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Formulas, Reactions, and Amounts

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

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Formulas, Reactions, and Amounts

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  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− SrF2: 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 • CO3−2, ClO4−

  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 + O2 → 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 × 1023 items • Abbreviation “mol” (save one whole letter!) • 6.02 × 1023 = “Avogadro’s number” = NA • Compare to • dozen • pair • gross • score

  17. Avogadro’s Number • Why 6.02 × 1023? • NA 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 1 mol Na 400 g Na = 17.40 mol Na 22.99 g 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(NO3)2 • Ca: 40.08 × 1 = 40.08 • N: 14.01 × 2 = 28.02 • O: 16.00 × 6 = 96.00 • Ca(NO3)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(NO3)2 1 mol Ca(NO3)2 164.10 g Ca(NO3)2 100 g Ca(NO3)2 = 0.609 mol Ca(NO3)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 CaCl2 = 1 mol 1 eq H2O = 2 mol Ca(OH)2 + 2HCl → CaCl2 + 2H2O

  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 → CaCl2 + 2H2O 1.80 eqHCl∙(2 mol/1 eq) = 3.60 molHCl 1.80 eqCaCl2∙(1 mol/1 eq) = 1.80 mol CaCl2 1.80 eqH2O∙(2 mol/1 eq) = 3.60 mol H2O

  31. Find Equivalent Moles • Activity XII

  32. Masses from Equivalent Moles If we use 1.80 mol Ca(OH)2 = 133.37gthat is (1.80 mol)(1 eq/1 mol) = 1.80 eq Ca(OH)2 + 2HCl → CaCl2 + 2H2O 1.80 eqHCl = 3.60 mol= 131.26 g 1.80 eqCaCl2 = 1.80 mol= 199.77 g 1.80 eqH2O = 3.60 mol= 64.85 g

  33. Find Masses from Equivalents • Activity XIII

  34. Equivalent Masses Ca(OH)2 + 2HCl → CaCl2 + 2H2O • 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 → CaCl2 + 2H2O (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 CaCl2 = 0.13496 mol = 14.98 g • 2N mol H2O = 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 → MgCl2 + 2H2O; 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 • MgCl2: make 0.686 mol × 95.21 g/mol = 65.31 g • H2O: make 1.371 mol × 18.016 g/mol = 24.70 g

  40. Find the Limiting Reagent and Yields • Activity XV

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