# Chapter 3. Stoichiometry: Ratios of Combination. Insert picture from First page of chapter. Copyright McGraw-Hill

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1 Chapter 3 Insert picture from First page of chapter Stoichiometry: Ratios of Combination Copyright McGraw-Hill

2 3.1 Molecular and Formula Masses Molecular mass - (molecular weight) The mass in amu s of the individual molecule Multiply the atomic mass for each element in a molecule by the number of atoms of that element and then total the masses Formula mass (formula weight)- The mass in amu s of an ionic compound Copyright McGraw-Hill

3 Calculating Molar Mass Calculate the molar mass for carbon dioxide, CO 2 Write down each element; multiply by atomic mass C = 1 x = amu O = 2 x = amu Total: = amu Copyright McGraw-Hill

4 Your Turn! Calculate the molar mass for each of the following: Sulfur trioxide Barium phosphate Silver nitrate Acetic acid Copyright McGraw-Hill

5 3.2 Percent Composition of Compounds Calculate by dividing the total mass of each element in a compound by the molecular mass of the compound and multiplying by 100 % composition allows verification of purity of a sample Copyright McGraw-Hill

7 % Composition Calculate the percent composition of iron in a sample of iron (III) oxide Formula: Fe 2 O 3 Calculate formula mass Fe = 2 x = amu O = 3 x = amu Total mass: = amu Copyright McGraw-Hill

8 % Composition % by mass = = 69.9% Fe What is the % oxygen in this sample? (hint :100%) Copyright McGraw-Hill

9 Your Turn! Calculate the percent oxygen in a sample of potassium chlorate Copyright McGraw-Hill

10 3.3 Chemical Equations Chemical equations represent chemical sentences Read the following equation as a sentence NH 3 + HCl NH 4 Cl ammonia reacts with hydrochloric acid to produce ammonium chloride Copyright McGraw-Hill

11 Chemical Equations Reactant: any species to the left of the arrow (consumed) Product: any species to the right of the arrow (formed) State symbols: (s) solid (l) liquid (g) gas (aq) water solution Copyright McGraw-Hill

12 Balancing Equations Balanced: same number and kind of atoms on each side of the equation Copyright McGraw-Hill

13 Balancing Equations Steps for successful balancing 1. Change coefficients for compounds before changing coefficients for elements.(never change subscripts!) 2. Treat polyatomic ions as units rather than individual elements. 3. Count carefully, being sure to recount after each coefficient change. Copyright McGraw-Hill

14 Balancing Equations Balance the equation representing the combustion of hexane C 6 H 14(l) + O 2(g) CO 2(g) + H 2 O (l) (Hint: Make a list of all elements and count to keep track) Copyright McGraw-Hill

15 Balancing Equations Balance the equation representing the combustion of hexane C 6 H 14(l) +19/2O 2(g) 6CO 2(g) + 7H 2 O (l) Or multiply through the entire equation to eliminate fractions 2C 6 H 14(l) +19O 2(g) 12CO 2(g) + 14H 2 O (l) Copyright McGraw-Hill

16 Chemical Equations Equations can represent physical changes: KClO 3(s) KClO 3(l) Or chemical changes: Note the symbol for heat above the arrow 2 KClO 3(s) 2 KCl (s) + 3 O 2(g) Copyright McGraw-Hill

17 3.4 The Mole and Molar Masses Balanced equations tell us what is reacting and in what relative proportions on the molecular level. However, chemists must work with the chemical reactions on a macroscopic level. Copyright McGraw-Hill

18 The Mole The unit of measurement used by chemists in the laboratory 1 mole = x (Avogadro s number represents the number of atoms that exist in exactly 12 grams of carbon-12) This is our counting number for atoms, molecules and ions much like a dozen is our counting number for cookies or doughnuts) Copyright McGraw-Hill

19 The Mole 2 H 2(g) + O 2(g) 2 H 2 O (l) 2 molecules H 2(g) + 1 molecule O 2(g) 2 molecules H 2 O (l) 2 moles H 2(g) + 1 mole O 2(g) 2 moles H 2 O (l) This relationship can be made because of Avogadro s number (N A ) Copyright McGraw-Hill

20 Moles and Atoms Calculate the number of atoms found in 4.50 moles of silicon. How many moles of silicon are in 2.45 x atoms? Copyright McGraw-Hill

21 Molar Mass Molar mass - the mass of one mole of a substance in grams Carbon = 12.0 grams/mole Sodium = 22.9 grams/mole What is the relationship between molar mass and atomic mass? Copyright McGraw-Hill

22 Molar Mass What is molar mass for each of the following? Copper metal = Helium gas = Calcium metal = Copyright McGraw-Hill

23 Molar Mass for Compounds Calculate the molar mass for each of the following: H 2 O H 2 x 1.01 g/mol = 2.02 O 1 x g/mol = Molar mass = g/mol Copyright McGraw-Hill

24 Your Turn! Calculate the molar mass for each of the following: Carbon dioxide Ammonia Oxygen gas (Don t forget the diatomics!) Copyright McGraw-Hill

25 Conversions between grams, moles and atoms Copyright McGraw-Hill

26 Interconverting mass, moles and number of particles Determine the number of moles in grams of sodium chlorate, NaClO 3 1 mole NaClO g NaClO mol NaClO g NaClO 3 3 Copyright McGraw-Hill

27 Another Determine the number of molecules in 4.6 moles of ethanol, C 2 H 5 OH. (1 mole = x ) molecules C H OH 4.6 mol C H OH molecules 2 5 1mol C H OH 2 5 Copyright McGraw-Hill

28 Another Determine how many H atoms are in 4.6 moles of ethanol. Begin with the answer to the last problem molecules C 2 H 5 OH 6 H atoms 1 molecule C 2 H 5 OH atoms H Copyright McGraw-Hill

29 Your Turn Solve the following conversions How many atoms of silver are in 3.50 moles of silver? Determine the number of moles of carbon disulfide in grams of CS 2. Determine the number of sulfur atoms in grams of CS 2. Copyright McGraw-Hill

30 Another How many grams of oxygen are present in 5.75 moles of aluminum oxide, Al 2 O 3? Strategy: Copyright McGraw-Hill

31 Challenge Determine the number of fluorine atoms in grams of sulfur hexafluoride. (hint: make a plan first!) Copyright McGraw-Hill

32 Empirical and Molecular Formulas Empirical- simplest whole-number ratio of atoms in a formula Molecular - the true ratio of atoms in a formula; often a whole-number multiple of the empirical formula We can determine empirical formulas from % composition data; a good analysis tool. Copyright McGraw-Hill

33 Empirical Formulas Steps for success Convert given amounts to moles Mole ratio (divide all moles by the smallest number of moles) The numbers represent subscripts. If the numbers are not whole numbers, multiply by some factor to make them whole. Copyright McGraw-Hill

34 Empirical Formula Determine the empirical formula for a substance that is determined to be 85.63% carbon and 14.37% hydrogen by mass. Copyright McGraw-Hill

35 3.5 Combustion Analysis Analysis of organic compounds (C,H and sometimes O) are carried using an apparatus like the one below Copyright McGraw-Hill

36 Combustion Analysis The data given allows an empirical formula determination with just a few more steps. The mass of products (carbon dioxide and water) will be known so we work our way back. Copyright McGraw-Hill

37 Combustion Analysis Suppose that 18.8 grams of glucose was burned in a combustion train resulting in 27.6 grams of carbon dioxide and 11.3 grams of water. Calculate the empirical and molecular formula of glucose. Molar mass = 180 g/mol (Assumptions: all C in CO 2 originates from glucose; all H in H 2 O originates from glucose; O is found by difference) Copyright McGraw-Hill

38 Combustion Analysis Steps: Convert 27.6 g CO 2 into g of C Convert 11.3 g H 2 O into g H Calculate g O = g sample - (g C + g H) Find empirical formula as before (g to moles, mole ratio) Copyright McGraw-Hill

39 Combustion Analysis Molecular formula = molecular mass/empirical mass = 180/30 = 6 Multiply through empirical formula to obtain new subscripts Molecular formula = C 6 H 12 O 6 Copyright McGraw-Hill

40 3.6 Calculations with Balanced Chemical Equations Balanced equations allow chemists and chemistry students to calculate various amounts of reactants and products. The coefficients in the equation are used as mole ratios. Copyright McGraw-Hill

41 Stoichiometry Stoichiometry- using balanced equations to find amounts How do the amounts compare in the reaction below? Copyright McGraw-Hill

42 Mole Ratios Many mole ratios can be written from the equation for the synthesis of urea Mole ratios are used as conversion factors 2molNH 3 1mol CO 2 or 1 molco 2 2 mol NH 3 Copyright McGraw-Hill

43 Calculations with Balanced Equations How many moles of urea could be formed from 3.5 moles of ammonia? 2NH 3(g) + CO 2(g) (NH 2 ) 2 CO (aq) + H 2 O (l) 3.5 mol NH 3 1 mol (NH 2) 2 CO 2 mol NH 3 = 1.8 mol (NH 2 ) 2 CO Copyright McGraw-Hill

44 Mass to Mass A chemist needs grams of urea, how many grams of ammonia are needed to produce this amount? Strategy: Grams moles mole ratio grams g (NH 2 ) 2 CO 1 mol(nh 2 ) 2 CO g(nh 2 ) 2 CO 2 mol NH 3 1 mol (NH 2 ) 2 CO g NH 3 1mol NH 3 = g NH 3 Copyright McGraw-Hill

45 Correction to the publisher s prevous slide: MW urea g/mol So mass is 33.34g. Copyright McGraw-Hill

46 You Try! How many grams of carbon dioxide are needed to produce 125 grams of urea? Copyright McGraw-Hill

47 3.7 Limiting Reactants Limiting reactant - the reactant that is used up first in a reaction (limits the amount of product produced) Excess reactant - the one that is left over Industry often makes the more expensive reactant the limiting one to ensure its complete conversion into products Copyright McGraw-Hill

48 Limiting Reactant If one loaf of bread contains 16 slices of bread and a package of lunchmeat contains 10 slices of turkey, how many sandwiches can be made with 2 pieces of bread and one slice of meat? Which is the limiting reactant? How much excess reactant is left? Copyright McGraw-Hill

49 Limiting Reactant How do you identify a limiting reactant problem? Example: If 5.0 moles of hydrogen react with 5.0 moles of oxygen, how many moles of water can be produced? Notice: both reactant amounts are given and a product amount is requested Copyright McGraw-Hill

50 Steps for Success Step 1: write a balanced equation Step 2: identify the limiting reactant Must compare in terms of moles Step 3: use a mole ratio to desired substance Step 4: convert to desired units Copyright McGraw-Hill

51 Limiting Reactant How many molecules of water are formed when 7.50 grams of hydrogen gas react with 5.00 grams of oxygen gas? Step 1: 2H 2(g) + O 2(g) 2 H 2 O (l) Step 2: 7.50 g H 2 /2.02 g/mol = mol 5.00 g O 2 / g/mol = mol Copyright McGraw-Hill

52 Limiting Reactant Step 2 continued: Decide which is limiting - look at the mole ratio of reactants--it takes twice as much H 2 as O 2 so O 2 limits in this case. Step 3 and step 4: molo 2 2 molh 2 O 1 molo molecules H 2 O 1 molh 2 O molecules H 2 O Copyright McGraw-Hill

53 Limiting Reactant In the previous example, how many grams of hydrogen were left in excess? Step 1: how much H 2 is used molo 2 2 molh 2 1 molo g H 2 1 molh g H 2 used Copyright McGraw-Hill

54 Limiting Reactant In the previous example, how many grams of hydrogen were left in excess? Step 2: initial H 2 - used H g g = 6.87 g excess Copyright McGraw-Hill

55 Your Turn! When grams of nitrogen react with grams of hydrogen, how many grams of ammonia are produced? How many grams of excess reagent remain in the reaction vessel? Copyright McGraw-Hill

56 Reaction Yield Theoretical yield: the maximum amount of product predicted by stoichiometry Actual yield: the amount produced in a laboratory setting Percent yield: a ratio of actual to theoretical (tells efficiency of reaction) Copyright McGraw-Hill

57 Percent Yield When a student reacted 3.75 grams of zinc with excess hydrochloric acid, 1.58 grams of zinc chloride were collected. What is the percent yield for this reaction? % yield = actual theoretical 100 Copyright McGraw-Hill

58 Percent Yield Step 1: Balanced equation Step 2: Calculate theoretical yield Step 3: Substitute into formula and solve Copyright McGraw-Hill

59 Percent Yield Zn (s) + 2 HCl (aq) ZnCl 2(aq) + H 2(g) Theoretical yield = 7.82 g ZnCl 2 Actual yield = 1.58 g ZnCl 2 Calculate % yield: (1.58g/7.82g)x100% = 20.2% Copyright McGraw-Hill

60 A Few Reaction Types Combination: one product is formed Decomposition: one reactant produces more than one product Combustion: a hydrocarbon reacts with oxygen to produce carbon dioxide and water Copyright McGraw-Hill

61 Combination Reaction General formula: A + B AB Sodium + chlorine sodium chloride 2Na + Cl 2 2 NaCl Sulfur dioxide + water sulfurous acid SO 2 + H 2 O H 2 SO 3 Copyright McGraw-Hill

62 Decomposition Reaction General formula: AB A + B Copper (II) carbonate decomposes with heat into copper (II) oxide and carbon dioxide CuCO 3 CuO + CO 2 Potassium bromide decomposes into its elements 2KBr 2K + Br 2 Copyright McGraw-Hill

63 Combustion (hydrocarbons) General formula: C x H y + O 2 CO 2 + H 2 O Methane gas burns completely CH 4 + 2O 2 CO 2 + 2H 2 O Butane liquid in a lighter ignites 2C 4 H O 2 8CO H 2 O Copyright McGraw-Hill

64 Review Molecular mass Percent composition Chemical equations Reactants Products State symbols Balancing Copyright McGraw-Hill

65 Review continued Mole concept and conversions Empirical and molecular formulas Combustion analysis Stoichiometry Limiting reactant % yield Types of reactions Copyright McGraw-Hill

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