Chapter 3 Stoichiometry: Calculations with Chemical Formulas and Equations

1. Chapter 3Stoichiometry:Calculations with Chemical Formulas and Equations

2. Physical Change • is one where the substance retains its identity. • Examples of physical reactions: • melting / freezing • boiling / condensing • subliming • subdivision • dissolving (solvation)

3. Chemical Change • the atoms of one or more substances are rearranged to produce new substances with new properties. • the atoms remain the same, but their arrangement changes: • H2O  H2 + O2 • water is broken down into two substances, oxygen gas and hydrogen gas. • The two new substances bear no resemblance to the water of which they are made.

4. How to Recognize a Chemical Change • a colour change • gas production • an energy change • precipitate formation

5. Stoichiometry • refers to balancing chemical equations and the associated mathematics. • is possible because of the law of conservation of mass.

6. Chemical Equations Concise representations of chemical reactions

7. Anatomy of a Chemical Equation CH4 (g) + 2 O2 (g) CO2 (g) + 2 H2O (g)

8. Anatomy of a Chemical Equation CH4 (g) + 2 O2 (g) CO2 (g) + 2 H2O (g) Reactants appear on the left side of the equation.

9. Anatomy of a Chemical Equation CH4 (g) + 2 O2 (g) CO2 (g) + 2 H2O(g) Products appear on the right side of the equation.

10. Anatomy of a Chemical Equation CH4 (g) + 2 O2 (g) CO2(g) + 2 H2O(g) The states of the reactants and products are written in parentheses to the right of each compound.

11. Anatomy of a Chemical Equation CH4 (g) + 2 O2 (g) CO2(g) + 2 H2O(g) Coefficients are inserted to balance the equation.

12. Subscripts and Coefficients Give Different Information • Subscripts tell the number of atoms of each element in a molecule

13. Subscripts and Coefficients Give Different Information • Coefficients tell the number of molecules

14. Balancing Chemical Equations • Matter cannot be lost in any chemical reaction. • all the atoms must balancein the chemical equation. • same number and type on each side of the equation • can only change the stoichiometric coefficients in front of chemical formulas. • Subscripts in a formula are never changed when balancing an equation.

15. Balancing Chemical Equations • Balance the following equation: Na(s) + H2O(l) H2 (g) + NaOH(aq) • count the atoms of each kind on both sides of the arrow: • The Na and O atoms are balanced (one Na and one O on each side) • there are two H atoms on the left and three H atoms on the right.

16. Balancing Chemical Equations • Balance the following equation: Na(s) + 2 H2O(l) H2 (g) + NaOH(aq) • the 2 increases the number of H atoms among the reactants. • O is now unbalanced. • have 2 H2O on the left; can balance H by putting a coefficient 2 in front of NaOH on the right:

17. Balancing Chemical Equations • Balance the following equation: Na(s) + 2 H2O(l) H2 (g) + 2NaOH(aq) • H is balanced • O is balanced • Na is now unbalanced, with one on the left but two on the right. • put a coefficient 2 in front of the reactant:

18. Balancing Chemical Equations • Balance the following equation: 2 Na(s) + 2 H2O(l) H2 (g) + 2 NaOH(aq) • Check our work: • 2 Na on each side • 4 H on each side • 2 O on each side • Balanced !!

19. CH4 (g) + O2 (g) CO2 (g) + H2O (g) • CH4 (g) + 2 O2 (g) CO2 (g) + 2 H2O (g)

20. Pb(NO3)2(aq)+ NaI(aq)  PbI2(aq) + NaNO3(aq) • Pb(NO3)2 (aq) + 2 NaI(aq) PbI2 (aq) + 2 NaNO3 (aq)

21. Balancing Chemical Equations • Remember: • when changing the coefficient of a compound, the amounts of all the atoms are changed. • do a recount of all the atoms every time you change a coefficient. • do a final check at the end. • fractions are not allowed. Multiply to reach whole numbers.

22. States of Reactants and Products • (s) - solid • (l) - liquid • (g) - gas • (aq) - aqueous (meaning that the compound is dissolved in water.) • (ppt) - precipitate (meaning that the reaction produces a solid which falls out of solution.) • Reaction conditions occasionally appear above or below the reaction arrow • e.g., "Δ" is often used to indicate the addition of heat).

23. Balancing Chemical Equations • Complete 3.11 to 3.14

24. Reaction Types

25. Combination Reactions • Two or more substances react to form one product • Examples: N2 (g) + 3 H2 (g) 2 NH3 (g) C3H6 (g) + Br2 (l)  C3H6Br2 (l) 2 Mg (s) + O2 (g)  2 MgO (s)

26. 2 Mg (s) + O2 (g)  2 MgO (s)

27. Decomposition Reactions • One substance breaks down into two or more substances • Examples: CaCO3 (s) CaO (s) + CO2 (g) 2 KClO3 (s)  2 KCl (s) + O2 (g) 2 NaN3 (s) 2 Na (s) + 3 N2 (g)

28. Combustion Reactions • Rapid reactions that produce a flame • Most often involve hydrocarbons reacting with oxygen in the air • Examples: CH4 (g) + 2 O2 (g) CO2 (g) + 2 H2O (g) C3H8 (g) + 5 O2 (g)  3 CO2 (g) + 4 H2O (g)

29. Balancing Chemical Reactions • Complete Problems 3.15 to 3.20

30. Formula Weights

31. Formula Weight (FW) • Sum of the atomic weights for the atoms in a chemical formula • So, the formula weight of calcium chloride, CaCl2, would be Ca: 1(40.08 amu) + Cl: 2(35.45 amu) 110.98 amu • These are generally reported for ioniccompounds

32. C: 2(12.01 g/mol) + H: 6(1.01 g/mol) 30.08 g/mol Molecular Weight (MW) • Sum of the atomic weights of the atoms in a molecule • For the molecule ethane, C2H6, the molecular weight would be

33. Molecular & Formula Weights • formula and molecular weights are often called molar mass. • The formula weight (in amu) is numerically equal to the molar mass (in g/mol). • Always report molar mass in g/mol. • Always round the atomic weights to 2 decimal places.

34. Formula & Molecular Weights • Complete questions 3.21 and 3.22

35. (number of atoms)(atomic weight) x 100 % element = (molar mass of the compound) Percent Composition Percentage composition is obtained by dividing the mass contributed by each element (number of atoms times AW) by the formula weight of the compound and multiplying by 100.

36. (2)(12.01 g/mol) %C = (2(12.01) + 6(1.01))g/mol x 100 = 24.02 g/mol 30.08 g/mol = 79.85% Percent Composition So the percentage of carbon in ethane (C2H6) is…

37. % Composition of H2SO4 • the formula weight is 98.09 amu (do calculation) • % H = 2(1.01 amu) x 100 = 2.06 % H 98.09 amu • % S = 32.07 amu x 100 = 32.69 % S 98.09 amu • % O = 4(16.00 amu) x 100 = 65.25 % O 98.09 amu the sum of the percent compositions will always equal 100 %

38. Percent Composition • Complete questions 3.24 and 3.26 • Calculate the % composition of each element in each compound.

39. Moles

40. Avogadro’s Number • 6.022 x 1023 • 1 mole of 12C has a mass of 12 g

41. Molar Mass • By definition, these are the mass of 1 mol of a substance (i.e., g/mol) • is the atomic weight on the periodic table • The formula weight (in amu’s) will be the same number as the molar mass (in g/mol)

42. Using Moles Moles provide a bridge from the molecular scale to the real-world scale

43. Molar Calculations • Using the mass and molar mass to calculate the number of MOLES of a compound or element. • m Mass grams g • n amount of matter moles mol • M Molar mass grams per mole g/mol number of = mass moles molar mass n = m M

44. Sample Calculation • If you have 20.2 grams of Potash, how many moles do you have? • Molar Mass KCl (calculated):74.55 g/mol • Mass KCl: 20.2 g 20.2 g = 0.271 moles KCl 74.55 g/mol • if the units work out, you are correct.

45. Example • How many moles are present in 0.750 g of ammonium phosphate? • Ammonium phosphate: (NH4)3PO4 • Molar mass: (3(14.01) + 12(1.01) + 30.98 + 4(16.00)) g/mol = 149.13 g/mol • Number of moles = 0.750 g = 5.03 x 10-3 mol 149.13 g/mol (NH4)3PO4

46. Example • What is the mass of 12.31 mol of NaOH ? • Step 1. Calculate molar mass • 1 x 22.00 g/mol • 1 x 16.00 g/mol • 1 x 1.01 g/mol • 40.00 g/mol • Step 2. Calculate mass • (12.31 mol)(40.00 g/mol) = 492.4 g NaOH

47. Example • What is the mass of 2.1 x 10-4 mol of diarsenic trioxide? • Diarsenic trioxide: As2O3 • Molar mass: (2(74.92) + 3(16.00)) g/mol = 197.84 g/mol • Mass = (2.1 x 10-4 mol)(197.84 g/mol) = 0.042 g As2O3

48. Example How many moles of caffeine (C8H10N4O2 ) is required to reach the 12.0 g lethal dose in a 140 pound female? • Step 1. Molar Mass • 8 x 12.01 g/mol • 10 x 1.01 g/mol • 4 x 14.01 g/mol • 2 x 16.00 g/mol • 194.22 g/mol • Step 2. Calculate Moles • 12.00 g = 0.06179 mol caffeine • 194.22 g/mol

49. Calculating Atoms… • We can also find the number of atoms of a substance using Avogadro’s Number: • 6.022 x 1023 atoms/mol • Example: • Calculate the number of atoms in 0.500 moles of silver? 0.500mol Ag x 6.022x1023 atoms = 3.01 x 1023 atoms mol of Ag

50. The Same Goes… • For molecules! • Avogadro’s number represents the number of particles, whether atoms or molecules. • Eg. Calculate the number of molecules in 4.99mol of methane. 4.99 mol x 6.022 x 1023 molecules = 3.00 x 1024 molecules mol of CH4