Some Basic Concepts of Chemistry

In this chapter: classification of matter, SI units, mass, volume, density and temperature, scientific notation, precision and accuracy, significant figures, dimensional analysis, laws of chemical combination, Dalton's atomic theory, atomic and molecular masses, the mole, percentage composition, empirical and molecular formulae, stoichiometry and limiting reagent, and concentration of solutions.

Matter and its classification

Anything that has mass and occupies space is matter. It exists as solid (definite shape and volume), liquid (definite volume, no definite shape) or gas (neither). The three states can be interconverted by changing temperature and pressure.

• Mixtures contain two or more substances in any ratio. A homogeneous mixture (sugar solution, air) has uniform composition throughout; a heterogeneous mixture (salt and sugar, grains with pulses) does not. Components can be separated by physical methods such as filtration, crystallisation and distillation.
• Pure substances have fixed composition. An element contains only one kind of atom (Na, Cu, or molecules like H₂, O₂). A compound is formed when atoms of different elements combine in a fixed ratio; its properties differ from those of its elements, and it can be broken down only by chemical methods.

Measurement and SI units

The International System of Units (SI) has seven base units. Every other unit is derived from these.

• Mass and weight: mass is the amount of matter and stays constant; weight is the force of gravity on it and can change from place to place. A balance measures mass.
• Volume: SI unit m³. In the lab we use litre: 1 L = 1 dm³ = 1000 cm³, and 1 mL = 1 cm³.
• Density = mass/volume. SI unit kg m⁻³; g cm⁻³ is more common in chemistry.
• Temperature: K = °C + 273.15 and °F = 95(°C) + 32. Values below zero are possible on the Celsius and Fahrenheit scales, never on the Kelvin scale.

Uncertainty in measurement

Very large and very small numbers are written in scientific notation, N × 10ⁿ, with N between 1 and 10. So 0.00016 is 1.6 × 10⁻⁴ and 232.508 is 2.32508 × 10².

Precision is how close repeated measurements are to each other. Accuracy is how close they are to the true value. A student who gets 1.94, 1.95, 1.96 g for a true mass of 2.00 g is precise but not accurate.

Significant figures

Significant figures are the meaningful digits in a measurement: all the certain digits plus the last, uncertain one.

• All non-zero digits are significant: 285 cm has three.
• Zeros before the first non-zero digit are not significant: 0.03 has one, 0.0052 has two.
• Zeros between non-zero digits are significant: 2.005 has four.
• Zeros at the end are significant only if there is a decimal point: 0.200 has three, 100 has one, but 100. has three and 100.0 has four.
• Counted numbers (2 balls, 20 eggs) and exact defined numbers have infinite significant figures.

In addition and subtraction, keep as many decimal places as the number with the fewest decimal places: 12.11 + 18.0 + 1.012 = 31.122, reported as 31.1. In multiplication and division, keep as many significant figures as the least precise number: 2.5 × 1.25 = 3.125, reported as 3.1.

Rounding off: if the digit to be dropped is more than 5, raise the previous digit by one (1.386 to 1.39); if less than 5, leave it (4.334 to 4.33). If it is exactly 5, leave the previous digit if it is even and raise it if it is odd (6.35 becomes 6.4, 6.25 becomes 6.2).

Dimensional analysis

To change units, multiply by a unit factor that equals one, such as 1 in / 2.54 cm. Units cancel like numbers. For 3 in: 3 in × (2.54 cm / 1 in) = 7.62 cm. Write the units at every step and you will rarely make a conversion mistake.

Laws of chemical combination

Dalton's atomic theory

• Matter consists of indivisible atoms.
• All atoms of a given element have identical properties, including identical mass. Atoms of different elements differ in mass.
• Compounds are formed when atoms of different elements combine in a fixed ratio.
• Chemical reactions involve reorganisation of atoms. Atoms are neither created nor destroyed in a chemical reaction.

The theory explained the laws of conservation of mass, definite proportions and multiple proportions, but it could not explain the law of gaseous volumes.

Atomic, molecular and formula mass

The atomic mass unit is defined as exactly one-twelfth of the mass of one carbon-12 atom: 1 u = 1.66056 × 10⁻²⁴ g. On this scale hydrogen is 1.0078 u and oxygen 15.995 u.

Most elements occur as a mix of isotopes, so the atomic mass used in calculations is an average atomic mass, weighted by natural abundance. For chlorine: ³⁵Cl (34.9689 u, 75.77%) and ³⁷Cl (36.9659 u, 24.23%) give 34.9689 × 0.7577 + 36.9659 × 0.2423 = 35.45 u.

Molecular mass is the sum of the atomic masses of the atoms in a molecule. Glucose, C₆H₁₂O₆: 6(12.011) + 12(1.008) + 6(16.00) = 180.162 u. Ionic solids like NaCl have no discrete molecules, so we use formula mass: 23.0 + 35.5 = 58.5 u.

The mole

One mole contains exactly 6.02214076 × 10²³ elementary entities (atoms, molecules, ions or electrons). This number is the Avogadro constant, N_A, and its unit is mol⁻¹. The mass of one mole of a substance in grams is its molar mass, numerically equal to its atomic, molecular or formula mass in u. So the molar mass of water is 18.02 g mol⁻¹ and one mole of water weighs 18.02 g.

Almost every problem in this chapter is a conversion through moles. Convert whatever you are given into moles first, use the balanced equation (which gives mole ratios), and convert back at the end.

Percentage composition, empirical and molecular formula

The empirical formula gives the simplest whole-number ratio of atoms; the molecular formula gives the actual number of each kind of atom. Molecular formula = n × empirical formula, where n = molar mass ÷ empirical formula mass.

The method: take 100 g of compound so that percentages become grams, divide each by the atomic mass to get moles, divide all by the smallest, and round to whole numbers (multiply through if you get something like 1.5).

Worked example: A compound contains 40.0% C, 6.7% H and 53.3% O. Its molar mass is 180 g mol⁻¹. Find its empirical and molecular formula.
Solution: In 100 g: C = 40.0/12 = 3.33 mol, H = 6.7/1 = 6.7 mol, O = 53.3/16 = 3.33 mol. Dividing by 3.33 gives C : H : O = 1 : 2 : 1, so the empirical formula is CH₂O (formula mass 12 + 2 + 16 = 30). n = 180/30 = 6, so the molecular formula is C₆H₁₂O₆.

Stoichiometry and the limiting reagent

A balanced equation gives the ratio of moles (and, for gases, volumes) in which substances react. For the combustion of methane,

Balance an equation by counting atoms of each element on both sides; balance the elements that appear in only one compound on each side first and leave H and O for the end. Never change a formula to balance an equation, only the coefficients.

When reactants are not taken in the exact ratio, the one that gets used up first is the limiting reagent. It decides how much product forms; the other reactant is left in excess.

Worked example: 3.0 g of H₂ reacts with 29.0 g of O₂ to form water. Which is the limiting reagent, how much water forms, and how much of the excess reagent is left? (H = 1, O = 16)
Solution: 2H₂ + O₂ → 2H₂O. Moles of H₂ = 3.0/2 = 1.5 mol; moles of O₂ = 29.0/32 = 0.906 mol. 1.5 mol H₂ needs only 0.75 mol O₂, so H₂ is the limiting reagent. Water formed = 1.5 mol = 1.5 × 18 = 27.0 g. O₂ used = 0.75 × 32 = 24.0 g, so 29.0 − 24.0 = 5.0 g O₂ is left over.

Concentration of solutions

Molarity depends on temperature because volume expands or contracts on heating and cooling; mass does not. On dilution the moles of solute stay the same, so M₁V₁ = M₂V₂.

Worked example: 4.0 g of NaOH is dissolved in water and the solution is made up to 250 mL. Find its molarity.
Solution: Molar mass of NaOH = 23 + 16 + 1 = 40 g mol⁻¹, so moles = 4.0/40 = 0.10 mol. Volume = 0.250 L. Molarity = 0.10/0.250 = 0.40 mol L⁻¹ (0.40 M).

Common mistakes: (1) Using volume of solvent instead of volume of solution for molarity, or mass of solution instead of mass of solvent for molality. (2) Forgetting to convert mL to L and g to kg. (3) Counting trailing zeros in 100 or 2500 as significant when there is no decimal point. (4) Deciding the limiting reagent by comparing masses directly; compare moles, divided by their coefficients in the balanced equation. (5) Using 22.4 L mol⁻¹ when the question defines STP as 1 bar.

JEE and NEET focus

• Mole conversions between mass, number of particles and gas volume, including counting atoms inside molecules (atoms in 0.5 mol CO₂ is 1.5 N_A).
• Limiting reagent and percentage yield problems from a balanced equation.
• Empirical and molecular formula from percentage composition.
• Molarity, molality and mole fraction, and converting one into another using the density of the solution.
• Significant figures in results of arithmetic operations, and the rounding rule for a dropped 5.
• Examples of each law of chemical combination, especially multiple proportions and gaseous volumes.

Practice questions