Some Basic Concepts of Chemistry
Laws of chemical combination, mole concept, stoichiometry & concentration terms
Laws of Chemical Combination and the Mole Concept
Matter and the Laws of Chemical CombinationTopic 1
Chemistry studies matter — anything that has mass and occupies space — and the changes it undergoes. Matter is classified physically into solids, liquids and gases, and chemically into pure substances (elements and compounds, with fixed composition) and mixtures (variable composition, separable by physical means). A compound, unlike a mixture, has its constituents combined in a fixed ratio with entirely new properties, an idea NEET often tests through examples.
The quantitative foundation of chemistry rests on the laws of chemical combination. The law of conservation of mass states that mass is neither created nor destroyed in a chemical reaction, so the total mass of reactants equals the total mass of products. The law of definite proportions states that a given compound always contains the same elements in the same fixed proportion by mass — pure water is always $11.1\%$ hydrogen and $88.9\%$ oxygen, whatever its source.
The law of multiple proportions states that when two elements form more than one compound, the masses of one element combining with a fixed mass of the other are in a simple whole-number ratio — as in $\text{CO}$ and $\text{CO}_2$, where the oxygen masses per fixed carbon are in the ratio $1:2$. Gay-Lussac's law of gaseous volumes adds that gases react in simple whole-number volume ratios, and Avogadro's law states that equal volumes of gases at the same temperature and pressure contain equal numbers of molecules.
These laws were unified by Dalton's atomic theory, which proposed that matter is made of indivisible atoms, that atoms of an element are identical in mass, and that compounds form when atoms combine in fixed whole-number ratios. Although later refined (atoms are divisible, isotopes exist), Dalton's theory remains the conceptual basis for stoichiometry and is a frequent NEET starting point.
| Law | Statement |
|---|---|
| Conservation of mass | mass of reactants = mass of products |
| Definite proportions | fixed mass ratio in a compound |
| Multiple proportions | simple whole-number mass ratios |
| Avogadro's law | equal volumes to equal molecules |
Carbon forms two oxides: $\text{CO}$ (with $1.33\ \text{g}$ oxygen per g carbon) and $\text{CO}_2$ (with $2.66\ \text{g}$ oxygen per g carbon). Which law does this illustrate?
Show solution
The masses of oxygen per fixed mass of carbon are in the ratio $1.33 : 2.66 = 1 : 2$, a simple whole-number ratio — this is the law of multiple proportions.
When $4\ \text{g}$ of hydrogen reacts completely with $32\ \text{g}$ of oxygen, what mass of water is formed?
Show solution
By the law of conservation of mass, mass of products = mass of reactants $= 4 + 32 = 36\ \text{g}$ of water.
The law of conservation of mass states that in a reaction, mass is:
$\text{CO}$ and $\text{CO}_2$ illustrate the law of:
Equal volumes of gases at the same T and P contain equal numbers of molecules. This is:
A pure compound always has its elements in a:
Which statement of Dalton's theory was later found incorrect?
NEET tip: Distinguish the laws by their key phrase — conservation (mass equal), definite (fixed ratio), multiple (whole-number ratios), Avogadro (equal volumes to equal molecules). These underpin all stoichiometry.
The Mole Concept and Molar MassTopic 2
Atoms and molecules are far too small to count individually, so chemists group them into a convenient unit called the mole. One mole is the amount of substance containing exactly Avogadro's number, $N_A = 6.022\times10^{23}$, of elementary entities — atoms, molecules, ions or electrons. The mole is to chemistry what a dozen is to eggs, just on an enormous scale, and it is the single most important quantity in this chapter.
The mass of one mole of a substance is its molar mass, numerically equal to its atomic or molecular mass but expressed in grams per mole. So one mole of carbon atoms has a mass of $12\ \text{g}$, and one mole of water has a mass of $18\ \text{g}$. The number of moles is found from $n = \dfrac{\text{given mass}}{\text{molar mass}}$, the workhorse relation of mole calculations.
Three quantities are linked through the mole: mass, number of particles, and (for gases) volume. The number of particles is $N = n\times N_A$, and at standard temperature and pressure (STP) one mole of any ideal gas occupies the molar volume of $22.4\ \text{L}$. Being able to convert quickly between grams, moles, molecules and litres is exactly what NEET numericals demand.
A subtle but examinable point is the difference between atomic mass, molecular mass and formula mass. Atomic mass is for a single atom; molecular mass is the sum for a molecule (e.g. $\text{H}_2\text{O} = 18\ \text{u}$); and formula mass is used for ionic compounds like $\text{NaCl}$, which do not exist as discrete molecules. Atomic masses themselves are averages weighted by isotopic abundance, which is why many are not whole numbers.
| Relation | Expression |
|---|---|
| Moles from mass | $n = \dfrac{\text{mass}}{\text{molar mass}}$ |
| Number of particles | $N = n\times N_A$ |
| Avogadro's number | $N_A = 6.022\times10^{23}$ |
| Molar volume (STP) | $22.4\ \text{L/mol}$ |
How many moles are present in $36\ \text{g}$ of water ($\text{H}_2\text{O}$, molar mass $18\ \text{g/mol}$)?
Show solution
$n = \dfrac{\text{mass}}{\text{molar mass}} = \dfrac{36}{18} = 2\ \text{mol}$.
How many molecules are there in $0.5\ \text{mol}$ of $\text{CO}_2$?
Show solution
$N = n\times N_A = 0.5\times(6.022\times10^{23}) = 3.011\times10^{23}$ molecules.
Avogadro's number is:
The number of moles equals:
The molar volume of an ideal gas at STP is:
One mole of carbon-12 atoms has a mass of:
Formula mass (rather than molecular mass) is used for:
NEET tip: Master the conversions: mass $\leftrightarrow$ moles ($n = $ mass/molar mass), moles $\leftrightarrow$ particles ($\times N_A$), and moles $\leftrightarrow$ volume ($\times 22.4\ \text{L}$ at STP).
Stoichiometry and Concentration
Empirical Formula, Molecular Formula and StoichiometryTopic 3
The composition of a compound is described by its formulae. The empirical formula gives the simplest whole-number ratio of atoms, while the molecular formula gives the actual number of atoms in a molecule. The two are related by a whole number $n$: molecular formula $= n\times$ empirical formula, where $n = \dfrac{\text{molecular mass}}{\text{empirical formula mass}}$. For example, the empirical formula of glucose is $\text{CH}_2\text{O}$ and its molecular formula $\text{C}_6\text{H}_{12}\text{O}_6$ ($n = 6$).
To find an empirical formula from percentage composition, divide each element's percentage by its atomic mass to get a mole ratio, then divide through by the smallest value to obtain the simplest whole-number ratio. This standard procedure — percent to moles to ratio — is a guaranteed NEET calculation type, so the method should be automatic.
Stoichiometry is the quantitative study of reactants and products in a balanced equation. The balanced equation gives the mole ratios in which substances react; these ratios are then used with $n = \text{mass}/\text{molar mass}$ to relate masses or volumes. The golden rule is to always work in moles, never directly in grams, when applying the coefficients.
In real reactions the reactants are rarely present in exact stoichiometric amounts, so one runs out first — the limiting reagent — and it decides how much product forms. The other reactant, present in excess, is left over. Identifying the limiting reagent (by comparing available moles against the required ratio) before calculating product is the most common stoichiometry pitfall NEET sets up, so always check it.
| Step | Action |
|---|---|
| Empirical formula | % to moles to simplest ratio |
| Molecular formula | $n\times$ empirical; $n = \dfrac{M}{\text{EFM}}$ |
| Stoichiometry | work in moles via the balanced equation |
| Limiting reagent | runs out first, decides product |
A compound contains $40\%$ carbon, $6.7\%$ hydrogen and $53.3\%$ oxygen by mass. Find its empirical formula.
Show solution
Moles: $\text{C} = 40/12 = 3.33$, $\text{H} = 6.7/1 = 6.7$, $\text{O} = 53.3/16 = 3.33$. Dividing by the smallest ($3.33$): $\text{C}:\text{H}:\text{O} = 1:2:1$, so the empirical formula is $\text{CH}_2\text{O}$.
In $2\text{H}_2 + \text{O}_2 \rightarrow 2\text{H}_2\text{O}$, if $4\ \text{mol}$ of $\text{H}_2$ react with $1\ \text{mol}$ of $\text{O}_2$, which is the limiting reagent?
Show solution
The ratio needed is $2:1$. For $1\ \text{mol}\ \text{O}_2$ only $2\ \text{mol}\ \text{H}_2$ are required, but $4\ \text{mol}$ are available. So $\text{O}_2$ runs out first — $\text{O}_2$ is the limiting reagent.
The simplest whole-number ratio of atoms in a compound is the:
The empirical formula of glucose ($\text{C}_6\text{H}_{12}\text{O}_6$) is:
In stoichiometric calculations one should work in:
The reactant that is completely consumed first is the:
Molecular formula equals empirical formula multiplied by:
NEET tip: For empirical formula: percent to moles to ratio. Always identify the limiting reagent (compare available moles with the equation's ratio) before computing product mass.
Concentration Terms and Significant FiguresTopic 4
The composition of a solution is expressed through several concentration terms, and NEET expects you to know each definition and when it is temperature-independent. Molarity ($M$) is moles of solute per litre of solution, $M = \dfrac{\text{moles of solute}}{\text{volume of solution in L}}$. It is the most common term but depends on temperature, because volume changes with temperature.
Molality ($m$) is moles of solute per kilogram of solvent, $m = \dfrac{\text{moles of solute}}{\text{mass of solvent in kg}}$. Because it is based on mass, molality is independent of temperature — a frequently tested distinction from molarity. Mole fraction ($x$) is the ratio of the moles of one component to the total moles, and the mole fractions of all components add up to one.
Two more terms appear for dilute or trace solutions. Mass percentage is the mass of solute per $100\ \text{g}$ of solution, and parts per million (ppm) expresses very low concentrations as parts of solute per million parts of solution — used for pollutants and trace impurities. A useful skill is the dilution relation $M_1V_1 = M_2V_2$, which keeps the moles of solute constant when a solution is diluted.
Finally, measurement precision is conveyed by significant figures. All non-zero digits are significant; leading zeros are not; trailing zeros after a decimal point are. In multiplication and division the answer carries as many significant figures as the least precise value, while in addition and subtraction it is fixed by the least number of decimal places. Reporting answers to the correct number of significant figures, and using consistent units via dimensional analysis, is part of scoring full marks in NEET numericals.
| Term | Definition |
|---|---|
| Molarity $M$ | moles solute / L solution (T-dependent) |
| Molality $m$ | moles solute / kg solvent (T-independent) |
| Mole fraction $x$ | moles of one / total moles |
| Dilution | $M_1V_1 = M_2V_2$ |
Find the molarity of a solution containing $0.5\ \text{mol}$ of $\text{NaCl}$ in $250\ \text{mL}$ of solution.
Show solution
$M = \dfrac{\text{moles}}{\text{volume in L}} = \dfrac{0.5}{0.25} = 2\ \text{M}$.
To what volume must $100\ \text{mL}$ of $2\ \text{M}$ HCl be diluted to make it $0.5\ \text{M}$?
Show solution
$M_1V_1 = M_2V_2$, so $V_2 = \dfrac{M_1V_1}{M_2} = \dfrac{2\times100}{0.5} = 400\ \text{mL}$.
Molarity is defined as moles of solute per:
Which concentration term is independent of temperature?
The sum of the mole fractions of all components in a solution is:
On dilution of a solution, the quantity that stays constant is the:
In multiplication, the result is reported to the significant figures of the:
NEET tip: Molarity is per litre of solution (T-dependent); molality is per kg of solvent (T-independent). Use $M_1V_1 = M_2V_2$ for dilution and report answers to the correct significant figures.
Quick Revision — Some Basic Concepts of Chemistry
- Laws: conservation of mass; definite proportions (fixed mass ratio); multiple proportions (whole-number ratios, e.g. $\text{CO}$, $\text{CO}_2$); Avogadro (equal volumes to equal molecules).
- Mole: $N_A = 6.022\times10^{23}$ entities. $n = \dfrac{\text{mass}}{\text{molar mass}}$; $N = n\times N_A$; $1\ \text{mol}$ gas $= 22.4\ \text{L}$ at STP.
- Formulae: empirical = simplest ratio (percent to moles to ratio); molecular $= n\times$ empirical.
- Limiting reagent is consumed first and fixes the product amount — always check it.
- Molarity (mol/L solution, temperature-dependent) vs molality (mol/kg solvent, temperature-independent); dilution $M_1V_1 = M_2V_2$.
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