General, Organic, and Biochemistry, 8e

1. General, Organic, and Biochemistry, 8e Bettelheim, Brown, Campbell, & Farrell

2. FormulaWeight • Formula weight: the sum of the atomic weights in atomic mass units (amu) of all atoms in a compound’s formula:

3. Formula Weight • Formula weight can be used for both ionic and molecular compounds; it tells nothing about whether a compound is ionic or molecular. • Molecular weight should be used only for molecular compounds. • In this text, we use formula weight for ionic compounds and molecular weight for molecular compounds.

4. The Mole • Mole (mol) • Mole; the amount of substance that contains as many atoms, molecules, or ions as are in exactly 12 g of carbon-12. • A mole, whether it is a mole of iron atoms, a mole of methane molecules, or a mole of sodium ions, always contains the same number of formula units. • The number of formula units in a mole is known as Avogadro’s number. • Avogadro’s number has been measured experimentally • Its value is 6.02214199 x 1023 formula units per mole.

5. The Mole • One Mole can equal three other measurements: • 1 Mole = 6.022 x 1023 atoms, molecules or formula units 1 Mole = the atomic mass (molar mass) expressed in grams 1 Mole = 22.4 L of a gas at standard temperature and pressure 1 Mole = 6.022x1023 units = molar mass = 22.4L

6. Avagadro’s Number • 1 Mole = 6.022x1023 units -- The unit depends on the type of substance you have. • Atom = a single element • Molecule = a molecular compound (covalent bonds), this includes the seven diatomic elements (H2, N2, O2, F2, Cl2, Br2, I2) • Formula unit = the simplest ratio of a compound, for an ionic compound this is the empirical formula, for a molecular compound it is the molecular formula (for a more detailed explanation see the next slide)

7. Avagadro’s Number • A formula unit is the exact make up of a molecule, or a group of bonded atoms, and is the smallest ratio of atoms in a compound, or group of loosely connected ions. The formula unit for any non-ionic molecule — one that has strong chemical bonds, like water — is called the molecular formula, which is H20 for water. In ionic compounds, no strong chemical bond is formed, and the formula units that are used to represent these compounds are called the empirical formula, or the smallest atomic ratio.

8. Molar Mass • Molar mass: the formula weight of a substance expressed in grams. Calculate by adding the atomic mass of all atoms (found on the periodic table). Don’t forget to multiply if there are subscripts! • Glucose, C6H12O6 • (C = 6 x 12.01) + (H = 12 x 1.01) + (O = 6 x 16.00) = • molecular weight: 180.18 amu • molar mass: 180.18 g/mol • one mole of glucose has a mass of 180.18 g RULE: always record two decimal places on molar mass!

9. Molar Volume • Molar Volume is 22.4 L of a gas at standard Temperature and Pressure (STP). (0 C and 1 atmosphere of pressure) This will hold true for any substance in its gas phase. So 1 mole of water (aka: steam) will fill 22.4L of space, and 1 mole of oxygen gas, O2 (g), will fill 22.4L of space. The key is to be sure the substance is a gas and the environment is at STP. Also, keep in mind the seven diatomics!

10. Dimensional Analysis with the Mole • There are four choices for creating the conversion factors necessary for all problems. • 1 mole 6.022x1023 units molar mass 22.4L • To create the conversion factors choose two of the four to use. In order to decide which two to use determine the amount given and the label needed in the word problem. Place the needed label over the given label to make the conversion factor.

11. Given moles and need atoms put these two together: • 1 mole 6.022x1023 units molar mass 22.4L • #moles given x 6.022x1023 atoms = #of atoms • 1 mole

12. Given atoms and need moles put these two together: • 1 mole 6.022x1023 units molar mass 22.4L • # of atoms given x 1 mole = #of moles • 6.022x1023 atoms

13. Given moles need grams put these two together: • 1 mole 6.022x1023 units molar mass 22.4L • #moles given x mass in grams = # of grams • 1 mole • (to find mass use periodic table)

14. Given grams need moles put these two together: • 1 mole 6.022x1023 units molar mass 22.4L • #of grams given x 1 mole = #of moles • mass in grams • (to find mass use periodic table)

15. Given moles need liters put these two together: • 1 mole 6.022x1023 units molar mass 22.4L • #moles given x 22.4 L = #of liters • 1 mole

16. Given liters need moles put these two together: • 1 mole 6.022x1023 units molar mass 22.4L • # of liters given x 1 mole = #of liters • 22.4 L

17. Molar Mass • We can use molar mass to convert from grams to moles, and from moles to grams • calculate the number of moles of water in 36.0 g water

18. Grams to Moles - Practice • Calculate the number of moles in 5.63 g of sodium sulfate, Na2SO4

19. Grams to Moles - Practice • Calculate the number of moles in 5.63 g of sodium sulfate, Na2SO4 • What is the given? Look for the term following “in”……5.63g of sodium sulfate is given. • What is needed? Look for the how much or how many term…..we are looking for moles. • Now choose the two terms that you will need to make the conversion factor. • 1 mole and molar mass remember needed over given so: 1 mole__ • molar mass

20. Grams to Moles - Practice • Calculate the number of moles in 5.63 g of sodium sulfate, Na2SO4 • the molar mass of Na2SO4 is: 2(23.0) + 32.1 + 4(16.0) = 142.1 amu • therefore, 1 mol of Na2SO4 = 142.1 g Na2SO4

21. Grams to Molecules • A tablet of aspirin, C9H8O4, contains 0.360 g of aspirin. How many molecules of aspirin are present?

22. Grams to Molecules • A tablet of aspirin, C9H8O4, contains 0.360 g of aspirin. How many molecules of aspirin are present? • What is the given? Look for the term following “in”……0.360 g of asprin sulfate is given. • What is needed? Look for the how much or how many term…..we are looking for molecules. • Now choose the two terms that you will need to make the conversion factor. • 6.022x1023 molecules and molar mass remember needed over given so: 6.022x1023 molecules__ • molar mass

23. Grams to Molecules • A tablet of aspirin, C9H8O4, contains 0.360 g of aspirin. How many molecules of aspirin are present? • the molar mass of asprin is 180.00g. 0.360 g aspirin x 6.022x1023 molecules = 1.20x1021 molecules 180.00g aspirin

24. Complete the practice in the packet: • Practice: Complete the following molar mass and volume questions. • (use conversion factors and unit cancellation) • grams in 2.5 moles of calcium • 2.5 moles x 40.08 grams = 100.2g  100 g (2 sig.figs) • 1 mole • grams in 4 moles of Al • 4 moles x 26.98g = 107.92g  100g (1 sig fig) • 1 mole • moles in 3.10 x 104 atoms of sulfur • 3.10 x 104atoms x 1 mole = 5.15 x 10-20moles (3 sf) • 6.022 x 1023atoms

25. Complete the practice in the packet: • grams in 6.0 L of oxygen gas • 6.0L x 32.00g = 8.6g (2 sf) **remember oxygen is diatomic O2 ** • 22.4 L • atoms in 35 grams of water • 35g x 6.022 x 1023molecules x 3 atoms = 3.8 x 1025 atoms • 18.02g 1 molecule • grams in 3.011 x 1016 atoms of zinc • 3.011 x 1016atoms x 65.38 g _____ = 3.269 x 10-6g • 6.022 x 1023atoms • liters in 0.500 moles of chlorine gas • 0.500 moles x 22.4 L = 11.2 L • 1 mole

26. PERCENT COMPOSITION: • Percent composition is an expression of content of each element in a compound in comparison to the total mass. • 1st determine the molar mass of compound. • 2nd determine the mass of the elements in the compound separately • 3rd divide the mass of the element by the total mass and multiply by 100% to get the percent. • Do this for each element.

27. PERCENT COMPOSITION: • Percent composition of water: H2O • 1st determine the molar mass of compound. (H 2 x 1.01g) + (O 1 x 16.00g) = 18.02 g • 2nd determine the mass of the elements in the compound separately. hydrogen accounts for 2.02g and oxygen for 16.00g • 3rd divide the mass of the element by the total mass and multiply by 100% to get the percent. • Hydrogen = 2.02 g x 100 = 11.2% hydrogen • 18.02g • Oxygen = 16.00 g x 100 = 88.8% oxygen • 18.02g

28. PERCENT COMPOSITION: • Determine the percent composition of sodium sulfate: Na2SO4 • 1st determine the molar mass of compound. • 2nd determine the mass of the elements in the compound separately. • 3rd divide the mass of the element by the total mass and multiply by 100% to get the percent.

29. PERCENT COMPOSITION: • Determine the percent composition of sodium sulfate: Na2SO4 • 1st determine the molar mass of compound. Na 2(23.00) + S 32.07 + O 4(16.00) = 142.07 grams • 2nd determine the mass of the elements in the compound separately. Na = 46.00g S 32.07g O 64.00g • 3rd divide the mass of the element by the total mass and multiply by 100% to get the percent. • Na = 46.00g x 100 = 32.38% sodium • 142.07g • S = 32.07g x 100 = 22.57% sulfur • 142.07g • O = 64.00g x 100 = 45.05% oxygen • 142.07g • Now complete the practice in the packet.

30. Complete the practice in the packet: • HCl = 36.46g • %H = (1.01/36.46) x 100 = 2.77% hydrogen • % Cl = (35.45/36.46) x 100 = 97.23% chlorine • AgNO3 = 169.88g • % Ag = (107.87/169.88) x 100 = 63.50% silver • % N = (14.01/169.88g) x 100 = 8.25% nitrogen • % O = (48.00/169.88g) x 100 = 28.26% oxygen • BaCrO4 = 253.33g • % Ba = (137.33/253.33) x 100 = 54.21% barium • % Cr = (52.00/253.33) x 100 = 20.53% chromium • % O = (64.00/253.33) x 100 = 25.26% oxygen

31. Complete the practice in the packet: • ZnSO4 • % Zn = • % S = • % O = • KClO4 • % K = • % Cl = • % O = • Fe(OH)3 • % Fe = • % O = • % H =

32. Homework: • Complete the homework pages that follow percent composition in your packet. (front and back  )