What do the Grade 10 Chemistry cards cover?

36 cards across 6 chapters of the national textbook: Energy changes and electrochemistry, Sollutions, Important inorganic compounds, Metals and nonmetals, Hydrocarbons and their natural sources and 1 more. You can take any chapter one card at a time on the page itself.

Is there a national exam in Grade 10?

No. Ethiopia sets national exams in Grade 6, Grade 8 and Grade 12 only. These cards are for your school's own exams, and for the national exam that comes a few years later.

Where do these cards come from?

They are drafted from Grade 10 Chemistry, the Ministry of Education textbook for this grade. A second pass that cannot see the chapter then re-derives every formula, constant and table row, and anything it cannot confirm is held back instead of published.

Yes. Every card here is free to read and the printable sheet is free to download. Neither needs an account.

When was this last checked?

6 September 2026. Cards arrive chapter by chapter, and the line under each one says when that card was last read through.

Free short notes and revision cards

Grade 10 Chemistry

Grade 10 Chemistry on Temari has 36 revision cards, arranged by the chapters of the Ethiopian national curriculum. Every card says when the rule applies, what each symbol in it stands for, and the mistake students most often make with it. They are free to read and need no account.

36
Cards: 6
Chapters: 11
Formulas: 15
Reference tables: Grade 10 Chemistry
Textbook: Free
Price: 6 September 2026
Last checked

Everything36Reference table15Formula11Common mistake4Rule4Method2

Energy changes and electrochemistry

Four Basic Types of Chemical ReactionsReference tableCombination · Decomposition · Single displacement · +1

Reaction Type	General Form
Combination	$A + B \rightarrow AB$
Decomposition	$AB \rightarrow A + B$
Single displacement	$A + BC \rightarrow B + AC$
Double displacement	$AB + CD \rightarrow AD + CB$

When you use it

Use when classifying chemical reactions based on how reactants transform into products.

Watch out

In single displacement, a free element only displaces another element if it is more reactive.

Drafted from Grade 10 Chemistry, pages 2-50, then checked twice before it went up

Inverting Oxidizing and Reducing AgentsCommon mistake

When you use it

Use when identifying the substance oxidized, substance reduced, oxidizing agent, and reducing agent in a redox reaction.

Watch out

The substance that loses electrons undergoes oxidation and is the reducing agent. The substance that gains electrons undergoes reduction and is the oxidizing agent.

Drafted from Grade 10 Chemistry, pages 2-50, then checked twice before it went up

Molecular Formula DeterminationFormula

n = \frac{M_{\text{molecular}}}{M_{\text{empirical}}}

n = \frac{M_{\text{molecular}}}{M_{\text{empirical}}}

$n$: Whole number multiplier relating empirical formula to molecular formula $dimensionless$
$M_{\text{molecular}}$: Molar mass of the molecular formula $g\,mol^{-1}$
$M_{\text{empirical}}$: Molar mass of the empirical formula $g\,mol^{-1}$

When you use it

Use when converting an empirical formula to a true molecular formula given the actual molar mass.

Watch out

The multiplier n must be a whole number. Multiply every subscript in the empirical formula by n.

Drafted from Grade 10 Chemistry, pages 2-50, then checked twice before it went up

Percent Composition by MassFormula

\% \text{ composition} = \frac{n \times M_{\text{element}}}{M_{\text{compound}}} \times 100

\% \text{ composition} = \frac{n \times M_{\text{element}}}{M_{\text{compound}}} \times 100

$n$: Number of moles of the element in 1 mole of the compound $dimensionless$
$M_{\text{element}}$: Molar mass of the element $g\,mol^{-1}$
$M_{\text{compound}}$: Molar mass of the compound $g\,mol^{-1}$

When you use it

Use when finding the percentage of a specific element present in a chemical compound.

Watch out

Multiply the molar mass of the element by its subscript subscript n in the formula before dividing.

Drafted from Grade 10 Chemistry, pages 2-50, then checked twice before it went up

Rules for Assigning Oxidation NumbersReference tableUncombined element · Monatomic ion · Group IA, Group IIA · +4

Species or Element	Oxidation Number Rule	Exceptions
Uncombined element	$0$	None
Monatomic ion	Equal to ion charge	None
Group IA, Group IIA	$+1$ for Group IA, $+2$ for Group IIA	None
Oxygen in compounds	$-2$	Peroxides: $-1$ , Superoxides: $-1/2$ , $OF_2$ : $+2$
Hydrogen in compounds	$+1$	Metal hydrides: $-1$
Neutral compound sum	Sum equals $0$	None
Polyatomic ion sum	Sum equals ion charge	None

When you use it

Use when determining the oxidation state of any element in a neutral molecule or polyatomic ion.

Watch out

Do not forget exceptions for oxygen in peroxides like Na2O2 and hydrogen in metal hydrides like NaH.

Drafted from Grade 10 Chemistry, pages 2-50, then checked twice before it went up

Sollutions

Classification of ColloidsReference tableLiquid · Solid · Gas

Dispersed Phase	Dispersing Medium	Colloid Type	Example
Liquid	Gas	Aerosol	Fog, clouds, mist
Solid	Gas	Aerosol	Smoke, exhaust
Gas	Liquid	Foam	Shaving cream
Liquid	Liquid	Emulsion	Milk, face cream
Solid	Liquid	Sol	Milk of magnesia
Gas	Solid	Foam	Sponge, pumice
Liquid	Solid	Gel	Jelly, cheese, butter
Solid	Solid	Solid sol	Gemstone, milky glass

When you use it

Use this table to identify the colloid type based on the physical states of dispersed phase and dispersing medium.

Watch out

Gas dispersed in gas does not form a colloid, because all gases mix completely to form a true solution.

Drafted from Grade 10 Chemistry, pages 51-110, then checked twice before it went up

Enthalpy of Solution for Ionic SolidsFormula

\Delta H_{\text{soln}} = \Delta H_{\text{lat}} + \Delta H_{\text{hydr}}

\Delta H_{\text{soln}} = \Delta H_{\text{lat}} + \Delta H_{\text{hydr}}

$\Delta H_{\text{soln}}$: molar enthalpy of solution $\text{kJ}\,\text{mol}^{-1}$
$\Delta H_{\text{lat}}$: lattice energy $\text{kJ}\,\text{mol}^{-1}$
$\Delta H_{\text{hydr}}$: enthalpy of hydration $\text{kJ}\,\text{mol}^{-1}$

When you use it

Use this to determine overall heat change when an ionic solid dissolves into water from its lattice and hydration energies.

Watch out

Lattice energy overcomes crystal bonds and is positive (endothermic), while hydration energy is always negative (exothermic).

Drafted from Grade 10 Chemistry, pages 51-110, then checked twice before it went up

Heat of SolutionFormula

q = m C_g \Delta T

q = m C_g \Delta T

$q$: heat released or absorbed $\text{J}$
$m$: mass of the solution $\text{g}$
$C_g$: specific heat capacity of the solution $\text{J}\,\text{g}^{-1}\,{^{\circ}\text{C}}^{-1}$
$\Delta T$: temperature change $^{\circ}\text{C}$

When you use it

Use this formula to find the total heat energy absorbed or released during the dissolution process in a calorimeter.

Watch out

The mass m must be the total mass of the solution (solute mass plus solvent mass), not just the solvent mass.

Drafted from Grade 10 Chemistry, pages 51-110, then checked twice before it went up

Henry's LawFormula

C = k P

C = k P

$C$: concentration of dissolved gas $\text{mol}\,\text{L}^{-1}$
$k$: Henry's law constant $\text{mol}\,\text{L}^{-1}\,\text{atm}^{-1}$
$P$: partial pressure of the gas $\text{atm}$

When you use it

Use this formula to calculate the solubility or concentration of a gas in a liquid at a specific partial pressure.

Watch out

Gases that react chemically with water or form hydrogen bonds such as HCl and NH3 do not obey Henry's law.

Drafted from Grade 10 Chemistry, pages 51-110, then checked twice before it went up

Important inorganic compounds

Acid Base IndicatorsReference tableLitmus · Phenolphthalein · Methyl orange · +1

Indicator	Colour in Acid	Colour in Base
Litmus	Red	Blue
Phenolphthalein	Colourless	Pink
Methyl orange	Red	Yellow
Universal Indicator	Red	Blue

When you use it

Use this reference table to determine whether a solution is acidic or basic based on the observed colour of added indicators.

Watch out

Phenolphthalein remains completely colourless in acidic solutions, it only turns pink in basic solutions.

Drafted from Grade 10 Chemistry, pages 111-168, then checked twice before it went up

Classification of OxidesReference tableAcidic · Basic · Amphoteric · +2

Oxide Type	Chemical Behavior	Examples
Acidic	Forms acid with water, neutralizes bases	$\mathrm{CO_2}$ , $\mathrm{SO_2}$ , $\mathrm{N_2O_5}$
Basic	Forms base with water, neutralizes acids	$\mathrm{Na_2O}$ , $\mathrm{CaO}$ , $\mathrm{CuO}$
Amphoteric	Reacts with both strong acids and strong bases	$\mathrm{Al_2O_3}$ , $\mathrm{ZnO}$ , $\mathrm{PbO}$
Neutral	Reacts neither with acids nor with bases	$\mathrm{CO}$ , $\mathrm{NO}$ , $\mathrm{N_2O}$ , $\mathrm{H_2O}$
Peroxide	Contains $-O-O-$ link with oxygen in $-1$ state	$\mathrm{H_2O_2}$ , $\mathrm{Na_2O_2}$ , $\mathrm{CaO_2}$

When you use it

Use this table to classify binary oxygen compounds and predict their reaction products with water, acids, or bases.

Watch out

Neutral oxides like CO, NO, and N2O are non-metal oxides that do not form salts.

Drafted from Grade 10 Chemistry, pages 111-168, then checked twice before it went up

Diluting Concentrated AcidCommon mistake

When you use it

Apply when answering questions about laboratory safety rules and procedures for acid dilution.

Watch out

Never add water directly into concentrated acid. Always pour the concentrated acid slowly into water to prevent violent exothermic splattering.

Drafted from Grade 10 Chemistry, pages 111-168, then checked twice before it went up

pH and pOH RelationshipFormula

\mathrm{pH} + \mathrm{pOH} = 14

\mathrm{pH} + \mathrm{pOH} = 14

$\mathrm{pH}$: pH value $1$
$\mathrm{pOH}$: pOH value $1$

When you use it

Use this equation at 25 °C to convert between pH and pOH, especially when finding the pH of a basic solution from hydroxide ion concentration.

Watch out

This relationship is valid strictly at 25 °C where the ionic product of water equals 1.0 x 10^-14.

Drafted from Grade 10 Chemistry, pages 111-168, then checked twice before it went up

pH CalculationFormula

\mathrm{pH} = -\log[\mathrm{H}^+]

\mathrm{pH} = -\log[\mathrm{H}^+]

$\mathrm{pH}$: pH value $1$
$[\mathrm{H}^+]$: Hydrogen ion concentration $\mathrm{mol\,L^{-1}}$

When you use it

Use this formula to calculate the pH of an acidic solution from the molar concentration of hydrogen ions, or to find hydrogen ion concentration from a known pH.

Watch out

For diprotic strong acids like H2SO4, the hydrogen ion concentration is twice the molar concentration of the acid.

Drafted from Grade 10 Chemistry, pages 111-168, then checked twice before it went up

Acids, bases and salts: definitions and neutralizationRuleArrhenius acid · Arrhenius base · Bronsted-Lowry · +4

Term	What it means	Examples
Arrhenius acid	Releases hydrogen ions in water, carried in the solution as the hydronium ion $\mathrm{H_3O^+}$	$\mathrm{HCl}$ , $\mathrm{H_2SO_4}$ , $\mathrm{HNO_3}$ , $\mathrm{CH_3COOH}$
Arrhenius base	Releases hydroxide ions in water. A base that dissolves is called an alkali, and an insoluble base such as $\mathrm{CuO}$ moves no indicator until it reacts with an acid	$\mathrm{NaOH}$ , $\mathrm{KOH}$ , $\mathrm{Ca(OH)_2}$
Bronsted-Lowry	The wider definition: an acid donates a proton and a base accepts one. Ammonia belongs here, since it takes a proton from water and leaves hydroxide behind without carrying an $\mathrm{OH}$ group of its own	$\mathrm{NH_3}$
Salt	The hydrogen of an acid replaced by a metal ion or by the ammonium ion $\mathrm{NH_4^+}$	$\mathrm{NaCl}$ , $\mathrm{KNO_3}$ , $\mathrm{CuSO_4}$ , $\mathrm{NH_4Cl}$
Neutralization	Acid plus base gives salt and water. Between the ions it is $\mathrm{H^+ + OH^- \rightarrow H_2O}$ , and the metal ion with the acid's other ion is left as the salt	$\mathrm{HCl + NaOH \rightarrow NaCl + H_2O}$
Acid on a carbonate	A related but different reaction, giving salt, water and carbon dioxide	$\mathrm{2HCl + CaCO_3 \rightarrow CaCl_2 + H_2O + CO_2}$
The pH scale	A continuous scale, written 0 to 14 for school work at about 25 degrees Celsius. Below 7 is acidic, exactly 7 is neutral, above 7 is basic	pH 6.5 is acidic; pH 7.4 is basic

When you use it

Use when a question asks you to classify a substance as an acid, a base or a salt, or to complete a neutralization equation. Name the definition you are working from, because Arrhenius describes what a substance releases in water while Bronsted-Lowry describes a proton given and taken.

Watch out

Students copy the one to one pattern of HCl with NaOH onto an acid that has two hydrogens, and write H2SO4 plus NaOH giving NaSO4 and H2O. Both hydrogens have to be replaced, so the salt is Na2SO4, two sodium ions against one sulphate ion, and the equation needs two NaOH and two H2O. Build the salt from the charges first, then balance. The same slip produces CaCl instead of CaCl2 with calcium hydroxide.

Common names of chemical compoundsReference tableQuicklime · Slaked lime · Limestone, marble, chalk · +19

Common name	Chemical name	Formula
Quicklime	Calcium oxide	$\mathrm{CaO}$
Slaked lime	Calcium hydroxide	$\mathrm{Ca(OH)_2}$
Limestone, marble, chalk	Calcium carbonate	$\mathrm{CaCO_3}$
Baking soda	Sodium hydrogen carbonate	$\mathrm{NaHCO_3}$
Washing soda	Sodium carbonate decahydrate	$\mathrm{Na_2CO_3} \cdot 10\mathrm{H_2O}$
Soda ash	Sodium carbonate	$\mathrm{Na_2CO_3}$
Caustic soda	Sodium hydroxide	$\mathrm{NaOH}$
Blue vitriol	Copper(II) sulphate pentahydrate	$\mathrm{CuSO_4} \cdot 5\mathrm{H_2O}$
White vitriol	Zinc sulphate heptahydrate	$\mathrm{ZnSO_4} \cdot 7\mathrm{H_2O}$
Oil of vitriol	Sulphuric acid	$\mathrm{H_2SO_4}$
Muriatic acid	Hydrochloric acid	$\mathrm{HCl}$
Plaster of Paris	Calcium sulphate hemihydrate	$\mathrm{2CaSO_4} \cdot \mathrm{H_2O}$
Chloroform	Trichloromethane	$\mathrm{CHCl_3}$
Glacial acetic acid	Ethanoic acid	$\mathrm{CH_3COOH}$
Spirit	Ethanol	$\mathrm{C_2H_5OH}$
Marsh gas	Methane	$\mathrm{CH_4}$
Laughing gas	Dinitrogen monoxide	$\mathrm{N_2O}$
Dry ice	Solid carbon dioxide	$\mathrm{CO_2(s)}$
Potash, pearl ash	Potassium carbonate	$\mathrm{K_2CO_3}$
Table sugar	Sucrose	$\mathrm{C_{12}H_{22}O_{11}}$
Silica	Silicon dioxide	$\mathrm{SiO_2}$
Bleaching powder	Calcium chloride hypochlorite	$\mathrm{Ca(OCl)Cl}$

When you use it

Reach for this when a question, a label or a shop name uses the traditional name and the answer needs the chemical one, or the other way round. Copy the formula column carefully, because several of these names belong to hydrated crystals rather than to the dry salt.

Watch out

Students mix up the three limes, because all three names carry the word. Limestone is the rock you start with, CaCO3. Heat it and it loses carbon dioxide to give quicklime, CaO. Add water to that and it becomes slaked lime, Ca(OH)2, and slaked means water has been added, so it is the only one of the three with OH in it. The vitriols and washing soda hold a second trap: those names belong to the crystals, which carry water. Blue vitriol is copper sulphate pentahydrate and is blue because of that water, so drive the water off and you get white anhydrous CuSO4, which nobody calls blue vitriol.

The five reactions of acidsReference tableMetal hydroxide · Metal above hydrogen · Metal carbonate · +2

Acid reacts with	What you get	Example
Metal hydroxide	A salt and water. This is neutralization.	$\mathrm{HCl(aq) + NaOH(aq) \rightarrow NaCl(aq) + H_{2}O(l)}$
Metal above hydrogen	A salt and hydrogen gas, and only with a dilute non-oxidizing acid such as dilute HCl or dilute H2SO4. Copper, silver and gold sit below hydrogen in the reactivity series and give no reaction with dilute HCl.	$\mathrm{Zn(s) + 2HCl(aq) \rightarrow ZnCl_{2}(aq) + H_{2}(g)}$
Metal carbonate	A salt, water and carbon dioxide. A metal hydrogencarbonate gives the same three products.	$\mathrm{CaCO_{3}(s) + 2HCl(aq) \rightarrow CaCl_{2}(aq) + H_{2}O(l) + CO_{2}(g)}$
Metal oxide	A salt and water. A metal oxide is a basic oxide, so this is neutralization as well.	$\mathrm{CuO(s) + 2HCl(aq) \rightarrow CuCl_{2}(aq) + H_{2}O(l)}$
Ammonia	An ammonium salt and no water. Ammonia is the base here, but it carries no hydroxide and no oxide, so there is nothing to build water from. The NH4Cl forms as a white solid.	$\mathrm{NH_{3}(g) + HCl(g) \rightarrow NH_{4}Cl(s)}$

When you use it

Use when you have to predict the products of an acid reaction or write the balanced equation for one. A salt forms whenever the hydrogen of an acid is replaced by a metal or by the ammonium ion, and every one of these five rows produces one.

Watch out

Students write the salt before checking the metal's valency, so they get Zn + HCl giving ZnCl + H2 and never find the 2 that belongs in front of HCl. Build the salt from the ion charges first, since Zn2+ with Cl- gives ZnCl2, then balance the acid to match it. The same habit of copying a pattern without checking produces the other favourite mistake, Cu + 2HCl giving CuCl2 + H2. Copper sits below hydrogen in the reactivity series, so nothing happens, and that is exactly why copper reaches row four as CuO instead.

Unit 4

Electrode polarity confusionCommon mistake

When you use it

Use when identifying the signs and reactions of anode and cathode in electrochemical cells.

Watch out

Oxidation ALWAYS occurs at the anode and reduction at the cathode in both cells. The anode is negative in a Galvanic cell, but positive in an electrolytic cell.

Drafted from Grade 10 Chemistry, pages 169-200, then checked twice before it went up

Galvanic and electrolytic cellsReference tableEnergy conversion · Redox reaction · Anode sign and process · +1

Feature	Galvanic (Voltaic) Cell	Electrolytic Cell
Energy conversion	Chemical energy to electrical energy	Electrical energy to chemical energy
Redox reaction	Spontaneous	Nonspontaneous
Anode sign and process	Negative (-), Oxidation	Positive (+), Oxidation
Cathode sign and process	Positive (+), Reduction	Negative (-), Reduction

When you use it

Use when distinguishing between battery operation and electrolysis setups.

Drafted from Grade 10 Chemistry, pages 169-200, then checked twice before it went up

Metallic versus electrolytic conductionReference tableCharge carriers · Matter transfer · Chemical change · +1

Property	Metallic Conduction	Electrolytic Conduction
Charge carriers	Electrons (delocalized)	Ions (cations and anions)
Matter transfer	No transfer of matter	Involves transfer of ions
Chemical change	No chemical change	Chemical decomposition occurs
Effect of temperature	Conductivity decreases as temperature rises	Conductivity increases as temperature rises

When you use it

Use when comparing conduction in solid metals to molten salts or aqueous solutions.

Drafted from Grade 10 Chemistry, pages 169-200, then checked twice before it went up

Quantity of heatFormula

q = c \cdot m \cdot \Delta T

q = c \cdot m \cdot \Delta T

$q$: Heat absorbed or released $\text{J}$
$c$: Specific heat capacity $\text{J}\,\text{kg}^{-1}\,\text{K}^{-1}$
$m$: Mass of the substance $\text{kg}$
$\Delta T$: Temperature change $\text{K}$

When you use it

Use to calculate heat transferred during a temperature change.

Watch out

Temperature change can use Kelvin or degrees Celsius since the temperature difference is numerically identical.

Drafted from Grade 10 Chemistry, pages 169-200, then checked twice before it went up

Metals and nonmetals

Direct Water Dissolution of Sulfur TrioxideCommon mistake

When you use it

Use when outlining the industrial steps of the Contact Process for sulfuric acid manufacturing.

Watch out

Never write sulfur trioxide reacting directly with water in the Contact Process. It must first be absorbed into concentrated sulfuric acid to form oleum, which is then diluted.

Drafted from Grade 10 Chemistry, pages 201-236, then checked twice before it went up

Gold Purity PercentageFormula

\text{Mass \% of Gold} = \frac{\text{Carat}}{24} \times 100

\text{Mass \% of Gold} = \frac{\text{Carat}}{24} \times 100

$\text{Carat}$: Carat rating representing parts of pure gold by mass in 24 parts of alloy $\text{dimensionless}$

When you use it

Use this formula to calculate the mass percent of pure gold from a given carat value.

Watch out

Pure gold is 24 carats. Do not divide the carat value by 100.

Drafted from Grade 10 Chemistry, pages 201-236, then checked twice before it went up

Industrial Extraction of ElementsReference tableAluminum · Iron · Copper · +2

Element	Chief Ore or Source	Industrial Extraction Method
Aluminum	$\text{Al}_2\text{O}_3\cdot 2\text{H}_2\text{O}$ (Bauxite)	Hall-Héroult process (Electrolysis with $\text{Na}_3\text{AlF}_6$ )
Iron	$\text{Fe}_2\text{O}_3$ (Hematite)	Blast furnace reduction using $\text{CO}$
Copper	$\text{CuFeS}_2$ (Chalcopyrite)	Roasting, smelting, and electrolytic refining
Sulfur	Underground elemental deposits	Frasch process using superheated water
Chlorine	$\text{NaCl}(\text{aq})$ (Brine)	Electrolysis of concentrated aqueous $\text{NaCl}$

When you use it

Use this table to identify raw ores and corresponding extraction processes for major metals and nonmetals.

Watch out

Cryolite is not the aluminum ore. It is an additive that lowers the melting point of alumina from 2045 °C to 1000 °C.

Drafted from Grade 10 Chemistry, pages 201-236, then checked twice before it went up

Hydrocarbons and their natural sources

Alkane Structural Isomers FormulaFormula

N = 2^{n-4} + 1

N = 2^{n-4} + 1

$N$: number of structural isomers of the alkane
$n$: number of carbon atoms, valid for 4 to 7 carbons

When you use it

Use this shortcut to find the exact number of structural isomers for alkanes containing between 4 and 7 carbon atoms.

Watch out

This formula applies strictly for n between 4 and 7. It fails for methane, ethane, propane, and alkanes with 8 or more carbons.

Drafted from Grade 10 Chemistry, page 237 onwards, then checked twice before it went up

Complete Combustion of AlkanesFormula

\mathrm{C}_n\mathrm{H}_{2n+2} + \left(\frac{3n+1}{2}\right)\mathrm{O}_2 \rightarrow n\,\mathrm{CO}_2 + (n+1)\,\mathrm{H}_2\mathrm{O}

\mathrm{C}_n\mathrm{H}_{2n+2} + \left(\frac{3n+1}{2}\right)\mathrm{O}_2 \rightarrow n\,\mathrm{CO}_2 + (n+1)\,\mathrm{H}_2\mathrm{O}

$n$: number of carbon atoms in the alkane

When you use it

Use this general equation to balance any complete combustion reaction of an alkane producing carbon dioxide and water.

Watch out

When n is an even number, multiply the entire equation by 2 to clear the fractional coefficient for oxygen.

Drafted from Grade 10 Chemistry, page 237 onwards, then checked twice before it went up

Hydrocarbon Homologous FamiliesReference tableAlkane · Alkene · Cycloalkane

Class	General Formula	Saturation	Characteristic Feature
Alkane	$C_n H_{2n+2}$	Saturated	Single $C-C$ bonds only
Alkene	$C_n H_{2n}$	Unsaturated	At least one $C=C$ bond
Cycloalkane	$C_n H_{2n}$	Saturated	Single-bonded carbon ring

When you use it

Use this to classify open-chain and cyclic hydrocarbons, deduce molecular formulas from carbon counts, or check degree of saturation.

Watch out

Cycloalkanes share the general formula CnH2n with alkenes, but cycloalkanes are saturated rings containing only single bonds.

Drafted from Grade 10 Chemistry, page 237 onwards, then checked twice before it went up

Laboratory Preparation of AlkanesReference tableHydrogenation · Wurtz synthesis · Decarboxylation

Reaction Type	Reactants	Reagent or Catalyst	Product
Hydrogenation	Alkene $+ \mathrm{H}_2$	$\mathrm{Pt}$ , heat	Alkane
Wurtz synthesis	$2\mathrm{R-X} + 2\mathrm{Na}$	Heat	$\mathrm{R-R} + 2\mathrm{NaX}$
Decarboxylation	$\mathrm{RCOONa} + \mathrm{NaOH}$	$\mathrm{CaO}$ , heat	$\mathrm{R-H} + \mathrm{Na}_2\mathrm{CO}_3$

When you use it

Use this summary to identify reagents and products for standard synthetic routes to alkanes in the laboratory.

Watch out

Decarboxylation yields an alkane with one fewer carbon atom than the parent carboxylate salt, because one carbon leaves as sodium carbonate.

Drafted from Grade 10 Chemistry, page 237 onwards, then checked twice before it went up

The first ten alkanesReference tableMethane · Ethane · Propane · +7

Name	Carbon atoms	Molecular formula
Methane	1	$\mathrm{CH_4}$
Ethane	2	$\mathrm{C_2H_6}$
Propane	3	$\mathrm{C_3H_8}$
Butane	4	$\mathrm{C_4H_{10}}$
Pentane	5	$\mathrm{C_5H_{12}}$
Hexane	6	$\mathrm{C_6H_{14}}$
Heptane	7	$\mathrm{C_7H_{16}}$
Octane	8	$\mathrm{C_8H_{18}}$
Nonane	9	$\mathrm{C_9H_{20}}$
Decane	10	$\mathrm{C_{10}H_{22}}$

When you use it

Reach for this when a question names an alkane and asks for its formula, or gives a formula and asks for the name. These rows are the straight chains, because from butane on the formula alone no longer names one compound: C4H10 is butane, and it is also methylpropane.

Watch out

Losing the plus two. A student doubles the carbon count and stops, so hexane comes out as C6H12, which is a different compound carrying a double bond or a ring. Each end of the chain has one spare bond where the chain stops, so every alkane holds two hydrogens more than twice its carbons, and hexane is C6H14. Check any answer by counting: the hydrogen number is always even, and always two above the doubled carbon count.

Addition reactions of alkenesReference tableHydrogenation · Halogenation · Hydration · +2

Reaction	What is added, and the conditions	Product
Hydrogenation	$\mathrm{H_2}$ over heated $\mathrm{Ni}$ , or over $\mathrm{Pt}$ or $\mathrm{Pd}$ at room temperature	an alkane, $\mathrm{RCH_2-CH_3}$
Halogenation	$\mathrm{Br_2}$ at room temperature, with no catalyst	$\mathrm{RCHBr-CH_2Br}$ , and the orange colour fades
Hydration	$\mathrm{H_2O}$ with an acid catalyst	a secondary alcohol, $\mathrm{RCH(OH)-CH_3}$
Adding a hydrogen halide	$\mathrm{HBr}$ or $\mathrm{HCl}$ at room temperature	$\mathrm{RCHBr-CH_3}$ , the H joining the carbon that already carries more hydrogens
Polymerization	many $\mathrm{CH_2=CH_2}$ molecules with a catalyst	polyethene, $\mathrm{(-CH_2-CH_2-)_n}$

When you use it

Use when something is added across a C=C: hydrogen, bromine, water or a hydrogen halide. Addition breaks only the weaker pi bond of the double bond and puts two new single bonds in its place, which is why an alkene reacts in the cold where an alkane does not.

Watch out

Students put the OH on the wrong carbon. Water added to propene gives propan-2-ol, CH3CH(OH)CH3, and not propan-1-ol: the H joins the CH2 end that already carries two hydrogens, so the OH is left on the middle carbon. The same trap catches HBr, where propene gives 2-bromopropane. Ethene and but-2-ene are the exception, because both of their carbons carry the same number of hydrogens and only one product is possible.

The bromine water test for unsaturationMethodAn alkene, such as ethene · An alkyne, such as ethyne · An alkane, out of direct sunlight · +2

What is in the tube	What you see	What it tells you
An alkene, such as ethene	The orange brown colour goes at once, in the cold and with no catalyst	Unsaturated. The bromine adds across the double bond, and the Grade 10 equation is $\mathrm{C_2H_4 + Br_2 \rightarrow C_2H_4Br_2}$
An alkyne, such as ethyne	The colour goes as well, and twice as much bromine is used up	Unsaturated too. $\mathrm{C_2H_2}$ takes $\mathrm{2Br_2}$ and gives $\mathrm{C_2H_2Br_4}$ , so the test cannot tell a double bond from a triple one
An alkane, out of direct sunlight	The colour stays orange brown	Saturated
An alkane, in sunlight	The colour fades slowly and hydrogen bromide is given off	A false positive from substitution, so keep the tube away from the window
Phenol, or an aldehyde	The colour goes, and phenol also leaves a white precipitate	The sample reacts with bromine without being unsaturated, so the test does not name the group on its own

When you use it

Use to tell a saturated hydrocarbon from an unsaturated one. Add bromine water drop by drop and shake: if the orange brown colour goes at once in the cold, the sample holds a double or a triple bond.

Watch out

Two slips, and both cost marks. Writing that the solution turns clear instead of colourless: bromine water is already clear, meaning you can see through it, so the answer is orange brown going to colourless. Then reading a colourless tube as proof of a C=C. An alkyne does exactly the same thing, and uses twice the bromine doing it, so the only safe conclusion is that the compound is unsaturated.

Substitution or addition: how a hydrocarbon reactsRuleAlkanes, C_nH_2n+2, saturated · Alkenes C_nH_2n and alkynes C_nH_2n-2, unsaturated · Complete combustion, any hydrocarbon · +1

Family	How it reacts	Worked example
Alkanes, $\mathrm{C_nH_{2n+2}}$ , saturated	Substitution, because there is no bond to open. A chlorine atom takes the place of a hydrogen and $\mathrm{HCl}$ leaves. Sunlight starts a chain, so the real products are a mixture of $\mathrm{CH_3Cl}$ (chloromethane), $\mathrm{CH_2Cl_2}$ , $\mathrm{CHCl_3}$ and $\mathrm{CCl_4}$	$\mathrm{CH_4 + Cl_2 \rightarrow CH_3Cl + HCl}$ , the first step
Alkenes $\mathrm{C_nH_{2n}}$ and alkynes $\mathrm{C_nH_{2n-2}}$ , unsaturated	Addition. The multiple bond opens and both added atoms join the one molecule, so nothing is released	$\mathrm{CH_2=CH_2 + H_2 \rightarrow CH_3-CH_3}$ , over heated $\mathrm{Ni}$
Complete combustion, any hydrocarbon	Burning in plenty of oxygen gives carbon dioxide and water. A limited supply gives carbon monoxide or soot instead	$\mathrm{CH_4 + 2O_2 \rightarrow CO_2 + 2H_2O}$
Esterification, a different family	A carboxylic acid condenses with an alcohol. Neither one is a hydrocarbon, the reaction is reversible so it is written with a two way arrow, and the concentrated $\mathrm{H_2SO_4}$ is a catalyst and a dehydrating agent rather than a reactant	$\mathrm{CH_3COOH + C_2H_5OH \rightarrow CH_3COOC_2H_5 + H_2O}$

When you use it

Use to decide which reaction a hydrocarbon does before you write a single equation. An alkane has no bond to open, so it substitutes and pushes out a small molecule; an alkene or an alkyne opens its multiple bond and takes everything in.

Watch out

Students mix the two mechanisms in one equation and write CH2=CH2 with Cl2 giving CH3CH2Cl and HCl, opening the double bond and releasing HCl as well. Count the atoms and the mistake shows at once: the left side holds C2H4Cl2 and the right side C2H6Cl2, two hydrogens that came from nowhere. An addition joins everything into one product and releases nothing, while a substitution always kicks out a small molecule such as HCl. So the alkene answer is CH2ClCH2Cl on its own, and the alkane answer is CH3Cl with HCl beside it.

Naming a branched-chain alkaneRuleFind the parent chain · Number the chain · Give every substituent its locant · +4

Step	What to do	Example
Find the parent chain	Take the longest continuous chain of carbon atoms, counting around corners rather than along the row it happens to be drawn in. If two chains tie on length, take the one carrying more substituents.	3-ethylhexane
Number the chain	Number the carbons from the end that reaches a branch first, so the first substituent gets the lowest number it can have.	2-methylhexane
Give every substituent its locant	Each substituent takes the number of the parent carbon it is bonded to. Two on the same carbon each keep their own number, so that number is written twice.	2,2-dimethylpropane
Order the substituents	Write them alphabetically in front of the parent name. Ignore di-, tri- and tetra- and the italic prefixes n-, sec- and tert- when alphabetising, so n-propyl files under p. The prefixes iso-, neo- and cyclo- are part of the name and do count.	4-ethyl-2,2-dimethylhexane
Punctuate the name	Put a hyphen (-) between a number and a word and a comma (,) between two numbers, then close with the parent chain and its suffix.	2,3-dimethylbutane
Break a tie in the numbering	When both ends give the same first locant, compare the two full sets of locants term by term and keep the set that is lower at the first point of difference. If the two sets are identical, the lower number goes to the substituent that comes first alphabetically.	3-ethyl-4-methylhexane
Name a branched substituent	Treat it as a chain of its own, number its carbon 1 at the atom bonded to the parent chain, and put its whole name in parentheses.	4-(1-methylethyl)heptane

When you use it

Use whenever you have to turn a drawn branched alkane into its IUPAC name, or read a name back into a structure. Every name is built in one fixed order: the locant that says where a substituent sits, the prefix that names it, the parent chain, and the family suffix, which is -ane for an alkane.

Watch out

Two slips cost the most marks. The first is taking the chain that happens to be drawn straight across the page as the parent chain. A five-carbon row with a propyl branch on carbon 3 is not 3-propylpentane, because the chain that turns down the branch is six carbons long, so the name is 3-ethylhexane. The second is alphabetising by the multiplier. Dimethyl files under m, so the correct name is 4-ethyl-2,2-dimethylhexane and 2,2-dimethyl-4-ethylhexane is wrong.

Primary, secondary, tertiary and quaternary carbonsRuleNone · Primary · Secondary · +2

Class	Other carbons bonded to it	Where you meet it
None	0	the carbon in methane, CH4, which is bonded to no other carbon
Primary	1	the two end carbons of butane, CH3-CH2-CH2-CH3, and both carbons in ethane
Secondary	2	the two middle carbons of butane
Tertiary	3	the central carbon of 2-methylpropane, (CH3)3CH, which holds three methyl groups and one hydrogen
Quaternary	4	the central carbon of neopentane, C(CH3)4, which carries no hydrogen at all

When you use it

Use when a question asks you to classify a carbon atom, or when a reaction depends on how substituted that carbon is. Count only the other carbon atoms bonded directly to it: one makes it primary, two secondary, three tertiary and four quaternary, and a hydrogen takes the class of the carbon it sits on.

Watch out

Students count hydrogens instead of carbons. A CH3 carbon carries three hydrogens, so they call it tertiary, when it is primary: it touches only one other carbon. Read the neighbours rather than the hydrogens, because the carbon with the most hydrogens is the least substituted and the carbon with none is the quaternary one. Two follow-on traps come after that. An oxygen, nitrogen or halogen on the carbon is never counted, so the carbon in (CH3)3C-OH holds three carbons and is tertiary. And amines break the pattern entirely, because there the class is counted on the nitrogen, so a primary amine RNH2 has nothing to do with a primary carbon.

Free-radical halogenation of an alkaneMethodChain initiation · Chain propagation, step 1 · Chain propagation, step 2 · +1

Stage	What happens	With methane and chlorine
Chain initiation	Ultraviolet light or strong heat splits the halogen molecule homolytically, one bonding electron going to each atom. It is the halogen bond that breaks, because it is the weakest bond present, and the free radical it leaves is neutral with one unpaired electron, which is what makes it so reactive.	Cl2 gives two Cl radicals
Chain propagation, step 1	A chlorine radical pulls a hydrogen atom off the alkane and leaves an alkyl radical behind.	CH4 + Cl radical gives CH3 radical + HCl
Chain propagation, step 2	The alkyl radical attacks a fresh halogen molecule, takes one atom from it and hands back a chlorine radical. That returned radical is why one initiation event can drive a long run of cycles, and why this is called a chain reaction.	CH3 radical + Cl2 gives CH3Cl + Cl radical
Chain termination	Two radicals meet and combine, and that chain stops. Finding a trace of ethane among the products of methane chlorination is the evidence that methyl radicals were present.	CH3 radical + CH3 radical gives ethane, CH3-CH3

When you use it

Use when a question asks how an alkane reacts with chlorine or bromine, or asks you to write the mechanism out. A substitution reaction replaces an atom or group in a molecule with a different atom or group, and an alkane will only do it once ultraviolet light or strong heat is supplied, through three stages in this order: chain initiation, chain propagation, chain termination.

Watch out

Students write CH4 + Cl2 giving CH3Cl + HCl and stop there, as though the product were pure chloromethane. A chlorine radical attacks CH3Cl as readily as it attacks CH4, so the real product is a mixture of CH3Cl, CH2Cl2, CHCl3 and CCl4, plus a trace of ethane from termination, and a large excess of methane is what pushes the mixture towards the mono-substituted product. The second slip is in the initiation step. Many write CH4 giving a methyl radical and a hydrogen radical, but the bond that breaks is Cl-Cl, the weakest bond in the mixture, and the radicals must be drawn neutral with a single dot, never as Cl+ and Cl-.

Physical properties of the alkanesReference tableState at 25 degrees C · Polarity · Solubility · +4

Property	What it does	Why
State at 25 degrees C	Methane to butane (C1 to C4) are gases, pentane to heptadecane (C5 to C17) are liquids, and octadecane (C18) and longer are waxy solids.	The liquid to solid line is set by melting point: heptadecane melts at about 22 degrees C and octadecane at about 28 degrees C.
Polarity	Alkanes are nonpolar and carry no significant dipole.	Carbon and hydrogen sit close together on the Pauling scale, 2.55 and 2.20 in current tables and 2.5 and 2.1 in older ones, so a C-H bond has almost no charge separation.
Solubility	Insoluble in water, and freely soluble in nonpolar solvents such as diethyl ether, benzene or carbon tetrachloride.	Like dissolves like.
Density	Liquid alkanes run about 0.62 to 0.79 g/mL and solid paraffin wax about 0.9 g/mL, all below water's 1.00 g/mL, so petrol, kerosene and paraffin oil spread as a film on the surface.	Density climbs with chain length and then levels off near 0.8 g/mL for the liquids, so it never reaches the density of water.
Boiling point	Rises steadily as the chain gets longer, by roughly 20 to 30 degrees C for each added CH2 at the short-chain end and less as the chain grows.	A bigger molecule has a larger, more easily distorted electron cloud, so the London dispersion forces between neighbouring molecules are stronger and more heat is needed to separate them.
Melting point	Rises with chain length overall, but zigzags on the way up instead of climbing in a straight line.	Even-numbered chains pack into the crystal lattice better than odd-numbered ones.
Branching	Lowers the boiling point. Pentane boils at 36 degrees C, 2-methylbutane at 28 degrees C and 2,2-dimethylpropane at 9.5 degrees C, and all three are C5H12.	A branched isomer is more compact and closer to spherical, so neighbouring molecules touch over less surface and the dispersion forces between them are weaker.

When you use it

Use when a question asks you to compare two alkanes, or to explain why one boils higher than another, floats on water or refuses to dissolve in it. Alkanes are nonpolar molecules held to each other only by London dispersion forces, and every property in this table follows from that.

Watch out

Students learn that branching lowers the boiling point and then apply it to the melting point as well. Melting depends on how neatly molecules stack in a crystal, and a compact symmetric branched molecule stacks better than a long floppy chain: 2,2-dimethylpropane melts at about -17 degrees C while straight-chain pentane melts at about -130 degrees C, even though the branched one boils 27 degrees C lower. Branching lowers the boiling point and often raises the melting point, so check which property the question is asking about before you reach for the rule.

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What do the Grade 10 Chemistry cards cover?: 36 cards across 6 chapters of the national textbook: Energy changes and electrochemistry, Sollutions, Important inorganic compounds, Metals and nonmetals, Hydrocarbons and their natural sources and 1 more. You can take any chapter one card at a time on the page itself.
Is there a national exam in Grade 10?: No. Ethiopia sets national exams in Grade 6, Grade 8 and Grade 12 only. These cards are for your school's own exams, and for the national exam that comes a few years later.
Where do these cards come from?: They are drafted from Grade 10 Chemistry, the Ministry of Education textbook for this grade. A second pass that cannot see the chapter then re-derives every formula, constant and table row, and anything it cannot confirm is held back instead of published.
Is this free?: Yes. Every card here is free to read and the printable sheet is free to download. Neither needs an account.
When was this last checked?: 6 September 2026. Cards arrive chapter by chapter, and the line under each one says when that card was last read through.

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