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The Mole

The Mole. Ch.8. (8-1) Mole. (mol) : amt. of substance # of atoms in 12g of carbon-12 Avogadro’s constant : 6.02 x 10 23 particles / mol Atoms, molecules (covalent) , formula units (ionic). Moles  # of Atoms. 2.66 mol x 6.02 x 10 23 atoms = 1.60 x 10 24 1 mol atoms

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The Mole

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  1. The Mole Ch.8

  2. (8-1) Mole • (mol): amt. of substance • # of atoms in 12g of carbon-12 • Avogadro’s constant: 6.02 x 1023 particles / mol • Atoms, molecules (covalent) , formula units (ionic)

  3. Moles  # of Atoms • 2.66 mol x 6.02 x 1023 atoms = 1.60 x 1024 1 mol atoms • # of Atoms  Moles • 2.54 x 1024 atoms x 1 mol = 4.22 mol 6.02 x 1023 atoms

  4. Moles to Molecules • 1.04 mol x 6.02 x 1023 molecules = 6.26 x 1023 1 mol molecules • Formula units to moles 3.49 x 1024 form.u. x 1 mol = 5.80 mol 6.02 x 1023 form.u.

  5. Moles to Atoms Practice How many atoms are in 2.5 mol of Si? • List known 2.5 mol Si • Set up conv. factor (what you’re going to is on top, what needs to cancel is on bottom) 2.5 mol Si x 6.02 x 1023 atoms 1 mol Si

  6. Moles to Atoms Practice • Cancel units, multiply top #’s & divide by bottom 2.5 mol Si x 6.02 x 1023 atoms = 1.5 x 1024 atoms Si 1 mol Si

  7. Atoms to Moles Practice Convert 3.01 x 1023 atoms of Si to mols Si • List known 3.01 x 1023 atoms Si • Set up conv. factor (what you’re going to is on top, what needs to cancel is on bottom) 3.01 x 1023 atoms Si x 1 mol Si = 6.02 x 1023 atoms

  8. Atoms to Moles Practice • Cancel units, multiply top #’s & divide by bottom 3.01 x 1023 atoms Si x 1 mol Si = 0.500 mol Si 6.02 x 1023 atoms Si

  9. Relative Atomic Mass • Weighted avg. of each isotope’s mass • Need mass & % abundance (62.94 amu)(.6917) + (64.93 amu)(.3083) = 63.55 amu

  10. Molar Mass • Sum of all the atomic masses of a substance • Ex: CO2 C = 12.01 g/mol, O = 16 g/mol 12.01 g/mol + (2)(16 g/mol) = 44.01 g/mol

  11. Molar Mass Practice Calculate the molar mass of Ba(NO3)2 • List known (at. masses from PT) Ba = 137.33 g/mol, N = 14.01 g/mol, O = 16 g/mol • Calculate # of each atom in the formula (outer subscript multiplies w/ inner subscripts) 1 Ba 1 N x 2 = 2 N 3 O x 2 = 6 O

  12. Molar Mass Practice • Multiply each at. mass by the # of atoms & add together (1 Ba)(137.33 g/mol) + (2 N)(14.01 g/mol) + (6 O)(16 g/mol) = 261.35 g/mol

  13. Moles  Mass • 2.49 mol KF x 58.1 g KF = 145 g KF 1 mol KF • Grams  Moles • 110 g KF x 1 mol KF = 1.89 mol KF 58.1 g KF

  14. Moles to Mass Practice What’s the mass in grams of 3.50 mol Cu? • List known 3.50 mol Cu 2. Find at. mass or calculate molar mass 63.55 g Cu 1 mol Cu

  15. Moles to Mass Practice 3. Set up conv. factor (what you’re going to is on top, what needs to cancel is on bottom) 3.50 mol Cu x 63.55 g Cu 1 mol Cu 4. Cancel units, multiply top #’s, & divide by bottom 3.50 mol Cu x 63.55 g Cu = 222 g Cu 1 mol Cu

  16. Mass to Moles Practice Determine the # of moles in 237 g of Cu • List known 237 g Cu 2. Find at. mass or calculate molar mass 63.55 g Cu 1 mol Cu

  17. Mass to Moles Practice 3.Set up conv. factor (what you’re going to is on top, what needs to cancel is on bottom) 237 g Cu x 1 mol Cu = 63.55 g Cu 4. Cancel units, multiply top #’s, & divide by bottom 237 g Cu x 1 mol Cu = 3.73 mol Cu 63.55 g Cu

  18. Atoms to Grams 1.2 x 1024 atoms B x 1 mol B x 10.81 g B = 21.55 g B 6.02 x 1023 atoms 1 mol B

  19. (8-2) Percentage Composition • % by mass of each element in a cmpd • Ex: H2O = 88.7% O, 11.3% H H O

  20. Determine % Comp. from Chemical Formula • % = mass of component x 100 mass of whole • To verify answers all components %, should = 100%

  21. % Comp. Practice Find the % Comp. of Cu2S • Find at. masses & multiply by # of atoms Cu = (63.55 g/mol) (2) = 127.10 g/mol S = (32.07 g/mol) (1) = 32.07 g/mol

  22. % Comp. Practice 2. Divide each mass by total mass of cmpd & multiply by 100 127.10 g/mol Cu x 100 = 79.85% Cu 159.17 g/mol Cu2S 32.07 g/mol S x 100 = 20.15% S 159.17 g/mol Cu2S • Add % to verify it’s close to 100% 79.85% + 20.15% = 100%

  23. Hydrates • To determine % comp. of a hydrate, use same format as before • % = mass water x 100 mass of whole

  24. Hydrate % Comp. Practice Determine the % water in Na2CO3•10H2O • Calculate molar mass of water (coef. is distributed) (20)(1.01 g/mol) + (10)(16 g/mol) = 180.2 g/mol H2O • Divide water mass by the total hydrate mass 180.2 g/mol H2O x 100 = 62.96%H2O 286.2 g/mol Na2CO3•10H2O

  25. Empirical Formula • Empirical Formula: simplest ratio among the elements of a cmpd • Ex: CH2O: CH2O, C2H4O2, C6H12O6

  26. Determining Emp. Formulas 63.0% Mn, 37.0% O by mass • Convert to g assuming you have 100 g 63.0 g Mn, 37.0 g O • Convert g to mols using molar mass 63.0 g Mn x 1 mol Mn = 1.15 mol Mn 54.94 g 37.0 g O x 1 mol O = 2.31 mol O 16 g

  27. Emp. Formula (cont.) 3. Divide by smallest amt (round to whole #) 1.15 mol Mn = 1 mol Mn 1.15 2.31 mol O = 2 mol O 1.15 4. Write empirical formula MnO2

  28. Molecular Formula • Actual # of atoms in a cmpd • Determined by emp.formula & molar mass

  29. Determining a Molecular Formula emp. formula = P2O5 molar mass of cmpd = 284 g/mol • Find emp. formula molar mass (2)(30.97 g/mol) + (5)(16 g/mol) = 141.94 g/mol • Solve for n (round to nearest whole #) n = molar mass of cmpd = 284 g/mol = 2 molar mass emp.141.94 g/mol

  30. Determining a Molecular Formula 3. Use n to get molecular formula n (emp.form.) = molecular form. 2 (P2O5) = P4O10 • To verify answer, calculate molar mass & compare to value given (4)(30.97 g/mol) + (10)(16 g/mol) = 284 g/mol

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