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Chemical Composition

Chemical Composition. Chapter 10. Chemical Packages - Moles. We use a package for atoms and molecules called a mole A mole is the number of particles equal to the number of Carbon atoms in 12 g of C-12 One mole = 6.022 x 10 23 units

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Chemical Composition

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  1. Chemical Composition Chapter 10

  2. Chemical Packages - Moles • We use a package for atoms and molecules called a mole • A mole is the number of particles equal to the number of Carbon atoms in 12 g of C-12 • One mole = 6.022 x 1023 units • The number of particles in 1 mole is called Avogadro’s Number • This is the magic number that allows us to turn atomic masses (amu) into grams

  3. Representative Particles • Elements – made up of atoms • Molecular compounds – made up of molecules • Ionic compounds – made up of formula units (FU’s)

  4. Atomic Masses • Unit is the amu. • atomic mass unit • 1 amu = 1.66 x 10-24g • We define the masses of atoms, molecules, and formula units in terms of atomic mass units. • 1 Carbon atom = 12.01 amu • 1 O2 molecule = 2(16.00 amu) = 32.00 amu • 1 MgCl2 F.U. = 24.31 + 2(35.45) = 95.21 amu • Atomic mass, molecular mass, and formula mass are ALL measured in amu.

  5. Molar Mass • The molar mass is the mass in grams of one mole of a compound • The molecular mass can be calculated from atomic masses: water = H2O = 2(1.01 amu) + 16.00 amu = 18.02 amu • 1 mole of H2O will weigh 18.02 g, therefore the molar mass of H2O is 18.02 g • C = 12.01 amu = 12.01 g • O2 = 32.00 amu = 32.00 g • MgCl2 = 95.21 amu = 95.21 g

  6. Example #1 Compute the number of moles and number of atoms in 10.0 g of Al • Use the Periodic Table to determine the mass of 1 mole of Al 1 mole Al = 26.98 g • Use this as a conversion factor for grams-to-moles

  7. Example #1 Compute the number of moles and number of atoms in 10.0 g of Al • Use Avogadro’s Number to determine the number of atoms in 1 mole 1 mole Al = 6.022 x 1023 atoms • Use this as a conversion factor for moles-to-atoms

  8. Mole Conversions – mol to g A .267 mol sample of magnesium nitride equals how many grams? Formula = ??? Mg+2 N-3 Mg3N2 Molar Mass = ??? Mg = 3(24.31) = 72.93 g N = 2(14.01) = 28.02 g Molar Mass = 100.95 g 100.95 g Mg3N2 = 1 mol Mg3N2 .267 mol Mg3N2 x 27.0 g Mg3N2

  9. Mole Conversions – g to mol A 56.51 g sample of potassium oxide contains how many moles? Formula = ??? K+1 O-2  K2O Molar Mass = ??? K = 2(39.10) = 78.20 g O = 1(16.00) = 16.00 g Molar Mass = 94.20 g 56.51 g K2O 1 mol K2O 94.20 g K2O x = .5999 mol K2O

  10. Mole Conversions – g to molecules If you have a 24.31 g sample of propanol, C3H8O, how many molecules are produced? Molar Mass = ??? C = 3(12.01) = 36.03 g H = 8(1.01) = 8.08 g O = 1(16.00) = 16.00 g Molar Mass = 60.11 g 1 mol C3H8O 60.11 g C3H8O 6.022 x 1023 molecules 1 mol C3H8O 24.31 g C3H8O x x = 2.435 x 1023 molecules

  11. Mole Conversions – g to atoms A 6.93 g sample of sodium oxalate contains how many carbon atoms? Formula = ??? Na+1 C2O4-2 Na2C2O4 Molar Mass = ??? Na = 2(22.99) = 45.98 g C = 2(12.01) = 24.02 g O = 4(16.00) = 64.00 g Molar Mass = 134.00 g 6.93 g Na2C2O4 1 mol Na2C2O4 134.00 g Na2C2O4 6.022 x 1023 F.U. 1 mol Na2C2O4 x x x 2 carbon atoms 1 F.U. = 6.23 x 1022 carbon atoms

  12. Mole Conversions – Molar Volume • At STP, 1 mol (6.02 x 1023 rep. part.) of ANY gas occupies a volume of 22.4 L. • The quantity 22.4 L is called the molar volume of a gas. • STP = standard temperature and pressure • 0o C and 1 atm • What volume will 0.375 mol of O2 gas occupy at STP?

  13. Mass Percent • Measures the mass (by percent) of a single atom in a compound. • Ex. – A 35.0 g sample contains 11.2 g of potassium. What is the mass percent of potassium? element mass Mass Percent = x 100 total mass

  14. Percent Composition • Percentage of each element in a compound • By mass • Can be determined from: -the formula of the compound or -the experimental mass analysis of the compound • The percentages may not always total to 100% due to rounding

  15. Examples 1. A 23.9 g sample contains 14.3 g C, 3.2 g H, and some O. What is the % composition? 2. What is the % composition of aluminum carbonate? 3. What is the mass of C in a 121.3 g sample of aluminum carbonate?

  16. Empirical Formulas • The simplest, whole-number ratio of atoms in a molecule is called the Empirical Formula • can be determined from percent composition or from mass data. • Poem to Calculate Empirical Formula Percent to Mass (optional) Mass to Mole Divide by Small Multiply ‘til Whole

  17. Molecular Formulas • The molecular formula is a multiple of the empirical formula; it is the actual formula of a compound. • To determine the molecular formula you need to know the empirical formula and the molar mass of the compound.

  18. Examples • A 16.77 g sample of Ca and F contains 8.55 g Ca. What is the empirical formula? • Find the empirical formula of a compound with the following % composition: 26.7% P, 12.1% N, and 61.2% Cl. • For the last compound, if the molar mass is 695 g/mol, what is the molecular formula.

  19. Example • A 12.50 g sample of vitamin C contains 5.11 g C, .573 g H, and some O. If the molar mass is 176.14 g/mol, what is the molecular formula?

  20. Hydrate Pre-Lab 2.00 g of hydrated copper (II) sulfate is heated until all the water is gone. The mass of the dehydrated substance is 1.28 g. What is the formula of the hydrate? What is the mass percent of water?

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