1. Daniel L. Reger Scott R. Goode David W. Ball http://academic.cengage.com/chemistry/reger Lecture 03 (Chapter 3) Equations, the Mole, and Chemical Formulas

2. Chemical Equation Stoichiometry is the study of the quantitative relationships involving the substances in chemical reactions. A chemical equation describes the identities and relative amounts of reactants and products in a chemical reaction. Just like a chemical formula, a chemical equation expresses quantitative relations.

3. A chemical equation is a shorthand notation to describe a chemical reaction. 2Mg + O 2  2MgO magnesium react to form magnesium and oxygen oxide Chemical Reactions

4. Reactants are the substances consumed. Products are the substances formed. Coefficients are numbers before the formula of a substance in an equation. A balanced equation has the same number of atoms of each element on both sides of the equation. Definitions

5. The Chemical Equation 2H 2 (g) + O 2 (g)  2H 2 O( l )  Identifies reactant(s) and product(s)  Identifies states of matter (g = gas, l = liquid, s = solid, aq = aqueous)  Demonstrates law of conservation of matter in balanced equation  Law of conservation of matter: Atoms are neither created nor destroyed in a chemical reaction, but rearranged to form different molecules.  This equation, as written, describes the reaction between individual molecules (2 molecules of H 2 react with 1 molecule of O 2 ) as well as molar relationships (2 moles of H 2 react with 1 mole of O 2 ). Reactant(s)Product(s)

6. Balancing The Chemical Equation N 2 (g) + H 2 (g)  NH 3 (g)  You cannot change the natural molecular formulas, you can only change the coefficients indicating number of molecules or moles. N 2 (g) + H 2 (g)  2NH 3 (g)  By multiplying NH 3 by coefficient of 2, you now have 2N and 6H. This balances with N 2 on reactant side, but not H 2. N 2 (g) + 3H 2 (g)  2NH 3 (g)  Now check your work. 2 N 6 H 2 N 6 H Reactant(s)Product(s)

7. Balancing The Chemical Equation NO(g) + O 2 (g)  NO 2 (g)  Multiply NO 2 by coefficient of 2, then multiply NO by 2.  Now check your work. 2 N 4 O 2 N 4 O Reactant(s)Product(s) nitrogen monoxide  Reactant side has 3O, but product side has 2O. 2NO(g) + O 2 (g)  2NO 2 (g)

8. Balancing the Chemical Equation Write the correct formula for each substance. H 2 + Cl 2  HCl (unbalanced) Add coefficients so the number of atoms of each element are the same on both sides of the equation. H 2 + Cl 2  2HCl (balanced)

9. Write the correct formula for each substance. C 5 H 12 + O 2  CO 2 + H 2 O Assume one molecule of the most complicated substance. Adjust the coefficient of CO 2 to balance C. C 5 H 12 + O 2  5CO 2 + H 2 O Adjust the coefficient of H 2 O to balance H. C 5 H 12 + O 2  5CO 2 + 6H 2 O Adjust the coefficient of O 2 to balance O. C 5 H 12 + 8O 2  5CO 2 + 5H 2 O Check the balance by counting the number of atoms of each element. Balancing Chemical Equations

10. Test Your Skill Balance the equation C 4 H 9 OH + O 2  CO 2 + H 2 O

11. Test Your Skill Balance the equation C 4 H 9 OH + O 2  CO 2 + H 2 O Answer: C 4 H 9 OH + 6O 2  4CO 2 + 5H 2 O

12. Balancing Equations Sometimes fractional coefficients are obtained. C 5 H 10 + O 2  CO 2 + H 2 O C 5 H 10 + O 2  5CO 2 + H 2 O C 5 H 10 + O 2  5CO 2 + 5H 2 O C 5 H 10 + 15/2O 2  5CO 2 + 5H 2 O Multiply all coefficients by the denominator. 2C 5 H 10 + 15O 2  10CO 2 + 10H 2 O

13. Neutralization Reactions Neutralization is the reaction of an acid with a base to form a salt and water. An acid is a compound that dissolves in water to produce hydrogen ions. A base dissolves in water to produce hydroxide ions. It is usually a soluble metal hydroxide. A salt is an ionic compound consisting of the cation of a base and the anion of an acid.

14. Examples of Neutralization HCl + NaOH  NaCl + H 2 O H 2 SO 4 + 2KOH  K 2 SO 4 + 2H 2 O 2HClO 4 + Ca(OH) 2  Ca(ClO 4 ) 2 + 2H 2 O acid base  salt water

15. Balancing Neutralization Reactions The number of hydrogen ions provided by the acid must equal the number of hydroxide ions provided by the base. H 3 PO 4 + 3NaOH  Na 3 PO 4 + 3H 2 O 3H + + 3OH -  3H 2 O

16. Combustion Reactions A combustion reaction is the process of burning. Most combustion reactions are the combination of a substance with oxygen. When an organic compound burns in oxygen, the carbon is converted to CO 2, and the hydrogen forms water, H 2 O.

17. Identify the type of the reaction, then balance it. (a) (C 2 H 5 ) 2 O + O 2  CO 2 + H 2 O (b) H 2 SO 4 + Ca(OH) 2  CaSO 4 + H 2 O Test Your Skill

18. Identify the type of the reaction, then balance it. (a) (C 2 H 5 ) 2 O + O 2  CO 2 + H 2 O (b) H 2 SO 4 + Ca(OH) 2  CaSO 4 + H 2 O Answer: (a) Combustion (C 2 H 5 ) 2 O + 6O 2  4CO 2 + 5H 2 O (b) Neutralization H 2 SO 4 + Ca(OH) 2  CaSO 4 + 2H 2 O Test Your Skill

19. Oxidation-Reduction (Redox) Reactions An oxidation-reduction reaction is one in which electrons are transferred from one species to another. A positive or negative number, called the oxidation number, can be assigned to each element in a substance based on a series of rules. This will be covered in greater detail in Chapter 18 (Gen Chem II, for those of you who survive this class!!!)

20. 1. An atom in its elemental state has an oxidation number of zero. 2. The oxidation number of monatomic ions is the charge of the ion. 3. In compounds, the oxidation number of F is -1, O is generally -2, H is generally +1 when combined with a nonmetal and -1 when combined with a metal, and the other halogens are generally -1. 4. The sum of the oxidation numbers in any species (uncharged compounds or polyatomic ions) must equal the charge of the species. Assigning Oxidation Numbers

21. Determining Oxidation Numbers K 2 CO 3 CO 2 2(O.N of K) + (O.N. of C) + 3(O.N. of O) = 0 2(+1) + 3(-2) = -4 2(+1) + (+4) + 3(-2) = 0 (O.N of C) + 2(O.N. of O) = 0 2(-2) = -4 (+4) + 2(-2) = 0 CH 2 O(O.N of C) + 2(O.N. of H) + (O.N. of O) = 0 2(+1) + (-2) = 0 (0) + 2(+1) + (-2) = 0 NO 3 - (O.N of N) + 3(O.N. of O) = -1 + 3(-2) = -6 (+5) + 3(-2) = -1

22. Example: Oxidation Numbers Assign the oxidation number to all elements in (a) CO(b) SO 2 (c) Na 2 S (d) NO 3 -

23. Oxidation-Reduction Reactions Oxidation is the loss of electrons by a substance in a reaction (OIL). Reduction is the gain of electrons by a substance in a reaction (RIG). 2Na(s) + Cl 2 (g)  2NaCl(s) Oxidation States Reactants  Products Na = 0  Na + = +1 Oxidation Cl in Cl 2 = 0  Cl - = -1 Reduction

24. Redox e- Transfer: Examples 4(O.N. of Al) + 6(O.N. of O) = 0 4Al(s) + 3O 2 (g)  2Al 2 O 3 (s) (O.N of uncombined element = 0) + 6(-2) = -12  Al is oxidized (combines with O and each Al loses 3 e- and increases O.N. by 3)  Al is reducing agent  Oxygen is reduced (each of 6 O gains 2 e- and decreases O.N. by 2)  O is oxidizing agent Mission: Determine O.N. for each atom in the following and identify oxidizing and reducing agents. 4(+3) + 6(-2) = 0

25. The Mole One mole is the amount of substance that contains as many entities as the number of atoms in exactly 12 grams of the 12 C isotope of carbon. Avogadro’s number is the experimentally determined number of 12 C atoms in 12 g, and is equal to 6.022 x 10 23.

26. Fe Al Na Cu HgHg One Mole of Several Elements Argon in green balloons

27. Converting Moles and Entities One mole of anything contains 6.022 x 10 23 entities. 1 mol H = 6.022 x 10 23 atoms of H 1 mol H 2 = 6.022 x 10 23 molecules of H 2 1 mol CH 4 = 6.022 x 10 23 molecules of CH 4 1 mol CaCl 2 = 6.022 x 10 23 formula units of CaCl 2

28. The Mole and Chemical Equations  The Mole can be used to calculate mass relationships in chemical reactions.  Stoichiometry: Study of mass relationships  Coefficients apply to moles or molecules, but not mass. CH 4 (g) + 2O 2 (g)  CO 2 (g) + 2H 2 O( l ) 1 mol CH 4 (g) + 2 mol O 2 (g)  1 mol CO 2 (g) + 2 mol H 2 O( l )  This relationship tells us we need 2 moles of O 2 and 1 mole of CH 4 to produce 1 mol CO 2 and 2 moles of water.  The equation above has the following molar relationships Example

29. Moles to Number of Entities Moles of substance Number of atoms or molecules Avogadro’s number Avogadro’s number allows the interconversion of moles and numbers of atoms or molecules.

30. Example: Convert Moles and Entities How many atoms are present in 0.35 mol of Na? How many moles are present in 3.00 x 10 21 molecules of C 2 H 6 ?

31. Molar Mass The molar mass ( M ) of any atom, molecule or compound is the mass (in grams) of one mole of that substance. The molar mass in grams is numerically equal to the atomic mass or molecular mass expressed in u. Molar mass converts from mass (in grams) to amount (in moles) or the reverse.

32. Molar and Atomic Masses

33. Molar Mass Conversion What is the mass of 0.25 moles of CH 4 ? Molar mass of CH 4 = 16.0 g/mol.

34. Test Your Skill What mass of compound must be weighed out to have a 0.0223 mol sample of H 2 C 2 O 4 ( M = 90.04 g/mol)?

35. Test Your Skill What mass of compound must be weighed out to have a 0.0223 mol sample of H 2 C 2 O 4 ( M = 90.04 g/mol)? Answer: 2.01 g H 2 C 2 O 4

36. Percent Composition Mass relationships can be used to determine percent compositions 1 mol of H 2 SO 4 = 98.1 g of H 2 SO 4 % H = 2.0 gx100=2.0 % 98.1 g % S = 32.1 gx100=32.7 % 98.1 g % O = 64.0 gx100=65.2 % 98.1 g

37. % Mass of element in compound % mass N in HNO 3 ?N = 14.01 g/mol H = 1.008 g/mol O = 16.00 g/mol HN0 3 % mass= part X 100 total Total= (1 x H) + (1 x N) + (3 + O) = (1 x 1.008) + (1 x 14.01) + (3 x 16.00) = 63.018 g/mol N % mass= 14.01 g/mol X 100 = 22.23% 63.018 g/mol

38. Determining empirical formula of compound The empirical formula is the simplest ratio of atoms present in compound. Ratio values must be positive integers. The molecular formula may be the same as the empirical formula, or it may be a multiple of the empirical formula. To solve these problems, you need to know either the % of each element, or the mass of each element. For hydrocarbons, if the percentages of all elements does not total 100%, assume that the remaining fraction is oxygen.

39. What is the empirical formula of a compound that is 48.38% C, 8.12% H, and 43.50% O? Solution Since percentages add up to 100%, we can base our solution on the assumption that we are working with 100 g of sample. In step 01, convert percent to mass (if necessary), then mass to moles. Calculate Empirical Formula

40. What is the empirical formula of a compound that is 48.38% C, 8.12% H, and 43.50% O? Solution In step 02, divide the number of moles of each element by the smallest number of moles calculated, and round appropriately. A third step may be necessary to arrive at whole integer values. In this step, determine an integer multiple that will convert all fractions to integers, then multiply all numbers by that value. Calculate Empirical Formula 1.5 × 2 = 3 3.0 × 2 = 6 1.0 × 2 = 2 The empirical formula is C 3 H 6 O 2

41. Molecular Formula The molecular formula must be a whole number multiple of the empirical formula. If the empirical formula is CH 2, the molecular formula is (CH 2 ) n where The molar mass must be measured experimentally.

42. Example: Molecular Formula A compound has the empirical formula C 4 H 8 O. A mass spectrometry experiment determines that the molar mass of the compound is 216 g/mol. What is the molecular formula? Molar mass of empirical formula = 72 g/mol

43. Mole Relationships in Equations Because we can calculate moles from mass and vice-versa, we can determine how much mass of a compound is needed to produce a certain mass of product, how much reactant will be consumed, etc. This is Reaction Stoichiometry.

44. Guidelines for Reaction Stoichiometry Write the balanced equation. Calculate the number of moles of the species for which the mass is given. Use the coefficients in the equation (molar relationship) to convert the moles of the given substance into moles of the substance desired. Calculate the mass of the desired species.

45. Stoichiometry Examples  How many moles of O 2 are required to react with 1.72 mol CH 4 ? CH 4 (g) + 2O 2 (g)  CO 2 (g) + 2H 2 O( l ) 1 mol CH 4 (g) + 2 mol O 2 (g)  1 mol CO 2 (g) + 2 mol H 2 O( l ) 1.72 mol CH 4 = ? mol O 2  How many grams of H 2 O will be produced when 1.09 mol of CH 4 reacts with an excess of O 2 ? 1.09 mol CH 4 = ? Grams H 2 O

46. Stoichiometry Examples CH 4 (g) + 2O 2 (g)  CO 2 (g) + 2H 2 O( l ) 1 mol CH 4 (g) + 2 mol O 2 (g)  1 mol CO 2 (g) + 2 mol H 2 O( l )  How many grams of O 2 must react with excess CH 4 to produce 8.42 g CO 2 ? 8.42 g CO 2 = ? Grams O 2

47. Test Your Skill Given the equation 4FeS 2 + 11O 2  2Fe 2 O 3 + 8SO 2 what mass of SO 2 is produced from reaction of 3.8 g of FeS 2 with excess oxygen? Answer: 4.1 g SO 2

48. Theoretical Yield The previous calculation showed that given the equation 4FeS 2 + 11O 2  2Fe 2 O 3 + 8SO 2 4.1 g SO 2 is produced from reaction of 3.8 g of FeS 2 with excess oxygen. The 4.1 g SO 2 is the theoretical yield – the maximum quantity of product that can be obtained from a chemical reaction, based on the amounts of starting materials.

49. Limiting Reactant Limiting reactant: the reactant that is completely consumed when a chemical reaction occurs.

50. Limiting Reactant In the below reaction of Cl 2 (green) and Na (purple), Cl 2 is the limiting reactant. Once the limiting reactant is consumed the reaction stops and no more product forms. The formed NaCl and the excess Na are present at the end of the reaction.

51. The Limiting Reactant CH 4 (g) + 2O 2 (g)  CO 2 (g) + 2H 2 O( l ) 1 mol CH 4 (g) + 2 mol O 2 (g)  1 mol CO 2 (g) + 2 mol H 2 O( l )  Reaction will only continue as long as all necessary reactants are present  Limiting reactant is the one that gives the least amount of product  In a combustion reaction of 20.0 g methane and 100.0 g oxygen, how much carbon dioxide (mass) would be produced? Molar ratio CH 4 (g):CO 2 (g) is 1:1 Molar ratio O 2 (g):CO 2 (g) is 2:1  In this reaction, how much O 2 would remain unreacted?  g O 2 used in reaction 100.0 g – 80.0 g = 20.0 g O 2 unused in reaction So, 55.0 g CO 2 can be produced (theoretical yield).

52. Actual Yield, Percent Yield As shown in the picture, in many reactions not all of the product formed can be isolated, so the Actual yield may be less than the Theoretical yield (from your calculations). In the previous example, 55.0 g of CO 2 was calculated as the theoretical yield. Let’s say you perform the experiment and your actual yield is 44.0 g. What was your percent yield?

53. Example: Limiting Reactant Calculate the mass of the NH 3 product formed (theoretical yield) when 7.0 g of N 2 reacts with 2.0 g of H 2.