1. Section 3 Chemical Formulas and Equations

2. 2 Material was developed by combining Janusa’s material with the lecture outline provided with Ebbing, D. D.; Gammon, S. D. General Chemistry, 8th ed., Houghton Mifflin, New York, NY, 2005. Majority of figures/tables are from the Ebbing lecture outline. Mass and Moles of a Substance Chemistry requires a method for determining the numbers of molecules in a given mass of a substance. –This allows the chemist to carry out “recipes” for compounds based on the relative numbers of atoms involved. –stoichiometry - the calculation involving the quantities of reactants and products in a chemical equation.

3. 3 Material was developed by combining Janusa’s material with the lecture outline provided with Ebbing, D. D.; Gammon, S. D. General Chemistry, 8th ed., Houghton Mifflin, New York, NY, 2005. Majority of figures/tables are from the Ebbing lecture outline. Molecular Weight and Formula Weight The molecular weight (covalent bonds/nm- nm) of a substance is the sum of the atomic weights of all the atoms in a molecule of the substance. –For, example, a molecule of H 2 O contains 2 hydrogen atoms (at 1.01 amu each) and 1 oxygen atom (16.00amu), giving a molecular weight of 18.02 amu.

4. 4 Material was developed by combining Janusa’s material with the lecture outline provided with Ebbing, D. D.; Gammon, S. D. General Chemistry, 8th ed., Houghton Mifflin, New York, NY, 2005. Majority of figures/tables are from the Ebbing lecture outline. Molecular Weight and Formula Weight The formula weight of a substance is the sum of the atomic weights of all the atoms in one formula unit of the compound, whether molecular or not. –For example, one formula unit (FU) of NaCl contains 1 sodium atom (22.99 amu) and one chlorine atom (35.45 amu), giving a formula weight of 58.44 amu.

5. 5 Material was developed by combining Janusa’s material with the lecture outline provided with Ebbing, D. D.; Gammon, S. D. General Chemistry, 8th ed., Houghton Mifflin, New York, NY, 2005. Majority of figures/tables are from the Ebbing lecture outline. Mass and Moles of a Substance The Mole Concept –A mole is defined as the quantity of a given substance that contains as many molecules or formula units as the number of atoms in exactly 12 grams of carbon–12. –The number of atoms in a 12-gram sample of carbon–12 is called Avogadro’s number (N a ). The value of Avogadro’s number is 6.022 x 10 23. 1 mole = 6.022 x 10 23 ? ions, particles, atoms, molecules, items, etc. 1 mole Na 2 CO 3  6.022 x 10 23 FU Na 2 CO 3 1 mole CO 2  6.022 x 10 23 molecules CO 2

6. 6 Material was developed by combining Janusa’s material with the lecture outline provided with Ebbing, D. D.; Gammon, S. D. General Chemistry, 8th ed., Houghton Mifflin, New York, NY, 2005. Majority of figures/tables are from the Ebbing lecture outline. Mass and Moles of a Substance The molar mass (M m ) of a substance is the mass of one mole of a substance. –For all substances, molar mass, in grams per mole, is numerically equal to the formula weight in atomic mass units. –That is, one mole of any element weighs its atomic mass in grams. 1 molecule of H 2 O - MW = 18.02 amu  1 mole of H 2 O - M m = 18.02 g H 2 O 1 formula unit of NaCl - FW = 58.44 amu  1 mole of NaCl - M m = 58.44 g NaCl

7. 7 Material was developed by combining Janusa’s material with the lecture outline provided with Ebbing, D. D.; Gammon, S. D. General Chemistry, 8th ed., Houghton Mifflin, New York, NY, 2005. Majority of figures/tables are from the Ebbing lecture outline. How is it possible that M m and FW/MW are the same value but different units? A.)What is the mass in grams of one Cl atom? B.)in one HCl molecule?

8. 8 Material was developed by combining Janusa’s material with the lecture outline provided with Ebbing, D. D.; Gammon, S. D. General Chemistry, 8th ed., Houghton Mifflin, New York, NY, 2005. Majority of figures/tables are from the Ebbing lecture outline. Mass and Moles of a Substance Mole calculations –Converting the number of moles of a given substance into its mass, and vice versa, is fundamental to understanding the quantitative nature of chemical equations. 58.44 g NaCl1 mole NaCl 1 mole NaCl58.44 g NaCl

9. 9 Material was developed by combining Janusa’s material with the lecture outline provided with Ebbing, D. D.; Gammon, S. D. General Chemistry, 8th ed., Houghton Mifflin, New York, NY, 2005. Majority of figures/tables are from the Ebbing lecture outline. Mass and Moles of a Substance Mole calculations –Suppose we have 5.75 moles of magnesium. What is its mass?

10. 10 Material was developed by combining Janusa’s material with the lecture outline provided with Ebbing, D. D.; Gammon, S. D. General Chemistry, 8th ed., Houghton Mifflin, New York, NY, 2005. Majority of figures/tables are from the Ebbing lecture outline. Mass and Moles of a Substance Mole calculations –Suppose we have 100.0 grams of H 2 O. How many moles does this represent?

11. 11 Material was developed by combining Janusa’s material with the lecture outline provided with Ebbing, D. D.; Gammon, S. D. General Chemistry, 8th ed., Houghton Mifflin, New York, NY, 2005. Majority of figures/tables are from the Ebbing lecture outline. Mass and Moles and Number of Molecules or Atoms The number of molecules or atoms in a sample is related to the moles of the substance: How many molecules are there in 56 mg HCN? HW 20-22

12. 12 Material was developed by combining Janusa’s material with the lecture outline provided with Ebbing, D. D.; Gammon, S. D. General Chemistry, 8th ed., Houghton Mifflin, New York, NY, 2005. Majority of figures/tables are from the Ebbing lecture outline. Determining Chemical Formulas The percent composition of a compound is the mass percentage of each element in the compound. –We define the mass percentage of “A” as the parts of “A” per hundred parts of the total, by mass. That is,

13. 13 Material was developed by combining Janusa’s material with the lecture outline provided with Ebbing, D. D.; Gammon, S. D. General Chemistry, 8th ed., Houghton Mifflin, New York, NY, 2005. Majority of figures/tables are from the Ebbing lecture outline. Mass Percentages from Formulas Let’s calculate the percent composition (%C, %H) of one molecule of butane, C 4 H 10. First, we need the molecular wt of C 4 H 10. Now, we can calculate the percents.

14. 14 Material was developed by combining Janusa’s material with the lecture outline provided with Ebbing, D. D.; Gammon, S. D. General Chemistry, 8th ed., Houghton Mifflin, New York, NY, 2005. Majority of figures/tables are from the Ebbing lecture outline. How many grams of carbon are there in 83.5 g of formaldehyde, CH 2 O, (40.0% C, 6.73% H, 53.3% O)?

15. 15 Material was developed by combining Janusa’s material with the lecture outline provided with Ebbing, D. D.; Gammon, S. D. General Chemistry, 8th ed., Houghton Mifflin, New York, NY, 2005. Majority of figures/tables are from the Ebbing lecture outline. An unknown acid contains only C, H, O. A 4.24 mg sample of acid is completely burned. It gives 6.21 mg of CO 2 and 2.54 mg H 2 O. What is mass% of each element in the unknown acid?

16. 16 Material was developed by combining Janusa’s material with the lecture outline provided with Ebbing, D. D.; Gammon, S. D. General Chemistry, 8th ed., Houghton Mifflin, New York, NY, 2005. Majority of figures/tables are from the Ebbing lecture outline. Determining Chemical Formulas Determining the formula of a compound from the percent composition. –The percent composition of a compound leads directly to its empirical formula. –An empirical formula (or simplest formula) for a compound is the formula of the substance written with the smallest integer (whole number) subscripts.

17. 17 Material was developed by combining Janusa’s material with the lecture outline provided with Ebbing, D. D.; Gammon, S. D. General Chemistry, 8th ed., Houghton Mifflin, New York, NY, 2005. Majority of figures/tables are from the Ebbing lecture outline. Determining Chemical Formulas Determining the empirical formula from the percent composition. –Benzoic acid is a white, crystalline powder used as a food preservative. The compound contains 68.8% C, 5.0% H, and 26.2% O by mass. What is its empirical formula? –In other words, give the smallest whole-number ratio of the subscripts in the formula Cx HyOzCx HyOz

18. 18 Material was developed by combining Janusa’s material with the lecture outline provided with Ebbing, D. D.; Gammon, S. D. General Chemistry, 8th ed., Houghton Mifflin, New York, NY, 2005. Majority of figures/tables are from the Ebbing lecture outline. Determining Chemical Formulas –For the purposes of this calculation, we will assume we have 100.0 grams of benzoic acid. –Then the mass of each element equals the numerical value of the percentage. –Since x, y, and z in our formula represent mole- mole ratios, we must first convert these masses to moles. Cx HyOzCx HyOz

19. 19 Material was developed by combining Janusa’s material with the lecture outline provided with Ebbing, D. D.; Gammon, S. D. General Chemistry, 8th ed., Houghton Mifflin, New York, NY, 2005. Majority of figures/tables are from the Ebbing lecture outline. Determining Chemical Formulas This isn’t quite a whole number ratio, but if we divide each number by the smallest of the three, a better ratio might emerge. Determining the empirical formula from the percent composition. –Our 100.0 grams of benzoic acid would contain:

20. 20 Material was developed by combining Janusa’s material with the lecture outline provided with Ebbing, D. D.; Gammon, S. D. General Chemistry, 8th ed., Houghton Mifflin, New York, NY, 2005. Majority of figures/tables are from the Ebbing lecture outline. Determining Chemical Formulas Determining the empirical formula from the percent composition. –Our 100.0 grams of benzoic acid would contain: now it’s not too difficult to see that the smallest whole number ratio is 7:6:2 (multiple everything by 2 to get whole number. The empirical formula is C 7 H 6 O 2. 2 x C 3.5 H 3 O 1  C 7 H 6 O 2

21. 21 Material was developed by combining Janusa’s material with the lecture outline provided with Ebbing, D. D.; Gammon, S. D. General Chemistry, 8th ed., Houghton Mifflin, New York, NY, 2005. Majority of figures/tables are from the Ebbing lecture outline. Determining Chemical Formulas –An empirical formula gives only the smallest whole-number ratio of atoms in a formula. –The molecular formula should be a multiple of the empirical formula (since both have the same percent composition). C 2 H 3 O 2 empirical formula (lowest whole #) C 4 H 6 O 4 molecular formula C 8 H 12 O 8 molecular formula Which is not an empirical formula? CH 4 CH 4 OC 2 H 4 O 2 C 2 H 6 O –To determine the molecular formula, we must know the molecular weight (molar mass) of the compound.

22. 22 Material was developed by combining Janusa’s material with the lecture outline provided with Ebbing, D. D.; Gammon, S. D. General Chemistry, 8th ed., Houghton Mifflin, New York, NY, 2005. Majority of figures/tables are from the Ebbing lecture outline. Determining Chemical Formulas Determining the molecular formula from the empirical formula. Molecular weight = n x empirical formula wt. where n is the multiple factor n = molecular wt empirical wt

23. 23 Material was developed by combining Janusa’s material with the lecture outline provided with Ebbing, D. D.; Gammon, S. D. General Chemistry, 8th ed., Houghton Mifflin, New York, NY, 2005. Majority of figures/tables are from the Ebbing lecture outline. Acetic Acid contains 39.9% C, 6.7% H, 53.4% O. Determine empirical formula. The molecular mass of acetic acid is 60.0 g/mol. What is the molecular formula? HW 23

24. 24 Material was developed by combining Janusa’s material with the lecture outline provided with Ebbing, D. D.; Gammon, S. D. General Chemistry, 8th ed., Houghton Mifflin, New York, NY, 2005. Majority of figures/tables are from the Ebbing lecture outline. Stoichiometry: Quantitative Relations in Chemical Reactions Stoichiometry is the calculation of the quantities of reactants and products involved in a chemical reaction. –It is based on the balanced chemical equation and on the relationship between mass and moles. –Such calculations are fundamental to most quantitative work in chemistry.

25. 25 Material was developed by combining Janusa’s material with the lecture outline provided with Ebbing, D. D.; Gammon, S. D. General Chemistry, 8th ed., Houghton Mifflin, New York, NY, 2005. Majority of figures/tables are from the Ebbing lecture outline. 28.02 g N 2 + 3(2.02 g) H 2  2 (17.04 g) NH 3 28.02 g + 6.06 g  34.08 g 34.08 g = 34.08 g 1 molecule N 2 + 3 molecules H 2 2 molecules NH 3 Molar Interpretation of a Chemical Equation The balanced chemical equation can be interpreted in numbers of molecules, but generally chemists interpret equations as “mole-to-mole” relationships.

26. 26 Material was developed by combining Janusa’s material with the lecture outline provided with Ebbing, D. D.; Gammon, S. D. General Chemistry, 8th ed., Houghton Mifflin, New York, NY, 2005. Majority of figures/tables are from the Ebbing lecture outline. Molar Interpretation of a Chemical Equation Suppose we wished to determine the number of moles of NH 3 we could obtain from 4.8 mol H 2 (assume N 2 in excess).

27. 27 Material was developed by combining Janusa’s material with the lecture outline provided with Ebbing, D. D.; Gammon, S. D. General Chemistry, 8th ed., Houghton Mifflin, New York, NY, 2005. Majority of figures/tables are from the Ebbing lecture outline. Mass Relationships in Chemical Equations How many grams of HCl are required to react with 5.00 grams MnO 2 according to this equation?

28. 28 Material was developed by combining Janusa’s material with the lecture outline provided with Ebbing, D. D.; Gammon, S. D. General Chemistry, 8th ed., Houghton Mifflin, New York, NY, 2005. Majority of figures/tables are from the Ebbing lecture outline. Mass Relationships in Chemical Equations How many grams of CO 2 gas can be produced from 1.00 kg Fe 2 O 3 ? Fe 2 O 3 (s) + 3 CO (g)  2 Fe (s) + 3 CO 2 (g) HW 24

29. 29 Material was developed by combining Janusa’s material with the lecture outline provided with Ebbing, D. D.; Gammon, S. D. General Chemistry, 8th ed., Houghton Mifflin, New York, NY, 2005. Majority of figures/tables are from the Ebbing lecture outline. Limiting Reagent The limiting reactant (or limiting reagent) is the reactant that is entirely consumed when the reaction goes to completion. The limiting reagent ultimately determines how much product can be obtained. –For example, bicycles require one frame and two wheels. If you have 20 wheels but only 5 frames, how many bicycles can be made? o o o o o o o o o o o o o o o o o o o o 20 wheels 1 frame + 2 wheels  1 bike

30. 30 Material was developed by combining Janusa’s material with the lecture outline provided with Ebbing, D. D.; Gammon, S. D. General Chemistry, 8th ed., Houghton Mifflin, New York, NY, 2005. Majority of figures/tables are from the Ebbing lecture outline. Limiting Reagent If 0.30 mol Zn is added to hydrochloric acid containing 0.52 mol HCl, how many moles of H 2 are produced?

31. 31 Material was developed by combining Janusa’s material with the lecture outline provided with Ebbing, D. D.; Gammon, S. D. General Chemistry, 8th ed., Houghton Mifflin, New York, NY, 2005. Majority of figures/tables are from the Ebbing lecture outline. If 7.36 g Zn was heated with 6.45 g sulfur, what amount of ZnS was produced?

32. 32 Material was developed by combining Janusa’s material with the lecture outline provided with Ebbing, D. D.; Gammon, S. D. General Chemistry, 8th ed., Houghton Mifflin, New York, NY, 2005. Majority of figures/tables are from the Ebbing lecture outline. Theoretical and Percent Yield The theoretical yield of product is the maximum amount of product that can be obtained from given amounts of reactants. –The percentage yield is the actual yield (experimentally determined) expressed as a percentage of the theoretical yield (calculated).

33. 33 Material was developed by combining Janusa’s material with the lecture outline provided with Ebbing, D. D.; Gammon, S. D. General Chemistry, 8th ed., Houghton Mifflin, New York, NY, 2005. Majority of figures/tables are from the Ebbing lecture outline. Theoretical and Percent Yield To illustrate the calculation of percentage yield, recall that the theoretical yield of ZnS in the previous example was 11.0 g ZnS. If the actual yield of the reaction had been 9.32 g ZnS, what is the %yield? HW 25

34. 34 Material was developed by combining Janusa’s material with the lecture outline provided with Ebbing, D. D.; Gammon, S. D. General Chemistry, 8th ed., Houghton Mifflin, New York, NY, 2005. Majority of figures/tables are from the Ebbing lecture outline. If 11.0 g CH 3 OH are mixed with 10.0 g CO, what is the theoretical yield of HC 2 H 3 O 2 in the following reaction? If the actual yield was 19.1 g, what is the %yield of HC 2 H 3 O 2 ? CH 3 OH (l) + CO (g)  HC 2 H 3 O 2 (l)