1. Chemical Equations Balancing equations and applications

2. Objectives  Write and balance chemical equations  Perform calculations using moles and molar masses  Determine percent composition from formula  Determine empirical formula from percent composition  Determine molecular formula

3. The chemical equation aA + bB = cC + dD The Law of Conservation of Matter states that matter is neither created nor destroyed All the atoms on the left must be the same as those on the right Reactant side Product side coefficient ELEMENT or COMPOUND

4. Chemical book-keeping  The key to writing correct equations is to ask the question, “Have I gained or lost any atoms?”  Another thing is to put down the correct formula for each reactant or product  Formulas cannot be changed in order to balance the equation

5.  In the reaction of hydrogen with oxygen to produce water, the reactants are the elements H 2 and O 2, and the product is H 2 O  Count the atoms: 4 H and 2 O 4 H and 2 O The big number multiplies every atom after it The subscript only multiplies the atom before it

6. Balance the equations  CH 4 + O 2 = CO 2 + H 2 O  CH 4 + 2O 2 = CO 2 + 2H 2 O  C 3 H 8 + O 2 = CO 2 + H 2 O  C 3 H 8 + 5O 2 = 3CO 2 + 4H 2 O  N 2 + H 2 = NH 3  N 2 + 3H 2 = 2NH 3  Do balancing equation exercises

7. Molecules or moles  The numbers (coefficients) in chemical equation can refer to molecules  But for practical applications, we need a more useful number: we cannot count molecules

8. The Mole  The mole is a unit of quantity used in chemistry to measure the number of atoms or molecules  DEFINITION:  The number of atoms in exactly 12 g of 12 C  A mole of anything always has the same number of particles: atoms, molecules or potatoes – 6.02 x 10 23 – Avogadro’s number

9. Mole conversions

10. Atomic and molecular weights  Two scales:  Atomic mass unit scale  The mass of an individual atom or molecule in atomic mass units (amu)  Molar mass scale  The mass of a mole of atoms or molecules in grams  Confusing?...

11. The Good News  The weight of an atom in amu has the same numerical value as its molar mass in grams  The atomic mass of carbon is 12 amu  The molar mass of carbon is 12 g  The formula mass of H 2 O is 18 amu  The molar mass of H 2 O is 18 g

12. Examples  How many atoms are in 6.94 g of lithium if the atomic mass of Li is 6.94 amu?  6.02 x 10 23  What is the molar mass of H 2 O if the atomic mass of H = 1 amu and O = 16 amu  18 g

13. How many H atoms in 1 mol of CH 4 ?  1 mol = 6.02 x 10 23 particles

14. How many moles of O 2 in 64 g of oxygen?  Atomic mass of O = 16 AMU

15. Calculate the formula mass of Na 2 SO 4 in AMU. Use atomic mass Na = 23 AMU, O = 16 AMU, S = 32 AMU

16. Significance of formula unit  Ionic compounds do not contain molecules. Simplest formula is the formula unit  Covalent compounds, the molecular formula is the formula unit

17. Percent composition and empirical formula  Chemical analysis gives the mass % of each element in the compound  Molar masses give the number of moles  Obtain mole ratios  Determine empirical formula

18. Determining percent composition  Percent composition is obtained from the actual masses. Example: Sample contained 0.4205 g of C and 0.0795 g of H. Total mass = 0.5000 g (0.4205 + 0.0795) Therefore: in 100 g there are: (84.10 %) (84.10 %) (15.90 %) (15.90 %) Percent composition: 84.10 % C, 15.90 % H

19. Percent composition from formula  What is percent composition of C 5 H 10 O 2 ?  1 mol C 5 H 10 O 2 contains 5 mol C, 10 mol H and 2 mol O atoms  Mass of each element  Total mass = 102.13 g

20. Convert masses into percents  Percent composition: 58.80 % C + 9.870 % H + 31.33 % O = 100.00%

21. Empirical formula from percent composition: 84.1 % C, 15.9 % H 1.Convert percents into moles 84.10 g of C ≡ 7.00 mol C 15.9 g of H ≡ 15.8 mol H 2.Determine mole ratio Mole ratio H:C =  Simplest formula (decimal form): C 1 H 2.26  Make smallest integers by multiplying C 4 H 9 May require rounding. Errors in real data cause problems  Do percent composition and empirical formula exercises

22. What is percent C content of C 2 H 6 ?  Molar mass C = 12 g/mol; molar mass H = 1 g/mol

23. Empirical formula with more than two elements  Percent composition of vitamin C is:  40.9 % C, 4.58 % H, 54.5 % O 1.Convert into moles 2.Determine mole ratios 3.Find lowest whole numbers

24. Inaccuracy can lead to ambiguous or incorrect formulas  What if H:C is 2.20 rather than 2.26? An error of only 3 %  Formula becomes C 5 H 11 rather than C 4 H 9  What if H:C is 2.30 rather than 2.26? An error of only 2 %  Formula becomes C 3 H 7  Sometimes chemical intuition is required: we know there is FeO, Fe 3 O 4 and Fe 2 O 3 ; so a formula FeO 3 would indicate an error

25. Rounding or not: the role of chemical intuition  Formulae are always written with integers  Experimental ratios are always fractions  Two choices:  Round to nearest whole number  Multiply top and bottom to find ratio of whole numbers with same value  Choice depends on the type of substance

26. Hydrocarbons: A case for not rounding  There are millions of different hydrocarbons  What if H:C is 2.20 rather than 2.26? An error of only 3 %  Formula becomes C 5 H 11 rather than C 4 H 9  What if H:C is 2.30 rather than 2.26? An error of only 2 %  Formula becomes C 3 H 7  All formulae are reasonable  So how do I know what the composition is?  Additional knowledge about the substance is helpful: melting point, boiling point, molar mass

27. Inorganic compounds: Rounding makes sense  Inorganic compounds tend to have few compositions  Iron forms three oxides: FeO, Fe 3 O 4 and Fe 2 O 3  Experimental formula FeO 1.75 would indicate Fe 2 O 3 not FeO 2

28. Practice empirical formula problem  A compound contains 62.1 % C, 5.21 % H, 12.1 % N and 20.7 % O. What is the empirical formula?

29. Empirical and molecular formula  Percent composition gives the empirical (simplest) formula. It says nothing about the molecular formula.  Molecular formula describes number of atoms in the molecule  May be much larger than the empirical formula in the case of molecular covalent compounds  For ionic compounds empirical formula = “molecular” formula

30. Elements and compounds can have molecular formula different from simplest formula Substance Empirical formula Molecular formula Substance Empirical formula Molecular formula SulphurS S8S8S8S8PhosphorousP P4P4P4P4 BenzeneCH C6H6C6H6C6H6C6H6AcetyleneCH C2H2C2H2C2H2C2H2 Ethylene CH 2 C2H4C2H4C2H4C2H4Cyclohexane C 6 H 12

31. Determination of molecular formula  Require: 1.Empirical formula from percent composition analysis 2.Molar mass from some other source  Number of empirical formula units in molecule:  There are n (A a B b C c ) in molecule:  Molecular formula is A na B nb C nc

32. Molecular formula of vitamin C  Empirical formula of vitamin C is C 3 H 4 O 3  Molar mass vitamin C is 176.12 g/mol  Mass of empirical formula = 88.06 g/mol  (3 x 12.01 + 4 x 1.008 + 3 x 16.00)  Number of formula units per molecule =  Molecular formula = 2(C 3 H 4 O 3 ) = C 6 H 8 O 6

33. Ionic solids do not have molecular formulas  Infinite lattices like ionic solids and covalently bonded lattices are not molecular.  The formula used for an ionic compound is the same as the empirical formula – with one or two exceptions  Hg 2 Cl 2 rather than HgCl

34. Structural formula provides more information  The molecular formula indicates the number of atoms in the molecule  The structural formula indicates how those atoms are arranged  C 2 H 6 O is the molecular formula for ethanol and dimethyl ether  Structural formula for ethanol is CH 3 CH 2 OH  Structural formula for ether is CH 3 OCH 3  In a recipe we would need to use the structural formula to identify the correct reagent

35. In some cases the mole contents of the compound are obtained indirectly  Analysis of the hydrocarbon is performed by combustion.  Mole ratios of the elements are derived from the mole ratios of the combustion products CO 2 + H 2 O (1 mol H 2 O ≡ 2 mol H) (1 mol CO 2 ≡ 1 mol C)

36. The molecular or empirical formula can be used to determine the percent composition  Formula of aspirin is C 9 H 8 O 4  Molar mass is 180 g  Mass of C = 108 g  Mass of H = 8 g  Mass of O = 64 g  % C = 108/180 x 100 %  % H = 8/180 x 100 %  % O = 64/180 x 100 %