P & L. Johnson

Transcription

1 P & L. Johnson 2012

2 Imagine you have been invited to invest in a company that have just discovered a new way of making petrol. Could this be an easy way to make your fortune? But how much do they need to sell it at to make a profit and will this be cheaper and therefore more attractive than normal petrol? In order to work out the minimum price, you need to know what its going to cost to make it. To find this you will need to know what your starting chemicals are and how they react to form your product. Chemical formulae, balanced chemical equations and mole calculations are some important tools that chemists use to work out how chemicals react and how much of each substance they form. 1. Naming Chemical Compounds Page 4 2. Covalent Chemical Formulae Page 5 3. Ionic Chemical Formulae Page Chemical equations Page A Chemical Measure Page Can You Make a Profit? Page 28 P & L. Johnson

3 Nat 4 outcomes O Nat 5 outcomes By the end of this unit you should know the following: A chemical reaction which can be described using word equations, can also be described using chemical symbol equations. How to use balanced chemical equations to work out the mass of product formed. Chemical and ionic formulae including compounds containing group ions are used. The chemical formula of a covalent molecular substance gives the number of atoms present in the molecule. The formula of a covalent network or ionic compound gives the simplest ratio of atoms/ions in the substance. The gram formula mass is defined as the mass of one mole of a substance. Using the chemical formula of any substance the gram formula mass can be calculated using relative atomic masses of its constituent elements. The concentration of solutions in moles per litre. Calculations to determine the concentration and volume and the mass of a substance through the number of moles present. P & L. Johnson

4 Naming Compounds If a compound contains just two elements the compound name consists of the names of both elements but the last part of the second name becomes ide. e.g. magnesium and oxygen make magnesium oxide. carbon and chlorine make carbon chloride. If a compound contains more than two elements and one of them is oxygen the compound name consists of the names of both the other elements but the last part of the second name becomes ate or ite depending on the number of oxygen atoms. e.g. copper, sulfur and oxygen make copper sulfate. sodium, phosphorus and oxygen make sodium phosphate. DISCUSS After discussion with your teacher and others make sure you know how to name simple two and three element compounds. ACTIVITY 4.1 Naming Compounds Game Your teacher may let you play this simple card game to practice naming simple two and three element compounds. NOTE Fill in the blanks in the following table: Elements silver & sulfur Phosphorus & hydrogen lithium & iodine magnesium, nitrogen & oxygen hydrogen, carbon & oxygen calcium, bromine & oxygen Compound Name Silver sulfide phosphorus hydride Lithium iodide Magnesium nitrate Hydrogen carbonate calcium bromate P & L. Johnson

5 Symbols can be used to represent atoms of the different elements. A chemical formula is used to represent the type and number of each element in a compound. E.g. water (hydrogen oxide) has the chemical formula H 2 O this means that in every water molecule there are 2 hydrogen atoms bonded to 1 oxygen atom. DISCUSS After discussion with your teacher and others make sure you know how to work out the number of atoms of each element in a compound from its chemical formula. NOTE Complete the following table: Name You will remember from the previous topic on bonding that atoms join together in order to form a complete outer shell. In covalent bonding they do this by sharing their unpaired outer (valence) electrons. DISCUSS After discussion with your teacher and others make sure you know how to work out the valency of elements. NOTE Complete the following table: Chemical formula Number and type of each element hydrogen chloride HCl 1 hydrogen 1 chlorine magnesium fluoride MgF 2 1 magnesium 2 fluorine Silicon oxide SiO 2 1 silicon 2 oxygen Carbon sulfide CS 2 1 carbon 2 sulfur Group Number Valency P & L. Johnson

6 If we look at an example of two non-metal elements, oxygen and hydrogen, joining together to form water: You can see that oxygen has two unpaired outer electrons, a valency of 2 and hydrogen just one unpaired electron, a valency of 1. The oxygen therefore needs two hydrogen atoms to fill its outer electron clouds. A ratio of 1:2 oxygen to hydrogen atoms, in other words the valency number of one element tells you the number of atoms of the other element you need. This is called crossing over the valencies. An easy way of working out the formulae of simple compounds is to do the following: Step 1 Write down the symbols: H O 1 2 Step 2 Write the valency above H O each element 1 2 Step 3 Cross over the valency H O using arrows 2 1 Step 4 Simplify the numbers No need to write the 1. H 2 O Another example would be carbon sulfide: Step 1 Write down the symbols: C S 4 2 Step 2 Write the valency above C S each element 4 2 Step 3 Cross over the valency C S using arrows 2 4 Step 4 Simplify the numbers CS 2 No need to write the 1. P & L. Johnson

7 DISCUSS After discussion with your teacher and others make sure you know how to work out the chemical formulae of simple two element compounds. NOTES Work out the chemical formulae of the following compounds using crossing over the valencies method: 1. hydrogen sulfide 2. phosphorous chloride S H S S P Cl V 1 2 V 3 1 S 2 1 S 1 3 D 2 1 D 1 3 F H 2 S F PCl 3 3. silicon oxide 4. nitrogen hydride S Si O S N H V 4 2 V 3 1 S 2 4 S 1 3 D 1 2 D 1 3 F SiO 2 F NH 3 Your teacher may get you to try some extra questions on this. Exceptions Unfortunately in chemistry there are lots of substances that don t have chemical formulae that can be worked out using the above method, sometimes two elements can combine in two or more different ways. However these exceptions are given names that allow us to simply work out the formula from the name. Prefixes such as mon, di, tri, tetra, etc. are used to give the number of atoms of each element in the compound. NOTES Complete the following table: Prefix Number of atoms Prefix Number of atoms mon 1 tetra 4 di 2 penta 5 tri 3 hexa 6 P & L. Johnson

8 Carbon can combine with oxygen in two ways forming compounds with the chemical formula CO and CO 2. In order to tell them apart they are given the following names: CO = carbon monoxide CO 2 = carbon dioxide Here the prefixes tell you the number of oxygen atoms in the chemical formula. Another set of similar compounds are sulfur dioxide SO 2 and sulfur trioxide SO 3. DISCUSS After discussion with your teacher and others make sure you know how to work out the chemical formulae of simple two element compounds when given a chemical name containing a prefix. NOTES Work out the chemical formulae of the following compounds with a chemical name containing a prefix. 1. nitrogen dioxide 2. carbon tetrachloride NO 2 CCl 4 3. xenon hexafluoride 4. dinitrogen tetroxide XeF 6 N 2 O 4 Your teacher may get you to try some extra questions on this. ACTIVITY 4.2 Covalent Bonding Pairs Game Your teacher may let you play this simple card game to practice working out chemical formulae of covalent compounds. P & L. Johnson

9 Quick Test 1 1. Complete the following table: NAME OF COMPOUND ELEMENTS PRESENT Potassium chloride Sodium nitrate Copper(II) sulfate Lithium fluoride Magnesium nitrate Silver(I) bromide Hydrogen chloride Calcium carbonate 2. Work out the chemical formulae of the following compounds. 2a-f Formula by SVSDF (no prefix) 2g-j Formula from prefix a. phosphorus chloride b. nitrogen iodide c. sulfur bromide PCl 3 NI 3 SBr 2 d. hydrogen fluoride e. silicon carbide f. germanium hydride HF SiC GeH 4 g. dihydrogen oxide h. silicon tetrafluoride H 2 O SiF 4 i. uranium hexafluoride j. phosphorus pentachloride UF 6 PCl 5 3. The compound N 2 O 4 is called A. nitrogen monoxide B. nitrogen dioxide C. dinitrogen tetroxide D. dinitrogen trioxide 4. The valency of phosphorus in P 2 S 3 is: S P S A. 2 V 3 2 B. 3 S 2 3 C. 5 D 2 3 D. 1 F P 2 S 3 Work SVSDF backwards (i.e. start on formula line & ignore divide) to get answer P & L. Johnson Potassium & chlorine Sodium, nitrogen & oxygen Copper, sulfur & oxygen Lithium & fluorine Magnesium, nitrogen & oxygen Silver & bromine Hydrogen & chlorine Calcium, carbon & oxygen

10 You will remember from the previous topic on bonding that atoms join together in order to form a complete outer shell. In ionic bonding they do this by transferring their unpaired outer (valence) electrons from the metal atoms to the non-metal atoms, as shown in the example below of sodium chloride. Na The valency of the elements is the same as the size of their charge on the resulting ion; metals in group 1 having a 1+ charge, non-metals in group 6 having a 2 charge, etc. In the same way as with covalent compounds we can work out the chemical formula by using the crossing over the valency method. Step 1 Write down the symbols: Na Cl 1 1 Step 2 Write the valency above Na Cl each element 1 1 Step 3 Cross over the valency Na Cl using arrows 1 1 Step 4 Simplify the numbers No need to write the 1. NaCl This is the chemical formula, if you are asked for the ionic formula you need to show the charges on the ions as well. Na + Cl - P & L. Johnson

11 DISCUSS After discussion with your teacher and others make sure you know how to work out the chemical formulae of simple two element ionic compounds. NOTES Work out the chemical and ionic formulae of the following compounds using crossing over the valencies method: 1. magnesium bromide 2. potassium oxide S Mg Br S K O V 2 1 V 1 2 S 1 2 S 2 1 D 1 2 D 2 1 F MgBr 2 F K 2 O IF Mg 2+ (Br - ) 2 IF (K + ) 2 O 2-3. aluminium oxide 4. calcium nitride S Al O S Ca N V 3 2 V 2 3 S 2 3 S 3 2 D 2 3 D 3 2 F Al 2 O 3 F Ca 3 N 2 IF (Al 3+ ) 2 (O 2- ) 3 IF (Ca 2+ ) 3 (N 3- ) 2 Your teacher may get you to try some extra questions on this. ACTIVITY 4.3 Ionic Bonding Pairs Game Your teacher may let you play this simple card game to practice working out chemical formulae of ionic compounds. Exceptions Unfortunately in chemistry there are lots of substances that don t have chemical formulae that can be worked out using the above method, this is because transition metals can have more than one valency and have a group all of their own with no group number. In order to let you know what the valency of the transition metal is chemists use roman numerals in the chemical name. Roman Numeral Valency Roman Numeral Valency I 1 IV 4 II 2 V 5 III 3 VI 6 P & L. Johnson

12 Iron can have two different valencies, 2 or 3. Therefore the chemical names of iron oxide can be given as iron(ii) oxide or iron(iii) oxide. The Roman numerals tell you the valency of the iron, oxygen has its usual valency of 2. In order to work out the chemical formula of each form of iron oxide you use the crossing over the valency method as usual so: Iron(II) oxide 2 2 Fe O 2 2 FeO or Fe 2+ O 2- Iron(III) oxide 3 2 Fe O 2 3 Fe 2 O 3 or Fe 2 3+ O 3 2- NOTES Work out the chemical and ionic formulae of the following compounds using crossing over the valencies method: 1. copper(i) oxide 2. vanadium(v) chloride S Cu O S V Cl V 1 2 V 5 1 S 2 1 S 1 5 D 2 1 D 1 5 F Cu 2 O F VCl 5 IF (Cu + ) 2 O 2- IF V 5+ (Cl - ) 5 3. Nickel(II) sulfide 4. cobalt(ii) chloride S Ni S S Co Cl V 2 2 V 2 1 S 2 2 S 1 2 D 1 1 D 1 2 F NiS F CoCl 2 IF Ni 2+ S 2- IF Co 2+ (Cl - ) 2 P & L. Johnson

13 Ionic compounds containing more than two atoms Some ionic compounds contain more than two elements, their chemical names tend to end in ate or ite as we have already seen, but there are a few others. On p 8 of your data booklet you will see a table containing all the ones you will need to know. Ions containing more than two elements are known as group ions. NOTES Use page 8 of your data booklet to complete the following table: Group ion Formula Valency Charge nitrate NO 3 - carbonate CO hydroxide OH ammonium NH 4 + phosphate PO As many of the formulae for the group ions contain numbers it is important not to get these mixed up with the final number of ions in the chemical formula. For this reason groups ions are always placed in brackets, as shown below: Magnesium nitrate: Step 1 Write down the symbols: Mg (NO 3 ) Note the brackets 2 1 Step 2 Write the valency above Mg (NO 3 ) each ion 2 1 Step 3 Cross over the valency Mg (NO 3 ) using arrows 1 2 Step 4 Simplify the numbers Mg(NO 3 ) 2 Note the brackets remain If you are showing the ionic formula remember the charge on the group ion must be inside the bracket: Mg 2+ (NO 3 - ) 2 P & L. Johnson

14 Calcium sulfate: Step 1 Write down the symbols: Ca (SO 4 ) Note the brackets 2 2 Step 2 Write the valency above Ca (SO 4 ) each ion 2 2 Step 3 Cross over the valency Ca (SO 4 ) using arrows 2 2 Step 4 Simplify the numbers CaSO 4 or Ca 2+ SO 4 - Note the brackets go as there is only one sulfate ion NOTES Work out the chemical and ionic formulae of the following compounds using crossing over the valencies method: 1. magnesium carbonate 2. potassium sulfate 2-2- S Mg CO 3 S K SO 4 V 2 2 V 1 2 S 2 2 S 2 1 D 1 1 D 2 1 F MgCO 3 F K 2 SO 4 IF Mg CO 3 IF (K + 2- ) 2 SO 4 3. aluminium chromate 4. calcium phosphate 2-3- S Al CrO 4 S Ca PO 4 V 3 2 V 2 3 S 2 3 S 3 2 D 2 3 D 3 2 F Al 2 (CrO 4 ) 3 F Ca 3 (PO 4 ) 2 IF (Al 3+ ) 2 (CrO 4 2- ) 3 IF (Ca 2+ ) 3 (PO 4 3- ) 2 P & L. Johnson

15 Quick Test 2 1. The correct formula for sodium oxide is A. SO 2 B. S 2 O C. NaO D. Na 2 O 2. The correct formula for copper(ii) nitrate is A. CuNO 3 B. Cu(NO 3 ) 2 C. Cu 2 NO 3 D. Cu 2 N 3 3. In the compound Cr 2 (SO 4 ) 3, chromium has a valency of A. 4 B. 3 Work backwards from formula by SVSDF C. 2 D. 1 4 X is a metal. It forms a compound with fluorine with the formula XF 2. The metal X must belong to group A. 1 B. 2 Work backwards from formula by SVSDF. All metals with valency of 2 C. 3 must be in group 2 D Z is a non-metal. It forms a compound with magnesium with the formula Mg 3 Z 2. The non-metal Z must belong to group A. 3 B. 5 Work backwards from formula by SVSDF. All non-metals with valency C. 6 of 3 must be in group 5 (Remember group valency = 1, 2, 3, 4, 3, 2, 1, 0) D Which of the following pairs of elements would form a compound with a formula of the type X 2 Y 3? Where X is a metal and Y is a non-metal Work backwards from formula by X Y SVSDF. Metal with valency of 3 A potassium chlorine (Al only metal with this valency). B calcium oxygen So answer must be D. Sulfur is non C magnesium nitrogen -metal with a valency of two. D aluminium sulfur P & L. Johnson

16 Word Equations During a chemical reaction reactants are turned into new substances called products. This can be shown by a word equation where the chemicals reacting are shown on the left hand side and the products formed on the right. For example, in the reaction between magnesium ribbon and oxygen gas, a white powder magnesium oxide is produced. The word equation for this is: magnesium + oxygen magnesium oxide Notice that only the chemical names are used in the equation and not descriptions. DISCUSS After discussion with your teacher and others make sure you know how to work out a word equation from a description of a chemical reaction. NOTES Work out the word equations from the following descriptions of chemical reactions: 1. Sodium metal reacts violently with water producing hydrogen gas and a solution of sodium hydroxide. Sodium + water hydrogen + sodium hydroxide 2. Carbon dioxide gas and black copper oxide powder are formed when green copper carbonate powder is heated in a test tube. Copper carbonate carbon dioxide +copper oxide 3. When iron is produced in a blast furnace from iron ore, the iron(iii) oxide in the ore reacts with carbon monoxide gas. Carbon dioxide is also formed in the reaction. Iron(III) oxide + carbon monoxide iron + carbon dioxide P & L. Johnson

17 Chemical Equations Chemists often use shorthand to write chemical equations. They do this by using the chemical formula of each reactant and product and then simply swap them for the words in a word equation. Extra information can also be included by using state symbols. Solid = (s) Liquid = (l) Gas = (g) Aqueous solution = (aq) E.g. magnesium + oxygen magnesium oxide 2 2 Mg (s) + O 2(g) MgO (s) Mg O = MgO 2 2 Remember which elements are diatomic gases (H 2, N 2, O 2, F 2, Cl 2, Br 2 & I 2 ) For compounds you will need to correctly work out their formula using the crossing over the valency method, unless they have a prefix like mon or di in their name. DISCUSS After discussion with your teacher and others make sure you know how to convert word equations into chemical equations. NOTES Convert the following word equations into chemical equations. Remember diatomic elements (HON 7). All compound formulae worked out by SVSDF. Phosphorus exists naturally as a tetratomic element. 1. sodium (s) + chlorine (g) sodium chloride (s) Na (s) + Cl 2 (g) NaCl(s) 2. phosphorus (s) + chlorine (g) phosphorus chloride (s) P 4 (s) + Cl 2 (g) PCl 3 (s) 3. hydrogen (g) + iodine (g) hydrogen iodide (g) H 2 (g) + I 2 (g) HI(g) Your teacher may get you to try some extra questions on this. P & L. Johnson

18 Balanced Chemical Equations ACTIVITY 4.4 Balanced Chemical Reactions Try making a molecule each of hydrogen H 2 and oxygen O 2. Use them to make a molecule of water H 2 O. The chemical equation for this reaction should be: H 2 + O 2 H 2 O But you should find that you have an atom of oxygen left over. In reality this cannot happen as we cannot create or destroy matter, so in order to use up this extra oxygen atom we need to react it with another hydrogen molecule. + 2H 2 + O 2 2H 2 O This is known as a balanced chemical equation where there are equal numbers of each type of atom on both sides of the equation. It is important when writing balanced chemical equations not to alter the actual chemical formulae of any of the reactants or products, just the number of molecules. ACTIVITY 4.5 Balanced Chemical Equation Activity Your teacher may let you play this activity which should help you to understand how to balance chemical equations. An easy way to help you balance chemical equations is to use the following accountancy method where you count the number of atoms of each element on both sides of the equation: Mg + 2HCl MgCl 2 + H 2 Mg = 1 Mg = 1 H = 1 x2=2 H = 2 Cl = 1 x 2=2 Cl = 2 By having 2 HCl molecules the equation is balanced. P & L. Johnson

19 DISCUSS After discussion with your teacher and others make sure you know how to balance chemical equations. NOTES Balance the following chemical equations: 1. C + 2Cl 2 CCl 4 Step 1 balance Cl C = 1 C = 1 Cl = 2 4 Cl = 4 2. Mg + 2AgNO 3 Mg(NO 3 ) 2 + 2Ag Mg = 1 Mg = 1 Step 1 balance N or O Ag = 1 2 Ag = 1 2 Step 2 balance Ag N = 1 2 N = 2 O = 3 6 O = 6 3. C 2 H 4 + 3O 2 2CO 2 + 2H 2 O C = 2 C = 1 2 Step 1 balance C H = 4 H = 2 4 Step 2 balance H O = 2 O = Step 3 balance O 4. Ca + 2H 2 O Ca(OH) 2 + H 2 Ca = 1 Ca = 1 Step 1 balance H or O H = 2 4 H = 4 O = 1 2 O = 2 5. N 2 + 3I 2 2NI 3 N = 2 N = 1 2 Step 1 balance N I = 2 6 I = 3 6 Step 2 balance I Your teacher may get you to try some extra questions on this. P & L. Johnson

20 Quick Test 3 1. Write word equations for the following reactions: a. When lithium metal is added to water, a gas is given off, called hydrogen. A solution is also formed called lithium hydroxide. lithium + water lithium hydride b. Carbon dioxide gas and black copper oxide are formed when green copper carbonate powder is heated. Copper carbonate carbon dioxide + copper oxide c. Hydrogen peroxide is an unstable liquid that spontaneously decomposes to form water and oxygen gas. Hydrogen peroxide water + oxygen 2. Convert the following into balanced chemical equations: a. C + 2S CS 2 Step 1 balance S C = 1 C = 1 S = 1 2 S = 2 b. 2Li + ZnSO 4 Zn + Li 2 SO 4 Li = 1 2 Li = 2 Step 1 balance Li Zn = 1 Zn = 1 S = 1 S = 1 O = 4 O = 4 FOR THE NEXT TWO: CHEMICAL EQUATION MUST BE WRITTEN FIRST c. sulfur dioxide + oxygen sulfur trioxide 2SO 2 + O 2 2SO 3 Formula from prefix S = 1 2 S = 1 2 Step 1 try balance O O = 4 6 O = 3 6 Step 2 rebalance S (and O) d. nitrogen + hydrogen nitrogen hydride N 2 + 3H 2 2NH 3 Formula from SVSDF N = 2 N = 1 2 Step 1 balance N H= 2 6 H= 2 6 Step 2 balance H P & L. Johnson

21 We have just seen with balanced chemical equations that it is important that we have the right number of molecules reacting, but molecules are unimaginably small and it is impossible to see them to count, so how do chemists measure quantities of chemicals? If we just compare masses we could have two things weighing the same. For example, 1kg of cricket balls and 1kg of ping pong balls. However would we have the same number of balls? The answer is obviously no, there would need to be many more ping pong balls as they are so light. Chemicals are the same they all weigh different amounts, some molecules are very light others much heavier in comparison. We need a way that we can compare the number of molecules rather than the mass. ACTIVITY 4.6 A Chemical Measure Weigh out a dozen marbles or other small object that are all the same. NOTES Complete the following without using the balance again: Mass of a dozen = 8.50 g Mass of one = 0.71 g Mass of 2 dozen = 17 g Mass of 0.5 dozen = 4.25 g A dozen is just a fixed number, 12, and so if we know the mass of one dozen we can use simple calculations to find the mass of other amounts of the object. In chemistry we use a fixed amount called a mole, like a dozen it is a fixed number it just happens to be very large 602,000,000,000,000,000,000,000! This is because atoms and molecules are so small we need a colossal number of them to be able to weigh them. P & L. Johnson

22 The good news is that we don t need to know how many atoms or molecules are in a mole we just need to know how much one mole of a particular substance weighs. Just like you did with the small objects you weighed earlier. The mass of one mole of atoms is the atomic mass expressed in grams so: Therefore 1 mole of carbon = 12g 2 moles of carbon = 24g 1 mole of sodium = 23g Therefore 0.1 mole of sodium = 2.3g When we are looking at compounds we need to add up the masses of all the atoms in the chemical formula. This gives us the gram formula mass or GFM, where the GFM is the mass of one mole of a substance expressed in grams. H 2 O Mg(NO 3 ) 2 1x16 2x1 18g 6x16 2x14 1x24 148g So now 2 moles of H 2 O = 36g And 0.1 moles of Mg(NO 3 ) 2 = 14.8g To calculate the mass of a substance we have simply multiplied the number of moles by the GFM, this leads to the expression: mass = number of moles x GFM Or m = n x GFM We can rearrange the formula to find the number of moles of a substance if we know its mass: number of moles = mass GFM Or n = m GFM m n GFM P & L. Johnson

23 DISCUSS After discussion with your teacher and others make sure you know how to work out the number of moles of a substance. NOTES Work out the following, showing your working out: 1. What is the GFM of the following substance? a. NaCl b. CaF 2 c. CuSO 4 d. (NH 4 ) 2 CO 3 (1 x 23) (1 x 40) (1 x 63.5) (2 x 14) + (8 x 1) + (1 x 35.5) + (2 x 19) + (1 x 32) + (1 x 12) + (2 x 16) = 58.5 = 78 + (4 x 16) = 80 = What is the mass of the following? GFM for each worked out in Qu 1 above. a. 2 moles of NaCl b. 0.2 moles of CaF 2 1 mole NaCl = 58.5g 1 mole CaF 2 = 78 g (x2) (x2) (x 0.2) (x 0.2) 2 moles NaCl = 117 g 0.2 moles CaF 2 = 15.6 g OR m= n x GFM OR m= n x GFM m = 2 x 58.5 = 117g m = 0.2 x 78 = 15.6g c. 3 moles of CuSO 4 d. 0.5 moles of (NH 4 ) 2 CO 3 1 mole CuSO 4 = g 1 mole (NH 4 ) 2 CO 3 = 80 g (x3) (x3) (x 0.5) (x 0.5) 3 moles CuSO 4 = 478.5g 0.5 moles (NH 4 ) 2 CO 3 = 40 g OR m= n x GFM OR m= n x GFM m = 3 x = 478.5g m = 0.5 x 80 =40g 3. What is the number of moles in the following? GFM for each worked out in Qu 1 above. a. 5.85g of NaCl b. 380g of CaF g = 1 mole NaCl 78 g = 1 mole CaF 2 (/58.5) (/58.5) (/78) (/78) 1g = moles NaCl 1g = moles CaF 2 (x5.85) (x5.85) (x 380) (x 380) 5.85 g = 0.1 moles NaCl 380g = 4.87 moles CaF 2 P & L. Johnson

24 3. What is the number of moles in the following? GFM for each worked out in Qu 1 above. a. 5.85g of NaCl b. 380g of CaF 2 Alternative Method Alternative Method n = m/gfm n = m/gfm n = 5.85/58.5 = 0.1 moles NaCl n = 380/78 = 4.87 moles CaF 2 c. 1.6g of CuSO 4 d. 192g of (NH 4 ) 2 CO g = 1 mole CuSO 4 80 g = 1 mole (NH 4 ) 2 CO 3 (/159.5) (/159.5) (/80) (/80) 1g = moles CuSO 4 1g = moles (NH 4 ) 2 CO 3 (x1.6) (x1.6) (x 192) (x 192) 1.6 g = 0.01 moles CuSO 4 192g = 2.4 moles (NH 4 ) 2 CO 3 Alternative Method n = m/gfm Alternative Method n = m/gfm n = 1.6/159.5 = 0.01 moles CuSO 4 n = 192/80 = 2.4 moles (NH 4 ) 2 CO 3 P & L. Johnson

25 Solutions When you make up an aqueous solution you are dissolving a solid (solute) in water (solvent). The more solid you dissolve the more concentrated the solution. Alternatively we could dissolve the same mass of solid but reduce the volume of water used which would also make the solution more concentrated. So concentration depends on the mass of solid dissolved and the volume of water used. We have already seen that we cannot compare masses in chemistry but can compare number of moles. In chemistry we also use the litre as a standard measurement of volume of liquid. This leads to the following formula: Concentration = number of moles of solid Volume of water in litres Or c = n which can be rearranged to n = cv V n c v The units of concentration are moles per litre written as moll -1 or mol/l For example if you dissolve 2 moles of sodium chloride in 0.5litres of water you will get a 4 moll -1 sodium chloride solution. c = n so c = 2 c = 4moll -1 V 0.5 You can also use your knowledge of how to calculate the number of moles of a solid in order to work out what mass of solid to dissolve in a particular volume of water: For the above example you would need to dissolve 117g of sodium chloride in 0.5litres to get a 4moll -1 solution. m = n x GFM so m = 2 x 58.5 m = 117g P & L. Johnson

26 ACTIVITY 4.7 Making Solutions Make up the following solutions as accurately as possible. Your teacher may show you how to use more accurate glssware: GFM NaCl worked out previously = 58.5g NB To convert cm 3 into litres divide by This must be done for calculations involving concentration since the units of concentration are moles per litre. A. Dissolve 5.85g of sodium chloride in 100cm 3 of water. (0.1 litres of water) B. Dissolve 7.2g of sodium chloride in 250cm 3 of water. (0.25 litres of water) Work out their concentrations. A. 58.5g = 1 mole NaCl 0.1 litres contains 0.1 moles NaCl (/58.5) (/58.5) (/0.1) (/0.1) 1g = 0.017moles NaCl 1 litre contains 1 mole NaCl (x5.85) (x5.85) i.e. concentration = 1 mol l g = 0.1 moles NaCl Alternative method ( still have to work out how many moles either by proportion as shown above or using n=m/gfm calculation as explained previously) n = m/gfm c=n/v n = 5.85/58.5 = 0.1 moles NaCl c=0.1/0.1 = 1 mol l -1 B. 58.5g = 1 mole NaCl 0.25 litres contains 0.12 moles NaCl (/58.5) (/58.5) (/0.25) (/0.25) 1g = 0.017moles NaCl 1 litre contains 0.49 mole NaCl (x7.2) (x7.2) i.e. concentration = 0.49 mol l g = 0.12 moles NaCl Alternative method ( still have to work out how many moles either by proportion as shown above or using n=m/gfm calculation as explained previously) n = m/gfm c=n/v n = 7.2/58.5 = 0.12 moles NaCl c=0.12/0.25 = 0.49 mol l -1 P & L. Johnson

27 Quick Test 4 1. Calculate the number of moles of solute that needs to be dissolved to make the following solutions. a. 500cm 3 of a 1.0 moll -1 potassium hydroxide solution. (NB: 500cm 3 = 0.5 litre) 1.0 moll -1 means 1 litre contains 1 mole (x0.5) (x0.5) so 0.5 Litres contains 0.5 moles Alternative method n = c x v n = 1.0 x 0.5 = 0.5 moles b. 200cm 3 of a 0.5 moll -1 magnesium chloride solution. (NB: 200cm 3 = 0.2 litre) 0.5 moll -1 means 1 litre contains 0.5 mole (x0.2) (x0.2) so 0.2 Litres contains 0.1 moles Alternative method n = c x v n = 0.5 x 0.2 = 0.1 moles c. 2L of a 2.0 moll -1 silver(i) nitrate solution. 2.0 moll -1 means 1 litre contains 2 mole (x2) (x2) so 2 Litres contains 4 moles Alternative method n = c x v n = 2 x 2 = 4 moles 2. Calculate the mass of solute required to make the following solutions. a. 250cm 3 of a 0.5 moll -1 sodium hydroxide (NaOH) solution. (NB: 250cm 3 = 0.25 litres) 0.5 moll -1 means 1 litre contains 0.5 mole (x0.25) (x0.25) so 0.25 Litres contains moles GFM of NaOH = (1x23) + (1x16) + (1x1) = 40g 1 mole = 40g NaOH (x0.125) (x 0.125) mole = 5g NaOH P & L. Johnson

28 Quick Test 4 2. Calculate mass of solute required to make the following solutions. a. 250cm 3 of a 0.5 moll -1 sodium hydroxide (NaOH) solution. Alternative method n = c x v n = 0.5 x 0.25 = moles NaOH in 250cm 3 GFM of NaOH = (1 x 23) + (1 x 16) + (1 x 1) = 40g m = n x GFM = x 40 = 5 g NaOH b. 50cm 3 of a 0.5 moll -1 copper sulfate (CuSO 4 ) solution. (NB: 50cm 3 = 0.05 litres) 0.5 moll -1 means 1 litre contains 0.5 mole (x0.05) (x0.05) so 0.25 Litres contains moles GFM of CuSO 4 = (1 x 63.5) + (1 x 32) + (4 x 16) = 159.5g 1 mole = 159.5g CuSO 4 (x0.025) (x 0.025) mole = 3.99g CuSO 4 Alternative method n = c x v n = 0.5 x 0.05 = moles CuSO 4 in 50cm 3 GFM of CuSO 4 = (1 x 63.5) + (1 x 32) + (4 x 16) = 159.5g m = n x GFM = x = 3.99g CuSO 4 P & L. Johnson

29 Quick Test 4 2. Calculate mass of solute required to make the following solutions. c. 1.5L of a 2.0 moll -1 magnesium nitrate (Mg(NO 3 ) 2 ) solution. 2.0 moll -1 means 1 litre contains 2.0 mole (x1.5) (x1.5) so 1.5 Litres contains 3 moles GFM of Mg(NO 3 ) 2 = (1 x 24.5) + (2 x 14) + (6 x 16) = 134.5g 1 mole = 134.5g Mg(NO 3 ) 2 (x3) (x3) 3 mole = 403.5g Mg(NO 3 ) 2 Alternative method n = c x v n = 2.0 x 1.5 = 3 moles CuSO 4 in 1.5 litres GFM of Mg(NO 3 ) 2 = (1 x 24.5) + (2 x 14) + (6 x 16) = 134.5g m = n x GFM = 3 x = 403.5g Mg(NO 3 ) 2 P & L. Johnson

30 Using Balanced Chemical Equations: It is possible to use a balanced chemical equation to work out how much product is produced during a chemical reaction or how much reactant is needed to produce a certain amount of product. If 32g of methane (CH 4 ) is burnt, how much water is formed? CH 4 + 2O 2 CO 2 + 2H 2 O 2H 2 O If 1 mole produces 2moles 2x16 16g 36g 4x1 1g CH 4 32g 36 x 32 4x1 16 1x12 72g 16 Another way of tackling this question is to use mole calculations: CH 4 + 2O 2 CO 2 + 2H 2 O 1 : 2 : 1 : 2 m= 32g m=? CH 4 n = m 4x1 GFM 1x12 n = n = 2 n = 2 x 2 H 2 O m = n x GFM 1x16 m = 4 x 18 2x1 m = 72g 18 DISCUSS Your teacher may want you to use one particular method. After discussion with your teacher and others make sure you know how to use balanced chemical equations to work out the mass of a substance produced in a chemical reaction. P & L. Johnson

31 NOTES Work out the following, showing your working out: a. What mass of water is produced when 8g of hydrogen burns in oxygen? 2H 2 + O 2 2H 2 O GFM of H 2 = (2x1) = 2g 2moles produces 2moles GFM of H 2 0 = (2x1) + (1x16)=18g 2 x 2g produces 2 x 18g 4g H 2 produces 36g H 2 O (/4) (/4) 1g H 2 produces 9g H 2 O (x8) (x8) 8g H 2 produces 72g H 2 O b. What mass of carbon dioxide is formed when 64g of ethene (C 2 H 4 ) burns in oxygen? C 2 H 4 + 3O 2 2CO 2 + 2H 2 O 1 mole produces 2moles GFM of C 2 H 4 = (2x12)+(4x1)=28g 1 x 28g produces 2 x 44g GFM of CO 2 = (1x12)+(2x16)=44g 28g C 2 H 4 produces 88g CO 2 (/28) (/28) 1g C 2 H 4 produces 3.14g CO 2 (x64) (x64) 64g C 2 H 4 produces g CO 2 c. What mass of oxygen is required to burn 32g of methane (CH 4 ) completely? CH 4 + 2O 2 CO 2 + 2H 2 O 1 mole reacts with 2moles GFM of CH 4 = (1x12)+(4x1)=16g GFM of O 2 = (2x16)=32g 1 x 16g reacts with 2 x 32g 16g CH 4 reacts with 64g O 2 (/16) (/16) 1g CH 4 reacts with 4g O 2 (x32) (x32) 32g CH 4 reacts with 128g O 2 P & L. Johnson

32 A small chemical company that has just developed a new process for turning vinegar into unleaded petrol. The process is described below: Carbon Dioxide Petrol Vinegar Distillation During the process vinegar, which is a solution of ethanoic acid (CH 3 CO 2 H (aq) ), is distilled to separate the ethanoic acid from the water. The ethanoic acid is then super heated and reacted with powdered carbon (C (s) ) along with a special, top secret, catalyst. A hydrocarbon similar to petrol (octene, C 8 H 16 ) is formed along with carbon dioxide (CO 2(g) ) as a waste product. NOTES Water Heater Reactor Carbon a. Below is the chemical equation for the reaction, can you balance it. 4CH 3 CO 2 H (g) + 4C (s) 4CO 2(g) + C 8 H 16(l) C = 3 9 C = 9 12 Step 1 balance H (LHS) H = 4 16 H = 16 Step 2 balance O (RHS) O = 2 8 O = 2 8 Step 3 rebalance C (LHS) b. What mass of petrol (C 8 H 16(l) ) will be produced if 240kg of ethanoic acid reacts with 48kg of carbon? GFM of petrol C 8 H 16 = (8x12)+(16x1) = 112g GFM of ethanoic acid, CH 3 CO 2 H = (2x12) + (4x1) + (2x16) = 60g GFM of carbon, C = (1x12g) = 12g P & L. Johnson

33 b. What mass of petrol (C 8 H 16(l) ) will be produced if 240kg of ethanoic acid reacts with 48kg of carbon? 4CH 3 CO 2 H (g) + 4C (s) 4CO 2(g) + C 8 H 16(l) Normally in NAT 5 calculations from balanced equations only 2 substances are mentioned in the question stem (the two necessary to carry out the calculation). In this example 3 substances are mentioned. In this example either ethanoic acid can be used to work out the mass of petrol OR carbon can be used to work out the mass of petrol as shown below. Please note that in this particular question it doesn t matter which of the pairs of substances you chose but this is not always the case! However, such examples are not covered until higher chemistry. Calculation Using Ethanoic Acid and Petrol 4CH 3 CO 2 H (g) + 4C (s) 4CO 2(g) + C 8 H 16(l) 4 moles CH 3 CO 2 H (g) 1 mole C 8 H 16(l) (4 x 60g) CH 3 CO 2 H (g) (1 x 112g) C 8 H 16(l) 240g CH 3 CO 2 H (g) 112g C 8 H 16(l) So 240kg CH 3 CO 2 H (g) 112kg C 8 H 16(l) Calculation Using Carbon and Petrol 4CH 3 CO 2 H (g) + 4C (s) 4CO 2(g) + C 8 H 16(l) 4 moles C (s) 1 mole C 8 H 16(l) (4 x 12g) C (s) (1 x 112g) C 8 H 16(l) 48g C (s) 112g C 8 H 16(l) So 48kg C (s) 112kg C 8 H 16(l) c. If Vinegar costs 1.00 per kg, coal costs 2.00 per kg and running the plant costs 1 per kg of petrol produced; what is the minimum price the company must sell their petrol at per kg to make a profit. Step 1 Calculate the quantity of reactants required to make 1kg petrol. From question b we know that: 112kg of petrol was made using 240kg of ethanoic acid (/112) (/112) So 1 kg of petrol will be made using 2.14kg of ethanoic acid From question b we also know that: 112kg of petrol was made using 48kg of carbon (/112) (/112) So 1 kg of petrol will be made using 0.43kg of carbon P & L. Johnson

34 c. If Vinegar costs 1.00 per kg, coal costs 2.00 per kg and running the plant costs 1 per kg of petrol produced; what is the minimum price the company must sell their petrol at per kg to make a profit. Step 2 Calculate cost to make 1 kg petrol Vinegar = 1 per kg x 2.14kg = 2.14 Coal = 2 per kg x 0.43kg = 0.86 Running Plant = 1 per kg = 1.00 Total cost to make 1kg petrol = 4.00 Minimum price per kg of petrol to make any profit = 4.01 d. If 1 litre of petrol weighs roughly 700g can the business make a profit with today s petrol prices? Minimum price is 4.01 per kg of petrol (1 litre of petrol= 700g) i.e. 1000g = 4.01 (/1000) (/1000) 1g = (x700) (x700) 700g = 2.81 Minimum price per litre of petrol No profit could therefore be made since today s petrol prices are lower than this (~ 1.15 per litre). e. Would you invest in the company? No, since the method is not cost effective/ not profit generating. P & L. Johnson