Chemistry-ch03_chemical Reactions And Stoichiometry

3

Chemical reactions and stoichiometry

3.1 Chemical equations – A chemical reaction is the mixing of two or more species to produce new substances – A chemical equation describes what happens when a chemical reaction occurs – Consider the following reaction: 2H2 + O2 → 2H2O reactants: hydrogen and oxygen

react to produce

products: water

3.1 Chemical equations – Stoichiometry is concerned with the relative amounts of reactants and products in a chemical reaction – Stoichiometric coefficients indicate the number of molecules, ions or atoms among the reactants and products

3.1 Chemical equations – The law of conservation of mass says that atoms cannot be created or destroyed during a chemical reaction – Stoichiometric coefficients are used to balance an equation to meet this condition

3.1 Chemical equations • Specifying states of matter – It is useful to specify the physical states of the reactants and products – This is done by writing (s) for solid, (l) for liquid or (g) for gas after the chemical formula – (aq), meaning ‘aqueous solution’, can also be used to indicate that a particular substance is dissolved in water

3.2 Balancing chemical equations • To balance an equation: – Step 1: Write the unbalanced ‘equation’

Al(s) + HCl(aq) → AlCl3(aq) + H2(g) – Step 2: Adjust the coefficients so that there are equal numbers of each kind of atom on each side of the arrow – Is this balanced? 2Al(s) + 6HCl(aq) total reactants: 2 × Al, 6 × H + 3H2(g) and 6 × Cl – Is this balanced?

Yes

→total 2AlCl 3(aq) products: 2 × Al, 6 × H and 6 × Cl

3.3 The mole – Every atom has a certain mass – This is referred to as its atomic mass – The mole (abbreviated mol) is the SI unit of amount of substance – A mole is the amount of any substance with the same number of entities as there are atoms in exactly 12 g of 12C – Avogadro’s constant (6.022 × 1023 mol-1) gives the number of entities in 1 mole

3.3 The mole – The number of specified entities in a mole is constant – The mass of 1 mole depends on the mass of the individual entities

3.3 The mole – The molar mass, M, is the mass of 1 mole of a substance – What is the molar mass of water (H2O)? • The chemical formula tells us it is made up of 2 hydrogen atoms and 1 oxygen atom • Hence the molar mass is the sum of twice the atomic mass of hydrogen (1.008 g) plus the atomic mass of oxygen (15.999 g) • MH2O = (2 × 1.008 g) + 15.999 g = 18.015 g

– The relationship between moles (n), molar mass (M) and mass (m) is given by M = m/n

3.4 Empirical formulae The empirical formula is the simplest whole-number ratio of atoms within that compound. – For example, butane has the formula: –

C4H10

molecular formula

– However this isn’t the simplest ratio as both 4 and 10 are divisible by 2. Hence butane has the empirical formula: C2H5

empirical formula

3.4 Empirical formulae • Mole ratios from chemical formulae – Consider water: chemical formula:

H2O – The chemical formula tells us the ratio of H atoms to O atoms is always 2 to 1 – For chemical compounds, mole ratios are the ratios of individual atoms

3.4 Empirical formulae – The relative masses of elements in a compound is usually given as a percentage – This is called the percentage composition OR percentage composition by mass – The percentage by mass of an element is calculated using the following: % element =

mass of element x 100% mass of whole sample

3.4 Empirical formulae – A molecular formula gives the chemical composition of 1 molecule, e.g. P4O10 – The empirical formula for this compound is P2O5 – This gives the simplest whole number ratio between atoms – The empirical formula of a compound can be obtained experimentally by determining the mass of each element in a compound

3.4 Empirical formulae – There are three steps necessary to determine the empirical formula: 1. Assume we are studying 100 g of the compound and therefore individual mass percentages become the actual masses 2. Convert the ratio of elements by mass to a ratio by amount, by dividing the mass of each element by its molar mass 3. Divide the resulting numbers by the smallest, which will give the smallest wholenumber ratios of each element

3.4 Empirical formulae – The formula for ionic compounds is the same as the empirical formula – For molecules, the molecular formula and empirical are usually different – If the experimental molecular mass is available, the empirical formula can be converted into the molecular – The molecular formula will be a common multiplier times all the coefficients in the empirical formula

3.5 Stoichiometry, limiting reagents & percentage yield • Stoichiometry:

– The critical link between substances involved in a reaction is the mole-to-mole ratio – Consider the following: • 2C8H18 (l) + 25O2(g)→16CO2(g) +18H2O(g)

– This is interpreted as: 2 moles of liquid octane reacts with 25 moles of oxygen gas to produce 16 moles of carbon dioxide gas and 18 moles of steam

3.5 Stoichiometry, limiting reagents and percentage yield – Mole-to-mole relationships can be used to solve stoichiometry problems – The stoichiometry coefficients in the equation can be used to calculate conversion factors – To use these relationships in a stoichiometry problem, the equation must be balanced

3.5 Stoichiometry, limiting reagents and percentage yield

• Limiting reagents

– Ethanol, C2H5OH, is prepared industrially as follows: C2H4 + H2O → C2H5OH – The equation tells us that 1 molecule of ethene will react with 1 molecule of water to give 1 molecule of ethanol

3.5 Stoichiometry, limiting reagents and percentage yield – As we have learned, the mole-to-mole ratio is constant. So, 3 molecules of ethene will react with 3 molecule of water to give 3 molecules of ethanol

3.5 Stoichiometry, limiting reagents and percentage yield – What happens if we mix 3 molecules of ethene with 5 molecules of water?

– The ethene will be completely used up before the water, and so the product will contain 2 unreacted water molecules

3.5 Stoichiometry, limiting reagents and percentage yield – All reactions eventually use up a reactant and seem to stop – The reactant that is consumed first is called the limiting reagent because it limits the amount of product formed – Any reagent not completely consumed during the reactions is said to be in excess and is called an excess reagent – The calculated amount of product is always based on the limiting reagent

3.5 Stoichiometry, limiting reagents and percentage yield • Percentage yield – In most experiments, the amount of a product isolated falls short of the maximum amount – The actual yield of the desired product is simply how much is isolated – The theoretical yield of the product is what would be obtained if no losses occurred – The percentage yield is the actual yield calculated as a percentage of the theoretical yield

3.5 Stoichiometry, limiting reagents and percentage yield – Percentage yield is calculated using the following formula: % yield =

actual yield x 100% theoretical yield

– The calculation may be done in either grams or moles, but both yields must be in the same units – The actual yield can never be more than the theoretical yield

3.6 Solution stoichiometry – Chemical reactions are nearly always carried out in solution to allow mixing – A solution is a homogeneous mixture – When a solution forms, at least two substances are involved, a solvent and one or more solutes – The solvent is the component present in largest amount – The solute is any substance dissolved in the solvent

3.6 Solution stoichiometry

3.6 Solution stoichiometry • The concentration of solutions – The concentration is defined as the amount of solute dissolved in a particular volume of solution – The concentration of a substance X is represented as [X] – When the amount is given in moles and the volume in litres, it is called the molarity or molar concentration

3.6 Solution stoichiometry – Molarity (or molar concentration) has the units mol L–1 (often abbreviated M) – Concentration is based on the ratio of the amount of solute to the volume of solution – The equation defining concentration is:

The units of c are mol L-1

n C= V

n is the number of moles V is the volume (in Litres)

3.6 Solution stoichiometry • Diluting a solution – If a solute is already dissolved into a solution of high concentration, it can be diluted to decrease the concentration – Dilution is accomplished by adding more solvent to the solution, which causes the concentration to decrease – The same pure solvent should be used – The choice of apparatus used depends on the precision required

3.6 Solution stoichiometry

– When solvent is added to a solution, the solute particles become more spread out – The concentration of the solute in the solution becomes smaller

3.6 Solution stoichiometry – Ionic compounds dissociate into their constituent ions when dissolved in water – This can have important consequences in stoichiometric calculations • e.g. CaBr2 undergoes complete dissociation into Ca2+ and Br- ions, according to the equation CaBr2(s) → Ca2+(aq) + 2Br– (aq) • 0.10 moles of CaBr2 yields 0.10 moles of Ca2+ and 0.20 moles of Br– • In 0.10M CaBr2, the concentration of Ca2+ is 0.10M and the concentration of Br– is 0.20M

3.6 Solution stoichiometry – Spectator ions don’t take part in reactions – Balanced ionic equations are written without the inclusion of spectator ions – These are called net ionic equations • e.g. Consider the following reaction: – 2AgNO3(aq) + CaBr2(aq)→2AgBr(s) +Ca(NO3)2(aq)

• The NO3– and Ca2+ ions don’t take part in the reaction

– Q. What is the net ionic equation? A. Ag+(aq) + Br-(aq) → 2AgBr(s)

Chapter Summary – A chemical reaction is the formation of new substances (products) upon mixing two or more chemical species (reactants) – Stoichiometry is concerned with relative amounts of products and reactants – A balanced chemical equation has the same number of entities of each kind in the products and reactants – Chemical equations show physical states of the reactants and products

Chapter Summary – To balance a chemical equation, we use stoichiometric coefficients – The mole is the unit of substance – Avogadro’s constant is: 6.022 × 1023 mol-1 and gives the number of specified entities in 1 mole of a substance – The molar mass is the mass of 1 mole of a substance – The actual composition of a molecule is given by its molecular formula

Chapter Summary – An empirical formula gives the smallest whole-number ratio of atoms – Percentage composition is used to describe the relative amount of each element in a compound – A reactant present in a quantity less than that required by another reactant is called the limiting reagent – The other reactant is called the excess reagent

Chapter Summary – The percentage yield is the actual yield calculated as a percentage of the theoretical yield – When a solution forms, at least two substances are involved – Concentration is the ratio of the amount of solute to the volume of solution – Ionic compounds dissociate into their constituent ions when dissolved in water – Spectator ions don’t take part in reactions