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The Mole I

The Mole I. Mathematical Relationships with Chemical Formulas. The Mole. mole (mol) - the amount of a substance that contains the same number of entities as there are atoms in exactly 12 g of carbon-12.

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The Mole I

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  1. The Mole I Mathematical Relationships with Chemical Formulas

  2. The Mole mole (mol) - the amount of a substance that contains the same number of entities as there are atoms in exactly 12 g of carbon-12. This amount is 6.022x1023. The number is called Avogadro’s number and is abbreviated asN. One mole (1 mol) contains 6.022x1023 entities (to four significant figures)

  3. Oxygen 32.0 g One mole of common substances. Water 18.0 g CaCO3 100.1 g Copper 63.5 g

  4. molar mass (MM) = mass of one mole of a substance • 1 mol Ca = 40.1 g = 6.02 x 1023 Ca atoms • 1 mol Cu = 63.5 g = 6.02 x 1023 Cu atoms • 1 mol Hg = 200.6 g = 6.02 x 1023 Hg atoms

  5. Molar Mass • 1 mol Cl2 = 71.0 g = 6.02 x 1023 Cl2 molecules = 2(6.02 x 1023) Cl atoms • 1 mol (NH4)2SO4 = 6.02 x 1023 units of (NH4)2SO4 = 15(6.02 x 1023) atoms = [2(14.0g) + 8(1.0g) + 1(32.1g) + 4(16.0g)] = 132.1 g

  6. Information Contained in the Chemical Formula of Glucose C6H12O6 ( M = 180.16 g/mol) Oxygen (O) Carbon (C) Hydrogen (H) Atoms/molecule of compound 6 atoms 12 atoms 6 atoms Moles of atoms/ mole of compound 6 moles of atoms 12 moles of atoms 6 moles of atoms Atoms/mole of compound 6(6.022 x 1023) atoms 12(6.022 x 1023) atoms 6(6.022 x 1023) atoms Mass/moleculeof compound 6(12.01 amu) =72.06 amu 12(1.008 amu) =12.10 amu 6(16.00 amu) =96.00 amu Mass/mole of compound 72.06 g 12.10 g 96.00 g

  7. Calculating molesAfrom gramsA • How many moles of Ca are in 22 g of Ca? 1 mol Ca = 40.1 g so

  8. Calculating gramsA from molesA • How many g of (NH4)2SO4 are in 0.0335 mol of (NH4)2SO4 ? 1 mol (NH4)2SO4 = 132.1 g so

  9. Mass % of element X = moles of X in formula x molar mass of X x 100 molecular (or formula) mass of compound Mass percent from the chemical formula

  10. Calculating the Mass Percents and Masses of Elements in a Sample of Compound Glucose (C6H12O6) is the most important nutrient in the living cell for generating chemical potential energy. What is the mass percent of each element in glucose? Per mole glucose there are: 6 moles of C 12 moles H 6 moles O

  11. 1.008 g H 12.01 g C 1 mol H 1 mol C 16.00 g O 1 mol O 72.06 g C 96.00 g 180.16 g glucose 180.16 g glucose 12.096 g H 180.16 g glucose Calculating the Mass Percents and Masses of Elements in a Sample of Compound 12 mol H x = 12.096 g H 6 mol C x = 72.06 g C M = 180.16 g/mol 6 mol O x = 96.00 g O x 100 = 39.99 mass %C mass percent of C = mass percent of H = x 100 = 6.714 mass %H mass percent of O = x 100 = 53.29 mass %O

  12. Empirical and Molecular Formulas Empirical Formula - The simplest formula for a compound that agrees with the elemental analysis and gives rise to the smallest set of whole numbers of atoms. C1H2O1 Molecular Formula - The formula of the compound as it exists, it may be a multiple of the empirical formula. C6H12O6

  13. 1 mol O 1 mol Na 1 mol Cl 35.45 g Cl 16.00 g O 22.99 g Na Determining the Empirical Formula from Masses of Elements Elemental analysis of a sample of an ionic compound showed 2.82 g of Na, 4.35 g of Cl, and 7.83 g of O. What are the empirical formula and name of the compound? 2.82 g Na = 0.123 mol Na Divide each mole value by the smallest value to get the whole number subscripts. 4.35 g Cl = 0.123 mol Cl 7.83 g O = 0.489 mol O Na1 Cl1 O3.98 NaClO4 NaClO4 is sodium perchlorate.

  14. Determining a Molecular Formula from Elemental Analysis and Molar Mass During physical activity, lactic acid (MM=90.08 g/mol) forms in muscle tissue and is responsible for muscle soreness. Elemental analysis shows that this compound contains 40.0 mass% C, 6.71 mass% H, and 53.3 mass% O. (a) Determine the empirical formula of lactic acid. (b) Determine the molecular formula.

  15. 1 mol C 1 mol H 1 mol O 16.00 g O 12.01g C 1.008 g H 3.33 3.33 3.33 90.08 g molar mass of lactic acid 30.03 g mass of CH2O Determining a Molecular Formula from Elemental Analysis and Molar Mass Assuming there are 100. g of lactic acid, the constituents are: 40.0 g C 6.71 g H 53.3 g O 3.33 mol C 6.66 mol H 3.33 mol O Divide each mole amount by the smallest value… C 3.33 H6.66 O3.33 CH2O empirical formula Divide the molar mass by the mass of the empirical formula C3H6O3 is the molecular formula 3

  16. Calculating moleB<->moleA • How many moles of hydrogen are contained in 2.8 mol of (NH4)2SO4? • 1 mol (NH4)2SO4 = 2 mol N, 8 mol H, 1 mol S, 4 mol O

  17. Calculating particles from moles • Particles can be molecules, formula units (ionic compounds), atoms or ions. • 1 mol particles = 6.02 x 1023 units Ne atom NH3 molecule NaCl formula unit O2 molecule

  18. Calculating particles <-> moles • How many atoms of oxygen are in 0.580 mol of iron (II) nitrate? • How many moles of sulfur trioxide can be made from 4.0 x 1023 oxygen atoms?

  19. Calculating particles <-> g • How many atoms of potassium are in 23.4 g of potassium carbonate? • How many g of sulfur hexachloride can be made from 3.33 x 1024 chlorine atoms?

  20. GramA <-> GramBwith Formulas • How many g of oxygen are contained in 458 g of C12H22O11?

  21. Summary of the mass-mole-number relationships in a chemical formula. MASS(g) of substance A MASS(g) of substance B MM (g/mol) of compound A MM (g/mol) of compound B molar ratio from formula subscripts AMOUNT(mol) of substance A AMOUNT(mol) of substance B Avogadro’s number (particles/mol) Avogadro’s number (particles/mol) MOLECULES or ATOMS (or formula units) of substance A MOLECULES or ATOMS (or formula units) of substance B Modified from Silberberg, Principles of Chemistry

  22. Questions to Ask When Solving a Problem • Are grams mentioned? If yes you will need to use a molar mass conversion. • Are you comparing two different substances? If yes you will need a conversion using the subscripts in a formula. • Are particles (atoms or molecules) involved? If yes you will need a conversion with Avogadro’s number. You may need any number of these conversions depending on the problem.

  23. How many atoms of oxygen are required to form 6.8 grams of aluminum nitrate? • Grams? Yes, need molar mass • Two substances? Yes, need subscripts • Particles? Yes, need Avogadro’s # • So at least 3 conversions are needed…

  24. How many moles of nitrogen are contained in 45.0 grams of aluminum nitrate? • Grams? Yes, need molar mass • Two substances? Yes, need subscripts • Particles? No • So, at least 2 conversions are needed…

  25. Molarity • Because many reactions happen between substances in solution it is helpful to have a concentration based on moles. http://fuelcell.com

  26. 0.715 mol glycine 1000mL 495 mL soln 1 L Calculating the Molarity of a Solution Glycine (H2NCH2COOH) is the simplest amino acid. What is the molarity of an aqueous solution that contains 0.715 mol of glycine in 495 mL? Molarity is the number of moles of solute per liter of solution. = 1.44 M glycine

  27. MASS (g) of compound in solution AMOUNT (mol) of compound in solution MOLECULES (or formula units) of compound in solution VOLUME (L) of solution Summary of mass-mole-number-volume relationships in solution. MM (g/mol) Avogadro’s number (molecules/mol) M (mol/L)

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