Some Basic Concepts of Chemistry • Topic 1 of 3

Laws of Chemical Combination & Mole Concept

Chemistry is the study of matter — anything that has mass and occupies space. Matter is first classified physically into solids, liquids and gases, and chemically into pure substances (elements and compounds) and mixtures (homogeneous and heterogeneous). An element contains only one kind of atom; a compound contains two or more elements combined in a fixed ratio by mass.

The way elements combine is governed by the laws of chemical combination, the experimental foundations of modern chemistry:

  • Law of Conservation of Mass (Lavoisier): mass is neither created nor destroyed in a chemical reaction. The total mass of reactants equals the total mass of products.
  • Law of Definite (Constant) Proportions (Proust): a given compound always contains the same elements in the same fixed proportion by mass — e.g. pure water always has H and O in the mass ratio $1:8$.
  • Law of Multiple Proportions (Dalton): when two elements form more than one compound, the masses of one element that combine with a fixed mass of the other are in small whole-number ratios. For $CO$ and $CO_2$, the masses of oxygen per $12\,g$ of carbon are $16\,g$ and $32\,g$, a ratio of $1:2$.
  • Gay-Lussac's Law of Combining Volumes: gases react in volume ratios that are simple whole numbers (at constant $T$ and $P$).
  • Avogadro's Law: equal volumes of all gases at the same temperature and pressure contain equal numbers of molecules.

To count the enormous numbers of atoms in any real sample, chemists use the mole. One mole is the amount of substance containing exactly $N_A = 6.022\times10^{23}$ entities (atoms, molecules or ions) — the Avogadro constant. The molar mass (in $g\,mol^{-1}$) is numerically equal to the atomic or molecular mass expressed in unified atomic mass units (u). Thus $1\,mol$ of carbon-12 weighs exactly $12\,g$, and $1\,mol$ of water weighs $18\,g$.

The mole links three measurable quantities: mass $\left(n=\dfrac{m}{M}\right)$, number of particles $\left(N=n\times N_A\right)$, and for gases at STP, volume $\left(V=n\times 22.4\,L\right)$. Atomic masses themselves are relative values measured on the $^{12}C$ scale, where one atom of carbon-12 is assigned exactly $12\,u$.

The Mole Bridge: mass, moles, particles and gas volumeMass (g)weigh sampleMOLES (n)central hubParticles (N)atoms / moleculesVolume (L)gas at STPdivide by Mx N_Ax 22.4
Solved Examples
1
Worked Example
Calculate the number of moles in $36\,g$ of water ($H_2O$).
Solution
  1. Molar mass of $H_2O = (2\times1) + 16 = 18\,g\,mol^{-1}$.
  2. Moles $n = \dfrac{m}{M} = \dfrac{36}{18}$.
  3. $n = 2\,mol$.

Answer: $2\,mol$ of water.

2
Worked Example
How many molecules are present in $8\,g$ of oxygen gas ($O_2$)?
Solution
  1. Molar mass of $O_2 = 32\,g\,mol^{-1}$.
  2. Moles $n = \dfrac{8}{32} = 0.25\,mol$.
  3. Number of molecules $N = n\times N_A = 0.25\times 6.022\times10^{23}$.
  4. $N = 1.5055\times10^{23}$ molecules.

Answer: $\approx 1.51\times10^{23}$ molecules.

3
Worked Example
Carbon forms two oxides: $CO$ (containing $42.9\%$ C) and $CO_2$ (containing $27.3\%$ C). Show that this data follows the law of multiple proportions.
Solution
  1. In $CO$: per $42.9\,g$ C there is $57.1\,g$ O, so O per $1\,g$ C $= 57.1/42.9 = 1.33$.
  2. In $CO_2$: per $27.3\,g$ C there is $72.7\,g$ O, so O per $1\,g$ C $= 72.7/27.3 = 2.66$.
  3. Ratio of oxygen masses for fixed carbon $= 1.33 : 2.66 = 1 : 2$.

Answer: The ratio $1:2$ is a simple whole number, confirming the law of multiple proportions.

4
Worked Example
What is the mass of $0.5\,mol$ of sodium carbonate, $Na_2CO_3$?
Solution
  1. Molar mass $= (2\times23) + 12 + (3\times16) = 46 + 12 + 48 = 106\,g\,mol^{-1}$.
  2. Mass $m = n\times M = 0.5\times106$.
  3. $m = 53\,g$.

Answer: $53\,g$.

5
Worked Example
Calculate the number of atoms present in $1\,mol$ of $CaCO_3$.
Solution
  1. One formula unit of $CaCO_3$ has $1 + 1 + 3 = 5$ atoms.
  2. $1\,mol$ contains $N_A = 6.022\times10^{23}$ formula units.
  3. Total atoms $= 5\times 6.022\times10^{23} = 3.011\times10^{24}$.

Answer: $3.011\times10^{24}$ atoms.

6
Worked Example
What volume (at STP) is occupied by $11\,g$ of carbon dioxide gas, $CO_2$?
Solution
  1. Molar mass of $CO_2 = 12 + 32 = 44\,g\,mol^{-1}$.
  2. Moles $n = \dfrac{11}{44} = 0.25\,mol$.
  3. At STP, $V = n\times 22.4 = 0.25\times 22.4$.
  4. $V = 5.6\,L$.

Answer: $5.6\,L$ at STP.

Key Points

  • Law of conservation of mass: total mass of reactants = total mass of products.
  • Definite proportions fix the mass ratio of elements in a compound; multiple proportions give small whole-number mass ratios across different compounds.
  • Avogadro's law: equal volumes of gases at same T and P contain equal numbers of molecules.
  • One mole contains N_A = 6.022 x 10^23 entities; molar mass in g/mol equals the atomic/molecular mass in u.
  • Key conversions: n = m/M, N = n x N_A, and V = n x 22.4 L for a gas at STP.
Quick Check — 4 MCQs
Tap an option to check your answer0 / 4
Q1.The number of atoms in $0.1\,mol$ of a triatomic gas (e.g. $O_3$) is:
Explanation: Molecules $= 0.1\times N_A = 6.022\times10^{22}$; atoms $= 3\times$ that $= 1.806\times10^{23}$.
Q2.Which law is illustrated by the fact that pure water from any source always contains H and O in the mass ratio 1:8?
Explanation: A fixed mass composition of a single compound is the law of definite (constant) proportions.
Q3.The mass of $0.25\,mol$ of $H_2SO_4$ is:
Explanation: Molar mass of $H_2SO_4 = 98\,g\,mol^{-1}$; mass $= 0.25\times98 = 24.5\,g$.
Q4.Equal volumes of $H_2$ and $O_2$ at the same temperature and pressure contain:
Explanation: By Avogadro's law, equal volumes (same T, P) contain equal numbers of molecules — not equal masses or atoms.
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