5 5-1 © 2006 Thomson Learning, Inc. All rights reserved Bettelheim, Brown, Campbell, & Farrell General, Organic, and Biochemistry, 8e.

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Presentation transcript:

5 5-1 © 2006 Thomson Learning, Inc. All rights reserved Bettelheim, Brown, Campbell, & Farrell General, Organic, and Biochemistry, 8e

5 5-2 © 2006 Thomson Learning, Inc. All rights reserved Formula Weight Formula weight Formula weight: the sum of the atomic weights in atomic mass units (amu) of all atoms in a compound’s formula:

5 5-3 © 2006 Thomson Learning, Inc. All rights reserved Formula Weight 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 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.

5 5-4 © 2006 Thomson Learning, Inc. All rights reserved The Mole Mole (mol) MoleMole; 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 x formula units per mole.

5 5-5 © 2006 Thomson Learning, Inc. All rights reserved The Mole One Mole can equal three other measurements: 1 Mole = x 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 = 6.022x10 23 units = molar mass = 22.4L

5 5-6 © 2006 Thomson Learning, Inc. All rights reserved Avagadro’s Number 1 Mole = 6.022x10 23 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 (H 2, N 2, O 2, F 2, Cl 2, Br 2, I 2 ) 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)

5 5-7 © 2006 Thomson Learning, Inc. All rights reserved 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 H 2 0 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.

5 5-8 © 2006 Thomson Learning, Inc. All rights reserved Molar Mass 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, C 6 H 12 O 6 (C = 6 x 12.01) + (H = 12 x 1.01) + (O = 6 x 16.00) = molecular weight: amu molar mass: g/mol one mole of glucose has a mass of g RULE: always record two decimal places on molar mass!

5 5-9 © 2006 Thomson Learning, Inc. All rights reserved 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, O 2 (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!

© 2006 Thomson Learning, Inc. All rights reserved There are four choices for creating the conversion factors necessary for all problems. 1 mole 6.022x10 23 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. Dimensional Analysis with the Mole

© 2006 Thomson Learning, Inc. All rights reserved Given moles and need atoms put these two together: 1 mole 6.022x10 23 units molar mass 22.4L #moles given x 6.022x10 23 atoms = #of atoms 1 mole

© 2006 Thomson Learning, Inc. All rights reserved 1 mole 6.022x10 23 units molar mass 22.4L # of atoms given x 1 mole = #of moles 6.022x10 23 atoms Given atoms and need moles put these two together:

© 2006 Thomson Learning, Inc. All rights reserved Given moles need grams put these two together: 1 mole 6.022x10 23 units molar mass 22.4L #moles given x mass in grams = # of grams 1 mole (to find mass use periodic table)

© 2006 Thomson Learning, Inc. All rights reserved Given grams need moles put these two together: 1 mole 6.022x10 23 units molar mass 22.4L # of grams given x 1 mole = #of moles mass in grams (to find mass use periodic table)

© 2006 Thomson Learning, Inc. All rights reserved Given moles need liters put these two together: 1 mole 6.022x10 23 units molar mass 22.4L #moles given x 22.4 L = #of liters 1 mole

© 2006 Thomson Learning, Inc. All rights reserved Given liters need moles put these two together: 1 mole 6.022x10 23 units molar mass 22.4L # of liters given x 1 mole = #of liters 22.4 L

© 2006 Thomson Learning, Inc. All rights reserved 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

© 2006 Thomson Learning, Inc. All rights reserved Grams to Moles - Practice Calculate the number of moles in 5.63 g of sodium sulfate, Na 2 SO 4

© 2006 Thomson Learning, Inc. All rights reserved Grams to Moles - Practice Calculate the number of moles in 5.63 g of sodium sulfate, Na 2 SO 4 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

© 2006 Thomson Learning, Inc. All rights reserved Grams to Moles - Practice Calculate the number of moles in 5.63 g of sodium sulfate, Na 2 SO 4 the molar mass of Na 2 SO 4 is: 2(23.0) (16.0) = amu therefore, 1 mol of Na 2 SO 4 = g Na 2 SO 4

© 2006 Thomson Learning, Inc. All rights reserved Grams to Molecules A tablet of aspirin, C 9 H 8 O 4, contains g of aspirin. How many molecules of aspirin are present?

© 2006 Thomson Learning, Inc. All rights reserved Grams to Molecules A tablet of aspirin, C 9 H 8 O 4, contains 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 x10 23 molecules and molar mass remember needed over given so: 6.022x10 23 molecules__ molar mass

© 2006 Thomson Learning, Inc. All rights reserved Grams to Molecules A tablet of aspirin, C 9 H 8 O 4, contains g of aspirin. How many molecules of aspirin are present? the molar mass of asprin is g g aspirin x 6.022x10 23 molecules = 1.20x10 21 molecules g aspirin

© 2006 Thomson Learning, Inc. All rights reserved 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 grams = 100.2g  100 g (2 sig.figs) 1 mole grams in 4 moles of Al 4 moles x 26.98g = g  100g (1 sig fig) 1 mole moles in 3.10 x 10 4 atoms of sulfur 3.10 x 10 4 atoms x 1 mole = 5.15 x moles (3 sf) x atoms

© 2006 Thomson Learning, Inc. All rights reserved 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 O 2 ** 22.4 L atoms in 35 grams of water 35g x x molecules x 3 atoms = 3.8 x atoms 18.02g 1 molecule grams in x atoms of zinc x atoms x g _____ = x g x atoms liters in moles of chlorine gas moles x 22.4 L = 11.2 L 1 mole

© 2006 Thomson Learning, Inc. All rights reserved PERCENT COMPOSITION: Percent composition is an expression of content of each element in a compound in comparison to the total mass. 1 st determine the molar mass of compound. 2 nd determine the mass of the elements in the compound separately 3 rd divide the mass of the element by the total mass and multiply by 100% to get the percent. Do this for each element.

© 2006 Thomson Learning, Inc. All rights reserved PERCENT COMPOSITION: Percent composition of water: H 2 O 1 st determine the molar mass of compound. (H 2 x 1.01g) + (O 1 x 16.00g) = g 2 nd determine the mass of the elements in the compound separately. hydrogen accounts for 2.02g and oxygen for 16.00g 3 rd 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 = g x 100 = 88.8% oxygen 18.02g

© 2006 Thomson Learning, Inc. All rights reserved PERCENT COMPOSITION: Determine the percent composition of sodium sulfate: Na 2 SO 4 1 st determine the molar mass of compound. 2 nd determine the mass of the elements in the compound separately. 3 rd divide the mass of the element by the total mass and multiply by 100% to get the percent.

© 2006 Thomson Learning, Inc. All rights reserved PERCENT COMPOSITION: Determine the percent composition of sodium sulfate: Na 2 SO 4 1 st determine the molar mass of compound. Na 2(23.00) + S O 4(16.00) = grams 2 nd determine the mass of the elements in the compound separately. Na = 46.00g S 32.07g O 64.00g 3 rd 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 g S = 32.07g x 100 = 22.57% sulfur g O = 64.00g x 100 = 45.05% oxygen g Now complete the practice in the packet.

© 2006 Thomson Learning, Inc. All rights reserved 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 AgNO 3 = g % 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 BaCrO 4 = g % 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

© 2006 Thomson Learning, Inc. All rights reserved Complete the practice in the packet: ZnSO 4 % Zn = % S = % O = KClO 4 % K = % Cl = % O = Fe(OH) 3 % Fe = % O = % H =

© 2006 Thomson Learning, Inc. All rights reserved Homework: Complete the homework pages that follow percent composition in your packet. (front and back )