1. 3 Stoichiometry Contents 3-1 Law of Conservation of Matter 3-2 Balancing Equation 3-3 Equations on a Macroscopic Scale 3-4 Mass Relationship in Chemical Reactions 3-5 Limiting Reactants 3-6 Theoretical Yield, Actual Yield, and Percent Yield 3.7 Quantitative analysis 3.8 Empirical Formulas from Percent Composition 3.9 Molecular and Structural Formulas 3.10 Percent Composition from Formulas

2. Stoichiometry is the study of quantitative relationships between substances involved in chemical changes. It is a very important subject from the point of view of both theory and practice. Analytical chemistry is based on Stoichiometry.

3. 3-1 Law of Conservation of Matter The Quantity of matter is not changed by chemical reactions; Matter is neither created nor destroyed by chemical reactions. The number of atoms of each kind must be the same after reaction as it was before reaction. In chemical reactions, the quantity of matter does not change. The total mass of the products equals the total mass of the reactants. Law of Conservation of Matter provided the foundation for modern chemistry.

4. H 2 + N 2 NH 3 3H 2 + N 2 = 2NH 3

5. 3-2 Balancing Equation A chemical reaction is a process in which one set of substances called reactants is converted to a new set of substances called products. In other words, a chemical reaction is the process by which a chemical change occurs. We need evidence before we can say that a reaction has occurred. Some of the types of physical evidence to look for are shown here.

6. a color change formation of a solid (precipitate) with a clear solution evolution of a gas evolution or absorption of heat When none of these signs of a chemical reaction appears, we need chemical evidence. This requires a detailed chemical analysis of the reaction mixture to discover whether any new substances are present.

7. Balancing Chemical Equations 1.Write the correct formula(s) for the reactants on the left side and the correct formula(s) for the product(s) on the right side of the equation. Ethane reacts with oxygen to form carbon dioxide and water C 2 H 6 + O 2 CO 2 + H 2 O 2.Change the numbers in front of the formulas (coefficients) to make the number of atoms of each element the same on both sides of the equation. Do not change the subscripts. 3.7 2C 2 H 6 NOT C 4 H 12

8. Balancing Chemical Equations 3.Begin with the compound that has the most atoms or the most kinds of atoms and use one of these atoms as a starting point. C 2 H 6 + O 2 CO 2 + H 2 O 3.7 start with C 2 H 6 start with C or H but not O

9. Balancing Chemical Equations 4. Balance those elements that appear in only once on each side of the arrow first. C 2 H 6 + O 2 CO 2 + H 2 O 3.7 2 carbon on left 1 carbon on right multiply CO 2 by 2 C 2 H 6 + O 2 2CO 2 + H 2 O 6 hydrogen on left 2 hydrogen on right multiply H 2 O by 3 C 2 H 6 + O 2 2CO 2 + 3H 2 O start with C or H but not O

10. Balancing Chemical Equations 5. Balance those elements that appear more than once on a side. Balance free elements last. 3.7 2 oxygen on left 4 oxygen (2x2) C 2 H 6 + O 2 2CO 2 + 3H 2 O + 3 oxygen (3x1) multiply O 2 by 7 2 = 7 oxygen on right C 2 H 6 + O 2 2CO 2 + 3H 2 O 7 2 remove fraction multiply both sides by 2 2C 2 H 6 + 7O 2 4CO 2 + 6H 2 O The smallest whole-number coefficients are preferred!

11. Balancing Chemical Equations 6. Check to make sure that you have the same number of each type of atom on both sides of the equation. 2C 2 H 6 + 7O 2 4CO 2 + 6H 2 O 4 C (2 x 2)4 C 12 H (2 x 6)12 H (6 x 2) 14 O (7 x 2)14 O (4 x 2 + 6) 3.7

12. Balancing Chemical Equations 3.7 2C 2 H 6 + 7O 2 4CO 2 + 6H 2 O ReactantsProducts 4 C 12 H 14 O 4 C 12 H 14 O 6. Check to make sure that you have the same number of each type of atom on both sides of the equation.

13. The system to balance an equation Begin with the compound that has the most atoms or the most kinds of atoms and use one of these atoms as a starting point. Balance elements that appear only once on each side of the arrow first. Then balance elements that appear more than once on a side. Balance free elements last.

14. Here are some useful strategies for balancing equations If an element occurs in only one compound on each side of the equation, try balancing this element first. When one of the reactants or products exists as the free element, balance this element last. In some reactions, certain groups of atoms(for example, polyatomic ions) remain unchanged. In such cases, balance these groups as a unit. It is permissible to use fractional as well as integral numbers as coefficients. At times, an equation can be balanced most easily by using one or more fractional coefficients and then, if desired, clearing the fractions by multiplying all coefficients by a common multiplier.

15. Example Writing and balancing an equation: The combustion of a carbon- Hydrogen-Oxygen Compound. Liquid triethylene glycol, C 6 H 14 O 4, is used as a solvent and plasticizer for vinyl and polyurethane plastics. Write a balanced chemical equation for its complete combustion. Carbon- hydrogen-oxygen compounds, like hydrocarbons, yield carbon dioxide and water when burned in oxygen gas. Starting expression : C 6 H 14 O 4 + O 2 CO 2 + H 2 O Balance C : C 6 H 14 O 4 + O 2 6CO 2 + H 2 O Balance H : C 6 H 14 O 4 + O 2 6CO 2 + 7H 2 O Solution

16. At this point, the right side of the expression has 19 O atoms(12 in six CO 2 molecules and 7 in seven H 2 O molecules). To get 19 atoms on the left, we start with 4 in a molecule of C 6 H 14 O 4 and need 15 more. This requires a fractional coefficient of 15/2 for O 2. Balance O : C 6 H 14 O 4 + (15/2) O 2 6CO 2 + 7H 2 O (balanced) To remove the fractional coefficient, multiply all coefficients by 2, the denominator of the fractional coefficient, 15/2. 2 C 6 H 14 O 4 + 15 O 2 12CO 2 +14H 2 O (balanced) Check: Left : (2×6) =12 C; (2 ×14) = 28 H; [(2×4) + (15×2)] = 38 O Right : (12×1) =12 C; (14 ×2) = 28 H; [(12×2) + (14×1)] = 38 O

17. The term stoichiometry means, literally, to measure the elements, but from a more practical standpoint, it includes all the quantitative relationships involving atomic and formula masses, chemical formulas, and the chemical equation. The coefficients in the chemical equation 2H 2 (g) + O 2 (g) 2H 2 O(l) Mean that 2x molecules H 2 + 1x molecules O 2 2x molecules H 2 O Suppose that we let x=6.02214×10 23. Then the chemical equation also means that 2 mol H 2 + 1 mol O 2 2 mol H 2 O 3-3 Equations on a Macroscopic Scale

18. The coefficients in the chemical equation allow us to make statements, such as: Two moles of H 2 O are produced for every two moles of H 2 consumed. Two moles of H 2 O are produced for every one mole of O 2 consumed. Two moles of H 2 are consumed for every one mole of O 2 consumed. A stoichiometric factor relates the amounts a mole basis, of any two substances involved in a chemical reaction; thus a stoichiometric factor is a mole ratio. The mole is the key to quantitative relationships between substances involved in chemical changes on a practical scale.

19. Example Relating the Mass of a reactant and a product. What mass of H 2 O is formed in the reaction of 4.16g H 2 with an excess of O 2 ? Solution The general strategy for reaction stoichiometry problems, outlined earlier and illustrated in below, suggests these three steps. 1. Convert the quantity of H 2 from grams to moles. (use the inverse of molar mass of H 2 ) 2. From the number of moles of H 2, calculate the number of moles of H 2 O formed. 3. Convert the quantity of H 2 O from moles to grams. (use the inverse of molar mass of H 2 O) 3-4 Mass Relationship in Chemical Reaction

20. Grams of H 2 GIVEN Use inverse of molar mass as conversion factor: 1 mol H 2 /2.016 g H 2 Moles of H 2 Use coefficients in balanced chemical Equation to find mole ratio: 2 mol H 2 O/2 mol H 2 Moles of H 2 O = (2/2) ×mol H 2 Use molar mass as conversion factor: 18.02 g H 2 O/1 mol H 2 Grams of H 2 O FOUND

21. ( g H 2 mol H 2 mol H 2 O g H 2 O ) 123 1 mol H 2 2.016 g H 2 ? g H 2 O =4.16 g H 2 × × × 2 mol H 2 2 mol H 2 O 18.02 g H 2 O 1 mol H 2 O = 37.2 g H 2 O Always begin solving problem about quantitative relationships in reaction by writing an equation for the reaction. The coefficients in the equation describe the relations between moles of products and moles of reactions. For example: 2H 2 O = 2H 2 + O 2 Moles 2 2 1 Formula Mass 18.0 2.02 32.0 Mass 36.0 4.03 32.0

22. When all the reactants are completely and simultaneously consumed in a chemical reaction, the reactants are said to be in stoichiometric proportions--in the mole ratios dictated by the coefficients in the balanced equation. At other times, the reactants are not usually present in the proportions shown by the equation. The quantity of one reactant controls the amount of products that can be formed. The reactant that is completely consumed: the limiting reactant--determines the quantities of products formed. 3-5 Limiting Reactants

23. Solution 1. The first step in a stoichiometric calculation is to write A balanced equation for the reaction. If the equation Is not given, you must supply your own. Determining the Limiting Reactant in a Reaction. Phosphorus trichloride, PCI 3, is a commercially important compound used in the manufacture of pesticides, gasoline additives, and a number of other products. It is made by the direct combination of phosphorus and chlorine. P 4 (s) + 6 Cl 2 (g) 4 PCl 3 (l) What mass of PCl 3 (l) forms in the reaction of 125 g P 4 with 323 g Cl 2 ? Example

24. Phosphorus trichloride ? Mol Cl 2 = 323g Cl 2 × = 4.56 mol Cl 2 ? Mol P 4 = 125g P 4 × = 1.01 mol P 4 1mol Cl 2 70.91 g Cl 2 1 mol P 4 123.9 g P 4 P 4 (s) + 6 Cl 2 (g) 4 PCl 3 (l) 125 g 323 g ?g 2. Note what’s given and asked for the above reaction. 3. Write the formula masses needed below the reaction. formula masses, u 123.9 70.91 137.3 4. Determine which reactant in limiting. Convert grams to moles.

25. ? g PCl 3 = 323 gCl 2 × × × = 417 g PCl 3 1 mol Cl 2 70.91 g Cl 2 4 mol PCl 3 6 mol Cl 2 137.3 g PCl 3 1 mol PCl 3 We see rather clearly that there is less than 6 mol Cl 2 per mole of P 4 - Chlorine is the limiting reactant. The remainder of the calculation is to determine the mass of PCl 3 formed in the reaction of 323 g Cl 2 with an excess of P 4. 4. Calculation and check.

26. If calculated mole tatio <6/1 chlorine is limiting If calculated mole tatio >6/1 phosphorus is limiting Grams of P 4 Use inverse of molar Mass as conversion factor : 1 mol P 4 /123.9g P 4 Moles of P 4 Grams of Cl 2 Use inverse of molar Mass as conversion factor : 1 mol Cl 2 /70.91g Cl 2 Moles of Cl 2 Calculate mole ratio of Cl 2 to P 4 Mole ratio = moles of Cl 2 /moles of P 4

27. Example Determining the quantity of excess reactant(s) remaining after a reaction. What mass of P 4 remains in excess following the reaction in example above? Solution The key to this problem is to calculate the mass of P 4 that is consumed, and we can base this calculation either on the mass of Cl 2 consumed ： ?g P 4 = 323 g Cl 2 × 1 mol Cl 2 70.91 g Cl 2 1 mol P 4 6 mol Cl 2 ×× 123.9 g P 4 1 mol P 4 = 94.1 g P 4

28. Or on the mass of PCl 3 produced. ?g P 4 = 417 g PCl 3 × 1 mol PCl 3 137.3gPCl 3 1 mol P 4 4 mol PCl 3 × × 123.9 g P 4 1 mol P 4 = 94.1 g P 4 The mass of P 4 remaining after the reaction is simply the difference between what was originally present and what was consumed; that is, 125 gP 4 initially – 94.1 g P 4 consumed = 31 gP 4 remaining

29. Theoretical Yield, Actual Yield, and Percent Yield The theoretical yield of a reaction is the amount of product calculated to be formed by the reaction. The amount of product that is actually produced is called the actual yield. The percent yield is defined as: percent yield = actual yield / theoretical yield × 100% 3-6 Theoretical Yield, Actual Yield, and Percent Yield

30. In few reactions the actual yield almost exactly equals the theoretical yield, and the reactions are said to be quantitative. On the other hand, in some reactions the actual yield is less than the theoretical yield, and the percent yield is less than 100%. The yield may be less than 100% for many reasons: (1) The product of a reaction rarely appears in a pure form, and in the necessary purification steps, some product may be lost through handling. This reduces the yield. (2) In many cases the reactants may participate in reactions other than the one of central interest. These are called side reactions, and the unintended products are called by-products. To the extent that side reactions occur, the yield of the main product is reduced. (3) Finally, if a reverse reaction occurs, some of the expected product may react to re-form the reactants, and again the yield is less than expected.

31. Example Determining Theoretical, Actual, and Percent Yields. Billions of pounds of urea, CO(NH 2 ) 2, are produced annually for use as a fertilizer. The reaction used is 2 NH 3 + CO 2 CO(NH 2 ) 2 + H 2 O The typical starting reaction mixture has a 3: 1 mole ratio of NH 3 to CO 2. If 47.7 g urea forms per mole of CO 2 that reacts, what is the (a) theoretical yield; (b) actual yield; and (c) percent yield in this reaction?

32. 60.1 g CO(NH 2 ) 2 Solution (a) The stoichiometric proportions are 2 mol NH 3 : 1 mol CO 2. Because the mole ratio of NH 3 to CO 2 used is 3: 1, NH 3 is in excess and CO 2 is the limiting reactant. Because the quantity of urea is given per mole of CO 2, we should base the calculation on 1.00 mol CO 2. Theoretical yield = 1.00mol CO 2 × × = 60.1g CO(NH 2 ) 2 1 mol CO(NH 2 ) 2 1 mol CO 2 1 mol CO(NH 2 ) 2 (b) Actual yield = 47.7 g CO(NH 2 ) 2 (c) %yield = × 100% = 79.4% 47.7 g CO(NH 2 ) 2 60.1 g CO(NH 2 ) 2

33. 3.7 Quantitative analysis Quantitative analysis is finding out how much of a given substance is present in a sample. Qualitative analysis is finding out what substances are present in a sample.

34. 3.8 Empirical Formulas from Percent Composition The empirical formulas for a compound is the simplest formula that shows the ratios of the numbers of atoms of each kind in the compound. Empirical formulas –give the relative numbers and types of atoms in a molecule. –That is, they give the lowest whole number ratio of atoms in a molecule. –Examples: H 2 O, CO 2, CO, CH 4, HO, CH 2.

35. Example The composition of a compound is 36.4% Mn, 21.2% S, and 42.4% O. All percents are mass percent. What is the empirical formulas of the compound? Solution: Suppose there is 100 g sample, then: Mn: 36.4/54.9=0.663mol; S: 21.1/32.1=0.660mol; O: 42.4/16.0=2.65mol 0.663:0.660:2.65≈1: 1: 4 MnSO 4

36. Molecular formulas –give the actual numbers and types of atoms in a molecule. –Examples: H 2 O, CO 2, CO, CH 4, H 2 O 2, O 2, O 3, and C 2 H 4. 3.9 Molecular and Structural Formulas

37. Most molecular substances that we will study in this class contain only nonmetals. Space-filling models

38. Molecular and empirical formulas do not show how atoms are arranged when bonded together.

39. Picturing Molecules Molecules occupy three dimensional space. However, we often represent them in two dimensions. The structural formula gives the connectivity between individual atoms in the molecule. The structural formula may or may not be used to show the three dimensional shape of the molecule. If the structural formula does show the shape of the molecule, then either a perspective drawing, ball-and- stick model, or space-filling model is used.

40. Representing Structure in Molecules Accurately represents the angles at which molecules are attached.

41. Class Practice Exercise The structural formula of propane and butane is What is the chemical and empirical formula for these molecules? C CC HH H H H HH H C CC HH H H H HH HC H H

42. 3.10 Percent Composition from Formulas Example What is the percent by mass of iron (III) to one decimal place of FeCl 3 ? Solution: