1. Mass Relationships in Chemical Reactions Chang, Chapter 3 Bylinkin et al, Chapter 1 & 2

2. Chapter 3 Outline Atomic Mass Molar Mass Molecular Mass Percent Composition by Mass Chemical Reactions and Equations Stoichiometry Reaction Yield

3. Atomic Masses Atomic mass depends on the number of protons, electrons and neutrons in the atom Individual atoms cannot be weighed (too small to measure) All atomic masses a referenced to the standard of Carbon-12 The mass of one C-12 atom is 12 amu (atomic mass units) Therefore 1 amu – 1/12 the mass of a single C- 12 atom

4. Atomic Masses Atomic masses on the Periodic Table are weighed averages of all naturally occurring isotopes (relative atomic mass A r ) Example: Boron has two stable isotopes, B-10 (19.78%) and B-11 (80.22%) –Mass of B-10 = 10.0129 amu –Mass of B-11 = 1.0093 amu Calculate the weighted average atomic mass of Boron.

5. Atomic Masses Example: The atomic mass of copper is 63.540amu. It is composed of two isotopes, Cu-63 and Cu-65, with atomic masses of 62.930amu and 64.928amu, respectively. What are the relative abundances (%) of these isotopes in naturally occurring samples of copper?

6. Molar Masses The mole is defined as 6.022x10 23 This number is known as Avogadro's Number (N A ) or the mole. IMPORTANT unit factor: 1 mole X _ 6.022x10 23 atoms X

7. Molar Masses Molar mass ( M ) is the mass of 1 mole of elementary units of a substance (units = g/ mol) –6.022x10 23 atoms of C-12 (1 mole) weigh 12 g, therefore the molar mass of C-12 is 12g/mol IMPORTANT unit factor: 1 mol X _ molar mass X

8. Molecular Mass To determine molecular mass, add up the molar masses of each individual atom in the molecule –Example: CO 2 –Example: Skatole (C 9 H 9 N)

9. Sample Problem: Number of Moles How many moles of hydrazine (N 2 H 4 ) are present in a 50.0 g sample?

10. Sample Problem: Number of Atoms How many atoms of gold are present in a pure gold ring weighing 7.25 g?

11. Sample Problem: Number of molecules How many molecules of water are in 8.0 ounces (HINT: 1 ounce = 29.5 g)?

12. Percent Composition by Mass The percent composition by mass is simply the percent by mass of an element in a compound –It is determined by multiplying the moles of element present by the molar mass of that element and dividing my the molecular mass of the molecule –%Composition by Mass= n ∙ M element M molecule

13. Sample Problem: Percent by Mass Calculate the percent by mass of H, S, and O in sulfuric acid.

14. Sample Problem: Percent by Mass Calculate the percent by mass of N and H is ammonia.

15. Empirical Formula The empirical formula of a compound can be determined experimentally if the percent composition by mass of each element in the compound is known. The process is always the same: –Assume a 100 g sample –Convert from grams to moles –Divide by the smallest mole number –Use the results to determine the subscripts on each element in the formula of the compound.

16. Sample Problem: Empirical Formula Determine the empirical formula of a compound with 24.75 % K, 34.77 % Mn, and 40.51 % O.

17. Sample Problem: Combustion A sample of 1.000 g of a compound containing C and H reacts with excess O 2 to yield 0.692 g H 2 O and 3.381 g of CO 2. The molar mass of the sample is around 40 g. Calculate the mass of C in the sample Calculate the mass of H in the sample Did the compound contain any other elements? Calculate the mass % for all elements in original sample Determine the Empirical Formula Determine the Molecular Formula

18. Chemical Reactions A chemical reaction (rxn) is a process in which one substance is turned into one or more different substances A chemical equation is a symbolic representation of what occurs in a chemical rxn

19. Chemical Reactions: Chemical Equations Example: 2CO (g) + O 2 (g)  2CO 2 (g) –Translation: “carbon monoxide reacts with oxygen to yield carbon dioxide” –The reactants are on left of the arrow and the products are on the right –Physical states of reactants and products should be given g=gas, s=solid, l =liquid, aq=aqueous –Chemical equations should be BALANCED

20. Chemical Reactions: Balancing Equations Rules for Balancing Chemical Equations –Write the correct formula for all reactants and products –Add coefficients to yield the same number of ATOMS of each element on each side of the reaction arrow Begin with compounds Leave single elements last –Eliminate any fractions

21. Chemical Reactions: Balancing Equations Balance the following chemical equations: – Li (s) + N 2(g)  Li 3 N (s) – La 2 O 3(s) + H 2 O ( l )  La(OH) 3(aq) – NH 4 NO 3(s)  N 2(g) + O 2(g) + H 2 O (g) – Ca 3 P 2(s) + H 2 O ( l )  Ca(OH) 2(aq) + PH 3(g) – Al (s) + Cl 2(g)  AlCl 3 – K 2 O (s) + H 2 O ( l )  KOH (aq)

22. Stoichiometry The quantitative study of reactants and products in a chemical reaction –The Stoichiometric coefficients in a chemical equation can be interpreted as the number of moles of each substance – C 3 H 8(g) + O 2(g)  CO 2(g) + H 2 O (g) “One mole of propane reacts with five moles of oxygen to yield 3 moles of carbon dioxide and four moles of water” Can also be interpreted in terms of molecules and mass

23. Stoichiometry: Calculations Steps for Stoichiometric Calculations: –Begin with a balanced equation –Convert given quantities to moles –Identify the limiting reagent –Use the limiting reagent to calculate the consumption of reactant and/or production of product as needed

24. Stoichiometry: Sample Problem Determine the amount of sulfuric acid I can synthesize if I begin with 400.0g of sulfur dioxide, 175.0g of oxygen gas, and 125.0g of water. How much of each reactant is consumed? Step 1: Balanced Equation

25. Stoichiometry: Sample Problem Step 2: Convert to moles

26. Stoichiometry: Sample Problem Step 3: Determine the Limiting Reagent –Once you have all reactant quantities converted to moles, simply divide each by its corresponding stoichiometric coefficient. The smallest result is the limiting reagent.

27. Stoichiometry: Sample Problem Step 4: Use the limiting reagent to calculate the amount of product produced.

28. Stoichiometry: Sample Problem Step 4: Use the limiting reagent to determine how much of each reactant was consumed.

29. Stoichiometry: Sample Problem

30. Reaction Yield The theoretical yield is the maximum amount of product that would be produced if all the limiting reagent reacted –This is determined via stoichiometric calculations The actual yield is the amount of product obtained from a reaction in the laboratory –The actual yield is almost always LESS than the theoretical yield

31. Reaction Yield To determine how efficient a reaction is, a percent yield is calculated. Percent yield = actual yield ∙ 100% theoretical yield

32. Reaction Yield: Sample Problem 150.0g of iron(III) oxide is reacted with an excess of carbon monoxide to produce iron metal and carbon dioxide. If the actual yield of iron was 87.9g, what is the percent yield?

33. Reaction Yield: Sample Problem

34. TY = 105 g Fe PY = 83.7%