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Lecture 5. Chapter 3. Chemical Equations:

Lecture 5. Chapter 3. Chemical Equations:. 2 H 2 + O 2 → 2 H 2 O. Stoichiometry: Calculations with Chemical Formulas and Equations. Chemical Equations: The equation for hydrogen burning in oxygen is represented as: 2 H 2 + O 2 → 2 H 2 O

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Lecture 5. Chapter 3. Chemical Equations:

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  1. Lecture 5. Chapter 3. Chemical Equations: 2 H2 + O2 → 2 H2O

  2. Stoichiometry: Calculations with Chemical Formulas and Equations. Chemical Equations: The equation for hydrogen burning in oxygen is represented as: 2 H2 + O2 → 2 H2O The ‘+’ sign means ‘reacts with’The arrow means ‘produces’

  3. 2 H2 + O2 → 2 H2O The chemical substances on the left of the equation are the ‘reactants’. The chemical substances on the right of the equation are the ‘products’. The numbers in front of the formulas are the coefficients.

  4. Balanced equations: Because of the Law of conservation of matter, the numbers of each type of atom must be the same on the left and the right side of the equation. The equation must be BALANCED. (4 H’s on each side, 2 O’s on each side). Note: The coefficient multiplies through everything in the substance that follows: 2 H2 + O2 → 2 H2O 2 x 2 = 4 H 2 O 2 x 2 = 4H and 2 O

  5. View of reaction at the molecular level: 4 H’s and 2 O’s 4 H’s and 2 O’s Before reaction after reaction ‘produces’ O H H O + H H O O two hydrogen one oxygen two water molecules molecule molecules

  6. Balancing equations: Once we know the reactants and products in the reaction, we can write the unbalanced equation. e.g. CH4 + O2 → CO2 + H2O To balance the equation, we must find the coefficients that will lead to a balanced equation.

  7. Balancing equations: Change only the coefficients: In balancing an equation, only coefficients can be changed. Changing subscripts is not allowed because this would lead to a change in the nature of the substances involved.

  8. It is usually best to start by balancing the element(s) that occur in the fewest chemical formulas on either side of the equation. There always has to be the same number of carbon atoms on either side of the equation. Each C atom needs one O2 molecule to form CO2. The 4 H’s of CH4 need two O atoms = one O2 molecule to form 2 H2O: CH4 + 2 O2 → CO2 + 2 H2O (Note spaces between coefficients and formulas for substances).

  9. Molecular view of reaction: one C atom four O atoms one C atom two O-atoms four H atoms two O-atoms four H-atoms + + one methane two oxygen one carbon two water molecule molecules dioxide molecule molecules ‘before’ ‘after’

  10. Combination and Decomposition reactions. Combination reaction: Here two or more substances react to form one product: 2 Mg + O2(g) → 2 MgO(s) In a decomposition reaction, one substance breaks down into two or more: CaCO3(s) → CaO(s) + CO2(g)

  11. Combustion in air: Combustion reactions are rapid reactions in air that produce a flame. Hydrocarbons consist only of C and H. When balancing an equation with combustion of these molecules, start with the C. The number of CO2 molecules produced is the same as in the hydrocarbon: C3H8 + O2 → 3 CO2 + H2O (not yet balanced)

  12. Next balance the H-atoms. The number of waters produced (4) is half the number of H’s (8) in the hydrocarbon: C3H8 + O2 → 3 CO2 + 4 H2O 8 H 4 x 2 = 8 H

  13. Balance the O-atoms Finally, add up the number of O-atoms needed on the r.h.s. = 6 + 4 = 10. Therefore need 5 O2 molecules. C3H8 + 5 O2→ 3 CO2 + 4 H2O 5 x 2 = 10 O’s 3 x 2 = 6 O’s 4 O’s 10 O 10 O

  14. Molecular view of reaction: 3 C atoms 10 O atoms 3 C atoms 4 O-atoms 8 H atoms 6 O-atoms 8 H-atoms + + C3H8 + 5 O2→ 3 CO2 + 4 H2O

  15. Combustion of O-containing derivatives: One needs to remember here that there is an O on the left hand side, so the numbers of O2’s needed is going to be less by that amount: C2H5OH + x O2→ 2 CO2 + 3 H2O On r.h.s. are 4 + 3 = 7 O-atoms. But ethanol already has one O-atom, so need only 3 O2 molecules: C2H5OH + 3 O2→ 2 CO2 + 3 H2O 1 O 6 O’s 4 O’s 3 O’s 7 O’s 7 O’s

  16. 3.3. Formula weights. Chemical formulas and chemical equations contain quantitative information. They tell us how much of each reactant is needed for the reaction. H = 1.0 amu O = 16.0 amu Weight of H2O = 16.0 + 1.0 + 1.0 = 18.0 amu

  17. Formula and molecular weights: The formula weight (F. Wt.) is the sum of the atomic weights of each atom in its chemical formula. If the chemical formula is that of a molecule, then the formula weight is referred to as a molecular weight (M. Wt.).

  18. Formula weights: e.g. M. Wt. of glucose: C6H12O6: C = 12.0 amu H = 1.0 amu, O = 16.0 amu So F. Wt. = (6 x 12.0) + (12 x 1.0) + (6 x 16.0) = 180 amu. glucose Exercise 3.5: F. Wt. of sucrose: C12H22O11: (342.0) Ca(NO3)2: 164.1 amu.

  19. Percentage composition from formulas: When a new compound is analyzed to obtain its elemental composition, this is compared with the percentage composition obtained from its formula: % element = (no. atoms of element) x (At. Wt. element) x 100 F. Wt. of compound C12H22O11 (sucrose): F. Wt. = 342.0 amu. % C = (12 x 12.0) x 100/342 = 42.1% % H = (22 x 1.0) x 100/342 = 6.4% % O = (11 x 16.0) x 100/342 = 51.5% check: 42.1 + 6.4 + 51.5 = 100.0%.

  20. The Mole. 3.4. Avogadro’s number and the Mole. Amu are far too small to be useful in everyday chemistry. We use instead the Mole(abbreviated mol) which is basically the F. Wt. or M. Wt. expressed in grams. e.g. Water has a formula weight of 18.0 amu, so 1 mole of water weighs 18.0 grams. Lorenzo Romano Amedeo Carlo Avogadro, conte di Quaregna e di Cerreto (1776 - 1856)

  21. One mole of water Water has a formula weight of 18.0 amu, so 1 mole of water weighs 18.0 grams. How do we arrive at this? Conversion factors: 1 amu = 1.6 x 10-24 g, 1 mole contains 6.022 x 1023 molecules, 1 molecule of H2O weighs 18.0 amu: 18.0 amu x 6.022 x 1023 molecules x 1.6 x 1024 g 1 molecule 1 mol 1 amu = 18.0 g/mol 18.0 grams of water

  22. Avogadro’s number Note: 1 mol of 12C contains 6.022 x 102312C atoms. 1 mol of C2H4 contains 6.022 x 1023 C2H4 molecules.1 mol of glucose contains 6.022 x 1023 molecules 1 mol of water = 18.0 grams of water 1 mol of water contains 6.022 x 1023 water molecules. How many H-atoms does it contain?

  23. Think about dozens instead of moles: One dozen water molecules….. …contains one dozen oxygen atoms, but two dozen hydrogen atoms. One mol of water contains one mol of oxygen atoms but two moles of Hydrogen atoms

  24. Numbers of atoms in molecules: One mol of ethylene (C2H4) contains 6.022 x 1023 ethylene molecules, but: One mol of ethylene (C2H4) contains 4 x 6.022 x 1023 = 2.4088 x 1024 H-atoms One mol of ethylene (C2H4) contains 2 x 6.022 x 1023 = 1.2044 x 1024 C-atoms ethylene C2H4

  25. Counting BB’s Suppose you sell BB’s for air-guns in boxes each containing 1000 BB’s. How would you put 1000 BB’s in each box? Box must contain 1000 BB’s (one BB weighs 0.34 g)

  26. Weighing as a means of counting: Avogadro’s number (6.022 x 1023) is a very large number, but it is still just a number, like a dozen is a number. Just as with BB’s, there are too many molecules in even a tiny amount of a substance for us to count them. We have to weigh them. And we know that in one mol of substance there are 6.022 x 1023 molecules. So chemists work in moles, knowing 1 mole contains 6.022 x 1023 molecules.

  27. Fractions of moles and numbers of atoms: Exercise 3.8. Calculate the number of H atoms in 0.350 moles of C6H12O6. Conversion: 1 mole = 6.022 x 1023 molecules 1 molecule = 12 H-atoms 0.35 moles x 6.022 x 1023 molecules x 12 atoms = 1 mole 1 molecule = 25.3 x 1023 = 2.53 x 1024 H-atoms

  28. Table 3.2: Mole Relationships: Substance F.Wt. Molar Mass No. of particles (in one mole) Atomic N 14.0 14.0 g/mol 6.022 x 1023 N2 gas 28.0 28.0 g/mol 6.022 x 1023 N atoms in 1 mole N2 1.2044 x 1024 N’s Ag(s) 107.9 107.9 g/mol 6.022 x 1023 Ag+ ions 107.9 107.9 g/mol 6.022 x 1023 BaCl2 208.2 208.2 g/mol 6.022 x 1023 Ba2+ 1.2044 x 1024 Cl- AlCl3133.3 133.3 g/mol 6.022 x 1023 Al3+ 1.806 x 1024 Cl-

  29. Some problems: How many moles of H atoms are there in 1.0 moles of NH4Cl ? How many H atoms are there in 1.0 moles of NH4Cl ? How many H atoms are there in 2 moles of NH4Cl ? How many H atoms are there in 0.3724 moles of NH4Cl ?

  30. Interconverting masses and numbers of particles: e.g. How many Cu atoms in a Cu penny, weight = 3.0 g (assume the penny is pure copper). copper penny Conversion: Cu = 63.5 g/mol; 1 mol = 6.022 x 1023 atoms (note units) 3.0 g x 6.022 x 1023atoms x 1 mol = 1 mol 63.5 g 2.8 x 1022atoms

  31. 3.5 Empirical Formulas from Analyses. When Chemists discover new compounds, they may analyze them to get their percentage elemental composition. Thus, if we analyze a compound and find that it contains 73.9% Hg and 26.1% Cl by weight, we can work out the molar ratio. Note that when we say something is 73.9% Hg, that means 100.0 g of the substance would contain 73.9 g Hg, and similarly 26.1 g Cl.

  32. Assume that we have 100g sample. Thus, the percentages mean that we have 73.9 g Hg and 26.1 g Cl. Conversion factors (from periodic table): Hg, 1 mol = 200.6 g (= atomic wt.) Cl, 1 mol = 35.5 g (= atomic wt.) So the molar ratio is given by: Hg: 73.9 g x 1 mol/200.6 g = 0.368 mol Cl: 26.2 g x 1 mol/35.5 g = 0.735 mol

  33. To get the molar ratio, divide through by the lowest molar ratio present: Hg: 0.368 mol/0.368 mol = 1.00 Cl: 0.735 mol/0.368 mol = 1.997 (~ 2.00) We can say that within experimental error, the empirical formula of the compound is HgCl2. Example 3.13: Ascorbic acid is 40.92% C, 4.58% H, and 54.40% O by mass. What is the empirical formula?

  34. Example 3.13: Ascorbic acid Conversion factors: C, 1 mol = 12.0 g; H, 1 mol = 1.0 g; O, 1 mol = 16.0 g. C = 40.92 g x 1 mol/12.0 g = 3.407 mol H = 4.58 g x 1 mol/1.0 g = 4.54 mol O = 54.40 g x 1 mol/16.0 g = 3.406 mol Divide through by lowest number of moles (3.406) C = 3.407 mol/3.406 mol = 1.00 H = 4.54 mol /3.406 mol = 1.33 O = 3.406 mol /3.406 mol = 1.00 Multiply through by 3 to get all near whole numbers. Empirical formula is C3H4O3.

  35. Molecular formula from empirical Formula. If we know the molecular weight of a compound, we can then convert the empirical formula into a molecular formula . We calculate the empirical formula weight, which is the formula weight calculated from the empirical formula. We then obtain a whole number multiple by dividing the molecular weight by the empirical formula weight, and it is this multiple that we use to multiply through the subscripts in the empirical formula. Mesitylene has an empirical formula of C3H4. The experimentally determined M.Wt. is 121.0 amu. What is the molecular formula?

  36. An example: mesitylene Mesitylene has an empirical formula of C3H4. The experimentally determined M.Wt. is 121.0 amu. What is the molecular formula of mesitylene? C3H4 empirical formula weight = 3 x 12 + 4 x 1 = 40 amu Multiple = Experimental M.Wt empirical formula weight = 121amu /40 amu = 3.02 (~ 3.0 ) Multiple = 3.0, so molecular formula = C9H12

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