Me cchapter 4
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1) The document provides an overview of key concepts in chemistry including the mole concept, chemical formulas and equations, and different types of chemical reactions. 2) It explains that the mole is a unit used to measure amounts of substances and is equal to 6.022x1023 particles. Molar mass refers to the mass of one mole of a substance. 3) Chemical equations are used to represent chemical reactions and must satisfy the law of conservation of mass by being balanced with the same number and type of atoms on each side of the reaction arrow.
![4.5 Calculations Using the Relating Masses of Reactants
Chemical Equation and Products
• Calculate grams C3H8 required to produce
36.0 grams of H2O
• Use 3 conversion factors
– Grams H2O to moles H2O
– Moles H2O to moles C3H8
– Moles of C3H8 to grams C3H8
grams moles moles grams
H2 O H2 O C3 H8 C3 H8
– Set up the equation and cancel units
36.0 g H2O x 1 mol H2O x 1 mol C3H8 x 44.0 g C3H8
18.0 g H2O 4 mol H2O 1 mol C3H8
– 36.0 x [1/18.0] x [1/4] x 44.0 g C3H8 = 22.0 g C3H8](https://image.slidesharecdn.com/mecchapter4-120815081521-phpapp02/85/Me-cchapter-4-59-320.jpg)
![Calculating a Quantity of Reactant
4.5 Calculations Using the
• Ca(OH)2 neutralizes HCl
Chemical Equation
• Calculate grams HCl neutralized by 0.500 mol
Ca(OH)2
– Write chemical equation and balance
• Ca(OH)2(s) + 2HCl(aq) CaCl2(s) + 2H2O(l)
– Plan the path
moles moles grams
Ca(OH)2 HCl HCl
– Set up the equation and cancel units
0.500 mol Ca(OH)2 x 2 mol HCl x 36.5 g HCl
1 mol Ca(OH)2 1 mol HCl
Solve equation 0.500 x [2/1] x 36.5 g HCl = 36.5 g HCl](https://image.slidesharecdn.com/mecchapter4-120815081521-phpapp02/85/Me-cchapter-4-60-320.jpg)
![Sample Calculation
4.5 Calculations Using the
If the theoretical yield of iron was 30.0 g
Chemical Equation
and actual yield was 25.0 g, calculate the
percent yield:
2 Al(s) + Fe2O3(s) → Al2O3(aq) + 2Fe(aq)
• [25.0 g / 30.0 g] x 100% = 83.3%
• Calculate the % yield if 26.8 grams iron
was collected in the same reaction](https://image.slidesharecdn.com/mecchapter4-120815081521-phpapp02/85/Me-cchapter-4-64-320.jpg)
Me cchapter 4
- 1. Copyright© The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Chapter 4 Calculations and the Chemical Equation Denniston Topping Caret 7th Edition
- 2. 4.1 The Mole Concept and Atoms • Atoms are exceedingly small – Unit of measurement for mass of an atom is atomic mass unit (amu) – unit of measure for the mass of atoms • carbon-12 assigned the mass of exactly 12 amu • 1 amu = 1.66 x 10-24 g • Periodic table gives atomic weights in amu
- 3. 4.1 The Mole Concept and Mass of Atoms • What is the atomic weight of one atom of fluorine? Answer: 19.00 amu Atoms • What would be −the 19.00 amu F 1.661× 10 -24 g 3.156 ×10 23 g F × mass= this one of F atom 1 amuatom in grams? F F atom • Chemists usually work with much
- 4. 4.1 The Mole Concept and The Mole and Avogadro’s Number • A practical unit for defining a collection of atoms is the mole Atoms 1 mole of atoms = 6.022 x 1023 atoms • This is called Avogadro’s number – This has provided the basis for the concept of the mole
- 5. The Mole 4.1 The Mole Concept and • To make this connection we must define the mole as a counting unit – The mole is abbreviated mol Atoms • A mole is simply a unit that defines an amount of something – Dozen defines 12 – Gross defines 144
- 6. 4.1 The Mole Concept and Atomic Mass • The atomic mass of one atom of an element corresponds to: – The average mass of a single atom in amu Atoms – The mass of a mole of atoms in grams – 1 atom of F is 19.00 amu 19.00 amu/atom F – 1 mole of F is 19.00 g 19.00 g/mole F 19.00 amu F 1.66 × 10 −24 g F 6.022 × 10 23 atom F × × 1 atom F 1 amu F 1 mol F =19.00 g F/mol F or 19.00 g/mol F
- 7. Molar Mass 4.1 The Mole Concept and • Molar mass - The mass in grams of 1 mole of atoms • What is the molar mass of carbon? Atoms 12.01 g/mol C • This means counting out a mole of Carbon atoms (i.e., 6.022 x 1023) they would have a mass of 12.01 g • One mole of any element contains the same number of atoms, 6.022 x 1023, Avogadro’s number
- 8. Calculating Atoms, Moles, and Mass 4.1 The Mole Concept and • We use the following conversion factors: • Density converts grams – milliliters Atoms • Atomic mass unit converts amu – grams • Avogadro’s number converts moles – number of atoms • Molar mass converts grams – moles
- 9. 4.1 The Mole Concept and Strategy for Calculations • Map out a pattern for the required conversion • Given a number of grams and asked for Atoms number of atoms • Two conversions are required • Convert grams to moles 1 mol S/32.06 g S OR 32.06 g S/1 mol S • Convert moles to atoms mol S x (6.022 x 1023 atoms S) / 1 mol S
- 10. 4.1 The Mole Concept and Practice Calculations 1. Calculate the number of atoms in 1.7 moles of boron. 2. Find the mass in grams of 2.5 mol Na Atoms (sodium). 3. Calculate the number of atoms in 5.0 g aluminum. 4. Calculate the mass of 5,000,000 atoms of Au (gold)
- 11. 4.1 The Mole Concept and Interconversion Between Moles, Particles, and Grams Atoms
- 12. 4.2 The Chemical Formula, Formula Weight, and Molar Mass • Chemical formula - a combination of symbols of the various elements that make up the compound • Formula unit - the smallest collection of atoms that provide two important pieces of information – The identity of the atoms – The relative number of each type of atom
- 13. 4.2 The Chemical Formula, Formula Weight and Molar Mass Chemical Formula Consider the following formulas: • H2 – 2 atoms of hydrogen are chemically bonded forming diatomic hydrogen, subscript 2 • H2O – 2 atoms of hydrogen and 1 atom of oxygen, lack of subscript means one atom • NaCl – 1 atom each of sodium and chlorine • Ca(OH)2 – 1 atom of calcium and 2 atoms each of oxygen and hydrogen, subscript outside parentheses applies to all atoms inside
- 14. 4.2 The Chemical Formula, Formula Weight and Molar Mass Chemical Formula Consider the following formulas: • (NH4)3SO4 – 2 ammonium ions and 1 sulfate ion – Ammonium ion contains 1 nitrogen and 4 hydrogen – Sulfate ion contains 1 sulfur and 4 oxygen – Compound contains 2 N, 8 H, 1 S, and 4 O • CuSO4.5H2O – This is an example of a hydrate - compounds containing one or more water molecules as an integral part of their structure – 5 units of water with 1 CuSO4
- 15. 4.2 The Chemical Formula, Comparison of Hydrated and Formula Weight and Molar Mass Anhydrous Copper Sulfate Hydrated copper sulfate Anhydrous copper sulfate Marked color difference illustrates the fact that these are different compounds
- 16. Formula Weight and Molar Mass Formula Weight and Molar Mass 4.2 The Chemical Formula, • Formula weight - the sum of the atomic weights of all atoms in the compound as represented by its correct formula – expressed in amu • What is the formula weight of H2O? – 16.00 amu + 2(1.008 amu) = 18.02 amu • Molar mass – mass of a mole of compound in grams / mole – Numerically equal to the formula weight in amu • What is the molar mass of H2O? – 18.02 g/mol H2O
- 17. 4.2 The Chemical Formula, Formula Formula Unit Weight and Molar Mass • Formula unit – smallest collection of atoms from which the formula of a compound can be established • When calculating the formula weight (or molar mass) of an ionic compound, the smallest unit of the crystal is used What is the molar mass of (NH4)3PO4? 3(N amu) + 12(H amu) + P amu + 4(O amu)= 3(14.01) + 12(1.008) + 30.97 + 4(16.00)= 149.10 g/mol (NH4)3PO4
- 18. 4.3 The Chemical Equation and the Information It Conveys A Recipe For Chemical Change • Chemical equation - shorthand notation of a chemical reaction – Describes all of the substances that react and all the products that form, physical states, and experimental conditions – Reactants – (starting materials) – the substances that undergo change in the reaction – Products – substances produced by the reaction
- 19. and the Information It Conveys 4.3 The Chemical Equation Features of a Chemical Equation 1. Identity of products and reactants must be specified using chemical symbols 2. Reactants are written to the left of the reaction arrow and products are written to the right 3. Physical states of reactants and products may be shown in parentheses 4. Symbol ∆ over the reaction arrow means that energy is necessary for the reaction to occur 5. Equation must be balanced
- 20. and the Information It Conveys 4.3 The Chemical Equation Features of a Chemical Equation ∆ 2HgO( s ) → 2Hg( l ) + O 2 (g ) Products and reactants must be specified using chemical symbols Reactants – written on the left of arrow Products – written on the right ∆ – energy is needed Physical states are shown in parentheses
- 21. and the Information It Conveys 4.3 The Chemical Equation The Experimental Basis of a Chemical Equation We know that a chemical equation represents a chemical change • One or more substances changed into new substances • Different chemical and physical properties
- 22. and the Information It Conveys Evidence of a Reaction Occurring 4.3 The Chemical Equation The following can be visual evidence of a reaction: •Release of a gas – CO2 is released when acid is placed in a solution containing CO32- ions •Formation of a solid (precipitate) – A solution containing Ag+ ions mixed with a solution containing Cl- ions •Heat is produced or absorbed – Acid and base are mixed together •Color changes
- 23. and the Information It Conveys 4.3 The Chemical Equation Subtle Indications of a Reaction • Heat or light is absorbed or emitted • Changes in the way the substances behave in an electrical or magnetic field before and after a reaction • Changes in electrical properties
- 24. and the Information It Conveys Writing Chemical Reactions 4.3 The Chemical Equation • We will learn to identify the following patterns of chemical reactions: – combination – decomposition – single-replacement – double-replacement • Recognizing the pattern will help you write and understand reactions
- 25. and the Information It Conveys 4.3 The Chemical Equation Combination Reactions • The joining of two or more elements or compounds, producing a product of different composition A + B → AB • Examples: 2Na(s) + Cl2(g) → 2NaCl(s) MgO(s) + CO2(g) → MgCO3(s)
- 26. and the Information It Conveys 4.3 The Chemical Equation Types of Combination Reactions 1. Combination of a metal and a nonmetal to form a salt 2. Combination of hydrogen and chlorine molecules to produce hydrogen chloride 3. Formation of water from hydrogen and oxygen molecules 4. Reaction of magnesium oxide and carbon dioxide to produce magnesium carbonate
- 27. and the Information It Conveys 4.3 The Chemical Equation Decomposition Reactions • Produce two or more products from a single reactant • Reverse of a combination reaction AB → A + B • Examples: 2HgO(s) → 2Hg(l) + O2(g) CaCO3(s) → CaO(s) + CO2(g)
- 28. Types of Decomposition and the Information It Conveys 4.3 The Chemical Equation Reactions 1. Heating calcium carbonate to produce calcium oxide and carbon dioxide 2. Removal of water from a hydrated material
- 29. and the Information It Conveys 4.3 The Chemical Equation Replacement Reactions 1. Single-replacement • One atom replaces another in the compound producing a new compound A + BC → B + AC • Examples: Cu(s)+2AgNO3(aq) → 2Ag(s)+Cu(NO3)2(aq) 2Na(s) + 2H2O(l) → 2NaOH(aq) + H2(g)
- 30. and the Information It Conveys 4.3 The Chemical Equation Types of Replacement Reactions 1. Replacement of copper by zinc in copper sulfate 2. Replacement of aluminum by sodium in aluminum nitrate
- 31. and the Information It Conveys 4.3 The Chemical Equation Replacement Reactions 2. Double-replacement • Two compounds undergo a “change of partners” • Two compounds react by exchanging atoms to produce two new compounds AB + CD → AD + CB
- 32. and the Information It Conveys 4.3 The Chemical Equation Types of Double-Replacement • Reaction of an acid with a base to produce water and salt HCl(aq)+NaOH(aq) →NaCl(aq)+H2O(l) • Formation of solid lead chloride from lead nitrate and sodium chloride Pb(NO3)2(aq) + 2NaCl(aq) → PbCl2(s) + 2NaNO3(aq) AB + CD → AD + CB
- 33. and the Information It Conveys 4.3 The Chemical Equation Types of Chemical Reactions Precipitation Reactions • Chemical change in a solution that results in one or more insoluble products • To predict if a precipitation reaction can occur it is helpful to know the solubilities of ionic compounds
- 34. and the Information It Conveys 4.3 The Chemical Equation Predicting Whether Precipitation Will Occur • Recombine the ionic compounds to have them exchange partners • Examine the new compounds formed and determine if any are insoluble according to the rules in Table 4.1 • Any insoluble salt will be the precipitate Pb(NO3)2(aq) + NaCl(aq) → PbCl2 (s) + NaNO3 ( ?) (?) (aq)
- 35. and the Information It Conveys 4.3 The Chemical Equation Predict Whether These Reactions Form Precipitates • Potassium chloride and silver nitrate • Potassium acetate and silver nitrate
- 36. and the Information It Conveys 4.3 The Chemical Equation Reactions with Oxygen • Reactions with oxygen generally release energy • Combustion of natural gas – Organic compounds CO2 and H2O are usually the products CH4+2O2→CO2+2H2O • Rusting or corrosion of iron 4Fe + 3O2 → 2Fe2O3
- 37. and the Information It Conveys 4.3 The Chemical Equation Acid-Base Reactions • These reactions involve the transfer of a hydrogen ion (H+) from one reactant (acid) to another (base) HCl(aq) + NaOH(aq) → NaCl(aq) + H2O(l) The H+ on HCl was transferred to the oxygen in OH-, giving H2O
- 38. and the Information It Conveys 4.3 The Chemical Equation Oxidation-Reduction Reactions • Reaction involves the transfer of one or more electrons from one reactant to another Zn(s) + Cu2+(aq)→ Cu(s) + Zn2+(aq) Two electrons are transferred from Zn to Cu2+
- 39. and the Information It Conveys 4.3 The Chemical Equation Writing Chemical Reactions Consider the following reaction: hydrogen reacts with oxygen to produce water • Write the above reaction as a chemical equation H2 + O2 → H2O • Don’t forget the diatomic elements
- 40. and the Information It Conveys 4.3 The Chemical Equation Law of Conservation of Mass • Law of conservation of mass - matter cannot be either gained or lost in the process of a chemical reaction – The total mass of the products must equal the total mass of the reactants
- 41. A Visual Example of the Law of 4.4 Balancing Chemical Equations Conservation of Mass
- 42. 4.4 Balancing Chemical Equations • A chemical equation shows the molar quantity of reactants needed to produce a particular molar quantity of products • The relative number of moles of each product and reactant is indicated by placing a whole-number coefficient before the formula of each substance in the chemical equation
- 43. 4.4 Balancing Chemical Balancing Coefficient - how many of that substance are in the reaction Equations ∆ 2HgO( s ) → 2Hg(l ) + O 2 ( g ) • The equation must be balanced – All the atoms of every reactant must also appear in the products • Number of Hg on left? 2 – on right 2 • Number of O on left? 2 – on right 2
- 44. Examine the Equation 4.4 Balancing Chemical H 2 + O 2 → H 2O • Is the law of conservation of mass obeyed Equations as written? NO • Balancing chemical equations uses coefficients to ensure that the law of conservation of mass is obeyed • You may never change subscripts! • WRONG: H2 + O2 → H2O2
- 45. 4.4 Balancing Chemical Steps in Equation Balancing H2 + O2 → H2O The steps to balancing: Equations Step 1. Count the number of moles of atoms of each element on both product and reactant sides Reactants Products 2 mol H 2 mol H 2 mol O 1 mol O
- 46. 4.4 Balancing Chemical Steps in Equation Balancing H2 + O2 → H2O Step 2. Determine which elements are not balanced – do not have same number on Equations both sides of the equation – Oxygen is not balanced Step 3. Balance one element at a time by changing the coefficients H2 + O2 → 2H2O This balances oxygen, but is hydrogen still balanced?
- 47. 4.4 Balancing Chemical Steps in Equation Balancing H2 + O2 → 2H2O How will we balance hydrogen? Equations 2H2 + O2 → 2H2O Step 4. Check! Make sure the law of conservation of mass is obeyed Reactants Products 4 mol H 4 mol H 2 mol O 2 mol O
- 48. 4.4 Balancing Chemical Equations Balancing an Equation
- 49. Practice Equation Balancing 4.4 Balancing Chemical Balance the following equations: Equations 1. C2H2 + O2 → CO2 + H2O 2. AgNO3 + FeCl3 → Fe(NO3)3 + AgCl 3. C2H6 + O2 → CO2 + H2O 4. N2 + H2 → NH3
- 50. 4.5 Calculations Using the Chemical Equation • Calculation quantities of reactants and products in a chemical reaction has many applications • Need a balanced chemical equation for the reaction of interest • The coefficients represent the number of moles of each substance in the equation
- 51. 4.5 Calculations Using the General Principles 1. Chemical formulas of all reactants and Chemical Equation products must be known 2. Equation must be balanced to obey the law of conservation of mass • Calculations of an unbalanced equation are meaningless 1. Calculations are performed in terms of moles • Coefficients in the balanced equation represent the relative number of moles of products and reactants
- 52. 4.5 Calculations Using the Using the Chemical Equation Chemical Equation • Examine the reaction: 2H2 + O2 → 2H2O • Coefficients tell us? – 2 mol H2 reacts with 1 mol O2 to produce 2 mol H2O • What if 4 moles of H2 reacts with 2 moles of O2? – It yields 4 moles of H2O
- 53. Using the Chemical Equation 4.5 Calculations Using the 2H2 + O2 → 2H2O Chemical Equation • The coefficients of the balanced equation are used to convert between moles of substances • How many moles of O2 are needed to react with 4.26 moles of H2? • Use the factor-label method to perform this calculation
- 54. Use of Conversion Factors 4.5 Calculations Using the Chemical Equation 2H2 + O2 → 2H2O __mol O 2 1 4.26 mol H 2 × = 2.13 mol O2 __ mol H 2 2 • Digits in the conversion factor come from the balanced equation
- 55. 4.5 Calculations Using the Conversion Between Moles Chemical Equation and Grams • Requires only the formula weight • Convert 1.00 mol O2 to grams moles of grams of – Plan the path Oxygen Oxygen – Find the molar mass of oxygen • 32.0 g O2 = 1 mol O2 – Set up the equation – Cancel units 1.00 mol O2 x 32.0 g O2 1 mol O2 – Solve equation 1.00 x 32.0 g O2 = 32.0 g O2
- 56. Conversion of Mole Reactants to 4.5 Calculations Using the Chemical Equation Mole Products • Use a balanced equation • C3H8(g) + 5O2(g) 3CO2(g) + 4H2O(g) • 1 mol C3H8 results in: – 5 mol O2 consumed 1 mol C3H8 /5 mol O2 – 3 mol CO2 formed 1 mol C3H8 /3 mol CO2 – 4 mol H2O formed 1 mol C3H8 /4 mol H2O • This can be rewritten as conversion factors
- 57. 4.5 Calculations Using the Chemical Equation Calculating Reacting Quantities • Calculate grams O2 reacting with 1.00 mol C3H8 • Use 2 conversion factors – Moles C3H8 to moles O2 – Moles of O2 to grams O2 moles moles grams C3 H8 Oxygen Oxygen – Set up the equation and cancel units – 1.00 mol C3H8 x 5 mol O2 x 32.0 g O2 = 1 mol C3H8 1 mol O2 – 1.00 x 5 x 32.0 g O = 1.60 x 102 g O
- 58. 4.5 Calculations Using the Calculating Grams of Product Chemical Equation from Moles of Reactant • Calculate grams CO2 from combustion of 1.00 mol C3H8 • Use 2 conversion factors – Moles C3H8 to moles CO2 – Moles of CO2 to grams CO2 moles moles grams C3 H8 CO2 CO2 – Set up the equation and cancel units – 1.00 mol C3H8 x 3 mol CO2 x 44.0 g CO2 = 1 mol C3H8 1 mol CO2 – 1.00 x 3 x 44.0 g CO = 1.32 x 102 g CO
- 59. 4.5 Calculations Using the Relating Masses of Reactants Chemical Equation and Products • Calculate grams C3H8 required to produce 36.0 grams of H2O • Use 3 conversion factors – Grams H2O to moles H2O – Moles H2O to moles C3H8 – Moles of C3H8 to grams C3H8 grams moles moles grams H2 O H2 O C3 H8 C3 H8 – Set up the equation and cancel units 36.0 g H2O x 1 mol H2O x 1 mol C3H8 x 44.0 g C3H8 18.0 g H2O 4 mol H2O 1 mol C3H8 – 36.0 x [1/18.0] x [1/4] x 44.0 g C3H8 = 22.0 g C3H8
- 60. Calculating a Quantity of Reactant 4.5 Calculations Using the • Ca(OH)2 neutralizes HCl Chemical Equation • Calculate grams HCl neutralized by 0.500 mol Ca(OH)2 – Write chemical equation and balance • Ca(OH)2(s) + 2HCl(aq) CaCl2(s) + 2H2O(l) – Plan the path moles moles grams Ca(OH)2 HCl HCl – Set up the equation and cancel units 0.500 mol Ca(OH)2 x 2 mol HCl x 36.5 g HCl 1 mol Ca(OH)2 1 mol HCl Solve equation 0.500 x [2/1] x 36.5 g HCl = 36.5 g HCl
- 61. 4.5 Calculations Using the Chemical Equation General Problem-solving Strategy
- 62. Sample Calculation 4.5 Calculations Using the Na + Cl2 → NaCl Chemical Equation 1. Balance the equation 2Na + Cl2 → 2NaCl 2. Calculate the moles Cl2 reacting with 5.00 mol Na 3. Calculate the grams NaCl produced when 5.00 mol Na reacts with an excess of Cl2 4. Calculate the grams Na reacting with 5.00 g Cl2
- 63. Theoretical and Percent Yield 4.5 Calculations Using the Chemical Equation • Theoretical yield - the maximum amount of product that can be produced – Pencil and paper yield • Actual yield - the amount produced when the reaction is performed – Laboratory yield actual yield • Percent yield: % yield = × 100% theoretical yield = 125 g CO2 actual x 100% = 97.4% 132 g CO2 theoretical
- 64. Sample Calculation 4.5 Calculations Using the If the theoretical yield of iron was 30.0 g Chemical Equation and actual yield was 25.0 g, calculate the percent yield: 2 Al(s) + Fe2O3(s) → Al2O3(aq) + 2Fe(aq) • [25.0 g / 30.0 g] x 100% = 83.3% • Calculate the % yield if 26.8 grams iron was collected in the same reaction