5 5-1 © 2006 Thomson Learning, Inc. All rights reserved Bettelheim, Brown, Campbell, & Farrell...

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

Transcript of 5 5-1 © 2006 Thomson Learning, Inc. All rights reserved Bettelheim, Brown, Campbell, & Farrell...

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

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5-1© 2006 Thomson Learning, Inc.All rights reserved

Bettelheim, Brown,Bettelheim, Brown,Campbell, & FarrellCampbell, & Farrell

General, Organic, General, Organic, and Biochemistry, 8eand Biochemistry, 8e

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FormulaFormula WeightWeight• Formula weightFormula weight: the sum of the atomic weights in atomic mass units (amu) of all atoms in a compound’s formula:

Ionic CompoundsSodium chloride (NaCl)23.0 amu Na + 35.5 amu Cl = 58.5 amu

Aspirin (C9H8O4) 9(12.0 amu C) + 8(1.0 amu H) + 4(16.0 amu O) = 180.0 amu

Water (H2O) 2(1.0 amu H) + 16.0 amu O = 18.0 amu

Nickel(II) chloride hydrate(NiCl2•6H2O)

58.7 amu Ni + 2(35.5 amu Cl) + 12(1.0) amu H) + 6(12.0 amu O) = 237.7 amu

Molecular Compounds

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Formula WeightFormula Weight• Formula weightFormula weight can be used for both ionic and molecular compounds; it tells nothing about whether a compound is ionic or molecular.

• Molecular weightMolecular weight should be used only for molecular compounds.

• In this text, we use formula weight for ionic compounds and molecular weight for molecular compounds.

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The MoleThe 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 6.02214199 x 1023 formula units per mole.

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The MoleThe 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

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Avagadro’s NumberAvagadro’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)

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Avagadro’s NumberAvagadro’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.

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Molar MassMolar Mass• Molar massMolar 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 gRULE: always record two decimal places on molar mass!

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Molar VolumeMolar 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!

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• 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.

Dimensional Analysis with the Dimensional Analysis with the MoleMole

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Given moles and need atoms Given moles and need atoms put these two together:put these two together:

• 1 mole 6.022x1023 units molar mass 22.4L

• #moles given x 6.022x1023 atoms = #of atoms

• 1 mole

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• 1 mole 6.022x1023 units molar mass 22.4L

• # of atoms given x 1 mole = #of moles

• 6.022x1023 atoms

Given atoms and need moles Given atoms and need moles put these two together:put these two together:

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Given moles need grams put Given moles need grams put these two together: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)

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Given grams need moles put Given grams need moles put these two together: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)

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Given moles need liters put Given moles need liters put these two together:these two together:

• 1 mole 6.022x1023 units molar mass 22.4L

• #moles given x 22.4 L = #of liters

• 1 mole

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Given liters need moles put Given liters need moles put these two together:these two together:

• 1 mole 6.022x1023 units molar mass 22.4L

• # of liters given x 1 mole = #of liters

• 22.4 L

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Molar MassMolar 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

Moles of AGrams of A

Use molar mass (g/mol)as the conversion factor

You are given one of theseand asked to find the other

36.0 g H2O1.00 mol H2O

18.0 g H2O= 2.00 mol H2Ox

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

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Grams to Moles - Grams to Moles - PracticePractice• 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

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Grams to Moles - Grams to Moles - PracticePractice• 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

5.63 g Na2SO4 x 1.00 mol Na2SO4

142.1 g Na2SO4= 0.0396 mol Na2SO4

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

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Grams to MoleculesGrams 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

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Grams to MoleculesGrams 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

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Complete the practice in the Complete the practice in the packet: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•  

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Complete the practice in the Complete the practice in the packet: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•  

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PERCENT COMPOSITION: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.

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PERCENT COMPOSITION: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

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PERCENT COMPOSITION: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.

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PERCENT COMPOSITION: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.

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Complete the practice in the Complete the practice in the packet: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

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Complete the practice in the Complete the practice in the packet:packet:

• ZnSO4

• % Zn = • % S =• % O =•  • KClO4

• % K =• % Cl =• % O =•  • Fe(OH)3

• % Fe =• % O =• % H =

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