GAS FORMULAE THE GENERAL GAS EQUATION PV = nrt Note Useful Conversions P = Pressure ( kpa ) 760 mmhg = 1 atmosphere V = Volume ( L ) = 101,35Pa = 101.35 kpa n = Amount (in mol) of gas ( mol) 1 1 R = Gas constant = 8.31 JK mol 3 1 dm = 1000 ml = 1 L T = Temperature ( K ) 3 1cm = 1 ml The following relationship may be used to calculate parameters for different conditions of the same gas. For example, when there is a change in mole, volume, pressure or temperature. V1P1 n T 1 1 VP = n T Note Subscript 1 represents the original conditions whereas subscript denotes the conditions following a change. STANDARD TEMPERATURE AND PRESSURE (STP) One mole of any gas occupies.4 L at STP ( 0 C / 1 Atmosphere). When a gas is at STP, the General Gas Equation is simplified to V n = or V m V n = STP ( V m = molar volume).4 STANDARD LABORATORY CONDITIONS (SLC) 1 mole of gas occupies 4.5 L at SLC ( 5 C / 1 Atmosphere). When a gas is at SLC, the General Gas Equation is simplified to V n = SLC. 4.5 The School For Excellence 016 The Essentials Unit 3 Chemistry Book 1 Page 1
QUESTION 19 Find the amount of gas that occupies.37 L at a temperature of 17.6 C and a pressure of 96.7 kpa. Solution PV = nrt PV n = RT 96.7.37 n= = 0.0948 mol 8.314 90.6 QUESTION 0 A student inhales 500 ml of air at STP. If air contains 1% oxygen by volume, the number of atoms of oxygen inhaled is A B C D.8 10 5.64 10 5.66 10 1.13 10 1 1 4 5 QUESTION 1 The graph shows the maximum dissolved oxygen concentration in water as a function of temperature at normal atmospheric pressure. What is the volume of oxygen gas that can dissolve in 10.0 L of water at 5 C and normal atmospheric pressure? A B C D 6.0 ml 63.5 ml 80.0 ml 14 ml The School For Excellence 016 The Essentials Unit 3 Chemistry Book 1 Page
QUESTION Equal masses of nitrogen gas and hydrogen gas are placed in separate vessels. Both vessels have the same volume and are at the same temperature. The pressure exerted by the hydrogen gas is 15 kpa. Find the pressure, in kpa, exerted by the nitrogen gas. Solution QUESTION 3 0.500 mole of gas is introduced into a vessel of fixed volume at STP. If the temperature is increased to 380 K, find the pressure exerted on the vessel as a result of the change. Solution Identify the constant parameters (parameters that do not undergo a change) n, R, V Eliminate the constant parameters from the gas equation P P = T T 1 PT 380 101.35 = = = 141.04 = 141 1 P kpa kpa T1 73 V1 P1 n T 1 1 VP = n T QUESTION 4 A 0.150 L flask contains a gas at 0.0 C and at atmospheric pressure. Calculate the pressure in the flask when the flask is cooled to 10.0 C. Solution The School For Excellence 016 The Essentials Unit 3 Chemistry Book 1 Page 3
EMPIRICAL FORMULAE The empirical formula of a compound is the smallest whole number ratio of the atoms of the different elements that make up the compound. Empirical formulae are determined experimentally, usually by determining the mass of each element present in a given mass of compound. Therefore, to determine the empirical formula of a compound, an experimentally determined ratio by mass must be converted to a ratio by numbers of atoms. This is done by calculating the amount of each element. METHOD Step 1 Set up the elements present as a ratio. Step Calculate the mass or percentage composition of each element and set up a ratio by mass. Step 3 Set up a ratio by mole (convert the answers to amounts in mole). Step 4 Simplify the ratio by dividing through by the smallest number. Step 5 If required, multiply answers by an appropriate number to obtain the simplest whole number ratio. MOLECULAR FORMULAE The molecular formula describes the actual number of atoms of each element present in a molecule of a compound. The molecular formula is always a whole number multiple of the empirical formula, and may be obtained from the empirical formula if the molar mass of a compound is known. METHOD Step 1 Find the molar mass of the empirical formula unit. Step Find the molar mass of the compound. Step 3 Divide the molar mass of the compound by the molar mass of the empirical formula unit to give the number of empirical formula units. Step 4 Multiply the empirical formula by the number of empirical formula units. Molecular Formula= ( Empirical Formula) n RMM Where n = Empirical Formula ( mass) The School For Excellence 016 The Essentials Unit 3 Chemistry Book 1 Page 4
QUESTION 5 A 50 g sample of an experimental catalyst used in the polymerisation of butadiene is made up of 11.65 g Co and 5.7 g Cl. If the molar mass of the compound is 759 g/mol, find the molecular formula of this catalyst. Solution Co Mo Cl Find n 11.65 58.9 50 11.65 5.7 95.94 5.7 35.45 Simplify 0.1978 0.1319 0.750 Divide by smallest ratio 1.5 3 1 5.5 11 The empirical formula of the compound is Co 3MoCl11. QUESTION 6 A compound of carbon, hydrogen and oxygen was found to contain 54.5% carbon and 9.1% hydrogen. If the relative molecular mass of the compound is 87.95 (a) Determine the empirical formula of the compound. C H O Find n 54.5 1 9.1 1 36.4 15.99 Simplify 4. 54 9. 1. 764 Divide by smallest ratio 4 1 The empirical formula of the compound is C H 4O. (b) Determine the molecular formula of the compound. M ( C H 4O) = 44 Therefore, there are units of C H 4 O in the compound. i.e. Molecular Formula = C 4H 8O The School For Excellence 016 The Essentials Unit 3 Chemistry Book 1 Page 5
QUESTION 7 A molecule has a molecular mass of 99 g/mol. Analysis of this sample shows that it contains 4.3% carbon, 4.1% hydrogen and 71.6% chlorine. What is its molecular formula? Solution The School For Excellence 016 The Essentials Unit 3 Chemistry Book 1 Page 6
QUESTION 8 The graph shows the mass and amount of carbon, fluorine and chlorine atoms in 1 mole of a compound. What is the molecular formula for this compound? A CFCl B CFCl C CFCl 3 3 D CFCl 4 The School For Excellence 016 The Essentials Unit 3 Chemistry Book 1 Page 7
REDOX CHEMISTRY Reduction Oxidation reactions (redox reactions) are those reactions that involve the transfer of electrons. Oxidation involves the loss of electrons. Reduction involves the gain of electrons. Reduction Oxidation reactions occur simultaneously i.e. one reactant is oxidised at the same time as the other reactant is reduced. OXIDANTS An oxidant or oxidising agent is a chemical which oxidises another substance i.e. it gains electrons. In the process, the oxidant is reduced to its conjugate reductant. For example Cl ( g ) + e Cl( aq) Oxidant Properties Oxidant Conjugate Reductant Oxidants gain electrons. Oxidants undergo reduction i.e. are reduced. Oxidants cause the oxidation of another substance, and are therefore called oxidisers or oxidising agents. REDUCTANTS A reductant or reducing agent is a chemical which reduces another substance i.e. it loses electrons. In the process, the reductant is oxidised to its conjugate oxidant. For example Na Na + ( s ) ( aq) + e Reductant Properties Reductant Conjugate Oxidant Reductants donate electrons. Reductants undergo oxidation i.e. are oxidised. Reductants cause the reduction of another substance and are therefore called reducers or reducing agents. The School For Excellence 016 The Essentials Unit 3 Chemistry Book 1 Page 8
SUMMARY OF REDOX TERMINOLOGY OILRIG Oxidation Involves Loss Reduction Involves Gain OUR RIO RAROAO ROAD Oxidants Undergo Reduction Reductants Undergo Oxidation Reductant Is Oxidised Oxidant Is Reduced Reductants Are Reducers (Reducing Agents) Oxidants Are Oxidisers (Oxidising Agents) Reductants Donate Oxidants Accept OXIDATION NUMBERS The oxidation number is an arbitrary system that is used to provide an indication of the number of electrons lost or gained by a species, and hence whether a redox reaction has taken place. RULES The following rules are to be applied in the order they are presented. Rules are presented in order of priority i.e. Rule 1 has a greater priority than Rule etc. Components of each rule display equal priority i.e. Components (a), (b) and (c) in Rule 1 have equal priority. 1. (a) The oxidation number of an uncombined element, or a compound made from only one element is 0. For example The oxidation number of P in P 4 is 0. (b) The sum of the oxidation numbers in an electrically neutral formula is 0. (c) The sum of the oxidation numbers in an ion is equal to the charge on that ion. For example Oxidation Numbers in NO3 = 1 Oxidation Numbers in SO. (a) The oxidation number of Group I species (except H ) in compounds is +1. (b) The oxidation number of Group II species in compounds is +. (c) The oxidation number of Group III species in compounds is +3. 3. The oxidation number of F in compounds is 1. 4. The oxidation number of H in compounds is +1. 5. The oxidation number of O in compounds is. 4 = The School For Excellence 016 The Essentials Unit 3 Chemistry Book 1 Page 9
6. (a) The oxidation number of Group V species in binary compounds with metals is 3. (b) The oxidation number of Group VI species in binary compounds with metals is. (c) The oxidation number of Group VII species in binary compounds with metals is 1. DETERMINING THE OXIDATION NUMBER OF A SPECIFIC ELEMENT Step 1 Let the oxidation number of the unknown element be x. Step Assign an oxidation number to the other elements. Step 3 Multiply each oxidation number by the number of atoms of each corresponding element. Step 4 Let the sum of the oxidation numbers equal the charge on the species. Step 5 Solve for x. Note If a species consists of more than different elements and represents an ionic compound reduce the formula to anions and cations. For example H SO 4 can be + written as H + SO 4. If the unknown is present in more than one part of the structural formula and the species is ionic in nature reduce the formula to anions and cations. Oxidation numbers can be fractions! QUESTION 9 Find the oxidation number of (a) O in OF (b) P atoms in H 4P O7 (c) Cr in Cr( H O ) + (d) N in HN 3 3 6 The School For Excellence 016 The Essentials Unit 3 Chemistry Book 1 Page 30
QUESTION 30 The oxidation number of nitrogen in NH 4 NO is A 3 B 3 C ± 3 D 0 E Cannot be determined The School For Excellence 016 The Essentials Unit 3 Chemistry Book 1 Page 31
DETERMINING THE OXIDATION NUMBERS OF ALL THE ELEMENTS IN A SUBSTANCE Step 1 Assign an oxidation number to each element in the order specified in the Rules. Step Let the oxidation number of the last remaining element equal x. Step 3 Multiply each oxidation number by the number of atoms of each corresponding element. Step 4 Let the sum of the oxidation numbers equal the charge on the species. Step 5 Solve for x. QUESTION 31 Assign oxidation numbers to each element in the following compounds (a) S (b) H CO 3 4 O 6 (c) HOF (d) KClO 4 (e) KO The School For Excellence 016 The Essentials Unit 3 Chemistry Book 1 Page 3
IDENTIFYING REDOX PROCESSES When the oxidation number of two reactant species changes, a redox reaction has occurred. Oxidation involves an increase in oxidation number. If the oxidation number of an element increases, that element has undergone oxidation i.e. there has been a loss of electrons. Reduction involves a decrease in oxidation number. If the oxidation number of an element decreases, that element has undergone reduction i.e. there has been a gain in electrons. Oxidants will experience a decrease in oxidation number (as they undergo reduction; which involves a gain in electrons). Reductants will experience an increase in oxidation number (as they undergo oxidation; which involves a loss of electrons). Important Notes If there are no changes in oxidation numbers, the reaction is not a redox process. There must be two changes in oxidation numbers for a redox reaction to occur (oxidation and reduction processes occur simultaneously). Furthermore, the total increase in the oxidation number of one species MUST be equal to the total decrease in the oxidation number of the other species (taking mole ratios into consideration). Any reaction that involves a pure element or a compound that consists of only one element is most likely to be a redox reaction. For example Ca( s) + F( g) CaF One exception to this rule involves reactions of sugars such as glucose ( C 6H1O6 ). This generalisation, however, may be used to eliminate options in multiple choice questions, but confirm your final answer by assigning oxidation numbers in full. A gain of hydrogen is reduction and a loss of hydrogen is oxidation. A gain of oxygen is oxidation and loss of oxygen is reduction. The School For Excellence 016 The Essentials Unit 3 Chemistry Book 1 Page 33
QUESTION 3 Which one of the following reactions is a redox reaction? A B C D HO (l) + CO3 - (aq) OH - (aq) + HCO3 - (aq) Mg (s) + AgNO3 (aq) Mg(NO3) (aq) + Ag (s) CaO (s) + HO (l) Ca(OH) (aq) H3PO4 (aq) + 3KOH (aq) K3PO4 (aq) + 3HO (l) QUESTION 33 Which one of the following reactions is a redox reaction? BaCl + H SO BaSO + HCl A ( aq) 4( aq) 4( s) ( aq) + B Cu( aq) + CO3( aq) CuCO3( s) C MgCO3( s) MgO( s) + CO( g) D HNO( aq) + HO( l) HNO3( aq) + H O( l) QUESTION 34 In which of the following reactions are the nitrogen atoms being oxidised only? N O + H O HNO + HNO A 4( g) ( l) 3( aq) ( aq) B NH 4NO3( s) NO( g) + HO( g) C NO( g) NO4( g) D N H4( l) + HO( l) N( g) + 4HO( g) The School For Excellence 016 The Essentials Unit 3 Chemistry Book 1 Page 34
QUESTION 35 The following reaction involves hydrogen peroxide. + + 5H O( aq) + MnO4( aq) + 6H ( aq) 5O( g ) + 8H O( l) + Mn( aq) The hydrogen peroxide is acting as A B C D An acid A base A reductant An oxidant QUESTION 36 Consider the following reaction NaOH ( aq) + Cl( g ) NaCl( aq) + NaClO( aq) + H O( l) In this reaction, Cl is acting as A B C D An oxidant only A reductant only Both an oxidant and reductant An acid only QUESTION 37 Concentrated sulfuric acid reacts with glucose. One of the chemical reactions that can occur may be represented as C + 6H 1O6( aq) + 6H SO4( l ) 6C ( s) + 6H 3O( aq) + 6HSO 4( aq) This reaction is best described as being A B C D Dehydration only. Acid-base and redox only. Dehydration and acid-base only. Dehydration, acid-base and redox. The School For Excellence 016 The Essentials Unit 3 Chemistry Book 1 Page 35