Chapter Notes

Atoms and Molecules

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Atoms and Molecules

Ancient Indian and Greek philosophers pondered the nature of matter. In India, around 500 BC, Maharishi Kanad suggested that dividing matter (padarth) would lead to smaller particles, eventually reaching indivisible particles called Parmanu. Pakudha Katyayama added that these particles usually exist in combined forms, creating various types of matter.

Around the same time, Greek philosophers Democritus and Leucippus proposed that dividing matter would lead to indivisible particles, which Democritus named atoms (meaning indivisible). These ideas were philosophical, and experimental validation was limited until the eighteenth century.

By the late eighteenth century, scientists distinguished between elements and compounds and sought to understand how and why elements combine, and what happens during combination. Antoine L. Lavoisier established the foundation of chemical sciences by formulating two important laws of chemical combination.

Laws of Chemical Combination

The following two laws were established after extensive experiments by Lavoisier and Joseph L. Proust.

Law of conservation of mass

The law of conservation of mass states that mass can neither be created nor destroyed in a chemical reaction. This means that in any chemical reaction, the total mass of the reactants (the starting materials) is equal to the total mass of the products (the substances formed).

Law of constant proportions

Lavoisier and other scientists observed that compounds are composed of two or more elements, and each compound has the same elements in the same proportions by mass, regardless of the source or preparation method. This is known as the law of constant proportions, also known as the law of definite proportions. Proust stated this law as "In a chemical substance the elements are always present in definite proportions by mass".

For example, in water, the ratio of the mass of hydrogen to the mass of oxygen is always 1:81:8, irrespective of the water source. Thus, decomposing 9 g of water always yields 1 g of hydrogen and 8 g of oxygen. Similarly, in ammonia, nitrogen and hydrogen are always present in the ratio 14:314:3 by mass, regardless of the method or source of its creation.

The next challenge for scientists was to explain these laws. British chemist John Dalton provided a basic theory about the nature of matter. Dalton revived the idea of the divisibility of matter, which was then just a philosophy. He adopted the term 'atoms' from the Greeks and stated that the smallest particles of matter are atoms. His theory was based on the laws of chemical combination and provided an explanation for the law of conservation of mass and the law of definite proportions.

Dalton's atomic theory proposed that all matter, whether an element, a compound, or a mixture, is composed of small particles called atoms. The postulates of this theory can be stated as follows:

(i) All matter is made of very tiny particles called atoms, which participate in chemical reactions. (ii) Atoms are indivisible particles, which cannot be created or destroyed in a chemical reaction. (iii) Atoms of a given element are identical in mass and chemical properties. (iv) Atoms of different elements have different masses and chemical properties. (v) Atoms combine in the ratio of small whole numbers to form compounds. (vi) The relative number and kinds of atoms are constant in a given compound.

Example
Example In a reaction, 5.3 g5.3 \text{ g} of sodium carbonate reacted with 6 g6 \text{ g} of acetic acid. The products were 2.2 g2.2 \text{ g} of carbon dioxide, 0.9 g0.9 \text{ g} water and 8.2 g8.2 \text{ g} of sodium acetate. Show that these observations are in agreement with the law of conservation of mass. sodium carbonate + acetic acid →\rightarrow sodium acetate + carbon dioxide + water

Given

  • Mass of sodium carbonate = 5.3 g5.3 \text{ g}
  • Mass of acetic acid = 6 g6 \text{ g}
  • Mass of carbon dioxide = 2.2 g2.2 \text{ g}
  • Mass of water = 0.9 g0.9 \text{ g}
  • Mass of sodium acetate = 8.2 g8.2 \text{ g}

To Find

Show that the total mass of reactants is equal to the total mass of products.

Formula

Law of Conservation of Mass: Mass of reactants = Mass of products

Solution

Mass of reactants = Mass of sodium carbonate + Mass of acetic acid

Mass of reactants = 5.3 g+6 g=11.3 g5.3 \text{ g} + 6 \text{ g} = 11.3 \text{ g}

Mass of products = Mass of sodium acetate + Mass of carbon dioxide + Mass of water

Mass of products = 8.2 g+2.2 g+0.9 g=11.3 g8.2 \text{ g} + 2.2 \text{ g} + 0.9 \text{ g} = 11.3 \text{ g}

Final Answer Since the mass of reactants is equal to the mass of products (11.3 g11.3 \text{ g}), the observations are in agreement with the law of conservation of mass.

Example
Example Hydrogen and oxygen combine in the ratio of 1:8 by mass to form water. What mass of oxygen gas would be required to react completely with 3 g3 \text{ g} of hydrogen gas?

Given

  • Ratio of hydrogen to oxygen = 1:81:8
  • Mass of hydrogen = 3 g3 \text{ g}

To Find

Mass of oxygen required to react completely with 3 g3 \text{ g} of hydrogen.

Formula

Mass of HydrogenMass of Oxygen=18\frac{\text{Mass of Hydrogen}}{\text{Mass of Oxygen}} = \frac{1}{8}

Solution

Let the mass of oxygen required be x gx \text{ g}.

3 gx g=18\frac{3 \text{ g}}{x \text{ g}} = \frac{1}{8}

x=3 g×8=24 gx = 3 \text{ g} \times 8 = 24 \text{ g}

Final Answer 24 g24 \text{ g} of oxygen gas would be required to react completely with 3 g3 \text{ g} of hydrogen gas.

Example
Example Which postulate of Dalton's atomic theory is the result of the law of conservation of mass?

Given

Dalton's Atomic Theory and Law of Conservation of Mass

To Find

Postulate of Dalton's Atomic Theory that reflects Law of Conservation of Mass

Solution

Dalton's atomic theory states that atoms are indivisible and cannot be created or destroyed in a chemical reaction. This postulate directly reflects the law of conservation of mass, as the total number of atoms remains constant during a chemical reaction, ensuring that mass is conserved.

Final Answer Atoms are indivisible particles, which cannot be created or destroyed in a chemical reaction.

Example
Example Which postulate of Dalton's atomic theory can explain the law of definite proportions?

Given

Dalton's Atomic Theory and Law of Definite Proportions

To Find

Postulate of Dalton's Atomic Theory that reflects Law of Definite Proportions

Solution

Dalton's atomic theory states that atoms combine in the ratio of small whole numbers to form compounds. Also, The relative number and kinds of atoms are constant in a given compound. This explains the law of definite proportions, as the same elements always combine in the same fixed ratio by mass in a given compound.

Final Answer Atoms combine in the ratio of small whole numbers to form compounds. The relative number and kinds of atoms are constant in a given compound.

What is an Atom?

Atoms are the building blocks of all matter.

How big are atoms?

Atoms are incredibly small, much smaller than anything we can see or easily imagine. Millions of atoms stacked together would only form a layer as thick as a sheet of paper.

The size of an atom is indicated by its atomic radius, which is measured in nanometres (nm).

1 nm=1109 m1 \text{ nm} = \frac{1}{10^9} \text{ m}

1 m=109 nm1 \text{ m} = 10^9 \text{ nm}

Even though atoms are so small, they are essential because our entire world is made up of them. We may not see them, but they are always there, influencing everything we do. Modern technology allows us to produce magnified images of element surfaces, showing individual atoms.

What are the modern day SYMBOLS OF ATOMS OF DIFFERENT ELEMENTS?

Dalton was the first scientist to use symbols for elements in a specific way. A symbol represented not only the element but also a definite quantity of that element, specifically one atom. Berzilius suggested that symbols should be made from one or two letters of the element's name.

Initially, element names came from the place of discovery. For example, copper's name comes from Cyprus. Some names originated from specific colors, like gold from the English word for yellow. Nowadays, the IUPAC (International Union of Pure and Applied Chemistry), an international scientific organization, approves element names, symbols, and units.

Many symbols are the first one or two letters of the element's English name. The first letter is always capitalized, and the second letter is lowercase.

For example: (i) hydrogen, H (ii) aluminium, Al (not AL) (iii) cobalt, Co (not CO)

Some element symbols use the first letter and another letter from later in the name. Examples include: (i) chlorine, Cl (ii) zinc, Zn

Other symbols are derived from the element's name in Latin, German, or Greek. For example, iron's symbol is Fe from its Latin name ferrum, sodium is Na from natrium, and potassium is K from kalium. Each element has a unique name and chemical symbol.

Atomic mass

Dalton's atomic theory introduced the concept of atomic mass, stating that each element has a characteristic atomic mass. This theory effectively explained the law of constant proportions, prompting scientists to measure atomic masses. Determining the mass of individual atoms was challenging, so relative atomic masses were found using the laws of chemical combinations and the compounds formed.

Consider carbon monoxide (CO), formed by carbon and oxygen. Experiments showed that 3 g3 \text{ g} of carbon combines with 4 g4 \text{ g} of oxygen to form CO. This means carbon combines with 43\frac{4}{3} times its mass of oxygen. If we define the atomic mass unit (formerly 'amu', now 'u' for unified mass) as equal to the mass of one carbon atom, then carbon would have an atomic mass of 1.0 u1.0 \text{ u} and oxygen an atomic mass of 1.33 u1.33 \text{ u}.

It is more convenient to have these numbers as whole numbers or close to whole numbers. Initially, scientists used 116\frac{1}{16} of the mass of a naturally occurring oxygen atom as the unit. This was because:

  • Oxygen reacts with many elements and forms compounds.
  • This unit gave masses of most elements as whole numbers.

However, in 1961, carbon-12 isotope was chosen as the standard reference for a universally accepted atomic mass unit. One atomic mass unit is exactly one-twelfth (112th\frac{1}{12}^{\text{th}}) the mass of one atom of carbon-12. The relative atomic masses of all elements have been determined relative to the mass of carbon-12.

The relative atomic mass of an element is defined as the average mass of the atom, compared to 112th\frac{1}{12}^{\text{th}} the mass of one carbon-12 atom.

How do atoms exist?

Atoms of most elements cannot exist independently. They form molecules and ions, which then aggregate in large numbers to form matter that we can see, feel, or touch.

Example
Example Define the atomic mass unit.

Given

Definition of atomic mass

To Find

Define atomic mass unit

Solution

One atomic mass unit is a mass unit equal to exactly one-twelfth (112th\frac{1}{12}^{\text{th}}) the mass of one atom of carbon-12.

Final Answer One atomic mass unit is a mass unit equal to exactly one-twelfth (112th\frac{1}{12}^{\text{th}}) the mass of one atom of carbon-12.

Example
Example Why is it not possible to see an atom with naked eyes?

Given

Size of atoms

To Find

Reason why atoms cannot be seen with naked eyes

Solution

Atoms are extremely small, having radii on the order of nanometers (10−9 m10^{-9} \text{ m}). This size is much smaller than the wavelength of visible light, making it impossible to see atoms with the naked eye or even with conventional microscopes.

Final Answer Atoms are extremely small, much smaller than the wavelength of visible light, making it impossible to see them with the naked eye.

What is a Molecule?

A molecule is a group of two or more atoms chemically bonded together by attractive forces. It is the smallest particle of an element or compound that can exist independently and retain all the properties of that substance. Molecules can be formed from the same element or different elements.

Molecules of elements

Molecules of an element consist of the same type of atoms. Many elements, like argon (Ar) and helium (He), have molecules made up of only one atom and are called monoatomic. However, most nonmetals form molecules with two or more atoms. For example, oxygen molecules consist of two oxygen atoms (O2\text{O}_2) and are called diatomic. If three oxygen atoms combine, they form ozone (O3\text{O}_3).

The atomicity of a molecule is the number of atoms it contains.

Metals and some other elements, such as carbon, do not have simple structures but consist of a very large and indefinite number of atoms bonded together.

Molecules of compounds

Molecules of compounds are formed when atoms of different elements join together in definite proportions.

What is an ion?

Ions are charged species formed from compounds composed of metals and nonmetals. They can be single charged atoms or groups of atoms with a net charge. Ions can be negatively charged (anions) or positively charged (cations).

For example, sodium chloride (NaCl) consists of positively charged sodium ions (Na+\text{Na}^+) and negatively charged chloride ions (Cl−\text{Cl}^-). A polyatomic ion is a group of atoms carrying a charge.

Writing Chemical Formulae

The chemical formula of a compound is a symbolic representation of its composition. To write chemical formulas, you need to know the symbols and combining capacity of the elements.

The valency of an element is its combining power or capacity. It indicates how many atoms of another element one atom of the element can combine with to form a chemical compound. You can think of valency as the number of "hands" or "arms" an atom has available for bonding.

Formulae of simple compounds

Binary compounds are the simplest compounds, made up of two different elements. To write their chemical formulas, follow these steps:

  1. Write the symbols of the constituent elements.
  2. Write their valencies below the symbols.
  3. Crossover the valencies of the combining atoms.
Example
Example Formula of hydrogen chloride

Given

  • Elements: Hydrogen (H) and Chlorine (Cl)
  • Valency of Hydrogen = 1
  • Valency of Chlorine = 1

To Find

Chemical formula of hydrogen chloride

Formula

Crossover valency method

Solution

H1Cl1\text{H}^1 \text{Cl}_1

By crossing over the valencies, we get H1Cl1\text{H}_1\text{Cl}_1.

Final Answer The formula of the compound would be HCl.

Example
Example Formula of hydrogen sulphide

Given

  • Elements: Hydrogen (H) and Sulphur (S)
  • Valency of Hydrogen = 1
  • Valency of Sulphur = 2

To Find

Chemical formula of hydrogen sulphide

Formula

Crossover valency method

Solution

H1S2\text{H}^1 \text{S}_2

By crossing over valencies, we get H2S1\text{H}_2\text{S}_1.

Final Answer Formula: H2S\text{H}_2\text{S}

Example
Example Formula of carbon tetrachloride

Given

  • Elements: Carbon (C) and Chlorine (Cl)
  • Valency of Carbon = 4
  • Valency of Chlorine = 1

To Find

Chemical formula of carbon tetrachloride

Formula

Crossover valency method

Solution

C4Cl1\text{C}^4 \text{Cl}_1

By crossing over valencies, we get C1Cl4\text{C}_1\text{Cl}_4.

Final Answer Formula : CCl4\text{CCl}_4

Example
Example Formula of magnesium chloride

Given

  • Ions: Magnesium (Mg2+\text{Mg}^{2+}) and Chloride (Cl−\text{Cl}^-)
  • Charge of Magnesium = +2
  • Charge of Chloride = -1

To Find

Chemical formula of magnesium chloride

Formula

Crossover valency method

Solution

Mg2+Cl−1\text{Mg}^{2+} \text{Cl}^{-1}

By crossing over charges, we get MgCl2\text{MgCl}_2.

Final Answer Formula : MgCl2\text{MgCl}_2

Example
Example Formula for aluminium oxide

Given

  • Ions: Aluminium (Al3+\text{Al}^{3+}) and Oxide (O2−\text{O}^{2-})
  • Charge of Aluminium = +3
  • Charge of Oxide = -2

To Find

Chemical formula of aluminium oxide

Formula

Crossover valency method

Solution

Al3+O2−\text{Al}^{3+} \text{O}^{2-}

By crossing over charges, we get Al2O3\text{Al}_2\text{O}_3.

Final Answer Formula : Al2O3\text{Al}_2\text{O}_3

Example
Example Formula for calcium oxide

Given

  • Ions: Calcium (Ca2+\text{Ca}^{2+}) and Oxide (O2−\text{O}^{2-})
  • Charge of Calcium = +2
  • Charge of Oxide = -2

To Find

Chemical formula of calcium oxide

Formula

Crossover valency method

Solution

Ca2+O2−\text{Ca}^{2+} \text{O}^{2-}

By crossing over charges, we get Ca2O2\text{Ca}_2\text{O}_2. Simplifying the formula, we get CaO.

Final Answer Formula : CaO

Example
Example Formula of sodium nitrate

Given

  • Ions: Sodium (Na+\text{Na}^{+}) and Nitrate (NO3−\text{NO}_3^{-})
  • Charge of Sodium = +1
  • Charge of Nitrate = -1

To Find

Chemical formula of sodium nitrate

Formula

Crossover valency method

Solution

Na+NO3−\text{Na}^{+} \text{NO}_3^{-}

By crossing over charges, we get NaNO3\text{NaNO}_3.

Final Answer Formula : NaNO3\text{NaNO}_3

Example
Example Formula of calcium hydroxide

Given

  • Ions: Calcium (Ca2+\text{Ca}^{2+}) and Hydroxide (OH−\text{OH}^{-})
  • Charge of Calcium = +2
  • Charge of Hydroxide = -1

To Find

Chemical formula of calcium hydroxide

Formula

Crossover valency method

Solution

Ca2+OH−\text{Ca}^{2+} \text{OH}^{-}

By crossing over charges, we get Ca(OH)2\text{Ca(OH)}_2.

Final Answer Formula : Ca(OH)2\text{Ca(OH)}_2

Example
Example Formula of sodium carbonate

Given

  • Ions: Sodium (Na+\text{Na}^{+}) and Carbonate (CO32−\text{CO}_3^{2-})
  • Charge of Sodium = +1
  • Charge of Carbonate = -2

To Find

Chemical formula of sodium carbonate

Formula

Crossover valency method

Solution

Na+CO32−\text{Na}^{+} \text{CO}_3^{2-}

By crossing over charges, we get Na2CO3\text{Na}_2\text{CO}_3.

Final Answer Formula : Na2CO3\text{Na}_2\text{CO}_3

Example
Example Formula of ammonium sulphate

Given

  • Ions: Ammonium (NH4+\text{NH}_4^{+}) and Sulphate (SO42−\text{SO}_4^{2-})
  • Charge of Ammonium = +1
  • Charge of Sulphate = -2

To Find

Chemical formula of ammonium sulphate

Formula

Crossover valency method

Solution

NH4+SO42−\text{NH}_4^{+} \text{SO}_4^{2-}

By crossing over charges, we get (NH4)2SO4\text{(NH}_4)_2\text{SO}_4.

Final Answer Formula : (NH4)2SO4\text{(NH}_4)_2\text{SO}_4

Example
Example Write down the formulae of (i) sodium oxide (ii) aluminium chloride (iii) sodium sulphide (iv) magnesium hydroxide

Given

Names of the compounds

To Find

(i) Chemical formula of sodium oxide (ii) Chemical formula of aluminium chloride (iii) Chemical formula of sodium sulphide (iv) Chemical formula of magnesium hydroxide

Formula

Crossover valency method

Solution

(i) Sodium Oxide Na+O2−→Na2O\text{Na}^+ \text{O}^{2-} \rightarrow \text{Na}_2\text{O}

Answer for part (i) = Na2O\text{Na}_2\text{O}

(ii) Aluminium Chloride Al3+Cl−→AlCl3\text{Al}^{3+} \text{Cl}^- \rightarrow \text{AlCl}_3

Answer for part (ii) = AlCl3\text{AlCl}_3

(iii) Sodium Sulphide Na+S2−→Na2S\text{Na}^+ \text{S}^{2-} \rightarrow \text{Na}_2\text{S}

Answer for part (iii) = Na2S\text{Na}_2\text{S}

(iv) Magnesium Hydroxide Mg2+OH−→Mg(OH)2\text{Mg}^{2+} \text{OH}^- \rightarrow \text{Mg(OH)}_2

Answer for part (iv) = Mg(OH)2\text{Mg(OH)}_2

Example
Example Write down the names of compounds represented by the following formulae: (i) Al2(SO4)3\text{Al}_2\text{(SO}_4)_3 (ii) CaCl2\text{CaCl}_2 (iii) K2SO4\text{K}_2\text{SO}_4 (iv) KNO3\text{KNO}_3 (v) CaCO3\text{CaCO}_3

Given

Chemical Formulae of the compounds

To Find

(i) Name of the compound Al2(SO4)3\text{Al}_2\text{(SO}_4)_3 (ii) Name of the compound CaCl2\text{CaCl}_2 (iii) Name of the compound K2SO4\text{K}_2\text{SO}_4 (iv) Name of the compound KNO3\text{KNO}_3 (v) Name of the compound CaCO3\text{CaCO}_3

Solution

(i) Al2(SO4)3\text{Al}_2\text{(SO}_4)_3 is Aluminium Sulphate

Answer for part (i) = Aluminium Sulphate

(ii) CaCl2\text{CaCl}_2 is Calcium Chloride

Answer for part (ii) = Calcium Chloride

(iii) K2SO4\text{K}_2\text{SO}_4 is Potassium Sulphate

Answer for part (iii) = Potassium Sulphate

(iv) KNO3\text{KNO}_3 is Potassium Nitrate

Answer for part (iv) = Potassium Nitrate

(v) CaCO3\text{CaCO}_3 is Calcium Carbonate

Answer for part (v) = Calcium Carbonate

Example
Example What is meant by the term chemical formula?

Given

Term chemical formula

To Find

Definition of chemical formula

Solution

The chemical formula of a compound is a symbolic representation of its composition. It indicates the elements present in the compound and the ratio in which their atoms combine.

Final Answer The chemical formula of a compound is a symbolic representation of its composition.

Example
Example How many atoms are present in a (i) H2S\text{H}_2\text{S} molecule and (ii) PO43−\text{PO}_4^{3-} ion?

Given

Chemical Formulae: H2S\text{H}_2\text{S} and PO43−\text{PO}_4^{3-}

To Find

(i) Number of atoms in H2S\text{H}_2\text{S} (ii) Number of atoms in PO43−\text{PO}_4^{3-}

Solution

(i) In H2S\text{H}_2\text{S}, there are 2 hydrogen atoms and 1 sulphur atom. Total atoms = 2+1=32 + 1 = 3

Answer for part (i) = 3 atoms

(ii) In PO43−\text{PO}_4^{3-}, there is 1 phosphorus atom and 4 oxygen atoms. Total atoms = 1+4=51 + 4 = 5

Answer for part (ii) = 5 atoms

Molecular Mass

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

Example
Example 3.1 (a) Calculate the relative molecular mass of water (H2O\text{H}_2\text{O}). (b) Calculate the molecular mass of HNO3\text{HNO}_3.

Given

Atomic masses: Hydrogen = 1 u1 \text{ u}, Oxygen = 16 u16 \text{ u}, Nitrogen = 14 u14 \text{ u}

To Find

(a) Molecular mass of H2O\text{H}_2\text{O} (b) Molecular mass of HNO3\text{HNO}_3

Formula

Molecular mass = Sum of atomic masses of all atoms in the molecule

Solution

(a) Molecular mass of H2O\text{H}_2\text{O} The molecule contains two atoms of hydrogen and one atom of oxygen.

2×Atomic mass of H+1×Atomic mass of O2 \times \text{Atomic mass of H} + 1 \times \text{Atomic mass of O}

=2×1 u+1×16 u=2 u+16 u=18 u= 2 \times 1 \text{ u} + 1 \times 16 \text{ u} = 2 \text{ u} + 16 \text{ u} = 18 \text{ u}

Answer for part (a) = 18 u18 \text{ u}

(b) Molecular mass of HNO3\text{HNO}_3 The molecule contains one atom of hydrogen, one atom of nitrogen, and three atoms of oxygen.

1×Atomic mass of H+1×Atomic mass of N+3×Atomic mass of O1 \times \text{Atomic mass of H} + 1 \times \text{Atomic mass of N} + 3 \times \text{Atomic mass of O}

=1 u+14 u+3×16 u=1 u+14 u+48 u=63 u= 1 \text{ u} + 14 \text{ u} + 3 \times 16 \text{ u} = 1 \text{ u} + 14 \text{ u} + 48 \text{ u} = 63 \text{ u}

Answer for part (b) = 63 u63 \text{ u}

Formula unit mass

The formula unit mass of a substance is the sum of the atomic masses of all atoms in a formula unit of a compound. It is calculated the same way as molecular mass. The term "formula unit" is used for substances whose constituent particles are ions. For example, sodium chloride (NaCl) has a formula unit mass of:

1×23 u+1×35.5 u=58.5 u1 \times 23 \text{ u} + 1 \times 35.5 \text{ u} = 58.5 \text{ u}

Example
Example 3.2 Calculate the formula unit mass of CaCl2\text{CaCl}_2.

Given

Atomic masses: Ca = 40 u40 \text{ u}, Cl = 35.5 u35.5 \text{ u}

To Find

Formula unit mass of CaCl2\text{CaCl}_2

Formula

Formula unit mass = Sum of atomic masses of all atoms in the formula unit

Solution

Atomic mass of Ca+(2×Atomic mass of Cl)\text{Atomic mass of Ca} + (2 \times \text{Atomic mass of Cl})

=40 u+(2×35.5 u)=40 u+71 u=111 u= 40 \text{ u} + (2 \times 35.5 \text{ u}) = 40 \text{ u} + 71 \text{ u} = 111 \text{ u}

Final Answer 111 u111 \text{ u}

Example
Example Calculate the molecular masses of H2,O2,Cl2,CO2,CH4,C2H6,C2H4,NH3,CH3OH\text{H}_2, \text{O}_2, \text{Cl}_2, \text{CO}_2, \text{CH}_4, \text{C}_2\text{H}_6, \text{C}_2\text{H}_4, \text{NH}_3, \text{CH}_3\text{OH}.

Given

Atomic masses: H = 1 u1 \text{ u}, O = 16 u16 \text{ u}, Cl = 35.5 u35.5 \text{ u}, C = 12 u12 \text{ u}, N = 14 u14 \text{ u}

To Find

Molecular masses of the given compounds

Formula

Molecular mass = Sum of atomic masses of all atoms in the molecule

Solution

(i) H2\text{H}_2

2×Atomic mass of H=2×1 u=2 u2 \times \text{Atomic mass of H} = 2 \times 1 \text{ u} = 2 \text{ u}

Answer for part (i) = 2 u2 \text{ u}

(ii) O2\text{O}_2

2×Atomic mass of O=2×16 u=32 u2 \times \text{Atomic mass of O} = 2 \times 16 \text{ u} = 32 \text{ u}

Answer for part (ii) = 32 u32 \text{ u}

(iii) Cl2\text{Cl}_2

2×Atomic mass of Cl=2×35.5 u=71 u2 \times \text{Atomic mass of Cl} = 2 \times 35.5 \text{ u} = 71 \text{ u}

Answer for part (iii) = 71 u71 \text{ u}

(iv) CO2\text{CO}_2

Atomic mass of C+2×Atomic mass of O=12 u+2×16 u=12 u+32 u=44 u\text{Atomic mass of C} + 2 \times \text{Atomic mass of O} = 12 \text{ u} + 2 \times 16 \text{ u} = 12 \text{ u} + 32 \text{ u} = 44 \text{ u}

Answer for part (iv) = 44 u44 \text{ u}

(v) CH4\text{CH}_4

Atomic mass of C+4×Atomic mass of H=12 u+4×1 u=12 u+4 u=16 u\text{Atomic mass of C} + 4 \times \text{Atomic mass of H} = 12 \text{ u} + 4 \times 1 \text{ u} = 12 \text{ u} + 4 \text{ u} = 16 \text{ u}

Answer for part (v) = 16 u16 \text{ u}

(vi) C2H6\text{C}_2\text{H}_6

2×Atomic mass of C+6×Atomic mass of H=2×12 u+6×1 u=24 u+6 u=30 u2 \times \text{Atomic mass of C} + 6 \times \text{Atomic mass of H} = 2 \times 12 \text{ u} + 6 \times 1 \text{ u} = 24 \text{ u} + 6 \text{ u} = 30 \text{ u}

Answer for part (vi) = 30 u30 \text{ u}

(vii) C2H4\text{C}_2\text{H}_4

2×Atomic mass of C+4×Atomic mass of H=2×12 u+4×1 u=24 u+4 u=28 u2 \times \text{Atomic mass of C} + 4 \times \text{Atomic mass of H} = 2 \times 12 \text{ u} + 4 \times 1 \text{ u} = 24 \text{ u} + 4 \text{ u} = 28 \text{ u}

Answer for part (vii) = 28 u28 \text{ u}

(viii) NH3\text{NH}_3

Atomic mass of N+3×Atomic mass of H=14 u+3×1 u=14 u+3 u=17 u\text{Atomic mass of N} + 3 \times \text{Atomic mass of H} = 14 \text{ u} + 3 \times 1 \text{ u} = 14 \text{ u} + 3 \text{ u} = 17 \text{ u}

Answer for part (viii) = 17 u17 \text{ u}

(ix) CH3OH\text{CH}_3\text{OH}

Atomic mass of C+4×Atomic mass of H+Atomic mass of O=12 u+4×1 u+16 u=12 u+4 u+16 u=32 u\text{Atomic mass of C} + 4 \times \text{Atomic mass of H} + \text{Atomic mass of O} = 12 \text{ u} + 4 \times 1 \text{ u} + 16 \text{ u} = 12 \text{ u} + 4 \text{ u} + 16 \text{ u} = 32 \text{ u}

Answer for part (ix) = 32 u32 \text{ u}

Example
Example Calculate the formula unit masses of ZnO,Na2O,K2CO3\text{ZnO}, \text{Na}_2\text{O}, \text{K}_2\text{CO}_3, given atomic masses of Zn = 65 u65 \text{ u}, Na = 23 u23 \text{ u}, K = 39 u39 \text{ u}, C = 12 u12 \text{ u}, and O = 16 u16 \text{ u}.

Given

Atomic masses: Zn = 65 u65 \text{ u}, Na = 23 u23 \text{ u}, K = 39 u39 \text{ u}, C = 12 u12 \text{ u}, O = 16 u16 \text{ u}

To Find

Formula unit masses of the given compounds

Formula

Formula unit mass = Sum of atomic masses of all atoms in the formula unit

Solution

(i) ZnO\text{ZnO}

Atomic mass of Zn+Atomic mass of O=65 u+16 u=81 u\text{Atomic mass of Zn} + \text{Atomic mass of O} = 65 \text{ u} + 16 \text{ u} = 81 \text{ u}

Answer for part (i) = 81 u81 \text{ u}

(ii) Na2O\text{Na}_2\text{O}

2×Atomic mass of Na+Atomic mass of O=2×23 u+16 u=46 u+16 u=62 u2 \times \text{Atomic mass of Na} + \text{Atomic mass of O} = 2 \times 23 \text{ u} + 16 \text{ u} = 46 \text{ u} + 16 \text{ u} = 62 \text{ u}

Answer for part (ii) = 62 u62 \text{ u}

(iii) K2CO3\text{K}_2\text{CO}_3

2×Atomic mass of K+Atomic mass of C+3×Atomic mass of O=2×39 u+12 u+3×16 u=78 u+12 u+48 u=138 u2 \times \text{Atomic mass of K} + \text{Atomic mass of C} + 3 \times \text{Atomic mass of O} = 2 \times 39 \text{ u} + 12 \text{ u} + 3 \times 16 \text{ u} = 78 \text{ u} + 12 \text{ u} + 48 \text{ u} = 138 \text{ u}

Answer for part (iii) = 138 u138 \text{ u}

Exercises

Example
Example A 0.24 g0.24 \text{ g} sample of compound of oxygen and boron was found by analysis to contain 0.096 g0.096 \text{ g} of boron and 0.144 g0.144 \text{ g} of oxygen. Calculate the percentage composition of the compound by weight.

Given

  • Mass of compound = 0.24 g0.24 \text{ g}
  • Mass of boron = 0.096 g0.096 \text{ g}
  • Mass of oxygen = 0.144 g0.144 \text{ g}

To Find

Percentage composition of boron and oxygen by weight.

Formula

Percentage of element=Mass of elementMass of compound×100\text{Percentage of element} = \frac{\text{Mass of element}}{\text{Mass of compound}} \times 100

Solution

(i) Percentage of Boron

Percentage of Boron=0.096 g0.24 g×100=40%\text{Percentage of Boron} = \frac{0.096 \text{ g}}{0.24 \text{ g}} \times 100 = 40\%

Answer for part (i) = 40%40\%

(ii) Percentage of Oxygen

Percentage of Oxygen=0.144 g0.24 g×100=60%\text{Percentage of Oxygen} = \frac{0.144 \text{ g}}{0.24 \text{ g}} \times 100 = 60\%

Answer for part (ii) = 60%60\%

Example
Example When 3.0 g3.0 \text{ g} of carbon is burnt in 8.00 g8.00 \text{ g}

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