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Unit 2

Unit 2. Moles, Chemical Reactions, and Stoichiometry. The Mole. A mole (mol) is the amount of a substance that contains as many particles as there are atoms in exactly 12 g of carbon-12.

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Unit 2

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  1. Unit 2 Moles, Chemical Reactions, and Stoichiometry

  2. The Mole • A mole(mol) is the amount of a substance that contains as many particles as there are atoms in exactly 12 g of carbon-12. • It is a counting unit, similar to a dozen. In a dozen, there are 12 things. In a mole, there are 6.02 x 1023things. Visual Concept

  3. Molar Mass • Molar Mass is the mass of one mole of a pure substance. • Molar mass units are g/mol. • The molar mass of an element is the same number as its atomic mass, only the units are different. • Examples: Molar mass of H = 1.0 g/mol Molar mass of O = 16.0 g/mol

  4. Formula Mass • The formula mass of any compound is the sum of the masses of all the atoms in its formula. Example:Formula mass of water, H2O: H2 = 1.0 amu x 2 = 2.0 amu. O = + 16.0 amu. 18.0 amu • A compound’s molar massis numerically equal to its formula mass. Only the units are different. (Ex: Molar mass of H2O = 18.0 g.)

  5. Molar MassesSample Problem Determine the molar mass of each of the following compounds: • Al2S3 • Ba(OH)2 Solution: Al2 = 27.0 x 2 = 54.0 g S3 = 32.1 x 3 = + 96.3 g 150.3 g Ba = 137.3 g O2 = 16.0 x 2 = 32.0 g H2 = 1.0 x 2 = + 2.0 g 171.3 g

  6. Percent Composition • The percentage by mass of each elementin a compound is known as the percent composition of the compound. mass of element in compound % of element = x 100 molar mass of compound

  7. Percent CompositionSample Problem Find the percentage composition of copper(I) sulfide, Cu2S. • Find the molar mass of Cu2S: • Find the percentage by mass of each element: Cu2 = 63.5 x 2 = 127.0 g Solution: S = + 32.1 g 159.1 g 127.0 g 79.80% Cu % Cu = x 100 = 159.1 g 32.1 g 20.2% S % S = x 100 = 159.1 g

  8. Molar Mass as a Conversion Factor • The molar mass of a compound can be used as a conversion factor to convert between moles and grams for a given substance. Example: • What is the mass of 2.5 moles of H2O? molar mass of H2O = 18.0 g/mol conversion factor given g H2O 18.0 x 2.5 mol H2O = 45 g H2O 1 mol H2O

  9. Molar Mass as a Conversion FactorSample Problem Calculate the moles in 1170 g of copper (II) nitrate. Solution: - 2+ Cu(NO3)2 Cu (NO3) 1. Determine the correct formula: 2 Cu = 63.5 g 2. Calculate the molar mass: N2 = 14.0 x 2 = 28.0 g 3. Convert from g to mol: O6 = 16.0 x 6 = + 96.0 g Conversion factor 187.5 g Given 1 mol Cu(NO3)2 x 1170 g Cu(NO3)2 = 6.24 mol Cu(NO3)2 187.5 g Cu(NO3)2

  10. Avogadro’s Number • Avogadro’s number: 6.022 1415 × 1023— the number of particles in exactly one mole of a pure substance. • Named for nineteenth-century Italian scientist Amedeo Avogadro. Visual Concept

  11. Conversions with Avogadro’s Number • Avogadro’s number can be used as a conversion factor between number of particles (atoms or molecules) and moles. Example: • How many moles of silver, Ag, are in 3.01  1023 atoms of silver? Conversion factor Given mol Ag 1 3.01 x 1023 atoms Ag x = 0.500 mol Ag 6.02 x 1023 atoms Ag

  12. Combining Conversion FactorsSample Problem What is the mass in grams of 1.20  108 atoms of copper, Cu? Solution: 2nd Conversion factor 1st Conversion factor Given 1 63.5 mol Cu g Cu x x 1.20 x 108 atoms Cu 6.02 x 1023 1 atoms Cu mol Cu 1.27 x 10-14 g Cu =

  13. Standard Molar Volume • Standard Temperature and Pressure (STP) is 0oC and 1 atm. • The Standard Molar Volumeof a gas is the volume occupied by one mole of a gas at STP. It has been found to be 22.4 L.

  14. Molar Volume Conversion Factor • Standard Molar Volume can be used as a conversion factor to convert from the number of moles of a gas at STP to volume (L), or vice versa.

  15. Molar Volume ConversionSample Problem a. What quantity of gas, in moles, is contained in 5.00 L at STP? b. What volume does 0.768 moles of a gas occupy at STP? 1 mol x 5.00 L 0.223 mol = 22.4 L 22.4 L x 0.768 mol 17.2 L = 1 mol

  16. The Mole Map • You can now convert between number of particles, mass (g), and volume (L) by going through moles.

  17. Chemical Reactions • Achemical reactionis the process by which one or more substances are changed into one or more different substances. • Signs that a chemical reaction is taking place: • Release of energy as heat and/or light. • Production of a gas. • Formation of a precipitate - a solid that separates from a liquid solution. • Color change.

  18. The Law of Conservation of Mass • In any chemical reaction, the original substances are called reactantsand the resulting substances are called products. • According to the law of conservation of mass, the total mass of reactants must equal the total mass of products for any given chemical reaction.

  19. Chemical Equations • A chemical equationrepresents a chemical reaction using symbols and formulas. Example: • 2H2O(l) 2H2(g) + O2(g) Reactant Products

  20. Diatomic Molecules • Oxygen gas (O2) is an example of an element that normally exists as a diatomic molecule. You need to memorize all seven:

  21. Balancing Equations • The final step in writing correct chemical equations is to make sure the law of conservation of mass is satisfied. • The numbers and types of atoms on both sides of the equation must be the same – this is called balancing an equation. • Equations are balanced by inserting coefficients - whole numbers that appear in front of formulas in a chemical equation.

  22. Balancing EquationsSample Problem A Balance the following equation: CH4(g) + O2(g) CO2(g) + H2O(g) Solution: • Start with the easiest element…carbon. • Carbon is already balanced. • Next count the hydrogen atoms. • Two more hydrogen atoms are needed on the right. • Finally, count oxygen atoms. • There are 4 oxygens on the right side of the equation, but only two on the left. • Add a coefficient 2 in front of the O2 on the left. 2 2

  23. Balancing EquationsSample Problem B Balance the following equation: Al4C3(s) + H2O(l) CH4(g) + Al(OH)3(s) Solution: • Let’s start with aluminum. • Add a coefficient 4 to Al(OH)3 on the right. • Next count the carbon atoms. • Add a coefficient 3 to CH4 on the right. • Balance the oxygen atoms. • Add a 12 to the H2O on the left. • Lastly, count the hydrogen atoms. • Hydrogen is already balanced. 4 12 3

  24. Writing Chemical EquationsSample Problem Write an equation for the reaction that occurs when solid copper metal reacts with aqueous silver nitrate to produce solid silver metal and aqueous copper(II) nitrate. Solution: • First, use correct formulas and symbols to write a chemical equation. • Then, balance your equation. Ag(s) 2 2 Cu(s) + AgNO3(aq) + Cu(NO3)2(aq)

  25. Types of Chemical Reactions • There are 5 basic types of chemical reactions: • Synthesis • Decomposition • Single-Displacement • Double-Displacement • Combustion

  26. Synthesis Reactions • In a synthesis reaction (also called a composition reaction) two or more substances combine to form a new compound. • This type of reaction is represented by the following general equation: A + X AX

  27. Decomposition Reactions • In a decomposition reaction, a single compound breaks apart to form 2 or more simpler substances. • Decomposition is the opposite of synthesis. • This type of reaction is represented by the following general equation: AX A + X

  28. Single-Displacement Reactions • In a single-displacement reaction(also called single-replacement) one element replaces a similar element in a compound. • They often take place in aqueous solution. • This type of reaction is represented by the following general equation: A + BX AX + B

  29. Double-Displacement Reactions • In double-displacement reactions, the ionsof 2 compounds exchange places in an aqueous solution to form 2 new compounds. • One of the compounds formed is usually either a precipitate, a gas, or water. • Represented by the following general equation: AX + BY AY + BX

  30. Combustion Reactions • In a combustion reaction,a hydrocarbon fuel combines with oxygen, releasing a large amount of energy in the form of light and heat. • Products of combustion reactions are always carbon dioxide and water vapor. • Example: Combustion of propane C3H8(g) + 5O2(g) 3CO2(g) + 4H2O(g)

  31. Types of ReactionsSample Problem Classify each of the following reactions as a synthesis, decomposition, single-displacement, double-displacement, or combustion reaction. • N2(g) + 3H2(g) → 2NH3(g) • 2Li(s) + 2H2O(l) → 2LiOH(aq) + H2(g) • 2NaNO3(s) → 2NaNO2(s) + O2(g) • 2C6H14(l) + 19O2(g) → 12CO2(g) + 14H2O(l) synthesis single-displacement decomposition combustion

  32. What is Stoichiometry? • Stoichiometryis the branch of chemistry that deals withthe mass relationships of elements in a chemical reaction. • Stoichiometry calculationsalways start with a balanced chemical equation.

  33. Mole Ratio • Amole ratio is a conversion factor that relates the amounts in moles of any two substances involved in a chemical reaction Example:2Al2O3(l) → 4Al(s) + 3O2(g) Mole Ratios:2 mol Al2O3 2 mol Al2O3 4 mol Al 4 mol Al 3 mol O2 3 mol O2

  34. Mole to Mole Conversions • Using the mole ratio, you can convert from moles of one substance to moles of any other substance in a chemical reaction. Example: • For the reaction N2 + 3H2 → 2NH3, how many moles of H2 are required to produce 12 moles of NH3? Use mole ratio as a conversion factor This is what we’re given: 3 mol H2 x 12 mol NH3 = 18 mol H2 2 mol NH3

  35. Mole to Mass Conversions • To convert from moles of one substance to grams of another, you need 2 conversion factors: 1. mole ratio. 2. molar mass of the unknown. • To set up your conversion factor, always put the units you have on the bottom and the units you need on the top.

  36. Mole to Mass Conversions (continued) Example: Given the equation: 2Mg(s) + O2(g)→ 2MgO(s), Calculate the mass in grams of magnesium oxide which is produced from 2.00 mol of magnesium. 2nd C.F. Molar Mass of the Unknown 1st C.F. Mole Ratio Given 2 40.3 mol MgO g MgO x x 2.00 mol Mg 80.6 g MgO = 2 1 mol Mg mol MgO

  37. Mass to Mole Conversions • To convert from grams of one substance to moles of another, you need 2 conversion factors: 1. molar mass of the given. 2. mole ratio. • Mass to mole conversion factors are the inverse of mole to mass conversion factors.

  38. Mass to Mole Conversions (continued) Example: Given the equation: 2HgO(s) → 2Hg(s) + O2(g), How many moles of HgO are needed to produce 125g of O2? 1st C.F. 2nd C.F. Mole Ratio Molar Massof the Given Given 1 2 mol O2 mol HgO x x 125 g O2 7.81 mol HgO = 32.0 g O2 1 mol O2

  39. Mass to Mass Conversions • To convert from grams of one substance to grams of another, you need 3 conversion factors: 1. molar mass of the given. 2. mole ratio. 3. molar mass of the unknown.

  40. Mass to Mass Conversions (continued) Example: Given the equation: 2HgO(s) → 2Hg(s) + O2(g), How many grams of HgO are needed to produce 45.0 g of O2? 3rd C.F. Molar Mass ofthe Unknown 1st C.F. 2nd C.F. Mole Ratio Molar Massof the Given Given 1 2 216.6 mol O2 mol HgO g HgO x x x 45.0 g O2 609 g HgO = 32.0 1 1 g O2 mol O2 mol HgO

  41. Volume Ratios • You can use the volume ratios as conversion factors just like you would use mole ratios. 2CO(g) + O2(g) → 2CO2(g) 2 molecules 1 molecule 2 molecules 2 mole 1 mole 2 mol 2 volumes 1 volume 2 volumes • Example: What volume of O2 is needed to react completely with 0.626 L of CO to form CO2? 1 L O2 x 0.626 L CO 0.313 L O2 = 2 L CO

  42. Gas StoichiometrySample Problem Assume that 5.61 L H2 at STP reacts with excess CuO according to the following equation: CuO(s) + H2(g) → Cu(s) + H2O(g) • How many moles of H2 react? • How many grams of Cu are produced? 1 mol H2 x 5.61 L H2 0.250 mol H2 = 22.4 L H2 1 mol H2 1 mol Cu 63.5 g Cu x x x 5.61 L H2 15.9 g Cu = 22.4 L H2 1 mol H2 1 mol Cu

  43. Limiting Reactants • When combining 2 or more different things to make a product, you have to stop when one of the things is used up. • For example, no matter how many tires there are, if there are only 8 car bodies, then only 8 cars can be made. 

  44. Limiting Reactants (continued) • The limiting reactant is the reactant that limits the amount of product formed. • The excess reactantis the substance that is not used up completely. • Once the limiting reactant is used up, a chemical reaction will stop.

  45. Limiting Reactants (continued) Example: Silicon dioxide reacts with hydrogen fluoride according to the following equation: SiO2(s) + 4HF(g) → SiF4(g) + 2H2O(l) If 6.0 mol HF is added to 4.5 mol SiO2, which is the limiting reactant? The limitingreactant makes the least product. Mole ratio C.F. 1 mol SiF4 Set up 2 equations, one for eachgiven: 6.0 mol HF 6.0 mol HF x = 1.5 mol SiF4 1.5 mol SiF4 4 mol HF Limiting Reactant is HF 1 mol SiF4 x 4.5 mol SiO2 = 4.5 mol SiF4 1 mol SiO2

  46. Percentage Yield • The theoretical yield is the maximum amount of product that can be produced from a given amount of reactant (found with stoichiometry). • The actual yieldof a product is the measured amount of that product obtained from a reaction (given in problem). • The percentage yieldis the ratio of the actual yield to the theoretical yield, multiplied by 100.

  47. Percentage YieldSample Problem A Given the following equation: C6H6 (l) + Cl2(g) → C6H5Cl(l) + HCl(g) When 36.8 g C6H6 react with excess Cl2, the actual yield of C6H5Cl is 38.8 g. What is the % yield of C6H5Cl? This is theTheoretical Yield Set up a mass to mass conversion 1 112.5 mol C6H5Cl g C6H5Cl 1 x x x mol C6H6 36.8 g C6H6 53.1 g C6H5Cl = g C6H6 78.0 1 1 mol C6H6 mol C6H5Cl Actual 38.8 g C6H5Cl Percent Yield 73.1% 100 100 x x = Now use the formula = = Theoretical 53.1 g C6H5Cl

  48. Percentage YieldSample Problem B According to the following reaction : CO(g) + 2H2(g) → CH3OH(l) If the typical yield is 80.0%, what mass of CH3OH should be expected if 75.0 g of CO reacts with excess H2 gas? This is theTheoretical Yield Set up a mass to mass conversion 1 32.0 mol CH3OH g CH3OH 1 x x x mol CO 75.0 g CO 85.7 g CH3OH = g CO 28.0 1 1 mol CO mol CH3OH Multiply theoreticalyield by percentageto get actual yield 68.6 g CH3OH 85.7 g CH3OH 80.0% x =

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