Lavoisier: Law of conservation of mass! Chemical Reaction Mass before = Mass after .

1. Lavoisier: Law of conservation of mass! Chemical Reaction Mass before = Mass after. Chapter 3: Stoichiometry Scientific contributions cut short – Antoine Lavoisier was guillotined in 1794 at the age of 60. Today, considered to be father of modern chemistry.

2. Chemical Equations Chemical equations are descriptions of chemical reactions. There are 2 parts to an equation: reactants and products: H2 + O2H2O • To balance: change only the stoichiometric coefficients cannot change the chemical species Stoichiometric coefficients: numbers in front of the chemical formulas; give ratio of reactants and products.

3. Consider the unbalanced chemical equation: H2 + O2 H2O The equation H2 + O2 H2O2 is a balanced equation but for a different reaction. DO NOT change subscripts! Balancing options H2 + 1/2 O2 H2O, or 2H2 + O2 2H2O, or 4H2 + 2O2 4H2O, etc. Preferably, no fractions and smallest possible common coefficients used.

4. Let’s consider combustionreactionswhich involve the burning/oxidation of hydrocarbons to produce CO2 and H2O always. Hydrocarbons are organic compounds made up only of carbon and hydrogen. For burning/oxidation to occur O2 has to be present! All combustion equations will include O2(g) as a reactant! Example 1 Write a balanced equation for the combustion of octane, C8H18. We know that our reactants include C8H18 and O2(g), and the products will be CO2 and H2O. But in what ratios will they react and produce? Let’s balance!

5. C8H18 + O2(g)  CO2 + H2O We need to increase C on the product side to 8, using a stoichiometric coefficient. C8H18 + O2(g)  8CO2 + H2O We need to increase H on the product side to 18, using a stoichiometric coefficient. Start with C & H. Do O last because it appears in 3 species.

6. C8H18 + O2(g)  8CO2 + 9H2O We now need to increase O2 on the reactant side. It needs to change from 2 to 25. This can only be done using a fraction! C8H18 + 12.5 O2(g)  8CO2 + 9H2O Now to remove the fraction, and keep coefficients as small as possible we multiply ALL coefficients by 2! 2C8H18 + 25O2(g)  16CO2 + 18H2O = = =

7. Here is another set up of the same example: HINT: Balance O last (occurs in 3 species) but C and H only appear in two! C8H18 + O2 CO2 + H2O • Balance C C8H18 + O2 8CO2 + H2O • Balance H C8H18 + O2 8CO2 + 9H2O • Balance O C8H18 + 25/2O2 8CO2 + 9H2O 2C8H18 + 25O2 16CO2 + 18H2O • We usually add the physical states of the reactants and products 2C8H18(l) + 25O2(g)  16CO2(g) + 18H2O(l)

8. In combustion reactions: Rapid reactions that produce a flame. Usually very “clean” and products are predictable, but incomplete combustion (leftover ash) can be very harmful.

9. Example 2 Write a balanced equation for the combustion of purine, C5H4N4. C  CO2H  H2O N  N2

10. Patterns of chemical reactivity Elements in the same group of the periodic table react in a similar manner. Potassium Lithium Sodium Alkali metal + water  2Na + 2H2O  2NaOH + H2 2K + 2H2O  2KOH + H2

11. Alkali earth metal + water  Mg + 2H2O  Mg(OH)2 + H2 Ca + 2H2O  Ca(OH)2 + H2 Note: These are the EXACT same products as for alkali metals, except for the stoichiometry!!! Always look for patterns in chemistry!

12. Combination Reactions They are characterized by having fewer products than reactants. 2Mg(s) + O2(g)  2MgO(s) Mg has combined with O2 to form MgO. Two reactants combine to form a single product. Metal + oxygen 

13. Non-metal + hydrogen  C(s) + 2H2(g)  CH4(g) Metal + halogen  Ca(s) + Cl2(g)  CaCl2(s) 2Li(s) + F2(g)  2LiF(s)

14. Decomposition Reactions They are characterized by having fewer reactants than products. 2NaN3(s)  2Na(s) + 3N2(g) Metal azide The NaN3 is ignited and rapidly decomposes into Na and N2 gas Metal carbonate  D CaCO3(s)  CaO(s) + CO2(g) limestone, seashells lime

16. Quantitative vs. Qualitative Chemical formulae and equations both have qualitative and quantitative significance. Chemical formula says what compounds you are working with and the equation gives a qualitative idea of how the reactants react, and what products they produce. N2 +3H22NH3(balanced) Quality:We have nitrogen and hydrogen as reactants. The equation tells us that they undergo a combination reaction to form ammonia. Subscripts in formulae and coefficients in balanced equations represent precise quantities. Quantity:The product, NH3, is composed of 3H for every 1N atom. The balanced equation shows that 2 molecules of NH3 are produced for every N2 and 3H2 that are used up.

17. Average Atomic Mass REVISION FROM ISOTOPES! Most elements occur as mixtures of isotopes. To determine the average atomic mass of an element we must consider each isotopes mass as well as their respective abundances. Before we consider real isotopes, lets create an analogy using popcorn  I have a bowl of popcorn where 93% of the pieces weigh 1.14 mg, 2% weigh 1.09 mg and 5% weigh 1.03 mg. Which popcorn piece mass is in the greatest and lowest abundance? Which mass will dominate in calculating the average mass of the popcorn pieces? So what do we expect the average mass to be?

18. If the bowl contains 1000 popcorn pieces in it. Then 930 of them would weigh 1.14 mg, 20 would weigh 1.09 mg and 50 would weigh 1.03 mg. Let’s add those abundances together = 930 + 20 + 50 = 1000. This shows that each isotope’s (popcorn-piece mass’) abundance must make up 100% , which is 1000 in this case. In isotopes be sure to check that your abundances ALWAYS add up to 100%

19. Now let us calculate the average weight of the popcorn pieces. (We are expecting it to be just a little less than 1.14 mg.)

20. Example 3 Three isotopes of silicon occur in nature, calculate the average atomic weight of silicon. 28Si (92.21%) mass = 27.97693 amu 29Si (4.70%) mass = 28.97659 amu 30Si (3.09%) mass = 29.97376 amu First make a logical prediction!

21. Molecular and Formula Weights Formula weight (Fr or FW) is the mass of a collection of atoms represented by a chemical formula. So it is the sum of atomic masses (Ar) of the atoms in the chemical formula. For example, K2Cr2O7 Fr (K2Cr2O7) = 2Ar(K) + 2Ar(Cr) + 7Ar(O) = 2(39.10) + 2(52.00) + 7(16.00) = 294.20 amu Significant figures!!

22. If the chemical formula of a substance is the same as its molecular formula, then the formula weight is also called the molecular weight. Molecular weight (Mr or MW) is the mass of a collection of atoms represented by a chemical formula for a molecule. What is the molecular formula for glucose? MW(C6H12O6) = Formula weight and molecular weight are virtually identical and are used interchangeably. When dealing with an ionic compound (3D lattice) we do not use molecular formulas to name them, and so we could not calculate a molecular weight. In this case we use the formula weight.

23. 5 water molecules per CuSO4 unit (Add them into molecular weight!) CuSO4•5H2O → CuSO4 + 5H2O blue grey anhydrous copper sulphate Water of Crystallization Some compounds have water of crystallization. These are water molecules associated with the solid as it crystallizes from solution. CuSO4•5H2O Can often be driven off:

24. The Mole Mole: a convenient measure of enormous numbers. Defined as the amount of matter that contains as many objects (atoms/molecules etc.) as the number of atoms in 12g of 12C 12 g = 6.02214 x 1023 atoms 6.022 x 1023 is known as Avogadro’s Number, NA NEW PUBLIC HOLIDAY! Mole Day: October 23 from 6:02am to 6:02pm, invented May 15, 1991.

25. The Mole 1 mole of anything = 6.022  1023 of that thing. 1 mole 12C atoms = 6.022  102312C atoms. 1 mole H2O molecules = 6.022  1023 H2O molecules. 1 mole NO3- ions = 6.022  1023 NO3- ions. Example 4 Calculate the number of carbon atoms in 0.350 mol of C6H12O6.

26. Example 4 Calculate the number of carbon atoms in 0.350 mol of C6H12O6. Let’s look at this more mathematically now! Avo’s number provides a conversion factor between the number of moles and molecules of a given compound. 1 mole sugar = 6.022 x 1023 molecules of sugar. So we can use the “GIVEN and DESIRED” formula! (Given: mol, Desired molecules!)

27. Molar Mass The mass in grams of 1 mol of substance (units g/mol). The molar mass (in g) of any substance is always numerically equal to its formula weight (in amu). One HCl molecule weighs 36.46 amu  One mol of HCl weighs 36.46 g. Can you see where this comes from? Remember we said that amu and g.mol-1 are equivalent and can be used interchangeably? So….. HCl weighs 36.46 amu = 36.46 g.mol-1. 36.46 g.mol-1 x 1 mol = 36.46 g.

28. 28 g 1 mol of N2 58.45 g 1 mol of H2O 18 g 1 mol of NaCl

29. MAKE SURE YOU UNDERSTAND THESE RELATIONSHIPS!!!

30. Molar Mass Inter-converting masses, moles and number of particles: • Converting from mass to moles and moles to mass is straight forward: • Calculate the molar mass of the substance • Use the mole concept • number of moles = mass / molar mass • n = m / Mr • Once the number of moles of the substance is calculated, use Avogadro’s number to attain number of molecules: • number of molecules = n x AVO • Now we use the number of molecules to solve for number of atoms, or ions within the substance.

31. Molar Mass TAKE NOTE OF THESE PARTICLES AND HOW THEY RELATE TO EACH OTHER! Sugar  C6H12O6 This is 1 MOLECULE of sugar. This molecule contains 24 ATOMS; 6 carbon, 12 hydrogen and 6 oxygen. Sulphuric Acid  H2SO4

32. Mass Mols Molecules Example 5 Calculate the number of H atoms in 20.00 g of C6H12O6. Strategy n = moles m = mass (grams) MW = molecular weight or molar mass (g/mol) n = moles NA = Avogadro’s Number

33. MUST know two things to figure this out… • mass (g) and • molecular weight (molar mass) (g.mol-1) • …of the compound in question.

34. Definitions AMOUNTnumber of moles (must always compare moles!) n = mass / molar mass, or n = concentration/volume. PERCENT COMPOSITION Calculate by dividing the atomic weight (Ar) for each element by the formula weight (Fr) of the compound, and express as a percentage:

35. Example 6 What is the % by mass (% composition) of O in K2Cr2O7? First determine atomic and formula weights (or MWs). Then apply equation (works the same in amu or g/mol).

36. Percent Composition If you are battling to understand what is meant by percentage (by mass) contributed by each element in the compound then just think about a box of colourful smarties! What percentage of the box of smarties is yellow? So it’s the same as ELEMENTAL composition, except that we need to consider mass when we use elements!!!

37. m n = Mr Empirical Formulas from Analyses Strategy Mass % of elements Empirical formula Assume have 100 g sample Moles of each element Mole ratio Grams of each element in sample Mr of that specific element!

38. Example 7(p. 85 8th Ed; p. 93 9th Ed; p. 98 10th Ed) Ethylene glycol, the substance used in automobile antifreeze, is composed of 38.7% C, 9.7% H, and 51.6% O by mass. Its molar mass is 62.1g/mol. Information given in problem Ethylene glycol: 38.7% C 9.7% H 51.6% O by mass Molar mass (from mass spectrometry): 62.1 g mol-1 Calculate: (a) empirical formula (b) molecular formula

39. 38.7g 9.7g 51.6g 12.01 g mol-1 1.008 g mol-1 16.00 g mol-1 C H O 38.7% 9.7% 51.6% assume 100 g 38.7g 9.7g 51.6g number of moles 3.22 mol 9.62 mol 3.23 mol mole ratio (divide by smallest no. mols) 3.22/3.22 9.62/3.22 3.23/3.22 1 2.99 1.003

40. C H O C H O 1 2.99 1.003 1 3 1 Experimental error is to be expected, so is likely to be The empirical formula is therefore CH3O which has the formula weight 1 x 12.01 + 3 x 1.008 + 1 x 16.00 = 31.03 g/mol But molecular weight = 62.1 g/mol, which is twice this value The molecular formula must be C2H6O2.

41. CxHyOz Catalysis: complete oxidation of C to CO2 Hydrogen in CxHyOz is oxidized to H2O and absorbed in the H2O absorber C in CxHyOz is absorbed here Combustion Analysis to Determine Experimental Data O in CxHyOzis determined by mass difference

42. Example 8 8th ed., p. 99, 3.54(a); 9th ed., p. 107, 3.48(a); 10th ed., p. 114, 3.52(a) Combustion of 2.78 mg of ethyl butyrate produced 6.32 mg CO2 and 2.58 mg of H2O. What is the empirical formula of this compound? Consider the given information separately. Let’s look at CO2 first. Recall that we can only determine C and H content from information given! Ask yourself: How much of the 6.32 mg of CO2 is due to C?

43. Example 8 Combustion of 2.78 mg of ethyl butyrate produced 6.32 mg CO2 and 2.58 mg of H2O. What is the empirical formula of this compound? Now let’s deal with H2O, noting we can only determine H content. How much of the 2.58 mg of H2O is due to H?

44. Example 8 Combustion of 2.78 mg of ethyl butyrate produced 6.32 mg CO2 and 2.58 mg of H2O. What is the empirical formula of this compound? Now, determine how much O is present from the difference in mass. Finally, same strategy as before…we have all three masses and can determine the empirical formula.

45. START HERE THIS TIME!

46. 1.44 x 10-4 mol 2.87 x 10-4 mol 4.75 x 10-5 mol 4.75 x10-5 mol4.75 x10-5 mol4.75 x10-5 mol 1.73 x 10-3 g 0.289 x 10-3 g 0.76 x 10-3 g 12.01 g mol-1 1.008 g mol-1 16.00 g mol-1 C H O Determine moles of each element Divide each by smallest # moles to determine ratios

47. Quantitative Information from Balanced Equations In a balanced equation the coefficients can be interpreted as the relative numbers of molecules involved in the reaction AND as the relative number of moles. When given any chemical equation you always ensure that you BALANCE it 1st! Then you use the given information (usually a mass in grams, or a number of moles in mol), together with your calculated molar masses, to solve for the unknown.

48. Quantitative Information from Balanced Equations Example 9 What mass of CO2 is produced when 1.00 g of propane, C3H8, is burned? First, write the balanced equation Then, look at what is given…grams of reactant. In order to determine how much CO2 is produced, you need to consider reaction STOICHIOMETRY.

49. STRATEGY

50. What mass of CO2 is produced when 1.00 g of propane, C3H8, is burned? C3H8 + 5O2→ 3CO2 + 4H2O 1.00 g ??? g Calculate moles of propane: Next, consider the MOLAR relationship between propane and carbon dioxide. It is NOT 1:1, but rather 1:3 (THIS is stoichiometry!). Stoichiometric coefficients are MOLE relationships!!