Chemical Formula Relationships Chapter 7 Chemical Formula Relationships Presentation Slides to Accompany Cracolice/Peters Introductory Chemistry: An Active Learning Approach, Third Edition Copyright © 2007 Brooks/Cole, a part of the Thomson Corporation.

Section 7.1 The Number of Atoms in a Formula Presentation Slides to Accompany Cracolice/Peters Introductory Chemistry: An Active Learning Approach, Third Edition Copyright © 2007 Brooks/Cole, a part of the Thomson Corporation.

Goal 1 Given the formula of a chemical compound (or a name from which the formula may be written), state the number of atoms of each element in the formula unit. Presentation Slides to Accompany Cracolice/Peters Introductory Chemistry: An Active Learning Approach, Third Edition Copyright © 2007 Brooks/Cole, a part of the Thomson Corporation.

In writing the formula of a substance, subscript numbers are used to indicate the number of atoms or groups of atoms of each element in the formula unit However, subscript numbers of 1 are omitted Presentation Slides to Accompany Cracolice/Peters Introductory Chemistry: An Active Learning Approach, Third Edition Copyright © 2007 Brooks/Cole, a part of the Thomson Corporation.

ammonium carbonate: (NH4)2CO3 2 nitrogen atoms 8 hydrogen atoms 1 carbon atom 3 oxygen atoms Presentation Slides to Accompany Cracolice/Peters Introductory Chemistry: An Active Learning Approach, Third Edition Copyright © 2007 Brooks/Cole, a part of the Thomson Corporation.

Section 7.2 Molecular Mass and Formula Mass Presentation Slides to Accompany Cracolice/Peters Introductory Chemistry: An Active Learning Approach, Third Edition Copyright © 2007 Brooks/Cole, a part of the Thomson Corporation.

Goal 2 Distinguish among atomic mass, molecular mass, and formula mass. Presentation Slides to Accompany Cracolice/Peters Introductory Chemistry: An Active Learning Approach, Third Edition Copyright © 2007 Brooks/Cole, a part of the Thomson Corporation.

Goal 3 Calculate the formula (molecular) mass of any compound whose formula is given (or known). Presentation Slides to Accompany Cracolice/Peters Introductory Chemistry: An Active Learning Approach, Third Edition Copyright © 2007 Brooks/Cole, a part of the Thomson Corporation.

The average mass of all atoms of an element Atomic mass: The average mass of all atoms of an element as they occur in nature. Based on 12C at a mass of 12 amu, and weighted by natural isotopic distribution. (Section 5.5) Presentation Slides to Accompany Cracolice/Peters Introductory Chemistry: An Active Learning Approach, Third Edition Copyright © 2007 Brooks/Cole, a part of the Thomson Corporation.

The sum of the atomic masses of the atoms that make up the molecule. Molecular mass: The sum of the atomic masses of the atoms that make up the molecule. Presentation Slides to Accompany Cracolice/Peters Introductory Chemistry: An Active Learning Approach, Third Edition Copyright © 2007 Brooks/Cole, a part of the Thomson Corporation. Presentation Slides to Accompany Cracolice/Peters Introductory Chemistry: An Active Learning Approach, Second Edition Copyright © 2004 Brooks/Cole, a division of Thomson Learning, Inc.

The mass of one formula unit of any compound. Formula mass: The mass of one formula unit of any compound. Term must be used for ionic compounds; may be used for any compound. Formula mass = Sum of the atomic masses in the formula unit Presentation Slides to Accompany Cracolice/Peters Introductory Chemistry: An Active Learning Approach, Third Edition Copyright © 2007 Brooks/Cole, a part of the Thomson Corporation.

What is the formula mass of calcium phosphate? Example: What is the formula mass of calcium phosphate? Phosphate ion: PO43– Calcium ion: Ca2+ Calcium phosphate: Ca3(PO4)2 3 x 40.08 amu = 120.24 amu Ca 2 x 30.97 amu = 61.94 amu P O 8 x 16.00 amu = 128.00 amu 310.18 amu Ca3(PO4)2 Presentation Slides to Accompany Cracolice/Peters Introductory Chemistry: An Active Learning Approach, Third Edition Copyright © 2007 Brooks/Cole, a part of the Thomson Corporation.

Section 7.3 The Mole Concept Presentation Slides to Accompany Cracolice/Peters Introductory Chemistry: An Active Learning Approach, Third Edition Copyright © 2007 Brooks/Cole, a part of the Thomson Corporation.

Goal 4 Define the term mole. Identify the number of objects that corresponds to one mole. Presentation Slides to Accompany Cracolice/Peters Introductory Chemistry: An Active Learning Approach, Third Edition Copyright © 2007 Brooks/Cole, a part of the Thomson Corporation.

Goal 5 Given the number of moles (or units) in any sample, calculate the number of units (or moles) in the sample. Presentation Slides to Accompany Cracolice/Peters Introductory Chemistry: An Active Learning Approach, Third Edition Copyright © 2007 Brooks/Cole, a part of the Thomson Corporation.

of atoms in exactly 12 grams of carbon-12. Mole (mol): One mole is the amount of any substance that contains the same number of units as the number of atoms in exactly 12 grams of carbon-12. The definition of a mole refers to a number of particles, but it doesn’t say what that number is. Experiments must be conducted to find the number of particles in a mole. Presentation Slides to Accompany Cracolice/Peters Introductory Chemistry: An Active Learning Approach, Third Edition Copyright © 2007 Brooks/Cole, a part of the Thomson Corporation.

Avogadro’s Number (N): The experimentally-determined number of atoms in a mole. 6.02214199 X 1023/mol (1998 value) Analogy: 12 items dozen 6.02 X 1023 items mole Presentation Slides to Accompany Cracolice/Peters Introductory Chemistry: An Active Learning Approach, Third Edition Copyright © 2007 Brooks/Cole, a part of the Thomson Corporation.

How many carbon dioxide molecules are in 2.0 moles of carbon dioxide? Example: How many carbon dioxide molecules are in 2.0 moles of carbon dioxide? 6.02 x 1023 molecules CO2 mol CO2 = 2.0 mol CO2 x 1.2 x 1024 molecules CO2 Presentation Slides to Accompany Cracolice/Peters Introductory Chemistry: An Active Learning Approach, Third Edition Copyright © 2007 Brooks/Cole, a part of the Thomson Corporation.

Section 7.4 Molar Mass Presentation Slides to Accompany Cracolice/Peters Introductory Chemistry: An Active Learning Approach, Third Edition Copyright © 2007 Brooks/Cole, a part of the Thomson Corporation.

Goal 6 Define molar mass or interpret statements in which the term molar mass is used. Presentation Slides to Accompany Cracolice/Peters Introductory Chemistry: An Active Learning Approach, Third Edition Copyright © 2007 Brooks/Cole, a part of the Thomson Corporation.

Goal 7 Calculate the molar mass of any substance whose chemical formula is given (or known). Presentation Slides to Accompany Cracolice/Peters Introductory Chemistry: An Active Learning Approach, Third Edition Copyright © 2007 Brooks/Cole, a part of the Thomson Corporation.

The mass in grams of one mole of a substance. Units: g/mol Molar mass: The mass in grams of one mole of a substance. Units: g/mol Molar mass is a bridge between a macroscopic quantity of matter, grams, and a particulate-level quantity of matter, moles Chemists count particles by weighing Presentation Slides to Accompany Cracolice/Peters Introductory Chemistry: An Active Learning Approach, Third Edition Copyright © 2007 Brooks/Cole, a part of the Thomson Corporation.

1 amu = 1/12 the mass of a carbon-12 atom 12 X X 12 1 amu = 1/12 the mass of a carbon-12 atom 12 amu = the mass of a carbon-12 atom 1 mol = # atoms in 12 g carbon-12 12 g = the mass of 1 mol of carbon-12 atoms 12 amu/atom 12C 12 g/mol 12C An atomic, molecular, or formula mass in amu is numerically equal to molar mass in g/mol Presentation Slides to Accompany Cracolice/Peters Introductory Chemistry: An Active Learning Approach, Third Edition Copyright © 2007 Brooks/Cole, a part of the Thomson Corporation.

1 mole of... C Cu Zn S8 Hg Br2 Ni Ca I2 Presentation Slides to Accompany Cracolice/Peters Introductory Chemistry: An Active Learning Approach, Third Edition Copyright © 2007 Brooks/Cole, a part of the Thomson Corporation. Presentation Slides to Accompany Cracolice/Peters Introductory Chemistry: An Active Learning Approach, Third Edition Copyright © 2007 Brooks/Cole, a part of the Thomson Corporation.

1 mole of ... CuSO4 Hg2I2 ZnS CaBr2 NiCO3 CuBr2 CuS Presentation Slides to Accompany Cracolice/Peters Introductory Chemistry: An Active Learning Approach, Third Edition Copyright © 2007 Brooks/Cole, a part of the Thomson Corporation.

What is the mass of an ammonia molecule? Example: What is the mass of an ammonia molecule? What is the mass of a mole of ammonia molecules? Ammonia is NH3 14.01 amu N + 3(1.008 amu H) = 17.03 amu NH3 An ammonia molecule has a mass of 17.03 amu. To change from molecular mass to molar mass, change the units from amu to g/mol. A mole of ammonia has a mass of 17.03 g/mol. Presentation Slides to Accompany Cracolice/Peters Introductory Chemistry: An Active Learning Approach, Third Edition Copyright © 2007 Brooks/Cole, a part of the Thomson Corporation.

Section 7.5 Conversion among Mass, Number of Moles, and Number of Units Presentation Slides to Accompany Cracolice/Peters Introductory Chemistry: An Active Learning Approach, Third Edition Copyright © 2007 Brooks/Cole, a part of the Thomson Corporation.

Goal 8 Given any one of the following for a substance whose formula is given (or known), calculate the other two: (a) mass, (b) number of atoms, (c) number of formula units, molecules, or atoms. Presentation Slides to Accompany Cracolice/Peters Introductory Chemistry: An Active Learning Approach, Third Edition Copyright © 2007 Brooks/Cole, a part of the Thomson Corporation.

Number of atoms, 6.02 X 1023 particles molecules, mole formula units Moles # grams mole Grams Presentation Slides to Accompany Cracolice/Peters Introductory Chemistry: An Active Learning Approach, Third Edition Copyright © 2007 Brooks/Cole, a part of the Thomson Corporation.

How many hydrogen atoms are in Example: How many hydrogen atoms are in a 1.0-L sample of water? 1000 mL H2O L H2O 1.00 g H2O mL H2O x 1.0 L H2O x 1 mol H2O 18.02 g H2O x 6.02 x 1023 molecules H2O mol H2O x 2 atoms H molecule H2O x = 6.7 x 1025 atoms H Presentation Slides to Accompany Cracolice/Peters Introductory Chemistry: An Active Learning Approach, Third Edition Copyright © 2007 Brooks/Cole, a part of the Thomson Corporation.

Section 7.6 Mass Relationships among Elements in a Compound: Percentage Composition Presentation Slides to Accompany Cracolice/Peters Introductory Chemistry: An Active Learning Approach, Third Edition Copyright © 2007 Brooks/Cole, a part of the Thomson Corporation.

Goal 9 Calculate the percentage composition of any compound whose formula is given (or known). Presentation Slides to Accompany Cracolice/Peters Introductory Chemistry: An Active Learning Approach, Third Edition Copyright © 2007 Brooks/Cole, a part of the Thomson Corporation.

Goal 10 Given the mass of a sample of any compound whose formula is given (or known), calculate the mass of any element in the sample; or, given the mass of any element in the sample, calculate the mass of the sample or the mass of any other element in the sample. Presentation Slides to Accompany Cracolice/Peters Introductory Chemistry: An Active Learning Approach, Third Edition Copyright © 2007 Brooks/Cole, a part of the Thomson Corporation.

The percentage composition of a compound is the percentage by mass of % of A = parts of A X 100 total parts The percentage composition of a compound is the percentage by mass of each element in the compound. Presentation Slides to Accompany Cracolice/Peters Introductory Chemistry: An Active Learning Approach, Third Edition Copyright © 2007 Brooks/Cole, a part of the Thomson Corporation.

Determine the percentage composition of sugar, C12H22O11 Example: Determine the percentage composition of sugar, C12H22O11 In one mole of C12H22O11: 12(12.01 g C) + 22(1.008 g H) + 11(16.00 g O) = 342.30 g C12H22O11 Presentation Slides to Accompany Cracolice/Peters Introductory Chemistry: An Active Learning Approach, Third Edition Copyright © 2007 Brooks/Cole, a part of the Thomson Corporation.

. 12(12.01 g C) X 100 = 42.10% C 342.30 g C12H22O11 . 22(1.008 g H) X 100 = 6.479% H 342.30 g C12H22O11 . 11(16.00 g O) X 100 = 51.42% O 342.30 g C12H22O11 Check: 42.10% + 6.479% + 51.42% = 100.00% Presentation Slides to Accompany Cracolice/Peters Introductory Chemistry: An Active Learning Approach, Third Edition Copyright © 2007 Brooks/Cole, a part of the Thomson Corporation.

In one mole of sugar: 12(12.01 g C) + 22(1.008 g H) + 11(16.00 g O) = 342.30 g C12H22O11 12(12.01 g C) PER 342.30 g C12H22O11 12(12.01 g C) PER 22(1.008 g H) and so on Presentation Slides to Accompany Cracolice/Peters Introductory Chemistry: An Active Learning Approach, Third Edition Copyright © 2007 Brooks/Cole, a part of the Thomson Corporation.

How many grams of sugar are needed Example: How many grams of sugar are needed to obtain 10.0 g carbon? 342.30 g C12H22O11 12(12.01 g C) 10.0 g C x = 23.8 g C12H22O11 Presentation Slides to Accompany Cracolice/Peters Introductory Chemistry: An Active Learning Approach, Third Edition Copyright © 2007 Brooks/Cole, a part of the Thomson Corporation.

Section 7.7 Empirical Formula of a Compound Presentation Slides to Accompany Cracolice/Peters Introductory Chemistry: An Active Learning Approach, Third Edition Copyright © 2007 Brooks/Cole, a part of the Thomson Corporation.

Goal 11 Distinguish between an empirical formula and a molecular formula. Presentation Slides to Accompany Cracolice/Peters Introductory Chemistry: An Active Learning Approach, Third Edition Copyright © 2007 Brooks/Cole, a part of the Thomson Corporation.

Goal 12 Given data from which the mass of each element in a sample of a compound can be determined, find the empirical formula of the compound. Presentation Slides to Accompany Cracolice/Peters Introductory Chemistry: An Active Learning Approach, Third Edition Copyright © 2007 Brooks/Cole, a part of the Thomson Corporation.

The lowest whole-number ratio of atoms of the elements in a compound Empirical formula: The lowest whole-number ratio of atoms of the elements in a compound The empirical formula of C2H4 is CH2 Likewise, the empirical formula of C3H6 is CH2 All compounds with the general formula CnH2n have the same empirical formula and the same percentage composition Presentation Slides to Accompany Cracolice/Peters Introductory Chemistry: An Active Learning Approach, Third Edition Copyright © 2007 Brooks/Cole, a part of the Thomson Corporation.

Write the empirical formula of benzene, C6H6, and of octane, C8H18. Example: Write the empirical formula of benzene, C6H6, and of octane, C8H18. C6H6: Both divisible by 6 EF: CH C8H18: Both divisible by 2 EF: C4H9 Presentation Slides to Accompany Cracolice/Peters Introductory Chemistry: An Active Learning Approach, Third Edition Copyright © 2007 Brooks/Cole, a part of the Thomson Corporation.

How to Find an Empirical Formula 1. Find the masses of different elements in a sample of the compound 2. Convert the masses into moles of atoms of the different elements 3. Express the moles of atoms as the smallest possible ratio of integers 4. Write the empirical formula, using the number for each atom in the integer ratio as the subscript in the formula Presentation Slides to Accompany Cracolice/Peters Introductory Chemistry: An Active Learning Approach, Third Edition Copyright © 2007 Brooks/Cole, a part of the Thomson Corporation.

What is the empirical formula of a compound that analyzes as Example: What is the empirical formula of a compound that analyzes as 79.95% carbon, 9.40% hydrogen, and 10.65% oxygen? Many empirical formula questions can be solved with the following table: Element Grams Moles Mole Ratio Formula Empirical Presentation Slides to Accompany Cracolice/Peters Introductory Chemistry: An Active Learning Approach, Third Edition Copyright © 2007 Brooks/Cole, a part of the Thomson Corporation.

Mole Ratio Formula Ratio Element Grams Moles C H O 79.95 9.40 10.65 10 Empirical Formula Element Grams Moles 79.95 g C 12.01 g/mol C 9.40 g H 1.008 g/mol H 10.65 g O 16.00 g O 6.657 0.6656 9.33 C H O 79.95 9.40 10.65 10 14 1 6.657 9.33 0.6656 10.00 14.0 1.000 C10H14O Presentation Slides to Accompany Cracolice/Peters Introductory Chemistry: An Active Learning Approach, Third Edition Copyright © 2007 Brooks/Cole, a part of the Thomson Corporation.

Section 7.8 Determination of a Molecular Formula Presentation Slides to Accompany Cracolice/Peters Introductory Chemistry: An Active Learning Approach, Third Edition Copyright © 2007 Brooks/Cole, a part of the Thomson Corporation.

Goal 13 Given the molar mass and empirical formula of a compound, or information from which they can be found, determine the molecular formula of the compound. Presentation Slides to Accompany Cracolice/Peters Introductory Chemistry: An Active Learning Approach, Third Edition Copyright © 2007 Brooks/Cole, a part of the Thomson Corporation.

The molecular formula of a compound is found by determination of the number of empirical formula units in the molecule For example, if the empirical formula of a compound is CH2, the molecular formula is CH2 or C2H4 or C3H6 or C4H8, etc Presentation Slides to Accompany Cracolice/Peters Introductory Chemistry: An Active Learning Approach, Third Edition Copyright © 2007 Brooks/Cole, a part of the Thomson Corporation.

Empirical formula units in one molecule: molar mass of compound molar mass of empirical formula Presentation Slides to Accompany Cracolice/Peters Introductory Chemistry: An Active Learning Approach, Third Edition Copyright © 2007 Brooks/Cole, a part of the Thomson Corporation.

How to Find the Molecular Formula of a Compound 1. Determine the empirical formula of the compound 2. Calculate the molar mass of the empirical formula unit 3. Determine the molar mass of the compound (which will be given at this time) 4. Divide the molar mass of the compound by the molar mass of the empirical formula unit to get n, the number of empirical formula units per molecule 5. Write the molecular formula Presentation Slides to Accompany Cracolice/Peters Introductory Chemistry: An Active Learning Approach, Third Edition Copyright © 2007 Brooks/Cole, a part of the Thomson Corporation.

What is the molecular formula of a compound with the empirical formula Example: What is the molecular formula of a compound with the empirical formula C5H10O and a molar mass of 258 g/mol? Molar mass of C5H10O: 5(12.01 g/mol C) + 10(1.008 g/mol H) + 16.00 g/mol O = 86.13 g/mol C5H10O 258 g/mol molecule 86.13 g/mol ef unit = 3 (C5H10O)3 C15H30O3 Presentation Slides to Accompany Cracolice/Peters Introductory Chemistry: An Active Learning Approach, Third Edition Copyright © 2007 Brooks/Cole, a part of the Thomson Corporation.