Chapter Notes

Atomic Foundations of Matter
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Atomic Foundations of Matter

When elements combine to form a compound, their individual properties are often lost. For example, hydrogen is a combustible gas and oxygen is a gas that supports combustion. However, when they combine, they form water, a liquid at ordinary temperatures that extinguishes fire. Interestingly, the mass of the water formed is exactly equal to the sum of the masses of the hydrogen and oxygen that reacted. This observation is fundamental to chemistry and leads to our first important law.

Law of Conservation of Mass

Proposed by Antoine Lavoisier in 1789, the Law of Conservation of Mass states that matter can neither be created nor destroyed in a chemical reaction. This means that for any chemical reaction, the total mass of the substances before the reaction (reactants) is equal to the total mass of the substances formed after the reaction (products).

Total Mass of Reactants = Total Mass of Products

This principle holds true for both physical changes (like dissolving salt in water) and chemical changes.

Example
For example, in a reaction between sodium sulfate and barium chloride, a white substance called a precipitate is formed. Sodium sulfate + Barium chloride → Barium sulfate + Sodium chloride

If you weigh the solutions before and after mixing them, you will find that the total mass remains unchanged, demonstrating the Law of Conservation of Mass.

Note
When a reaction produces a gas, like the reaction between vinegar and baking soda, it must be performed in a closed system (like a flask with a balloon over the top) to prove this law. If the gas is allowed to escape into the air, the final mass will appear to be less than the initial mass.
Example
Example In a group activity, students place 4.0 g of calcium carbonate with 2.92 g of hydrochloric acid in a closed container. After the reaction is over, they measured 1.76 g of carbon dioxide, 0.72 g of water, and 4.44 g of calcium chloride. Verify whether the Law of Conservation of Mass is obeyed or not.

Given

  • Mass of calcium carbonate = 4.0 g4.0 \text{ g}
  • Mass of hydrochloric acid = 2.92 g2.92 \text{ g}
  • Mass of carbon dioxide = 1.76 g1.76 \text{ g}
  • Mass of water = 0.72 g0.72 \text{ g}
  • Mass of calcium chloride = 4.44 g4.44 \text{ g}

To Find

Verify if the Law of Conservation of Mass is obeyed.

Solution

First, calculate the total mass of the reactants. Total mass of reactants = Mass of calcium carbonate + Mass of hydrochloric acid 4.0 g+2.92 g=6.92 g4.0 \text{ g} + 2.92 \text{ g} = 6.92 \text{ g}

Next, calculate the total mass of the products. Total mass of products = Mass of carbon dioxide + Mass of water + Mass of calcium chloride 1.76 g+0.72 g+4.44 g=6.92 g1.76 \text{ g} + 0.72 \text{ g} + 4.44 \text{ g} = 6.92 \text{ g}

Comparing the two totals: Mass of reactants = Mass of products 6.92 g=6.92 g6.92 \text{ g} = 6.92 \text{ g}

Final Answer Hence, the Law of Conservation of Mass is obeyed.


Law of Constant Proportions

Following Lavoisier's work, Joseph Proust proposed another fundamental law. The Law of Constant Proportions (also known as the Law of Definite Proportions) states that in any compound formed by two or more elements, the elements always combine in a fixed ratio by mass, regardless of its source or how it was prepared.

Example
Water (H2OH_2O) is a classic example. Whether it comes from a river, an ocean, or is made in a lab, purified water always contains hydrogen and oxygen in a mass ratio of 1:8. This means that if you decompose 9 g9 \text{ g} of water, you will always get 1 g1 \text{ g} of hydrogen and 8 g8 \text{ g} of oxygen.
Example
Example 12 g of carbon combines with 32 g of oxygen to form 44 g of carbon dioxide as per the given equation. Carbon + Oxygen → Carbon dioxide If 2.4 g of carbon reacts completely with oxygen, how much carbon dioxide will be produced?

Given

  • 12 g of carbon produces 44 g of carbon dioxide.

To Find

The mass of carbon dioxide produced from 2.4 g of carbon.

Solution

From the given information, we can find the amount of carbon dioxide produced by 1 g of carbon. So, 1 g of carbon will give=4412 g of carbon dioxide\text{So, } 1 \text{ g of carbon will give} = \frac{44}{12} \text{ g of carbon dioxide} Now, we can calculate the amount for 2.4 g of carbon. Thus, 2.4 g of carbon will give=4412×2.4 g\text{Thus, } 2.4 \text{ g of carbon will give} = \frac{44}{12} \times 2.4 \text{ g} =8.8 g of carbon dioxide= 8.8 \text{ g of carbon dioxide}

Final Answer Hence, 8.8 g of carbon dioxide will be produced.


Example
Example Sodium chloride (NaCl) contains sodium and chlorine in the mass ratio of 23:35.5. If 46 g of sodium reacts completely, how much chlorine is needed to form NaCl?

Solution

The mass ratio of Na to Cl is 23:35.5. Mass of chlorine required=(35.523)×46=71 g\text{Mass of chlorine required} = (\frac{35.5}{23}) \times 46 = 71 \text{ g}

Final Answer 71 g of chlorine is needed.


Dalton's Atomic Theory

The Law of Conservation of Mass and the Law of Constant Proportions provided the foundation for John Dalton's Atomic Theory in 1808. This theory was a major step in understanding matter and explained why these laws work. A postulate is a fundamental assumption accepted as true to build further ideas.

John Dalton's Postulates:

  • All matter is made up of very tiny particles called atoms, which participate in chemical reactions.
  • Atoms are indivisible particles, which cannot be created or destroyed in a chemical reaction. (This explains the Law of Conservation of Mass).
  • Atoms of a given element are identical in mass and chemical properties.
  • Atoms of different elements have different masses and chemical properties.
  • Atoms combine in the ratio of simple whole numbers to form compounds. (This explains the Law of Constant Proportions).
  • The relative number and kinds of atoms are constant in a given compound.

How Atoms Combine?

Atoms combine to form more stable arrangements. An atom is considered stable when its outermost electron shell (valence shell) is full, which usually means having 8 electrons (an octet). To achieve this stability, atoms form a chemical bond, which is the force that holds them together.

A molecule is an electrically neutral entity made of more than one atom bonded together. It can exist independently and exhibits all the properties of that substance.

Atoms combine in two main ways:

  1. Sharing of electrons: Atoms share their valence electrons with other atoms.
  2. Transfer of electrons: One atom gives one or more electrons to another atom.

Bonding by sharing of electrons - Covalent Bond

A covalent bond is a chemical bond formed when two atoms share one or more pairs of electrons.

Molecules of Elements

  • Hydrogen (H2H_2): A hydrogen atom has one electron and needs one more to complete its first shell (which holds 2 electrons). Two hydrogen atoms share their single electrons to form a stable H2H_2 molecule. This involves one shared pair of electrons, forming a single bond, represented as H-H.
  • Chlorine (Cl2Cl_2): A chlorine atom has seven valence electrons and needs one more to complete its octet. Two chlorine atoms each share one electron, forming a single covalent bond, represented as Cl-Cl.
  • Oxygen (O2O_2): An oxygen atom has six valence electrons and needs two more. Two oxygen atoms each share two electrons with each other. This involves two shared pairs of electrons, forming a double bond, represented as O=O.

Molecules of Compounds

  • Hydrogen Chloride (HCl): A hydrogen atom needs one electron, and a chlorine atom also needs one electron. They share one pair of electrons, forming a single covalent bond, H-Cl.
  • Water (H2OH_2O): An oxygen atom needs two electrons, while each hydrogen atom needs one. The oxygen atom shares one electron with one hydrogen atom and a second electron with another hydrogen atom. This results in a water molecule with two single bonds.

Naming Covalent Compounds A prefix system is used to indicate the number of atoms of each element.

  • Prefixes: mono- (1), di- (2), tri- (3), tetra- (4), penta- (5), hexa- (6).
  • The first element keeps its name. The second element's name ends in -ide.
  • 'Mono-' is usually omitted for the first element.
  • Examples:
    • CO: Carbon monoxide
    • CO2CO_2: Carbon dioxide
    • PCl3PCl_3: Phosphorus trichloride
    • SF6SF_6: Sulfur hexafluoride
    • N2O5N_2O_5: Dinitrogen pentoxide
Note
Some compounds are known by common names, like H2OH_2O (water) and NH3NH_3 (ammonia).

Bonding by electron transfer - Ionic Bond

An ionic bond is formed by the complete transfer of one or more electrons from one atom to another. This typically occurs between a metal (which loses electrons) and a non-metal (which gains electrons).

When an atom loses electrons, it has more protons than electrons, resulting in a positively charged ion called a cation.

  • Example: A sodium atom (Na) has 11 protons and 11 electrons. It loses one valence electron to become stable. The resulting sodium ion (Na+Na^+) has 11 protons and 10 electrons, giving it a +1 charge.

When an atom gains electrons, it has more electrons than protons, resulting in a negatively charged ion called an anion.

  • Example: A chlorine atom (Cl) has 17 protons and 17 electrons. It gains one electron to become stable. The resulting chloride ion (ClCl^−) has 17 protons and 18 electrons, giving it a -1 charge.

The ionic bond is the strong electrostatic force of attraction between these oppositely charged ions (Na+Na^+ and ClCl^−) that holds them together in a compound like sodium chloride (NaCl).

Note
Ionic compounds don't exist as single molecules but form large, ordered, three-dimensional structures called crystal lattices.

Naming Ionic Compounds

  • The name of the cation (usually the metal) is written first.
  • The name of the anion (usually the non-metal) is written second, with its ending changed to -ide.
  • Examples: Sodium chloride (NaCl), Calcium oxide (CaO).
  • Some ions, called polyatomic ions, are made of multiple atoms bonded together with an overall charge (e.g., Sulfate, SO42SO_4^{2-}). Their names generally do not end in -ide.

Here are some common ions:

Name of ionFormulaValency
SodiumNa+Na^+1
PotassiumK+K^+1
CalciumCa2+Ca^{2+}2
MagnesiumMg2+Mg^{2+}2
AluminiumAl3+Al^{3+}3
Iron (Ferrous)Fe2+Fe^{2+}2
Iron (Ferric)Fe3+Fe^{3+}3
Copper (Cupric)Cu2+Cu^{2+}2
ChlorideClCl^−1
OxideO2O^{2-}2
SulfideS2S^{2-}2
HydroxideOHOH^−1
NitrateNO3NO_3^−1
CarbonateCO32CO_3^{2-}2
SulfateSO42SO_4^{2-}2
AmmoniumNH4+NH_4^+1

Writing Chemical Formulae

A chemical formula represents the composition of a compound. The "criss-cross" method is a quick way to determine the formula.

Writing chemical formulae of covalent compounds

  1. Write the symbols of the elements.
  2. Write the valency of each element below it.
  3. Criss-cross the valencies and write them as subscripts. Subscript '1' is not written.
  • Hydrogen sulfide:

    • Symbols: H S
    • Valencies: 1 2
    • Criss-cross: H2S1H_2S_1H2SH_2S
  • Carbon tetrachloride:

    • Symbols: C Cl
    • Valencies: 4 1
    • Criss-cross: C1Cl4C_1Cl_4CCl4CCl_4

Writing chemical formulae of ionic compounds

  1. Write the symbol of the cation first, then the anion.
  2. Write their charges below them (without the +/- signs).
  3. Criss-cross the charge numbers and write them as subscripts.
  4. If there is a common factor in the subscripts, simplify them to the simplest whole-number ratio.
  • Aluminium oxide:

    • Ions: Al3+Al^{3+} O2O^{2-}
    • Charges: 3 2
    • Criss-cross: Al2O3Al_2O_3
  • Magnesium oxide:

    • Ions: Mg2+Mg^{2+} O2O^{2-}
    • Charges: 2 2
    • Criss-cross: Mg2O2Mg_2O_2
    • Simplify by dividing by 2: MgO
Note
When a formula contains more than one polyatomic ion, the ion is placed in brackets with the subscript outside.
  • Magnesium hydroxide:
    • Ions: Mg2+Mg^{2+} OHOH^−
    • Charges: 2 1
    • Criss-cross: Mg(OH)2Mg(OH)_2 (The '2' applies to the entire OH ion).

Properties of the Ionic and the Covalent Compounds

PropertyIonic CompoundsCovalent Compounds
BondingTransfer of electronsSharing of electrons
Melting/Boiling PointsGenerally highGenerally low
Solubility in WaterGenerally solubleGenerally insoluble
Solubility in Petrol/KeroseneGenerally insolubleGenerally soluble
Electrical ConductivityConduct electricity in molten state or when dissolved in water. Do not conduct as solids.Do not conduct electricity in any state.

Explanation of Conductivity: In solid ionic compounds, the ions are locked in a fixed crystal lattice and cannot move. When melted or dissolved, the ions are free to move and carry an electric charge, allowing the substance to conduct electricity. Covalent compounds do not contain ions, so they cannot conduct electricity.

Molecular Mass of Covalent Compounds

The molecular mass of a substance is the sum of the atomic masses of all the atoms in a single molecule of that substance. It is expressed in atomic mass units (u).

Example
Example Calculate the molecular mass of water (H2OH_2O). (Atomic mass: H = 1 u; O = 16 u)

Solution

The formula H2OH_2O shows 2 atoms of Hydrogen and 1 atom of Oxygen. Molecular mass of H2O=(mass of H×2)+(mass of O×1)H_2O = ( \text{mass of H} \times 2) + (\text{mass of O} \times 1) (1 u×2)+(16 u×1)=2 u+16 u=18 u(1 \text{ u} \times 2) + (16 \text{ u} \times 1) = 2 \text{ u} + 16 \text{ u} = 18 \text{ u}

Final Answer The molecular mass of water is 18 u18 \text{ u}.


Example
Example Calculate the molecular mass of carbon dioxide (CO2CO_2). (Atomic mass: C = 12 u; O = 16 u)

Solution

The formula CO2CO_2 shows 1 atom of Carbon and 2 atoms of Oxygen. Molecular mass of CO2=(mass of C×1)+(mass of O×2)CO_2 = (\text{mass of C} \times 1) + (\text{mass of O} \times 2) (12 u×1)+(16 u×2)=12 u+32 u=44 u(12 \text{ u} \times 1) + (16 \text{ u} \times 2) = 12 \text{ u} + 32 \text{ u} = 44 \text{ u}

Final Answer The molecular mass of carbon dioxide is 44 u44 \text{ u}.


Formula Unit Mass of Ionic Compounds

Since ionic compounds form crystal lattices instead of individual molecules, we use the term formula unit mass. A formula unit is the simplest whole-number ratio of ions in the compound. The formula unit mass is calculated in the same way as molecular mass: by summing the atomic masses of all atoms in the formula unit.

Example
Example Calculate the formula unit mass of sodium oxide (Na2ONa_2O). (Atomic mass: Na = 23 u; O = 16 u)

Solution

The formula unit Na2ONa_2O contains 2 atoms of Sodium and 1 atom of Oxygen. Formula unit mass of Na2O=(mass of Na×2)+(mass of O×1)Na_2O = (\text{mass of Na} \times 2) + (\text{mass of O} \times 1) (23 u×2)+(16 u×1)=46 u+16 u=62 u(23 \text{ u} \times 2) + (16 \text{ u} \times 1) = 46 \text{ u} + 16 \text{ u} = 62 \text{ u}

Final Answer The formula unit mass of sodium oxide is 62 u62 \text{ u}.


Example
Example Calculate the formula unit mass of calcium nitrate, Ca(NO3)2Ca(NO_3)_2. (Atomic mass: Ca = 40 u; N = 14 u; O = 16 u)

Solution

The formula unit Ca(NO3)2Ca(NO_3)_2 contains:

  • 1 atom of Calcium (Ca)
  • 2 atoms of Nitrogen (N) (since NO3NO_3 is taken twice)
  • 6 atoms of Oxygen (O) (since O3O_3 is taken twice, 3×2=63 \times 2 = 6)

Formula unit mass = (mass of Ca×1)+[(mass of N×1)+(mass of O×3)]×2(\text{mass of Ca} \times 1) + [(\text{mass of N} \times 1) + (\text{mass of O} \times 3)] \times 2 (40 u×1)+[(14 u×1)+(16 u×3)]×2(40 \text{ u} \times 1) + [ (14 \text{ u} \times 1) + (16 \text{ u} \times 3) ] \times 2 40 u+(14 u+48 u)×240 \text{ u} + (14 \text{ u} + 48 \text{ u}) \times 2 40 u+(62 u)×240 \text{ u} + (62 \text{ u}) \times 2 40 u+124 u=164 u40 \text{ u} + 124 \text{ u} = 164 \text{ u}

Final Answer The formula unit mass of calcium nitrate is 164 u164 \text{ u}.

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