-mass check.
- Balance the equation before any arithmetic — every later step depends on the coefficients
- Write molar masses under the formulae they belong to
- Convert to moles and label the values with mol
- Show the mole ratio explicitly, such as
- Identify the limiting reactant in words whenever two amounts are given
- **Use only for gases at STP
- For empirical formulae, set out four columns — percentage, divide by atomic mass, divide by smallest, whole numbers
- Multiply, never round**, a ratio near , or
- **State and write the molecular formula when a molecular mass or vapour density is given
- Check conservation of mass at the end — it catches most arithmetic errors
The misconception to name. The coefficients of an equation are ratios of moles, not of grams.** does not mean of hydrogen reacts with of oxygen — it means moles with mole, which is with . Reading coefficients as masses is the commonest single error in stoichiometry.
A second trap. Stopping at the empirical formula when a vapour density is given. The vapour density is there to be used, and a question that supplies it expects the molecular formula as the final answer.
- Balance the equation before any arithmetic — every later step depends on the coefficients
- Write molar masses under the formulae they belong to
- Convert to moles and label the values with mol
- Show the mole ratio explicitly, such as
- Identify the limiting reactant in words whenever two amounts are given
- **Use only for gases at STP
- For empirical formulae, set out four columns — percentage, divide by atomic mass, divide by smallest, whole numbers
- Multiply, never round**, a ratio near , or
- **State and write the molecular formula when a molecular mass or vapour density is given
- Check conservation of mass at the end — it catches most arithmetic errors
The misconception to name. The coefficients of an equation are ratios of moles, not of grams.** does not mean of hydrogen reacts with of oxygen — it means moles with mole, which is with . Reading coefficients as masses is the commonest single error in stoichiometry.
A second trap. Stopping at the empirical formula when a vapour density is given. The vapour density is there to be used, and a question that supplies it expects the molecular formula as the final answer.
Did you know
How can a few grams of solid fill an airbag in an instant?
An airbag has to go from folded fabric to a full cushion in far less than the blink of an eye. No pump could work that fast. Some airbag designs have instead relied on a small charge of a solid called sodium azide, which decomposes almost instantly when triggered, releasing a large volume of nitrogen gas.
Stoichiometry decides how much solid to pack. The molar mass of sodium azide is .
- ** of sodium azide** is mol
- **The equation gives mol of nitrogen** for every mol of azide
- At STP, that is litres of gas
So a solid no heavier than a small cup of rice releases enough gas at STP to fill more than thirty two-litre bottles. Half the mass, , gives half the volume, litres — exactly the proportional reasoning of this lesson.
Checking the balance: sodium ; nitrogen on the left and on the right. Balanced — and without that balance, the volume would be wrong by a large factor.
The equation also reveals a problem that designers had to solve. One of the products is sodium metal, which reacts violently with water, as the periodic table chapter showed. Other ingredients are therefore added to the charge to convert the sodium into harmless compounds — a second set of reactions, each sized by the same mole calculations.
Real conditions complicate the numbers a little. The gas is released hot rather than at , so it occupies more than the STP volume, and the bag is not at exactly one atmosphere. But the starting point for every such design is the calculation above — moles from mass, a mole ratio from the equation, and a volume from the moles.
Stoichiometry decides how much solid to pack. The molar mass of sodium azide is .
- ** of sodium azide** is mol
- **The equation gives mol of nitrogen** for every mol of azide
- At STP, that is litres of gas
So a solid no heavier than a small cup of rice releases enough gas at STP to fill more than thirty two-litre bottles. Half the mass, , gives half the volume, litres — exactly the proportional reasoning of this lesson.
Checking the balance: sodium ; nitrogen on the left and on the right. Balanced — and without that balance, the volume would be wrong by a large factor.
The equation also reveals a problem that designers had to solve. One of the products is sodium metal, which reacts violently with water, as the periodic table chapter showed. Other ingredients are therefore added to the charge to convert the sodium into harmless compounds — a second set of reactions, each sized by the same mole calculations.
Real conditions complicate the numbers a little. The gas is released hot rather than at , so it occupies more than the STP volume, and the bag is not at exactly one atmosphere. But the starting point for every such design is the calculation above — moles from mass, a mole ratio from the equation, and a volume from the moles.
Exam relevance
How do empirical formulae and stoichiometry appear in JEE and NEET?
This is foundation work for Class 11 Some Basic Concepts of Chemistry, examined in both JEE Main and NEET Chemistry, and for the quantitative analysis of organic compounds in JEE Main.
Where empirical and molecular formulae lead. Class 11 treats percentage composition, empirical formula and molecular formula in the same way as this page, and both exams set direct numericals on them. The multiply-don't-round rule is tested deliberately, with percentages chosen so that one element comes out near or .
Where combustion analysis leads. JEE Main includes quantitative analysis of organic compounds, in which the masses of carbon dioxide and water from burning a sample give the masses of carbon and hydrogen in it. From there the calculation is exactly worked example 3 of the combining-masses section — masses to moles to ratio to empirical formula, then molecular mass to molecular formula.
Where stoichiometry leads. Limiting reagent problems, percentage yield and reactions in solution using molarity are central to Class 11 and recur in both exams. Worked example 2 of the last section — finding the limiting reactant and the excess left over — is the template for all of them, with concentrations replacing masses in the solution versions.
Where the gas volumes lead. Stoichiometry questions in both exams frequently ask for the volume of a gas released, in electrolysis, in decomposition or in combustion. The mole-to-litres step is the same, though the conditions may be given as STP or as a specified temperature and pressure.
Where water of crystallisation leads. Formulae of hydrated salts, and the mass lost on heating, appear in objective questions on percentage composition and in practical chemistry on the preparation of crystalline salts.
Question types to expect. At this level: empirical and molecular formula calculations, and mass-mass, mass-volume and limiting-reactant problems. In competitive papers: formula from combustion data, limiting reagent with percentage yield, solution stoichiometry with molarity, and gas volumes under stated conditions — mostly as numerical-answer questions.
The single trap that costs marks. Reading equation coefficients as mass ratios. Every coefficient is a mole ratio, and a candidate who skips the conversion to moles gets a plausible but wrong number that often appears among the options.
A second trap. Choosing the limiting reactant by comparing masses or raw moles. It must be found by comparing the moles present with the ratio the equation demands, and both exams set data where the lighter or the less abundant reactant is in excess.
Board versus competitive emphasis. The ICSE paper marks the balanced equation, the laid-out steps and correct units; a competitive paper marks a final numerical value, often after a formula determination and a stoichiometric step combined. The transferable habit is moles first and a conservation-of-mass check last — the pair that turns long calculations into reliable ones.
Where empirical and molecular formulae lead. Class 11 treats percentage composition, empirical formula and molecular formula in the same way as this page, and both exams set direct numericals on them. The multiply-don't-round rule is tested deliberately, with percentages chosen so that one element comes out near or .
Where combustion analysis leads. JEE Main includes quantitative analysis of organic compounds, in which the masses of carbon dioxide and water from burning a sample give the masses of carbon and hydrogen in it. From there the calculation is exactly worked example 3 of the combining-masses section — masses to moles to ratio to empirical formula, then molecular mass to molecular formula.
Where stoichiometry leads. Limiting reagent problems, percentage yield and reactions in solution using molarity are central to Class 11 and recur in both exams. Worked example 2 of the last section — finding the limiting reactant and the excess left over — is the template for all of them, with concentrations replacing masses in the solution versions.
Where the gas volumes lead. Stoichiometry questions in both exams frequently ask for the volume of a gas released, in electrolysis, in decomposition or in combustion. The mole-to-litres step is the same, though the conditions may be given as STP or as a specified temperature and pressure.
Where water of crystallisation leads. Formulae of hydrated salts, and the mass lost on heating, appear in objective questions on percentage composition and in practical chemistry on the preparation of crystalline salts.
Question types to expect. At this level: empirical and molecular formula calculations, and mass-mass, mass-volume and limiting-reactant problems. In competitive papers: formula from combustion data, limiting reagent with percentage yield, solution stoichiometry with molarity, and gas volumes under stated conditions — mostly as numerical-answer questions.
The single trap that costs marks. Reading equation coefficients as mass ratios. Every coefficient is a mole ratio, and a candidate who skips the conversion to moles gets a plausible but wrong number that often appears among the options.
A second trap. Choosing the limiting reactant by comparing masses or raw moles. It must be found by comparing the moles present with the ratio the equation demands, and both exams set data where the lighter or the less abundant reactant is in excess.
Board versus competitive emphasis. The ICSE paper marks the balanced equation, the laid-out steps and correct units; a competitive paper marks a final numerical value, often after a formula determination and a stoichiometric step combined. The transferable habit is moles first and a conservation-of-mass check last — the pair that turns long calculations into reliable ones.
Key takeaways
What must you be able to do from this part?
Two kinds of formula, two starting points and one method for every equation.
- Empirical formula: simplest whole-number ratio of atoms. Molecular formula: actual atoms per molecule, times the empirical formula
- From percentages: divide by atomic mass, divide by the smallest, convert to whole numbers
- ** C, H, O** gives ; ** Na, S, O** gives
- ** Fe, O** gives , multiplied to — **multiply, never round, near , or
- Find a missing percentage by difference**: C leaves H, giving
- From combining masses: the same method — Mg with O gives
- ** molecular mass empirical formula mass**, with molecular mass VD
- ** with VD ** gives ; ** with VD ** gives ; ** with ** gives
- Water of crystallisation: and water give , so
- Ionic compounds have only an empirical formula
- Equation method: balance, convert to moles, use the coefficients as a mole ratio, convert to grams or litres, check mass
- ** ** gives CaO and
- ** of oxygen** needs of
- ** of zinc** gives of hydrogen and uses of HCl
- Limiting reactant: compare moles present with moles required — of and of give water with hydrogen left
- ** of lead nitrate** gives PbO, and
- Coefficients are mole ratios, not mass ratios
The sharpest self-test is a two-step problem you build yourself. Take a compound that is carbon and hydrogen with vapour density , find its molecular formula, write the balanced equation for burning of it, and calculate the volume of carbon dioxide at STP — then check the masses balance.
- Empirical formula: simplest whole-number ratio of atoms. Molecular formula: actual atoms per molecule, times the empirical formula
- From percentages: divide by atomic mass, divide by the smallest, convert to whole numbers
- ** C, H, O** gives ; ** Na, S, O** gives
- ** Fe, O** gives , multiplied to — **multiply, never round, near , or
- Find a missing percentage by difference**: C leaves H, giving
- From combining masses: the same method — Mg with O gives
- ** molecular mass empirical formula mass**, with molecular mass VD
- ** with VD ** gives ; ** with VD ** gives ; ** with ** gives
- Water of crystallisation: and water give , so
- Ionic compounds have only an empirical formula
- Equation method: balance, convert to moles, use the coefficients as a mole ratio, convert to grams or litres, check mass
- ** ** gives CaO and
- ** of oxygen** needs of
- ** of zinc** gives of hydrogen and uses of HCl
- Limiting reactant: compare moles present with moles required — of and of give water with hydrogen left
- ** of lead nitrate** gives PbO, and
- Coefficients are mole ratios, not mass ratios
The sharpest self-test is a two-step problem you build yourself. Take a compound that is carbon and hydrogen with vapour density , find its molecular formula, write the balanced equation for burning of it, and calculate the volume of carbon dioxide at STP — then check the masses balance.