Metals

Metal Reactivity Metals display a wide range of reactivity with other substances, varying from very reactive to no reaction at all. Question: Why is knowing the reactivity of a metal useful to us? The other substances that most influence the choice of metal for a particular purpose are oxygen, water and acids. The order of metal reactivity is called the activity series of metals.

Reaction of metals with oxygen Most metals react with oxygen to form metallic oxides. All the oxides formed are ionic compounds. Why? Metal + Oxygen Metal oxide

Reactions of metals with oxygen For the reaction of iron with oxygen: Step 1: Write the word equation Iron + oxygen Iron oxide Step 2: Write the forumla Fe + O2 Fe 2O 3 Where do the subscripts come from?

Reactions of metals and oxygen Step 3: Balance the equation 4Fe + 3O2 2Fe2O3 Step 4: Add states 4Fe (s) + 3O2 (g) 2Fe2O3 (s)

Reaction of metals with water Most metals undergo no change when placed in cold water. Some exceptions to this are: lithium, potassium, sodium, calcium. These react with cold water to form hydrogen and a metal hydroxide. Sodium + water sodium hydroxide + hydrogen This reaction involves the transfer of electrons from sodium atoms to hydrogen atoms in the water.

Reaction of a metal and water For the reaction of sodium with water: Step 1: Write the word equation Sodium + water sodium hydroxide + hydrogen Step 2: Write the formula Na + H2O NaOH + H2

Reaction of a metal and water Step 3: Balance the equation 2Na + 2H2O 2NaOH + H2 Step 2: Add states 2Na (s) + 2H2O (l) 2NaOH (aq) + H2 (g)

Reaction of metals with water Some less reactive metals (Al, Zn, Fe) will NOT react with cold water but will react with steam to produce steam to produce hydrogen and a metal oxide. Zinc + steam zinc oxide + hydrogen

Reaction of metals with acids Some metals will react with acids to produce salt and hydrogen. Metal + acid salt + hydrogen During the reaction between a metal and an acid the metal loses electrons and becomes positively charge ions. Hydrogen ions from the acid gain electrons to form hydrogen gas.

Reaction of metals and acids For the reaction of zinc and hydrochloric acid: Step 1: Write the word equation Zinc + hydrochloric zinc + hydrogen acid chloride Step 2: Write the formula Zn + HCl ZnCl2 + H2

Reaction of metals with acids Step 3: Balance the equation Zn + 2HCl ZnCl2 + H2 Step 4: Add states Zn (s) + 2HCl (aq) ZnCl2 (aq) + H2 (g)

A common feature A common feature of all reactions of metals with oxygen, water and dilute acids is that atoms of the metals lose electrons to become positive ions. Reactions with oxygen: ionic oxides are formed MgO = ionic compound containing Mg 2+ and O 2- Reactions with water: ionic hydroxides are formed LiOH = ionic compound containing Li+ and H-

A common feature MgCl2 = ionic compound containing Mg2+ and Cl22- Reactions with acids: ionic metallic chlorides and sulfates are formed FeSo4 = ionic compound containing Fe2+ and SO42- MgCl2 = ionic compound containing Mg2+ and Cl22-

Zn (s) + 2HCl (aq) ZnCl2 (aq) + H2 (g) Ionic equations Zn (s) + 2HCl (aq) ZnCl2 (aq) + H2 (g) Species of the reactants are? Zn atoms, H+ ions and Cl- ions Species of the products are? Zn 2+ ions, H2 molecules and Cl- ions So we could write the complete ionic equation: Zn(s) + 2H+(aq)+ 2Cl-(aq) Zn 2+(aq) + 2Cl-(aq) + H2(g)

Zn(s) + 2H+(aq)+ 2Cl-(aq) Zn 2+(aq)+ 2Cl-(aq)+ H2(g) there are two Cl- ions on the right and two on the left. Zn atoms have changed to Zn 2+ ions. For this to happen zinc atoms must have given up 2 electrons. We can write this as: Zn Zn 2+ + 2e- Hydrogen ions have changed to hydrogen ions have changed to hydrogen molecules. Therefore each hydrogen ion must have gained one electron. 2H+ + 2e- H2

Zn (s) + 2H+ (aq) Zn 2+ (aq) + H2 (g) Net ionic equations Notice that this reaction is really between Zn and H because they are the only species that undergo a change. A net ionic equation only shows the ionic species that undergo a CHANGE in the reaction. Zn (s) + 2H+ (aq) Zn 2+ (aq) + H2 (g)

Spectator ions Note that in the reaction: Zn(s) + 2H+(aq)+ 2Cl-(aq) Zn 2+(aq) + 2Cl-(aq) + H2(g) the Cl- ions do not undergo a chemical change: there are two Cl- ions on the right and two Cl- ions on the left. Ions that do not undergo a chemical change during the reaction are called spectator ions

Working out net ionic equations Ionic equation for the reaction of Al and HCl Step 1: Write the word equation aluminium + hydrochloric Al + hydrogen acid chloride Step 2: Write the formula Al + HCl AlCl3 + H2

Working out net ionic equations Step 3: Balance the equation 2Al + 6HCl 2AlCl3 + 3H2 Step 4: Add states 2Al (s) + 6HCl (aq) 2AlCl3 (aq) + 3H2 (g)

Working out net ionic equaitons Step 5: Determine the species of the reactants and products Reactants: Al atoms, H+ ions and Cl- ions Products: Al 3+ atoms, H2 molecules and Cl- ions Step 6: write the complete ionic equation 2Al (s) + 6H+ (aq) + 6Cl- (aq) 2Al 3+ (aq) + 6Cl- (aq) + 3H2 (g)

Complete ionic equations Step 7: Write the net ionic equation 2Al (s) + 6H+ (aq) 2Al 3+ (aq) + 3H2 (g)

Ionisation energy The reactivity of a metal is related to the ease with which it loses valence electrons to form ions. Ionisation energy is a measure of the energy needed to remove the most loosely bound electron from an atom in the gaseous state. In general reactive metals have low ionisation energies and less reactive metals have high ionisation energies

Oxidation – reduction reactions Reactions which involve the transfer of electrons are called oxidation-reduction reactions. When an atom LOSES one or more electrons we say it has been oxidised. When an atom GAINS one or more electrons we say it has been reduced. OXIDATION = LOSS OF ELECTRONS REDUCTION = GAIN OF ELECTRONS

Oxidation – reduction reactions In normal chemical reactions there can be no overall loss or gain of electrons. Hence oxidation and reduction occur simultaneously. We call these reactions REDOX reactions. Half equations can be used to describe the oxidation and reduction processes separately in terms of electrons lost or gained.

Half equations Reaction between magnesium and oxygen Oxidation: Mg (s) Mg2+(s) + 2e- Reduction: O2(g) + 4e- 2O2-(s)

Relative atomic mass Because atoms are so small it is difficult to measure their actual individual masses. Relative atomic mass is NOT the mass of an atom of that element. It is just a relative mass – relative to the mass of a carbon atom. It is not a mass at all – just a number with no units. Example: a titanium atom is 4x the mass of a carbon atom so relative atomic mass of Ti is 48

Isotopes Most elements in nature consist of several isotopes with slightly different masses. This is because isotopes have a different number of neutrons in the nucleus. Example: 75% of Cl atoms have 18 neutrons 25% of Cl atoms have 20 neutrons There are two isotopes of Cl one with an atomic mass of 35 (Cl-35) and one with an atomic mass of 38 (Cl-37). Remember all Cl atoms have 17 protons.

Isotopes Therefore strictly speaking when we say the ‘mass of an atom’ we actually mean the ‘average mass of the atoms in the naturally occurring element’. Average mass = 75 x 35 + 25 x 37 = 35.5 100 Note: average mass is closest to the atomic mass of the most abundant isotope

Relative molecular mass Relative atomic mass (Ar) is used to describe the mass of atoms. Relative molecular mass (Mr) is used to describe the mass of molecules. (Mr) = the mass of a molecule of a substance or compound relative to the mass of an atom of the carbon-12 isotope taken exactly as 12

Relative molecular mass The relative molecular mass of a substance is found by adding the relative atomic masses of the constituent elements. Example: molecular mass of H2O = 2 x Ar(H) + 1 x Ar(O) = 2 x 1 + 16 = 18

Relative formula mass Many compounds, particularly ionic compounds (eg: NaCl) exist as an array of ions or atoms bound to each other but with no recognisable molecules. The formula NaCl instead tells us that throughout a sample of NaCl sodium and chlorine atoms are present in the ratio 1:1. Because ionic compounds do not contain molecules the sum of the relative atomic masses of the atoms in the formula is called the relative formula mass (still given the symbol Mr).

The mole Chemists measure the amount of any substance in moles. Mole: the quantity or amount of a substance that contains the same number of particles as there are atoms in exactly 12 grams of carbon-12 Avogadro’s number (NA): the number of atoms in exactly 12 grams of carbon-12 = 6.022 x 1023 Therefore one mole of a substance contains 6.022 x 1023 particles of that substance.

Working with moles When working in moles the particle or units being counted should be stated as atoms, molecules or ions. Eg: 1 mole Fe = 6.022 x 1023 atoms of Fe 1 mole of Pb = 6.022 x 1023 atoms of Pb 1 mole of H2SO4 = 6.022 x 1023 molecules of H2SO 1 mole N2 gas = 6.022 x 1023 molecules of nitrogen

Molar mass 6.022 x 1023 atoms of carbon has a mass of 12 grams Since all relative atomic masses are measured against the standard carbon 12 it follows that the atomic mass in grams of an element (or the formula mass in grams of any compound) is one mole of that substance and this one mole contains avogadro’s number of particles.

Consider: Titanium (atomic mass 48) 1 mole of titanium has 6.022 x 1023 atoms (each of which have a mass 4x that of carbon) Therefore mass of 1 mole of titanium = 4x12 =48 1 mole of titanium has a mass equal to its relative atomic mass.

Molar mass 1 mole of a substance has a mass equal to its: relative atomic mass (expressed in grams) relative molecular mass (expressed in grams) relative formula mass (expressed in grams) This is called the molar mass (M) of a substance. Molar mass has units – usually grams per mole (g mol -1).

Calculating molar mass Write down the symbol or formula of the substance Add up the relative atomic masses of the elements involved This is the symbol mass or formula mass In grams this is one mole of the substance This is made of avogadro’s number of particles

Summary There are therefore two ways of looking at a mole: 6.022 x 1023 = 1 mole of Ti = 48 grams of Ti atoms of Ti A number of particles (atoms, ions, molecules) 6.022 x 1023 A mass (the relative atomic, molecular or formula mass in grams) MOLE

Moles and numbers of particles Relationship between moles and number of particles (eg: atoms, molecules): Number of moles (n) = number of particles (N) number of particles in one mole (NA) n = N 6.022 x 1023 We often need to calculate the number of moles present in a given mass of a substance (either element or compound). At other times we need to calculate the number of atoms or molecules in a sample. Use the following formulas to do this.

Moles and numbers of particles Using this formula it is possible to calculate: Number of moles of a substance from the number of particles or basic units of a substance Number of particles or basic units of a substance (atoms, formula units) from the number of moles

Moles and mass The relationship between the number of moles and mass of a substances is: Number of moles (n) = mass (m) in grams mass of 1 mole (M) n = m M

Using this formula it is possible to calculate: Number of moles of any substance in a given mass Mass of a substance in a given number of moles

Summary n = m M n = N 6.022 x 1023 Therefore: N = m 6.022 x 1023 M

Percentage composition Percentage composition of a compound is simply the percentage by mass of each element present in the compound. To determine percentage composition you need two things : formula of the compound Relative atomic masses of the elements present

Steps in determining percentage composition Calculate the percentage composition of iron in Fe2O3 Step 1: determine mass of one mole of Fe2O3 2 x 55.9 + 3 x 16 = 159.8 grams Step 2: One mole of Fe2O3 contains 2 moles of Fe. 2 moles of Fe : 2 x 55.9 = 111.8 grams Step 3: % composition of Fe = 111.8 x 100 = 70% 159.8

Summary Therefore: % A in a compound = mass of A in 1 mole of the compound x 100 mass of 1 mole of the compound

Molecular v’s empirical formula Molecular formula: specifies the actual number of atoms of each element in a molecule hydrogen peroxide: H2O2 Empirical formula: specifies the simplest whole number ratio of each element hydrogen peroxide: HO

Determining empirical formula The empirical formula of a substance can be established by first determining the percentage composition of a substance by chemical analysis. Ex: Calculate the empirical formula of glucose with a chemical composition of 40% carbon, 6.6% hydrogen and 53.3% chlorine

Step 2: Determine the moles of each element in 100 grams Step 1: List the elements and the mass of each element in 100grams of freon-12 Element: C H O Mass in 100g: 40 6.6 53.3 Step 2: Determine the moles of each element in 100 grams n = m 40 6.6 53.3 M 12.01 1.008 16.00 = 3.33 = 6.55 = 3.33 Mole ratio 1 : 2 : 1 Step 3: Empirical formula CH2O

Determining molecular formula If the molecular mass of glucose =180.16 what is its molecular formula? Step 1: Determine the relative empirical formula mass CH2O = 1 x12.01 + 2 x 1.008 + 1 x 16.00 = 30.03 Step 2: Relative molecular mass = molecular formula Relative empirical formula mass empirical formula

molecular formula = 180.16 x CH2O 30.03 Relative molecular mass = molecular formula Relative empirical formula mass empirical formula 180.16 = molecular formula 30.03 CH2O molecular formula = 180.16 x CH2O 30.03 molecular formula = 6 x CH2O molecular formula = C6H12O6

Joseph Gay-Lussac Gay Lussac’s law of combining gas volumes states that when two gaseous elements combine: ‘the ratios of the volumes of gases involved, if measured at the same temperature and pressure, are expressed by small whole numbers’

Step 1: write the balanced equation for the reaction Example: calculate the volume of hydrogen that will combine with 6 L of nitrogen to form ammonia Step 1: write the balanced equation for the reaction N2 (g) + 3H2 (g) 2NH3 (g) Step 2: determine the ratios y 1 mole N2 3 mole H2 2 mole NH3 1 volume 3 volume 2 volume Step 3: calculate volume V (H2 ) = 3 x V(N2) = 3 x 6.0L = 18 L

Mass-mass calculations Chemical equations show the number of moles of reactants and products in a chemical reaction. They can also be used to determine the relationship between the masses of the reactants and products: N2 (g) + 3H2 (g) 2NH3 (g) 1 mole 3 mole 2 mole 28g + 6g 34g

Remember this! When carrying out mass-mass calculations remember the following step by step method: From the equation N = m/N N = m/N Mass of the known Moles of the known Moles of the unknown