1. Formulas, Reactions, and Amounts PHYS 1090 Unit 10

2. Why? Fine structure of matter not obvious –Formulas force us to mind atomic composition Materials react in definite proportions –Simple ratios emphasized in formulas Explain quantitative results –Understand the measurements

3. Molecular Formulas Molecule: group of connected atoms of definite composition Formula: tells how many atoms of each element per molecule –Often more information is necessary to unambiguously specify molecule

4. Formulas Elements represented by symbols (One capital letter or one cap + one lowercase) –H, Li, Na, C, N, etc. Subscripts tell how many atoms of each –No subscript means “1” LiBr: 1 Li + + 1 Br − SrF 2 : 1 Sr +2 + 2 F − Can subscript groups, e.g. B(OH) 3

5. Count Atoms Activity I

6. Ionic Compounds Ion: electrically-charged object Ionic compound: composed of ions, each containing one or more atoms, connected only by electrostatic attraction –Charges balance to zero

7. Charges of Ions Many atoms have one preferred charge –Na +, Ca +2, Br − Charges specified for others –Iron(II), lead(IV) Ions can be groups of atoms –CO 3 −2, ClO 4 −

8. Ionic Compound Formulas Formula unit: fewest positive + negative ions to balance charge Li + + Br − : 1 Li + + 1 Br − Sr +2 + F − : 1 Sr +2 + 2 F −

9. Balance Charges Activity II

10. Identify and Balance Activity III

11. Reaction Equations Reactants → Products “+” btw different reactants and products Coefficients: how many formula units of each species –No coefficient means “1” 2 C + O 2 → 2 CO Conservation of atoms from reactants to products

12. Counting Atoms Multiply number in each formula unit by coefficient Add together atoms of each type in all reactants Add together atoms of each type in all products These are the same in a balanced equation

13. Count Atoms Activity IV

14. Balance Equations Adjust coefficients to balance equations Activities V, VI

15. Moles How many?

16. Mole A counting unit: 6.02 × 10 23 items –Abbreviation “mol” (save one whole letter!) –6.02 × 10 23 = “Avogadro’s number” = N A Compare to –dozen –pair –gross –score

17. Avogadro’s Number Why 6.02 × 10 23 ? N A of Carbon-12 atoms has a mass of exactly 12 g.

18. Atomic Mass A sample of 1 mol of atoms of an element has a mass in grams equal to its atomic mass –More correctly “molar mass of the element” (Unstated) units = g/mol

19. Finding Atomic Mass On the periodic table –After the atomic number It’s that number that I warned you is not the mass number –Now you know what it’s for Depends on isotopic abundances –Generally similar for different sources

20. Find Masses Activity VII –Mass of 1 mole Activity VIII –Mass of arbitrary numbers of moles –Multiply atomic mass by moles –E.g. (2.0 mol B)(10.81 g B /mol B) = 21.62 g B

21. Finding Moles Divide sample mass by molar mass E.g. (400 g Na) / (22.99 g Na/mol Na) = 17.40 mol Na Or think of it as 400 g Na= 17.40 mol Na 22.99 g Na 1 mol Na

22. Find Moles Activity IX

23. Formula Mass Mass in grams of a mole of formula units –Mass of a mole of molecules Molar mass of compound –“molecular mass” –“molecular weight” –“formula weight”

24. Finding Formula Mass Multiply each element’s molar mass by its number in the formula unit Add products together Example: Ca(NO 3 ) 2 –Ca: 40.08 × 1 = 40.08 –N: 14.01 × 2 = 28.02 –O: 16.00 × 6 = 96.00 Ca(NO 3 ) 2 : 164.10 g/mol

25. Find Formula Masses Activity X

26. Other Way Around Given mass, how many moles are there? Divide sample mass by molar mass –Just like atomic masses Example: 100 g Ca(NO 3 ) 2 100 g Ca(NO 3 ) 2 = 0.609 mol Ca(NO 3 ) 2 164.10 g Ca(NO 3 ) 2 1 mol Ca(NO 3 ) 2

27. Find Moles Activity XI

28. Reactions Recipes and equivalents

29. Mole Equivalents 1 eq Ca(OH) 2 = 1 mol 1 eq HCl = 2 mol 1 eq CaCl 2 = 1 mol 1 eq H 2 O = 2 mol Ca(OH) 2 + 2HCl → CaCl 2 + 2H 2 O

30. Equivalent moles If we use 1.80 mol Ca(OH) 2, that is (1.80 mol)(1 eq/1 mol) = 1.80 eq Ca(OH) 2 + 2HCl → CaCl 2 + 2H 2 O 1.80 eq HCl ∙ (2 mol/1 eq) = 3.60 mol HCl 1.80 eq CaCl 2 ∙ (1 mol/1 eq) = 1.80 mol CaCl 2 1.80 eq H 2 O ∙ (2 mol/1 eq) = 3.60 mol H 2 O

31. Find Equivalent Moles Activity XII

32. Masses from Equivalent Moles If we use 1.80 mol Ca(OH) 2 = 133.37g that is (1.80 mol)(1 eq/1 mol) = 1.80 eq Ca(OH) 2 + 2HCl → CaCl 2 + 2H 2 O 1.80 eq HCl= 3.60 mol= 131.26 g 1.80 eq CaCl 2 = 1.80 mol= 199.77 g 1.80 eq H 2 O= 3.60 mol= 64.85 g

33. Find Masses from Equivalents Activity XIII

34. Equivalent Masses Ca(OH) 2 + 2HCl → CaCl 2 + 2H 2 O 1 mol Ca(OH) 2 74.096 g/mol 74.096 g 2 mol HCl 36.458 g/mol 72.916 g 1 mol CaCl 2 110.98 g/mol 110.98 g 2 mol H 2 O 18.016 g/mol 36.032 g 74.096 + 72.916 = 147.012 110.98 + 36.032 = 147.012

35. Finding Equivalent Masses Find moles of all reactants and products Convert to masses

36. Finding Equivalent Masses Example: –Ca(OH) 2 + 2HCl → CaCl 2 + 2H 2 O (previous) –10 g Ca(OH) 2 N mol Ca(OH) 2 = 10 g/74.096 g/mol = 0.13496 mol –2N mol HCl = 0.26992 mol = 9.84 g –N mol CaCl 2 = 0.13496 mol = 14.98 g –2N mol H 2 O = 0.26992 mol = 4.86 g

37. Find Equivalent Masses Activity XIV

38. Limiting Reagents Reactants may not be present in equivalent amounts! The one with the fewest equivalents limits the outcome.

39. Limiting Reagents Example: Mg(OH) 2 + 2HCl → MgCl 2 + 2H 2 O; 50 g Mg(OH) 2 + 50 g HCl –Mg(OH) 2 : 58.326 g/mol  50 g = 0.857 mol = 0.857 eq –HCl: 36.458 g/mol  50 g = 1.371 mol = 0.686 eq HCl is limiting –Mg(OH) 2 : use 0.686 mol × 58.326 g/mol = 40.01 g –MgCl 2 : make 0.686 mol × 95.21 g/mol = 65.31 g –H 2 O: make 1.371 mol × 18.016 g/mol = 24.70 g

40. Find the Limiting Reagent and Yields Activity XV