mole concept and stoichiometry

subatomic particles

• protons and neutrons (nucleons) → form dense nucleus of atoms

• electrons → surround the nucleus and are arranged in electron shells

relative charge and relative mass of particles

particle	actual mass/kg	charge relative to proton	relative mass (amu)
proton	1.671.671.67 x 10−2710^{-27}10−27	+1	1
neutron	1.671.671.67 x 10−2710^{-27}10−27	0	1
electron	9.119.119.11 x 10−3110^{-31}10−31	-1	1/1840

atomic number

number of protons in the nucleus of an atom

nucleon number (mass number)

total number of protons and neutrons in the nucleus of an atom

isotope

atoms of the same element having same number of protons (same atomic number) but different number of neutrons (different mass numbers)

carbon-12 scale

• one atom of ¹²C is assigned a mass of 12 units

• relative atomic masses have no units since they are not actual masses of atoms but masses relative to the arbitrary standard

relative atomic mass

average mass of one atom of an element compared to 1/12 the mass of one ¹²C atom

• due to presence of isotopes, the word ‘average’ is necessary

relative isotopic mass

• mass of one atom of an element compared to 1/12 the mass of one ¹²C atom

• relative abundance of isotopes: the percentage of the isotopes in the naturally occurring element

relative molecular mass

• for elements and contends that exist in the form of molecules

• average mass of one molecule of an element or compound compared to 1/12 the mass of one ¹²C atom

relative formula mass

• for ionic compounds, as they do not consist of individual molecules ⇒ cannot refer to their relative molecular mass

• average mass of one formula unit of an ionic compound compared to 1/12 the mass of one ¹²C atom

• addition of Ar (1 d.p.) of all elements present in the formula gives the Mr (1 a.p.)

avogadro constant (L)

is the number of particles per mole of substance. it has a value of 6.02 × 10²³

mole

defined as the amount of substance that contains 6.02 x 10²³ (or Avogadro Constant) elementary entities

molar mass

• is the mass of one mole of the substance

• it has the unit of g mol^-1 and is numerically equal to the relative atomic mass

• a mole of any substance has a mass numerically equal to its Ar or Mr in grams

relationship between amount, mass, molar mass and number of particles

• amount = mass of substance/molar mass of substance

  • n = m/Ar or Mr (g)

• number of particles = amount x avogadro constant

  • N = nL

molar volume

• the volume occupied by one mole of any gas at a particular temperature and pressure

• regardless of the identity of the gas, the molar volumes of gases are all the same when measured at the same temperature and pressure

avogadro’s law

equal volumes of all gases at the same temperature and pressure contain equal number of moles irrespective of the type of gases

to calculate volume of gas/amount of gas

• for gases in a balanced equation, volume ratio = mole ratio

• amount of gas = volume of gas (dm³) /molar volume (dm³mol^-1)

s.t.p. and r.t.p.

• for 1 mole of any gas [consists of L number of particles and has a mass equal to its Ar or Mr in grams]

  • standard temperature and pressure (s.t.p.) = 273K (0°C) and 10⁵ Pa (1 bar) → 22.7 dm³

  • room temperature and pressure (r.t.p.) = 293K (20°C) and 101325 Pa (1 atm) → 24.0 dm³

ideal gas equation

• volume can also be calculated using the ideal gas equation: PV = nRT

  • P: pressure (Nm^-2 or Pa)

  • V: volume (m³)

  • T: temperature (K)

  • m: number of moles

  • R: molar gas constant (8.31 J K^-1 mol^-1)

empirical and molecular formula

• empirical formula: the formula that shows the simplest whole number ratio of atoms of each element present in the compound

• molecular formula: the formula that shows the actual number of atoms of each element present in one molecule of the compound

calculation of empirical formula by composition by mass

• the empirical formula is not always the same as the molecular formula ⇒ usually a whole integer multiple of the empirical formula

mass/g OR % by mass
no. of moles/mol
simplest ratio

calculation of formula using combustion data

• when a gaseous hydrocarbon combusts in the presence of excess oxygen under room conditions, the reactions follow a general equation

  • C_xH_y (g) + (x+y/4) O₂ (g) → xCO₂ (g) + (y/2) H₂O (l)

• using the change in volume from the initial volume to the final volume to calculate the reacting ratios

calculations using reacting masses

• stoichiometric coefficients give information on the molar ratio

• stoichiometric amount: amount used in the exact proportion as shown in equation

• excess reagent: not completely used up in a chemical reaction

• limiting reagent: completely used up in a chemical reaction

  • decides how much product can be formed

percentage yield

• theoretical yield: quantity of product that will be formed if all limiting reagent reacts

• actual yield: quantity of product that is actually obtained during an experiment

• percentage yield = (actual yield/theoretical yield) x 100

calculations involving volumes and concentrations of solutions

• concentration can be measured in terms of

  1. amount of solute dissolved in 1 dm³ of solution → mol dm^-3

  2. mass of solute dissolved in 1 dm³ of solution → g dm^-3

• calculations of moles and concentrations

  • amount (mol) = concentration (mol dm^-3) x volume (dm³)

  • convert mol dm^-3 to g dm^-3, multiply by Ar/Mr of solute

  • to convert g dm^-3 to mol dm^-3, divide by Ar/Mr of solute

standard solution

a solution whose concentration is known accurately

calculate concentration of a diluted solution

• when a solution is diluted (adding more solvent), the concentration of the solution decreases but the number of moles of the solute in the diluted solution remains unchanged

• the ‘dilution factor’ method can also be used ⇒ how many times a solution is diluted compared to the original solution

parts per million

• for very small concentrations of substances, parts per million can be used as the unit of concentration (similar to percentage)

• ppm is the number of parts out of a total of one million parts (10⁶)

• when applied to substances in aqueous solutions, parts per million refers to grams of substances per million grams of solution → ppm = (mass of substance/total mass of mixture) x 10⁶

• when dealing with gases in gaseous mixture, parts per million refers to number of volume units of gas per million volume units of gaseous mixture → ppm = (volume of substance/total volume of mixture) x 10⁶