# Chapter 3 Stoichiometry Mole - Mass Relationships in Chemical Systems

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1 Chapter 3 Stoichiometry Mole - Mass Relationships in Chemical Systems 3.1 Atomic Masses 3.2 The Mole 3.3 Molar Mass 3.4 Percent Composition of Compounds 3.5 Determining the Formula of a Compound 3.6 Chemical Equations 3.7 Balancing Chemical equations 3.8 Stoichiometric Calculations: Amounts of Reactants and Products 3.9 Calculations Involving a Limiting Reactant 1 Atomic Definitions Atomic Mass Unit (amu) = 1/12 the mass of a carbon-12 atom; on this scale hydrogen has a mass of amu (unified atomic mass unit... u ) Dalton (D) = another name for the atomic mass unit; on this scale 12 C has a mass of Daltons;1 Dalton = 1 amu Isotopic Mass = The relative mass of an isotope compared to the 12 C isotope standard = amu Atomic Mass or Atomic Weight of an element = the average of the masses of its naturally occurring isotopes weighted according to their abundances. 2

2 How Do We Determine Atomic Mass? GRAVIMETRY weighing and analyzing an element or compound MASS SPECTROMETRY ionization of elements and deflection of their path by applied magnetic field; RATIO of atomic masses determined; accurate comparison of masses of elements and their isotopes. 3 4

3 Operation of a Mass Spectrometer 3.1) E A B C D A. Sample vaporized (made into a gas) B. Electron gun removes an electron, forming a positively charged species C. Positive ion accelerated through an electric field D. Magnetic deflection Lighter ions deflected more than heavier ones. E. Observe impact on detector 5 Figure 3.2a: Peaks of neon from injected sample THREE PEAKS, so THREE ISOTOPES! Figure 3.2b: Relative peak areas 91% 20 Ne 0.3% 21 Ne 9% 22 Ne This gives the isotopic composition of Ne. 6

4 What is the average atomic mass of the three stable isotopes of neon: 20 Ne ( 90.70%), 21 Ne (0.30%), and 22 Ne (9.00%)? amu amu amu amu amu 7 Calculating the Average Atomic Mass of an Element Problem: Calculate the average atomic mass of the three stable isotopes of magnesium: 24 Mg ( 78.7%), 25 Mg (10.2%), and 26 Mg (11.1%). 24 Mg (78.7%) amu x = amu 25 Mg (10.2%) amu x = amu 26 Mg (11.1%) amu x = amu Total = amu With appropriate significant digits 24.3 amu 8

5 What is the abundance of the two bromine isotopes, 79 Br = amu and 81 Br = amu, given that the average mass of bromine is amu? Br = %, 81 Br = % Br = 50.67%, 81 Br = 49.33% Br = 50.7%, 81 Br = 49.3% 9 Problem: Calculate the abundance of the two bromine isotopes, 79 Br = amu and 81 Br = amu, given that the average mass of bromine is amu. Plan: Let the abundance of 79 Br = X and of 81 Br = Y and X + Y = 1.0 Solution: X( ) + Y( ) = X + Y = 1.00, therefore X = 1.00 Y ( Y)( ) + Y( ) = Y Y = Y = Y = X = Y = = %X = % 79 Br = x 100 = % = 79 Br %Y = % 81 Br = x 100 = % = 81 Br 10

6 Isotopes of Hydrogen Natural Abundance 1 1H 1 Proton 0 Neutrons % amu 2 1H (D) 1 Proton 1 Neutron % amu 3 1H (T) 1 Proton 2 Neutrons The average mass of Hydrogen is amu 3 H radioactive with a half life of 12 years H 2 O normal water light water mass = 18.0 g/mole, BP = C D 2 O heavy water mass = 20.0 g/mole, BP = C 11 What will happen to the ball of H 2 O? D 2 O? 1. H 2 O : sink, D 2 O : float 2. H 2 O : float, D 2 O : sink 3. H 2 O : float, D 2 O : float 4. H 2 O : sink, D 2 O : sink 12

7 The mole NOT this kind of mole 13 MOLE just a number, like a dozen The amount of substance that contains as many elementary particles as there are atoms in exactly 12 grams of carbon -12 (i.e., 12 C). 1 mole = x particles (atoms, molecules, ions, marbles, apples, etc ) ~ 100 million x 100 million x 100 million 14

8 Avogadro s Number (N A ) N A = x = # of particles (atoms, molecules, ions, ) in one mole of that substance. Conversion factor: x whatever = 1 mol 15 One mole of common substances CaCO 3 :100.09g Oxygen gas: 32.00g Note: one mole of oxygen gas (O 2 ) would occupy 22.4 L at room temperature. Copper: 63.55g Water:18.02g 16

9 Dozen mass Mole mass relationships relationships 12 red marbles (7g each) = 84g 12 yellow marbles (4g each) = 48g x atoms Fe = 55.85g Fe x atoms S = 32.07g S 17 2 H 2 (g) + O 2 (g) 2 H 2 O(g) 2 dozen H 2 molecules react with exactly 1 dozen O 2 molecules to give exactly 2 dozen H 2 O molecules. 2 moles of H 2 molecules react with exactly 1 mole of O 2 molecules to give exactly 2 moles of H 2 O molecules. Why do we do this? Because these last amounts are in the gram range and easy to weigh. Conventions: 1 mole of 12 C atoms weighs exactly 12 g 1 atom of 12 C weighs exactly 12 amu (amu = atomic mass unit mass of a proton or neutron) 18

10 Mole - Mass Relationships of Elements Element Atomic Mass 1 Mole of Atoms Molar Mass 1 atom of H = amu x atoms = g H 1 atom of Fe = amu x atoms = g Fe 1 atom of S = amu x atoms = g S 1 atom of O = amu x atoms = g O Molecular mass: amu and g/mol are numerically the same! 1 molecule of O 2 = x 2 = amu 1 mole of O 2 = x molecules = g 1 molecule of S 8 = x 8 = amu 1 mole of S 8 = x molecules = g 19 What is the molar mass of water (H 2 O)? g/mol g/mol g/mol g/mol 20

11 Molecular Mass - Molar Mass (MM) The molecular mass of a compound expressed in amu is numerically the same as the mass of one mole of the compound expressed in grams, called its molar mass. For water: H 2 O Molecular mass = (2 x atomic mass of H ) + (1 x atomic mass of O) = 2 (1.008 amu) amu = amu Mass of one molecule of water = amu Molar mass = (2 x molar mass of H ) + (1 x molar mass of O) = 2 (1.008 g ) g = g per mole H 2 O g H 2 O = x molecules of water = 1 mole H 2 O 21 Calculating the Number of Moles and Atoms in a Given Mass of Element Problem: Tungsten (W) is the element used as the filament in incandescent light bulbs and has the highest melting point of any element, 3695 K (3422 o C). How many moles of tungsten, and atoms of the element, are present in a 35.0 mg sample of the metal? Plan: Convert Mass to Moles Molar Mass Convert Moles to Atoms Avogadro s Number From the Periodic Table, we find that the molar mass of W is g/mol. 22

12 Calculating the Number of Moles and Atoms in a Given Mass of Element Problem: Tungsten (W) is the element used as the filament in incandescent light bulbs and has the highest melting point of any element, 3680 o C. How many moles of tungsten, and atoms of the element, are present in a 35.0 mg sample of the metal? Solution: Moles of W = 35.0 x 10-3 g W 1 mol W = mol g W = 1.90 x 10-4 mol No. of W atoms = 1.90 x 10-4 mol W x atoms = 1 mol W 1.15 x atoms of tungsten

13 Calculating the Moles in a Given Mass of a Compound Problem: Sodium phosphate is a component of some detergents. How many moles are in a 38.6 g sample? Plan: We need to determine the formula and the molecular mass from the atomic masses of each element multiplied by the subscripts. 25 What is the chemical formula for sodium phosphate? 5 1. NaPO 3 2. NaPO 4 3. Na 2 PO 4 4. Na 3 PO 4 5. None of the above 26

14 What is the molecular mass for sodium phosphate (Na 3 PO 4 )? g/mol g/mol g/mol g/mol g/mol 27 Calculating the Moles in a Given Mass of a Compound Problem: Sodium phosphate is a component of some detergents. How many moles are in a 38.6 g sample? Plan: We need to determine the formula and the molecular mass from the atomic masses of each element multiplied by the subscripts. Solution: The formula is Na 3 PO 4. Calculating the molar mass: MM = 3 x sodium + 1 x phosphorous + 4 x oxygen = 3 x g/mol + 1 x g/mol + 4 x g/mol = g/mol g/mol g/mol = g/mol 28

15 How many moles of sodium phosphate are in a 38.6 g sample? MM= g/mol mol mol mol mol 29 Calculating the Moles in a Given Mass of a Compound Problem: Sodium phosphate is a component of some detergents. How many moles are in a 38.6 g sample? Plan: We need to determine the formula and the molecular mass from the atomic masses of each element multiplied by the subscripts. Solution: The formula is Na 3 PO 4. The molar mass is g/mol. Calculating the moles in a 38.6 g sample: moles of Na 3 PO 4 = 38.6 g sample 1 mol g sample = mol 30

16 MM (g/mol) MM (g/mol) 31 Atoms Molecular Formula Molecules x mol Moles x mol 32

17 33 Mass Fraction and Mass % Mass of Red Balls = 3 balls x 3.0 g/ball = 9.0 g Mass Fraction Red = 9.0 g / 16.0 g total = 0.56 Mass % Red = 0.56 x 100 = 56% red Mass Fraction Purple and Mass = % Purple = 2 balls x 2.0 g/ball / (16.0 g total) = 0.25 = 25% Similarly, mass fraction yellow = 3x1.0/16.0 = 0.19 Check: 56% + 25% + 19% = 100% 34

18 Calculate the Percent Composition of Sulfuric Acid H 2 SO 4 Molar mass of sulfuric acid = 2(1.008 g) + 1(32.07 g) + 4(16.00 g) = g/mol %H = 2(1.008 g H) x 100% = g %S = 1(32.07 g S) x 100% = g %O = 4(16.00 g O) x 100% = g 2.06% H 32.69% S 65.25% O Check = % 35 Calculating Mass Percentage and Masses of Elements in a Sample of a Compound Problem: Sucrose (C 12 H 22 O 11 ) is common table sugar. ( a) What is the mass percent of each element in sucrose? ( b) How many grams of carbon are in g of sucrose? 36

19 What is the mass percent of each element in sucrose (C 12 H 22 O 11 )? 1. %C=26.471%, %H=8.824%, %O=64.71% 2. %C=26.47%, %H=8.83%, %O=64.71% 3. %C=42.10%, %H=6.48%, %O=51.42% 4. %C=42.103%, %H=6.479%, %O=51.42% 37 Calculating Mass Percentage and Masses of Elements in a Sample of a Compound - I Problem: Sucrose (C 12 H 22 O 11 ) is common table sugar. ( a) What is the mass percent of each element in sucrose? ( b) How many grams of carbon are in g of sucrose? (a) Determining the mass percent of each element: mass of C per mole sucrose = 12 x g C/mol = g C/mol mass of H / mol = 22 x g H/mol = g H/mol mass of O / mol = 11 x g O/mol = g O/mol total mass per mole = g/mol Finding the mass fraction of C in sucrose & % C : Mass fraction of C = mass of C per mol sucrose = mass of 1 mol sucrose g C/mol g sucrose/mol = To find mass % of C = x 100% = % 38

20 Calculating Mass Percents and Masses of Elements in a Sample of Compound - II (a) continued Mass % of H = mol H x M of H x100% = 22 x g H x 100% mass of 1 mol sucrose g = 6.479% H Mass % of O = mol O x M of O x100% = 11 x g O x 100% mass of 1 mol sucrose g = 51.42% O 39 How many grams of carbon are in g of sucrose? g g g g 40

21 Calculating Mass Percentage and Masses of Elements in a Sample of a Compound Problem: Sucrose (C 12 H 22 O 11 ) is common table sugar. ( a) What is the mass percent of each element in sucrose? ( b) How many grams of carbon are in g of sucrose? (a) Determining the mass percent of each element: mass % of C = % (b) Determining the mass of carbon in g sucrose: Mass (g) of C = mass of sucrose x ( mass fraction of C in sucrose) Mass (g) of C = g sucrose g C 1 g sucrose = g C 41 Empirical and Molecular Formulas Empirical Formula - The simplest formula for a compound that agrees with the elemental analysis! The smallest set of whole numbers of atoms. Molecular Formula - The formula of the compound as it really exists. It must be a multiple of the empirical formula. 42

22 Some Examples of Compounds with the Same Elemental Ratios Empirical Formula Molecular Formula CH 2 (unsaturated hydrocarbons) C 2 H 4 C 3 H 6 C 4 H 8 OH or HO H 2 O 2 S S 8 P P 4 Cl Cl 2 CH 2 O (carbohydrates) C 6 H 12 O

23 Condensed structural formula CH 3 CH 2 OH CH 3 OCH 3 45 Elemental Analysis Decomposition or combustion analysis is used to determine the mass of each type of element present in a compound. Figure 3.5: Schematic Diagram of a Combustion Analysis Device 46

24 Steps for Determining Empirical Formulas Mass (g) of each element in sample MM (g/mol ) for that element Moles of each element Use # of moles as subscripts. Preliminary formula Empirical formula Change to integer subscripts: by smallest subscript, convert to whole # subscripts. 47 Problem: The elemental analysis of a sample gave the following results: 5.677g Na, g Cr, and g O. What is the empirical formula and name of the compound? Plan: Determining Empirical Formulas from Measured Masses of Elements - I Convert mass to moles Moles = mass/mm Construct preliminary formula Convert to empirical formula Divide each moles by smallest number of moles 48

25 Determining Empirical Formulas from Measured Masses of Elements - I Problem: The elemental analysis of a sample gave the following results: 5.677g Na, g Cr, and g O. What is the empirical formula and name of the compound? Solution: Finding the moles of the elements: Moles of Na = g Na Moles of Cr = g Cr Moles of O = g O 1 mol Na = mol Na g Na 1 mol Cr = mol Cr g Cr 1 mol O = mol O g O 49 Determining Empirical Formulas from Masses of Elements - II Constructing the preliminary formula: Na Cr O Converting to integer subscripts: (dividing all by smallest subscript) Na Cr O Rounding off to whole numbers: Na 2 CrO 4 sodium chromate 50

26 Determining the Molecular Formula from Elemental Composition and Molar Mass - I Problem: The sugar burned for energy in cells of the body is glucose (MM = g/mol), elemental analysis shows that it contains mass % C, mass % H, and mass % O. (a) Determine the empirical formula of glucose. (b) Determine the molecular formula. Plan: We are only given mass %, and no weight of the compound so we will assume we have 100 g of the compound. Convert mass to moles Moles = mass/mm Construct preliminary formula Convert to empirical formula Divide each moles by smallest number of moles 51 Determining the Molecular Formula from Elemental Composition and Molar Mass - I Problem: The sugar burned for energy in cells of the body is glucose (MM = g/mol), elemental analysis shows that it contains mass % C, mass % H, and mass % O. (a) Determine the empirical formula of glucose. (b) Determine the molecular formula. Remember: We are only given mass % and no weight of the compound so we will assume 100 g of the compound. Solution: Mass carbon = 40.00% x 100 g/100% = g C Mass hydrogen = 6.729% x 100 g/100 = g H Mass oxygen = 53.27% x 100 g/100% = g O g cmpd 52

27 What is the empirical formula of glucose? mass % C, mass % H, and mass % O 1. C 3 H 6 O 3 2. CH 2 O 3. C 3 H 7 O 3 4. CH 7 O 53 Determining the Molecular Formula from Elemental Composition and Molar Mass - II Converting from grams of elements to moles: Moles of C = g C 1 mole C = moles C g C Moles of H = g H 1 mol H = moles H g H Moles of O = g O 1 mol O = moles O g O Constructing the preliminary formula C 3.33 H 6.67 O 3.33 Converting to integer subscripts, all subscripts by the smallest: C 3.33/3.33 H / 3.33 O 3.33 / 3.33 = C 1 H 2 O 1 = CH 2 O Empirical Formula!! 54

28 Determining the Molecular Formula from Elemental Composition and Molar Mass - I Problem: The sugar burned for energy in cells of the body is glucose (MM = g/mol), elemental analysis shows that it contains mass % C, mass % H, and mass % O. (a) Determine the empirical formula of glucose: CH 2 O (b) Determine the molecular formula. Plan: We are only given mass %, and no weight of the compound so we will assume we have 100 g of the compound. Determine the empirical formula Determine molar mass of empirical formula Divide molecular formula mass by empirical formula mass MOLECULAR FORMULA Multiply empirical formula by n n 55 What is the molecular formula of glucose? CH 2 O, MM= g/mol 1. CH 2 O 2. C 3 H 6 O 3 3. C 6 H 12 O 6 4. C 9 H 18 O 9 56

29 Determining the Molecular Formula from Elemental Composition and Molar Mass - III (b) Determining the molecular formula: The molar mass of the empirical formula is: (1*C) + (2*H) + (1*O) = (1*12.01) + (2*1.008) + (1*16.00) = g/mol whole-number multiple = Therefore the molecular formula is: MM of the compound empirical molar mass g/mol = 6.00 = g/emp. mol Stated in the problem C 1 x 6 H 2 x 6 O 1 x 6 = C 6 H 12 O 6 57 Ascorbic acid (Vitamin C) - I contains only C, H, and O Upon combustion in excess oxygen, a 6.49 mg sample yielded 9.74 mg CO 2 and 2.64 mg H 2 O. Calculate the empirical formula of ascorbic acid. Plan: Calculate mass of C and H in CO 2 and H 2 O Generic: Calculate mass of all elements (except common element) in respective compounds Subtract masses of C and H from compound mass to determine mass of O Subtract known masses of all elements from compound mass to determine mass of common element Calculate empirical formula Calculate empirical formula 58

30 Ascorbic acid ( Vitamin C ) - I (contains only C, H, and O) Upon combustion in excess oxygen, a 6.49 mg sample yielded 9.74 mg CO 2 and 2.64 mg H 2 O. Calculate the empirical formula of ascorbic acid. Mass of C and H in the CO 2 and H 2 O, respectively: 9.74 x10-3 g CO g C = 2.65 x 10-3 g C g CO x10-3 g H 2 O g H = 2.95 x 10-4 g H g H 2 O Mass of O: 6.49 x10-3 g sample x10-3 g C x10-4 g H = 3.55 x 10-3 g O 59 What is the empirical formula for ascorbic acid? 2.65 x 10-3 g C, 2.95 x 10-4 g H, 3.55 x 10-3 g O 1. CH 1.33 O 2. C 2 HO 3. C 3 H 4 O 3 60

31 Vitamin C combustion - II Now we convert to moles: C: 2.65 x 10-3 g C 1 mol C = 2.21 x 10-4 mol C g C H: 2.95 x 10-3 g H 1 mol H = 2.93 x 10-4 mol H 1.008g H O: 3.55 x 10-3 g O 1 mol O = 2.22 x 10-4 mol O 16.00g O Divide each by smallest (2.21 x 10-4 ): C = 1.00 Multiply each by 3: C = 3.00 = 3.0 H = 1.33 (to get ~integers) H = 3.99 = 4.0 O = 1.00 C 3 H 4 O 3 O = 3.00 = If the empirical formula of ascorbic acid is C 3 H 4 O 3, and the molecular mass of ascorbic acid is 176 g/mol, what is the molecular formula? 1. C 3 H 4 O 3 2. C 6 H 8 O 6 3. C 9 H 12 O 9 4. C 12 H 16 O None of the above 62

32 Chemical Equations Qualitative information about a chemical reaction: Reactants Products States of Matter: (s) solid (l) liquid (g) gaseous (aq) aqueous 2 H 2 (g) + O 2 (g) 2 H 2 O (g) But also Quantitative Information! 63 (Like Table 3.2) 64

33 How to Balance Equations Mass Balance (or Atom Balance)- same number of each element on each side of the equation: (1) start with largest/most complicated molecule (2) progress to other elements, leaving lone elements for last (3) make all whole numbers (4) re-check atom balance 1 CH 4 (g) + O 2 (g) 1 CO 2 (g) + H 2 O (g) 1 CH 4 (g) + O 2 (g) 1 CO 2 (g) + 2 H 2 O (g) 1 CH 4 (g) + 2 O 2 (g) 1 CO 2 (g) + 2 H 2 O (g) 1 CH 4 (g) + 2 O 2 (g) 1 CO 2 (g) + 2 H 2 O (g) 65 Balancing Chemical Equations - I Problem: The hydrocarbon hexane is a component of gasoline that burns in an automobile engine to produce carbon dioxide and water, as well as energy. Write the balanced chemical equation for the combustion of hexane (C 6 H 14 ). Plan: Write the skeleton equation, converting the words into compounds, with blanks before each compound. Begin element balance, putting 1 on the most complex compound first, and save oxygen until last! Solution: C 6 H 14 (l) + O 2 (g) CO 2 (g) + H 2 O (g) + Energy Begin with one C 6 H 14 molecule which says that we will get 6 CO 2 s! 1 C 6 H 14 (l) + O 6 2 (g) CO 2 (g) + H 2 O (g) + Energy 66

34 Balancing Chemical Equations - II The H atoms in the hexane will end up as H 2 O, and we have 14 H atoms. Since each water molecule has two H atoms, we will get a total of 7 water molecules. 1 C 6 H 14 (l) + O 6 2 (g) CO 2 (g) + 7 H 2 O (g) + Energy Since oxygen atoms only come as diatomic molecules (two O atoms, O 2 ),we must have even numbers of oxygen atoms on the product side. We do not since we have 7 water molecules! Therefore, multiply everything by 2, giving a total of 2 hexane molecules, 12 CO 2 molecules, and 14 H 2 O molecules. 2 C 6 H 14 (l) + O 12 2 (g) CO 2 (g) + 14 H 2 O (g) + Energy This now gives 24 O from the carbon dioxide, and 14 O atoms from the water, which will be a total of = 38 O or 19 O 2! 2 C 6 H 14 (l) + 19 O 2 (g) 12 CO 2 (g) + 14 H 2 O (g) + Energy 67 Which of the following is always true for a balanced chemical equation? 1. Number of atoms in reactants equal number of atoms in products. 2. Volume of reactants equals volume of products. 3. Moles of reactants equal moles of products. 4. Mass of reactant equal mass of product. 5. More than one of the above. 68

35 Atoms (Molecules) Avogadro s Number Reactants Chemical Equations & Calculations x mol Moles Molar Mass Mass g/mol Products 69 MM (g/mol) of compound A MM (g/mol) of compound B 70

36 Calculating Reactants and Products in a Chemical Reaction Problem: Given the following chemical reaction between aluminum sulfide and water, if we are given g of Al 2 S 3 : a) How many moles of water are required for the reaction? b) What mass of H 2 S & Al(OH) 3 would be formed? Plan: _ Al 2 S 3 (s) + _ H 2 O (l) _ Al(OH) 3 (s) + _ H 2 S (g) Balance the chemical reaction Calculate the moles of aluminum sulfide Calculate moles of water needed, using the numbers in front of the compounds Convert to masses using molar mass of compounds Calculate moles of H 2 S and Al(OH) 3 produced, using the numbers in front of the compounds 71 What are the coefficients in the balanced equation? _ Al 2 S 3 (s) + _ H 2 O (l) _ Al(OH) 3 (s) + _ H 2 S (g) 1. 1,3,2,3 2. 1,3,2,6 3. 1,6,2,3 4. 2,6,4,3 72

37 Calculating Reactants and Products in a Chemical Reaction Problem: Given the following chemical reaction between aluminum sulfide and water, if we are given g of Al 2 S 3 : a) How many moles of water are required for the reaction? b) What mass of H 2 S & Al(OH) 3 would be formed? Al 2 S 3 (s) + 6 H 2 O (l) 2 Al(OH) 3 (s) + 3 H 2 S (g) Solution: a) molar mass of aluminum sulfide = g/mol moles Al 2 S 3 = g Al 2 S 3 mol Al 2 S 3 = moles Al 2 S g Al 2 S 3 Now think back to unit conversion and the information the chemical equation provides.. 73 Calculating Reactants and Products in a Chemical Reaction a) cont moles Al 2 S 3 6 moles H 2 O = moles H 2 O 1 mole Al 2 S 3 b) moles Al 2 S 3 3 moles H 2 S = moles H 2 S 1 mole Al 2 S 3 molar mass of H 2 S = g/mol mass H 2 S = moles H 2 S g H 2 S 1 mole H 2 S = g H 2 S moles Al 2 S 3 2 moles Al(OH) 3 = moles Al(OH) 3 1 mole Al 2 S 3 molar mass of Al(OH) 3 = g/mol mass Al(OH) 3 = moles Al(OH) g Al(OH) 3 = 1 mole Al(OH) 3 74 = g Al(OH) 3

38 Calculating the Amounts of Reactants and Products in a Reaction Sequence - I Problem: Calcium phosphate could be prepared from phosphorus in the following reaction sequence: POW!! 3 P 4 (s) + 10 KClO 3 (s) 3 P 4 O 10 (s) + 10 KCl (s) P 4 O 10 (s) + 6 H 2 O (l) HISSSSS 4 H 3 PO 4 (aq) 2 H 3 PO 4 (aq) + 3 Ca(OH) 2 (aq) 6 H 2 O (aq) + Ca 3 (PO 4 ) 2 (s) Given g P 4 and sufficient KClO 3, H 2 O and Ca(OH) 2. What mass of calcium phosphate could be formed? Plan: (1) Calculate moles of P 4 provided. (2) Use molar ratios to get moles from P 4 to Ca 3 (PO 4 ) 2. (3) Convert the moles of product back into mass using the molar mass of calcium phosphate. 75 How many moles of P 4 are in g? MM P = g/mol mol g mol g 76

39 Calculating the Amounts of Reactants and Products in a Reaction Sequence - II Solution: moles of phosphorous = g P 4 1 mole P 4 = mol P g P 4 For Reaction #1: 3 P 4 (s) + 10 KClO 3 (s) 3 P 4 O 10 (s) + 10 KCl (s) For Reaction #2: 1 P 4 O 10 (s) + 6 H 2 O (l) For Reaction #3: 2 H 3 PO Ca(OH) 2 4 H 3 PO 4 (aq) 1 Ca 3 (PO 4 ) H 2 O moles P 4 3 moles P 4 O 10 4 x moles H 3 PO 4 1 mole x Ca 3 (PO 4 ) 2 3 moles P 4 1 mole P 4 O 10 2 moles H 3 PO 4 = moles Ca 3 (PO 4 ) 2 Molar mass of Ca 3 (PO 4 ) 2 = g/mol mass of product = moles Ca 3 (PO 4 ) g Ca 3 (PO 4 ) 2 = g Ca 3 (PO 4 ) 1 mole Ca 3 (PO 4 ) P 4 (s) + 10 KClO 3 (s) 3 P 4 O 10 (s) + 10 KCl(s) Heat of reaction = - 9,425 kj The head of "strike anywhere" matches contain an oxidizing agent such as potassium chlorate together with tetraphosphorus trisulfide (P 4 S 3 ), glass and binder. The phosphorus sulfide is easily ignited, the potassium chlorate decomposes to give oxygen, which in turn causes the phosphorus sulfide to burn more vigorously. The head of safety matches are made of an oxidizing agent such as potassium chlorate, mixed with sulfur, fillers and glass powder. The side of the box contains red phosphorus, binder and powdered glass. The heat generated by friction when the match is struck causes a minute amount of red phosphorus to be converted to white phosphorus, which ignites spontaneously in air. This sets off the decomposition of potassium chlorate to give oxygen and potassium chloride. The sulfur catches fire and ignites the wood. 78

40 LIMITING REACTANT The amount of reagent that limits the amount of product that can be formed is the limiting reactant. When chemicals are mixed in exact stoichiometric quantities, none of the reagents is limiting. So far, we have been assuming that most of the reactants are available in excess of the amount needed to react with a fixed amount of one of the reactants. Read about the HABER process for producing ammonia (section 3.9 in the book) as an example of a limiting reactant. 79 LIMITING REAGENT 3N 2 (g) + 9 H 2 (g) 6 NH 3 (g) 80 Figure 3.9: Hydrogen and nitrogen reaction to form ammonia

41 81 Limiting Reactant Problem: A Sample Problem Problem: A fuel mixture used in the early days of rocketry is composed of two liquids, hydrazine (N 2 H 4 ) and dinitrogen tetraoxide (N 2 O 4 ). They ignite on contact to form nitrogen gas and water vapor. How many grams of nitrogen gas form when exactly 1.00 x 10 2 g N 2 H 4 and 2.00 x 10 2 g N 2 O 4 are mixed? Plan: First write the balanced equation. Since amounts of both reactants are given, this is a limiting reactant problem! Calculate the moles of each reactant. Pick one of the reactants and figure out how much of the other you would need. If you have more than enough of the second reagent, then the first is limiting. Use the moles of the limiting reactant to calculate the moles of nitrogen gas that will be formed. Finally, calculate the mass using the molecular weight of nitrogen gas. Solution: 2 N 2 H 4 (l) + N 2 O 4 (l) 3 N 2 (g) + 4 H 2 O (g) + Energy 82

42 Sample Problem cont. 2 N 2 H 4 (l) + N 2 O 4 (l) 3 N 2 (g) + 4 H 2 O (g) molar mass N 2 H 4 = ( 2 x x ) = g/mol molar mass N 2 O 4 = ( 2 x x ) = g/mol 1.00 x 10 Moles N 2 H 4 = 2 g = 3.12 moles N 2 H g/mol 4 Limiting! Moles N 2 O 4 = 2.00 x 10 2 g = 2.17 moles N 2 O g/mol For 3.12 moles N 2 H 4 we would need: 3.12 mol N 2 H 4 1 mol N 2 O 4 = 1.56 moles N 2 O 4 we have 2.17moles 2 mol N 2 H 4 Nitrogen yielded = 3.12 mol N 2 H 4 3 mole N 2 = 4.68 moles N 2 2 mol N 2 H 4 Mass of nitrogen = 4.68 moles N 2 x g N 2 /mol = 131 g N 2 83 Which of these is always true for limiting reagent questions? 1. It will say in the question that it is a limiting reactant question. 2. The equation provided will be balanced. 3. There will be quantitative information about more than one of the reactants. 4. There will be information about only products. 84

43 Acid - Metal Reaction 2 Al (s) + HCl (g) 2 AlCl 3(s) + H 2(g) Consider the reaction above. Is it balanced? (An unbalanced reaction won t help you much.) 2 Al (s) + 6 HCl (g) 2 AlCl 3(s) + 3 H 2(g) If we react 30.0 g Al and 20.0 g HCl, how many moles of aluminum chloride will be formed? Yes 2. No Is this a Limiting Reagent question? 86

44 Acid - Metal Limiting Reactant - I 2 Al (s) + 6 HCl (g) 2 AlCl 3(s) + 3 H 2(g) If we react 30.0 g Al and 20.0 g HCl, how many moles of aluminum chloride will be formed? A shortcut calculate the moles of each reactant and divide each # of moles by the reactant coefficient in the balanced reaction to find which reactant is limiting. Use the moles of that reactant to calculate the moles of product formed g Al * (mol Al/26.98 g Al) = 1.11 mol Al 1.11 mol Al / 2 = equivalents of Al 20.0 g HCl * (mol HCl/36.5 g HCl) = mol HCl mol HCl / 6 = equivalents of HCl Fewer equivalents of HCl, so HCl is the limiting reactant! 87 Acid - Metal Limiting Reactant - II 2 Al (s) + 6 HCl (g) 2 AlCl 3(s) + 3 H 2(g) Since 6 moles of HCl yield 2 moles of AlCl 3, moles of HCl will yield: mol HCl 2 moles of AlCl 3 = mol of AlCl 3 6 mol HCl 88

45 a A + b B + c C Steps for solving: Limiting Reactant Problems d D + e E + f F 1) Identify it as a limiting reactant problem - Information about the mass, number of moles, number of molecules, or volume and molarity of a solution is given for more than one reactant! 2) Calculate moles of each reactant! 3) Determine which reactant is limiting: either 1) calculate the amount of other reactants needed and compare with what is available or 2) divide the moles of each reactant by stoic. coefficient (a, b, etc...) and whichever is smallest, that reactant is the limiting reactant! 4) Use the limiting reactant to calculate the moles of product(s) that will form or the moles of other reactants that will be used up, then convert to the units needed (moles, mass, volume, number of atoms, etc...)! 89 Solving for Masses of Reactants and Products Start here 90

46 Titrations as Examples of Limiting Reactant Problems Before the endpoint the titrant is limiting and the sample is in excess. At the endpoint there are stoichiometric quantities (equivalent amounts) of both reactants. Past the endpoint the sample is limiting and the titrant is in excess. (endpoint equivalence point) You ll see this titration terminology in Lab #2 and we ll revisit it in Chapter Chemical Reactions in Practice: Theoretical, Actual, and Percent Yields Theoretical yield: The amount of product expected based on amounts of reactants and stoichiometry (molar ratios) in the balanced equation. Actual yield: The actual amount of product that is obtained. Side Reactions: These form smaller amounts of different products that reduce the actual yield of the product of interest. Percent yield (%Yield): Actual Yield (mass or moles) % Yield = x 100% Theoretical Yield (mass or moles) 92

47 93 What is the actual yield of a chemical reaction if the percent yield was 50%, and the theoretical yield was 8 g product? g product 2. 1 g product 3. 2 g product 4. 4 g product g product 6. None of the above 94

48 Percent Yield Calculation Problem: Given the chemical reaction between iron and water to form the iron oxide Fe 3 O 4 and hydrogen gas, if 4.55 g of iron is reacted with sufficient water to convert all of the iron to rust, what is the percent yield if only 6.02 g of the oxide forms? Plan: Balance the chemical reaction, calculate the theoretical yield and use it and the actual yield to calculate the percent yield. Solution: _ Fe (s) + _ H 2 O (l) _ Fe 3 O 4 (s) + _ H 2 (g) 95 What are the coefficients of the balanced chemical reaction? _ Fe (s) + _ H 2 O (l) _ Fe 3 O 4 (s) + _ H 2 (g) 1. 1,4,1,2 2. 3,2,1,2 3. 3,4,1,2 4. 3,4,1,4 96

49 Percent Yield Calculation Problem: Given the chemical reaction between iron and water to form the iron oxide Fe 3 O 4 and hydrogen gas, if 4.55 g of iron is reacted with sufficient water to convert all of the iron to rust, what is the percent yield if only 6.02 g of the oxide forms? Plan: Balance the chemical reaction, calculate the theoretical yield and use it and the actual yield to calculate the percent yield. Solution: 3 Fe (s) + 4 H 2 O (l) Fe 3 O 4 (s) + 4 H 2 (g) 97 What is the theoretical yield of Fe 3 O 4 (s)? Fe (s) = 4.55 g, MM Fe = g/mol, MM O = g/mol g mol g mol 98

50 Problem: Given the chemical reaction between iron and water to form the iron oxide Fe 3 O 4 and hydrogen gas, if 4.55 g of iron is reacted with sufficient water to convert all of the iron to rust, what is the percent yield if only 6.02 g of the oxide forms? Plan: Balance the chemical reaction, calculate the theoretical yield and use it and the actual yield to calculate the percent yield. Solution: 3 Fe (s) + 4 H 2 O (l) Fe 3 O 4 (s) + 4 H 2 (g) 4.55 g Fe g Fe/mol Percent Yield Calculation = mol Fe = mol Fe mol Fe 1 mol Fe 3 O 4 = mol Fe 3 O 4 3 mol Fe mol Fe 3 O g Fe 3 O 4 = 6.30 g Fe 3 O 4 1 mol Fe 3 O 4 Percent Yield = Actual Yield 6.02 g Fe x 100% = 3 O 4 x100%= 95.6 % Theoretical Yield 6.30 g Fe 3 O 4 99 Percent Yield Problem - I Problem: Ammonia is produced by the Haber process using nitrogen and hydrogen gas. If g of nitrogen are reacted with g hydrogen and the reaction yielded g of ammonia what was the percent yield of the reaction. N 2 (g) + 3 H 2 (g) 2 NH 3 (g) Plan: Since we are given the masses of both reactants, this is a limiting reactant problem. First determine which is the limiting reagent then calculate the theoretical yield, and finally the percent yield. 100

51 What is the limiting reactant? N 2 (g) = g, H 2 (g) = g, MM N = g/mol, MM H = g/mol 1. N 2 (g) 2. H 2 (g) 3. NH 3 (g) 4. N 2 (l) 5. H 2 (l) 101 Percent Yield Problem - I Problem: Ammonia is produced by the Haber process using nitrogen and hydrogen gas. If g of nitrogen are reacted with g hydrogen and the reaction yielded g of ammonia what was the percent yield of the reaction. N 2 (g) + 3 H 2 (g) 2 NH 3 (g) Plan: Since we are given the masses of both reactants, this is a limiting reactant problem. First determine which is the limiting reagent then calculate the theoretical yield, and finally the percent yield. Solution: g N moles N 2 = 2 = mol N g N 2 2 /mol g H moles H 2 = 2 = mol H g H 2 2 /mol Divide by coefficient to get eqs. of each: mol N 2 = eqs. 1 LR mol H 2 = eqs

52 What is the theoretical yield of NH 3 (g)? N 2 (g)= mol, H 2 (g) =10.74 mol, MM N = g/mol, MM H = g/mol N 2 (g) + 3 H 2 (g) 2 NH 3 (g) g mol g mol 103 Percent Yield Problem - II Solution Cont. N 2 (g) + 3 H 2 (g) 2 NH 3 (g) We have moles of nitrogen, and it is limiting, therefore the theoretical yield of ammonia is: mol N 2 2 mol NH 3 = mol NH 3 1 mol N 2 (Theoretical Yield) mol NH g NH = g NH mol NH 3 (Theoretical Yield) Actual Yield Percent Yield = x 100% Theoretical Yield g NH Percent Yield = 3 x 100% = % g NH 3 104

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