2. The Mole & Formulas Dr. Ron Rusay Spring 2008 © Copyright 2008 R.J. Rusay

3. Stoichiometry Mole - Mass Relationships Chemical Reactions The Mole % Composition: Determining the Formula of an Unknown Compound Writing and Balancing Chemical Equations Calculating the amounts of Reactant and Product Limiting Reactant

4. The Mole The number of carbon atoms in exactly 12 grams of pure 12 C. The number equalsThe number of carbon atoms in exactly 12 grams of pure 12 C. The number equals 6.02  10 23  1 mole of anything = 6.02  10 23 units 6.02  10 23 “units” of anything: atoms, people, stars, $s, etc., etc. = 6.02  10 23 “units” of anything: atoms, people, stars, $s, etc., etc. = 1 mole

5. Avogadro’s Number Avogadro’s number equals 1 mole ….which equals 6.022  10 23 “units” 6.022  10 23 “units”

6. Counting by Weighing 12 red marbles @ 7g each = 84g 12 yellow marbles @4g each=48g 55.85g Fe = 6.022 x 10 23 atoms Fe 32.07g S = 6.022 x 10 23 atoms S

7. CaCO 3 100.09 g Oxygen 32.00 g Copper 63.55 g Water 18.02 g Relative Masses of 1 Mole

8. Atomic and Molecular Weights Mass Measurements 1 H weighs 1.6735 x 10 -24 g and 16 O 2.6560 x 10 -23 g. 1 H weighs 1.6735 x 10 -24 g and 16 O 2.6560 x 10 -23 g. DEFINITION: mass of 12 C = exactly 12 amu. DEFINITION: mass of 12 C = exactly 12 amu. Using atomic mass units: Using atomic mass units: 1 amu = 1.66054 x 10 -24 g1 amu = 1.66054 x 10 -24 g 1 g = 6.02214 x 10 23 amu1 g = 6.02214 x 10 23 amu

9. QUESTION

10. ANSWER  –22  23  –22 5)1.06 10 g Section 3.3 The Mole (p. 82) A mole of copper atoms has a mass of 63.55 g. The mass of 1 copper atoms is 63.55 g/(6.022 10) = 1.06 10 g.

11. Atomic and Molecular Weights Formula Weight a.k.a. Molecular WeightFormula Weight a.k.a. Molecular Weight Formula weights (FW): sum of Atomic Weights (AW) for atoms in formula.Formula weights (FW): sum of Atomic Weights (AW) for atoms in formula. FW (H 2 SO 4 ) = 2 AW(H) + AW(S) + 4 AW(O)FW (H 2 SO 4 ) = 2 AW(H) + AW(S) + 4 AW(O) = 2(1.0 amu) + (32.0 amu) + 4(16.0)= 2(1.0 amu) + (32.0 amu) + 4(16.0) = 98.0 amu= 98.0 amu

12. Atomic and Molecular Weights Molecular weight (MW) is the weight of the molecular formula in amu.Molecular weight (MW) is the weight of the molecular formula in amu. MW of sugar (C 6 H 12 O 6 ) = ?MW of sugar (C 6 H 12 O 6 ) = ? MW = 6(12.0 amu) + 12(1.0 amu) + 6(16.0 amu)MW = 6(12.0 amu) + 12(1.0 amu) + 6(16.0 amu) = 180 amu= 180 amu

13. Molar Mass A substance’s molar mass (equal to the formula weight: atomic or molecular weight in grams) is the mass in grams of one mole of the element or compound.A substance’s molar mass (equal to the formula weight: atomic or molecular weight in grams) is the mass in grams of one mole of the element or compound. ð C = 12.01 grams per mole (g/mol) ð CO 2 = ?? ð 44.01 grams per mole (g/mol) 12.01 + 2(16.00) = 44.01

14. QUESTION

15. ANSWER )  3)46.07 Section 3.4 Molar Mass(p. 86 The molar mass is the sum of masses of all the atoms in the molecule. 2 12.01 + 6 1.008 + 1 16.00 = 46.07

16. QUESTION

17. ANSWER 3 ) 3)CHCl Section 3.4 Molar Mass(p. 86 The molar mass has units of g/mol. (12.8 g/0.256 mol) = 50.0 g/mol. The molecule with the closest molar mass is CH 3 Cl.

18. QUESTION

19. ANSWER  2)278 g Section 3.4 Molar Mass (p. 86) The molar mass of sodium hydroxide, NaOH, is 22.99 g/mol + 16.00 g/mol + 1.008 g/mol = 40.00 g/mol. Convert to grams: 6.94 mol (40.00 g/mol) = 278 g.

20. The Mole & Stoichiometry Stoichiometry - Mole - Mass Relationships in Chemical Reactions Stoichiometry: Writing and Balancing Chemical Equations Calculating the amounts of Reactant and Product

21. Chemical Reactions Atoms, Mass & Balance: eg. Zn + S -->

22. Chemical Equation Representation of a chemical reaction:Representation of a chemical reaction: _ C 2 H 5 OH + _ O 2  _ CO 2 + _ H 2 O reactants products reactants products C=2; H =5+1=6; O=2+1C=1; H=2; O=2+1C=2; H =5+1=6; O=2+1C=1; H=2; O=2+1 1 C 2 H 5 OH + 3 O 2  2 CO 2 + 3 H 2 O1 C 2 H 5 OH + 3 O 2  2 CO 2 + 3 H 2 O

23. Chemical Equation C 2 H 5 OH + 3 O 2  2 CO 2 + 3 H 2 OC 2 H 5 OH + 3 O 2  2 CO 2 + 3 H 2 O The equation is balanced and the reaction can be completely stated as: ð 1 mole of ethanol reacts with 3 moles of oxygen ð to produce 2 moles of carbon dioxide and 3 moles of water.

24. QUESTION Agriculturally, the following equation is important because it is used to make millions of tons of urea. When the equation is balanced, how many hydrogen atoms will be present on both sides of the equation? Also, how many moles of NH 3 would be needed to react completely with 0.5 moles of CO 2 ? NH 3 + CO 2  (NH 2 ) 2 CO + H 2 O 1.Three; two 2.Three; one 3.Six; one 4.Six ; two

25. ANSWER Choice 3. Six; one 3 answers both stoichiometry questions correctly. By placing a coefficient of 2 in front of NH 3 six hydrogen atoms would be represented on the left side of the equation. This equals the six on the right side. 2 NH 3 + CO 2  (NH 2 ) 2 CO + H 2 O Next, the balanced equation shows a 2:1 relationship between NH 3 and CO 2. So, if only 0.50 mole of CO 2 were present then one mole of NH 3 would be needed to maintain the 2:1 reacting ratio. 0.50 mole CO 2  2 mol NH 3 /1mol CO 2 = 1.0 mol NH 3 Section 3.7: Balancing Chemical Equations Section 3.8: Stoichiometric Calculations: Amounts of Reactants and Products

26. The Chemical Equation: Mole & Masses C 2 H 5 OH + 3 O 2  2 CO 2 + 3 H 2 OC 2 H 5 OH + 3 O 2  2 CO 2 + 3 H 2 O ð 46g (1 mole) of ethanol reacts with 3 moles of oxygen to produce 2 moles of carbon dioxide and 3 moles of water. ð How many grams of carbon dioxide and water are respectively produced?

27. The Chemical Equation: Moles & Masses C 2 H 5 OH + 3 O 2  2 CO 2 + 3 H 2 OC 2 H 5 OH + 3 O 2  2 CO 2 + 3 H 2 O ð 46g (1 mole) of ethanol requires 3 moles of oxygen to produce 2 moles of carbon dioxide and 3 moles of water. ð How many grams of oxygen are needed to react with all of the 46g (1 mole) of ethanol ? ð How many grams of oxygen are needed to react with 15.3g of ethanol in a 12oz. beer ?

28. Chemical Stoichiometry ð Stoichiometry is the study of chemicals and their quantities that are consumed and produced in chemical reactions. ð It quantitatively relates the behavior of atoms and molecules to observable chemical change and measurable mass effects.

29. QUESTION The fuel in small portable lighters is butane (C 4 H 10 ). Suppose after using such a lighter for a few minutes (perhaps to encourage your favorite concert performer to play one more encore) you had used 1.0 gram of fuel. How many moles of butane would this be? 1.58 moles 2.0.077 moles 3.1.7  10 –24 moles 4.0.017 moles

30. ANSWER 4) 0.017 mol is the answer for a proper grams to mole conversion. You need to know the molar mass of butane: 4  12g/mol = 48 for carbon + 10  1g/mol = 10.0 for hydrogen. Total = 48.0 + 10.0 = 58.0 g/mol. Next; 1.0 gram of butane  1 mol/58.0 g = 0.017 mol Section 3.3: Molar Mass

31. Mass Calculations: Products Reactants © Copyright 1995-2002 R.J. Rusay

32. Mass Calculations: Products Reactants © Copyright 1995-2008 R.J. Rusay

33. QUESTION The fuel in small portable lighters is butane (C 4 H 10 ). Suppose after using such a lighter for a few minutes (perhaps to encourage your favorite concert performer to play one more encore) you had used 1.0 gram of fuel.. How many moles of butane would this be? 0.017 moles How many grams of carbon dioxide would this produce? 1.) 750 mg2.) 6.0 g 3) 1.5 g4.) 3.0 g

34. ANSWER Choice 4.) 3.0 g is the answer for the correct grams to moles to grams conversion. 1.0 g C 4 H 10 ?g CO 2 C 4 H 10 mol C 4 H 10 1.0 g 58 g C 4 H 10 mol CO 2 44 g CO 2 ?g CO 2 2 mol C 4 H 10 8 mol CO 2 3.0 g CO 2