Fundamental Concepts of Chemistry notes

Atomic Structure, Atomic Number and Mass Number

An atom contains protons and neutrons in its nucleus, while electrons move around the nucleus. The atomic number identifies an element and is equal to the number of protons in one atom of that element.

The mass number is the total number of protons and neutrons. In a neutral atom, the number of electrons is equal to the number of protons. Isotopes have the same atomic number but different mass numbers.

A mass spectrometer is used to determine the relative masses of isotopes and their relative abundance. These data are used to calculate the relative atomic mass of an element.

• Atomic number, Z = number of protons.
• For a neutral atom: number of protons = number of electrons.
• Mass number, A = number of protons + number of neutrons.
• Number of neutrons = mass number minus atomic number.
• For an atom with mass number 23 and atomic number 11: protons = 11, electrons = 11 and neutrons = 12.
• Isotopes have the same number of protons but different numbers of neutrons.
• Mass spectrometer determines isotopic masses and relative abundance.

Mole and Avogadro's Number

A mole is the amount of a substance that contains the same number of elementary entities as the number of atoms present in 0.012 kg of carbon-12. This number is called Avogadro's number, represented by NA.

The elementary entities may be atoms, molecules, ions, formula units or electrons. The type of entity must always be stated or understood from the formula.

One mole of carbon atoms, one mole of water molecules and one mole of sodium ions contain the same number of entities, but their masses are different.

• One mole contains 6.022 × 10^23 elementary entities.
• The standard definition refers to the atoms present in 0.012 kg, or 12 g, of carbon-12.
• 1 mol of Na contains 6.022 × 10^23 sodium atoms.
• 1 mol of H2O contains 6.022 × 10^23 water molecules.
• 1 mol of Cl2 contains 6.022 × 10^23 chlorine molecules and 2 × 6.022 × 10^23 chlorine atoms.
• One mole of diamond and one mole of gold contain the same number of atoms, but their masses are different.
• Number of particles = number of moles × Avogadro's number.
• Number of moles = number of particles ÷ Avogadro's number.

Molar Mass, Moles and Number of Particles

Molar mass is the mass of one mole of a substance. Its unit is g mol−1. Numerically, molar mass is equal to the relative atomic mass, relative molecular mass or formula mass, but its unit is g mol−1.

For calculations, first find the molar mass from the chemical formula. Then use the mass to calculate moles. The same method applies to atoms, molecules, ions and ionic formula units.

• Moles from mass: n = mass in grams ÷ molar mass.
• Mass: m = n × molar mass.
• For silica, SiO2, molar mass = 28 + 2(16) = 60 g mol−1.
• 0.6 g of SiO2 contains 0.6 ÷ 60 = 0.01 mol.
• Mass of sulphur in H2SO4: one mole of H2SO4 contains 32 g sulphur. Therefore, 24.5 g contains 24.5 × 32 ÷ 98 = 8 g sulphur.
• Number of particles = moles × 6.022 × 10^23.
• For ionic compounds, count formula units. For example, 1 mol of NaCl contains 1 mol of Na+ ions and 1 mol of Cl− ions.

Molar Volume of Gases at STP

At standard temperature and pressure, one mole of any ideal gas occupies 22.4 dm3. This is called the molar volume at STP. Equal moles of different gases at STP occupy equal volumes.

The gases do not have equal masses because their molar masses are different. Gas volume calculations can therefore be performed using either the mole relationship or the molar volume relationship.

• At STP: 1 mol gas = 22.4 dm3.
• At STP: number of moles = gas volume in dm3 ÷ 22.4.
• At STP: gas volume = moles × 22.4 dm3.
• 0.5 mol of a gas occupies 0.5 × 22.4 = 11.2 dm3 at STP.
• 5.6 dm3 of O2 at STP represents 5.6 ÷ 22.4 = 0.25 mol, or approximately 1.5 × 10^23 O2 molecules.
• 4.4 g CO2 is 0.1 mol because its molar mass is 44 g mol−1. Its volume at STP is 2.24 dm3.
• Equal moles of all gases at STP have the same volume but different masses.

Counting Atoms, Ions, Electrons and Protons

A chemical formula shows the number of atoms of each element in one molecule or formula unit. To count atoms in a given sample, first calculate the number of moles of the substance, then multiply by Avogadro's number and by the subscript of the required element.

The same approach is used for protons and electrons. First determine the number of protons or electrons in one particle, then multiply by the number of particles.

• One molecule of CO2 contains 1 carbon atom and 2 oxygen atoms.
• 4.4 g CO2 contains 0.1 mol CO2 molecules. Oxygen atoms = 0.2 mol, approximately 1.2 × 10^23 atoms.
• One formula unit of CaCO3 contains 20 + 6 + 3(8) = 50 protons.
• 10 g CaCO3 is 0.1 mol. Total protons = 0.1 × 6.023 × 10^23 × 50 = 3.01 × 10^24 protons.
• One molecule of CH4 has 6 + 4(1) = 10 electrons in a neutral molecule.
• 1.6 g CH4 is 0.1 mol. Its electrons equal 0.1 × 10 × NA = 6.02 × 10^23 electrons.
• One formula unit of NH4NO3 has two nitrogen atoms. A mass of 160 amu represents two NH4NO3 formula units because one formula unit has a mass of 80 amu. Therefore, nitrogen atoms = 4.
• If a question says 2 mol of chlorine atoms, the number of atoms is 2 × 6.022 × 10^23. If it says 2 mol of Cl2 molecules, the number of molecules is 2 × 6.022 × 10^23, while chlorine atoms are twice this number.

Chemical Equations and Stoichiometric Ratios

A balanced chemical equation gives the mole ratio between reactants and products. The coefficients, not the subscripts, are used for stoichiometric calculations.

A correct calculation normally follows these steps: write and balance the equation, convert the given quantity into moles, use the coefficient ratio, and convert the answer into the required unit such as grams, dm3 or particles.

• For 2H2 + O2 → 2H2O, the mole ratio H2:O2:H2O is 2:1:2.
• One mole of CaCO3 contains 1 mol Ca, 1 mol C and 3 mol O atoms.
• On heating: CaCO3 → CaO + CO2.
• 100 g CaCO3 is 1 mol because its molar mass is 100 g mol−1. It produces 1 mol CO2, equal to 44 g.
• Combustion of methane: CH4 + 2O2 → CO2 + 2H2O.
• 32 g CH4 is 2 mol. It requires 4 mol O2, which occupies 4 × 22.4 = 89.6 dm3 at STP.
• For ionic compounds, Ca(OH)2 contains two hydroxide groups. One mole, or 74 g, contains 2 mol OH− ions, having a combined mass of 34 g.

Limiting and Excess Reactants

The limiting reactant is the reactant that is completely consumed first. It limits the amount of product formed. The other reactant remains partly unreacted and is called the excess reactant.

To identify the limiting reactant, convert the given masses or volumes of all reactants into moles. Compare the available mole ratio with the balanced equation. The reactant that produces the smaller amount of product is limiting.

• Limiting reactant: the reactant consumed completely and responsible for limiting product formation.
• Excess reactant: the reactant left over after the limiting reactant has been consumed.
• For CaCO3 + 2HCl → CaCl2 + H2O + CO2, 20 g CaCO3 = 0.2 mol and 20 g HCl is approximately 0.548 mol.
• The equation requires 0.4 mol HCl for 0.2 mol CaCO3. Since 0.548 mol HCl is available, HCl is in excess and CaCO3 is the limiting reactant.
• The limiting reactant determines the maximum or theoretical amount of product.
• The excess reactant does not determine the maximum product amount.
• Masses must be converted into moles before comparing reactants because coefficients represent moles, not grams.

Theoretical Yield, Actual Yield and Percentage Yield

Theoretical yield is the maximum amount of product predicted by stoichiometric calculations when the limiting reactant reacts completely. Actual yield is the amount of product obtained experimentally.

Actual yield is usually less than theoretical yield because reactions may be incomplete, side reactions may occur, products may be lost during filtration or transfer, or the reactants may contain impurities.

• Theoretical yield is calculated from the balanced equation and the limiting reactant.
• Actual yield is measured in the laboratory.
• Percentage yield = (actual yield ÷ theoretical yield) × 100.
• Actual yield cannot normally be greater than theoretical yield in a correctly performed calculation. An apparent value above 100 percent may result from impurities, wet product or experimental error.
• Reasons for actual yield being less than theoretical yield include incomplete reaction, reversible reactions, side reactions and loss of product during separation or transfer.
• If theoretical yield is 50 g and actual yield is 40 g, percentage yield = (40 ÷ 50) × 100 = 80%.
• Theoretical yield is not necessarily obtained even when the equation is correctly balanced.

Empirical Formula, Molecular Formula and Percentage Composition

The empirical formula gives the simplest whole-number ratio of atoms in a compound. The molecular formula gives the actual number of atoms in one molecule. The molecular formula is a whole-number multiple of the empirical formula.

Percentage composition can be used to determine a formula. Assume 100 g of compound, convert each element's mass into moles, divide all mole values by the smallest value, and convert the ratios into simple whole numbers.

• Empirical formula is the simplest ratio of atoms.
• Molecular formula = empirical formula × n, where n is a whole number.
• n = molecular mass ÷ empirical formula mass.
• C6H12O6 has empirical formula CH2O.
• CH3COOH has molecular formula C2H4O2 and empirical formula CH2O. Therefore, C6H12O6 and CH3COOH have the same empirical formula.
• For a compound containing 85.45% carbon and 14.55% hydrogen, a proposed formula must reproduce these percentages. C2H6 contains 80.0% carbon and 20.0% hydrogen, so it does not obey the stated composition.
• Percentage of an element = mass of that element in one mole of compound ÷ molar mass of compound × 100.