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Chemistry 101 : Chap. 3

Chemistry 101 : Chap. 3. Stoichiometry. Chemical Equations (2) Some Simple Patterns of Chemical Reactivity (3) Formula Weight (4) Avogadro’s Number and the Mole (5) Empirical Formulas from Analyses (6) Quantitative Information from Balanced Equations (7) Limiting Reactants. Stoichiometry.

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Chemistry 101 : Chap. 3

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  1. Chemistry 101 : Chap. 3 Stoichiometry • Chemical Equations (2) Some Simple Patterns of Chemical Reactivity (3) Formula Weight (4) Avogadro’s Number and the Mole (5) Empirical Formulas from Analyses (6) Quantitative Information from Balanced Equations (7) Limiting Reactants

  2. Stoichiometry A study of mass relationships that exist between substances consumed (reactants) and produced (products) in chemical reactions. Stoichiometry is build upon an understanding of atomic masses, chemical formulas and law of conservation of mass. “… nothing is created: an equal quantity of matter exists both before and after the experiment.” Antoine Lavoisier (1743 ~ 1794) With the Dalton’s atomic theory, law of conservation of mass is known as law of conservation of atom

  3. + Balancing Chemical Equations  Chemical Equation: 2H2 + O2 = 2H2O coefficient reactants products • Balanced Chemical Equation: Equations that represent the correct amount of reactants and products in chemical reaction. Balanced chemical equations satisfy the law of conservation of atoms

  4. Balancing Chemical Equations • Balancing Chemical Equations: Determining the coefficients that provide equal numbers of each type of atom on each side of the equation (1) You only determine the coefficients, not the subscripts. (2) Subscripts should never be changed in balancing equations.  It changes the identities of reactants or products (3) Balanced equation should contain the smallest possible whole-number coefficients (4) It is usually best to balance first those elements that occur in the fewest chemical formulas on each side of equation

  5. → + + + → Balancing Chemical Equations 2 CO + O2 → CO2 2 CH4 + Cl2 → CCl4 + HCl 4 4

  6. 2 3 2 3 x 3 1 2 x 2 Balancing Chemical Equations 2 3 2 C2H4 + O2 → CO2 + H2O 2 Al + HCl → AlCl3 + H2 2 Al + 6 HCl → 2 AlCl3 + 3 H2 2 NH4NO3 → N2 + O2 + H2O 2 NH4NO3 → 2 N2 + O2 + 4 H2O

  7. Balancing Chemical Equations Example: Law of conservation of atom (or mass) How many NH3 molecules should be shown in the container on the right?

  8. Three Basic Reactions Types Combination Reactions:A + B  C 2Mg (s) + O2 (g)  2MgO (s) Two reactants combine to form a single product NOTE: The symbols (s), (l), (g), (aq) specify the physical state of each substance Examples of combination reactions [ Not balanced!] (1) N2 + H2 NH3 (2) Fe + O2 Fe2O3 Iron (III) oxide (3) S + O2 SO3

  9. Three Basic Reactions Types Decomposition Reactions:A  B + C 2 HgO (s)  2 Hg (l) + O2 (g) A single reactant decomposes to form two or more products Examples of decomposition reactions [not balanced!] (1) NaN3 (s) Na (s) + N2 (g) (2) CaCO3 (s) CaO + CO2 (g) NOTE: Over 22 million tons of limestone (CaCO3) are converted to lime (CaO) each year for use in making glass, mortar, and for extracting iron from its ore

  10. Three Basic Reactions Types • Combustion Reactions : Organic compounds (or hydrocarbons) is burned (combusted) in oxygen to produce CO2 and H2O C3H8 (g) + O2 (g) CO2 (g) + H2O (g) 5 3 4 NOTE: When balancing combustion reaction, it is best to balance the oxygen atoms last. Examples of combustion reactions [not balanced] (1) CH4(g) + O2 (g) CO2 (g) + H2O (g) (2) C4H10 (g) + O2 (g) CO2 (g) + H2O (g)

  11. Formula Weight Chemical equations tell us the exact numbers of molecules and atoms involved in chemical reactions. But, in laboratory, what we usually measure is the weight of each compound. How can we relate them? • Formula Weight (FW) : The sum of the atomic weight of each atom in its chemical formula FW of H2SO4 = 2 x (AW of H) + (AW of S) + 4 x (AW of O) = 2 x 1.0 + 32.1 + 4 x 16.0 = 98.1 (amu) NOTE: If the chemical formula is that of a molecule, the formula weight is also calledmolecular weight (MW). NOTE: The masses used in the formula weight are from the periodic table (weighted average of all isotope’s masses)

  12. Formula Weight Examples : Determined the formula weights of following substances (1) FW of Mg(OH)2 = (2) FW of Ca(NO3)2 =

  13. Formula Weight Percentage Composition : The percentage by mass contributed by each element in the substance Example : Calculate the % weight of C, H and O in C12H22O11

  14. The Mole The mole is the SI unit for amount (number) of substance. It was originally proposed by Wilhelm Ostward in 1896 • Mole (mol):A quantity of objects equal to the number of atoms of carbon in exactly 12 g of 12C. Wilhelm Ostward (1853 ~ 1932) He was awarded the novel prize for chemistry for his work on catalysis in 1909. A mole of particles is an extremely large number of objects and it is approximately equal to 6.022  1023 particles. This number is referred to as Avogadro’s Number (NA) in honor of Amadeo Avogadro. 1 mol = 6.022  1023 particles Amadeo Avogadro (1776 ~ 1856)

  15. The Mole The mole is simply a unit of counting 1 dozen eggs = 12 individual eggs ½ dozen eggs = 6 individual eggs 1 mol of eggs = 6.022  1023 individual eggs ½ mol of eggs = 3.011  1023 individual eggs 1 mol of oxygen atoms = 6.022 1023 oxygen atoms (O) 1 mol of oxygen molecules = 6.022 1023 oxygen molecules (O2) = 2 mol of oxygen atoms

  16. The Mole • Example : How many moles of eggs are in an egg carton holding 12 eggs? Example : How many moles of oxygen atoms are in 1.5 moles of Ca(NO3)2 ?

  17. Molar Mass Molar mass = mass of 1 mole of a substance in gram (1) Molar mass is nothing to do with Molecule. (2) 1 mole of He atoms includes the same number of particles as 1 mole of Ne atoms. However, 1 mole of He atoms and 1 mole of Ne atoms have different weights because He and Ne have different molar mass. • 12 amu = mass of a single 12C atom (exact) 12 gram = mass of 1 mole of 12C atoms (exact) molar mass of 12C = 12 g/mol FW (or MW) in amu = molar mass in g/mol

  18. Molar Mass  1 mol of Cu = 6.02  1023 copper atoms  AW of Cu = 63.5 amu = average mass of naturally occurring (FW or MW) single Cu atom (This number is from the periodic table)  molar mass of Cu = 63.5 g/mol = mass of 6.02  1023 copper atoms amount of a substance in gram amount of a substance in mol  molar mass  molar mass

  19. Molar Mass Example: Calculate the mass of 0.433 mole of calcium nitrate. Example: How many grams of oxygen are in 0.433 mole of Ca(NO3)2

  20. Molar Mass  Problem: How many hydrogen atoms are in 4.5g of CH3OH ? • Compute the molecular weight or molar mass of CH3OH: 2.Determine the amount of CH3OH in mol. 3. Determine the amount of H

  21. Molar Mass  Problem: How many grams of carbon are in 42g of CH3CH2OH ? • Compute the molecular weight or molar mass of CH3CH2OH: 2.Determine the amount of CH3CH2OH in mol. 3. Determine the amount of C

  22. Molar Mass and Avogadro’s Number 1.08  1024 Copper atoms 1.8 mol of Copper 114.3 g of Copper Avogadro’s number molar mass

  23. Empirical Formulas from Analyses • Empirical Formula : Indicate the relative number of atoms of each type in a molecule If we know the relativenumber of atoms in mol, we can determine the compound’s empirical formula • Example : A compound made of Hg and Cl has 73.9% mercury (Hg) and 26.1% chlorine (Cl) by mass. What is the empirical formula for this compound?

  24. Empirical Formulas from Analyses Example : A compound is found to contain by mass 47.7% C, 10.5% H and 42.1% O. What is the empirical formula of the compound?

  25. Empirical Formulas from Analyses Example: A compound is found to have by mass 53.5% C, 11.1% H and 35.6% O. The experimentally determined molecular weight is 90 amu. What is the molecular formula of this compound?

  26. Quantitative Information from Balanced Chemical Equations Conservation of atom FOUR H atoms TWO O atoms FOUR H atoms + TWO O atoms 2H2 + O2 2H2O TWO hydrogen molecules TWO water molecules ONE oxygen molecule  6.021023 2 mol H2 + 1 mol O2 2 mol H2O  molar mass Conservation of mass 4g H2 + 32 g O2 36g H2O

  27. Quantitative Information from Balanced Chemical Equations  Example: How many moles of carbon dioxide would be produced by burning 3 moles of carbon monoxide? • Write down the reactants and products: (2) Balance the chemical equation : (3) Compute the amount of CO2 :

  28. Quantitative Information from Balanced Chemical Equations  Example: How many grams of H2O are formed from the complete combustion of 3.0 g of C2H6? • Write down the reactants and products : (2) Balance the chemical equation : (3) Convert the mass (gram) to mol (4) Compute the amount of water in mol : (5) Convert the mol to mass (gram) :

  29. Quantitative Information from Balanced Chemical Equations Stoichiometric relations between compounds A and B

  30. Quantitative Information from Balanced Chemical Equations • Example: Octane (C8H18), which is liquid in room temperature, has a density of 0.692 g/ml at 20oC. How many grams of O2 are required to completely burn 1.00 gal of octane? • Balance the combustion reaction: (2) Compute the mass of octane : (3) Convert the mass of octane to mol (4) Compute the amount of oxygen to react (in mol) (5) Compute the amount of oxygen to react (in gram)

  31. Limiting Reactants • Limiting reactant : The reactant that is completely consumed in a reaction. 2H2 + O2 2H2O Suppose we have a mixture of 10 mol H2 and 7 mol O2. Then, how many moles of O2 will be used? Since 2 moles of H2 consume 1 mole of O2, 10 moles of H2 will consume 5 moles of O2 to produce 10 moles of H2O  We have 7 mol – 5 mol = 2 mol of excess oxygens. Hydrogen is the limiting reactant !

  32. Limiting Reactants 2H2 + O2 2H2O Initial quantities : 10 mol 7 mol 0 mol Changes : 10 mol 5 mol 10 mol Final quantities : 0 mol 2 mol 10 mol

  33. Limiting Reactants What if we had 4 moles of oxygens to start the same reaction? 2H2 + O2 2H2O Initial quantities : 10 mol 4 mol 0 mol Changes : 8 mol 4 mol 8 mol Final quantities : 2 mol 0 mol 8 mol Oxygens are completely used up. Therefore, O2 is the limiting reactant

  34. Limiting Reactants How many compete cars can be built from these parts?

  35. Limiting Reactants Example : How many grams of P4O10 can be produced by the reaction of 1.0g of phosphorous and 3.0g of oxygens? P4 + 5O2 P4O10 MW=124 MW=32 MW=284 • Convert grams to mol : (2) Find the limiting reactant : (3) Set up the stoichiometic table (if you want) (4) Convert mol to grams

  36. Yields • Theoretical Yield : The quantity of product that is calculated to form when all of the limiting reactants react.  This is from balanced chemical equations • Actual Yield : The amount of product actually obtained in a reaction.  This is from actual experiments Actual yield ≤ theoretical yield • Percent Yield : Ratio between the theoretical yield and the actual yield

  37. Yields  Example : 2.50 g of copper is reacted with excess sulfur and 3.00 g of copper(I) sulfide is produced. What is the %yield of the reaction? 16Cu(s) + S8(s)  8Cu2S (s) AW = 63.5 FW=159 • What is the limiting reactant? (2) Calculate the theoretical yield of copper sulfide (3) Calculate % yield :

  38. Yields  Example : Lithium and nitrogen react to produce lithium nitride. If 5.00 g of each reactant undergoes a reaction with a 88.5% yield, how many grams of Li3N are obtained? 6 Li (s) + N2 (g)  2 Li3N (s) (1) Convert grams to mol: (2) Determine the limiting reactant (3) Compute the theoretical yield (4) Compute the actual yield

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