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William L Masterton Cecile N. Hurley Edward J. Neth University of Connecticut Chapter 3 Mass Relations.

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Presentation on theme: "William L Masterton Cecile N. Hurley Edward J. Neth University of Connecticut Chapter 3 Mass Relations."— Presentation transcript:

1 William L Masterton Cecile N. Hurley http://academic.cengage.com/chemistry/masterton Edward J. Neth University of Connecticut Chapter 3 Mass Relations in Chemistry; Stoichiometry

2 3.1 Atomic Masses Atomic mass – (atomic weight) – The atomic mass of an element indicates how heavy, on average, an atom of an element is when compared to an atom of another element Atomic mass units – (amu) – the units for atomic masses on the periodic table

3 The Carbon-12 Scale Mass of one 12 C atom = 12 amu (exactly) Note that 12 C and C-12 mean the same thing

4 Atomic Masses and Isotopic Abundances: Mass spectrometer – a device used to experimentally determine the atomic mass of an atom Isotopic abundances – the percentage of each isotope that exists in nature (also, determined using the mass spec.)

5 Figure 3.1 – Mass Spectrometer A mass spectrometer is used to determine atomic masses

6 Chlorine (NOT IN NOTES) Two isotopes Cl-35 Cl-37

7 Figure 3.2 – Mass Spectrum of Cl (N in N) The area under the peak in the mass spectrogram gives the isotopic abundance

8 Atomic Mass Calculations:

9 Example 3.1

10 Masses of Individual Atoms; Avogadro’s Number: Avogadro’s Number – The number of atoms that is equal to the atomic mass of any element N A = 6.022 X 10 23 For Example: 6.022x10 23 H atoms in 1.008 grams of H (atomic mass of H = 1.008)

11 Figure 3.3 – One Mole of Several Substances

12 Example 3.2

13 3.2 The Mole Mole – equal to Avogadro’s Number, equal to 6.022x10 23 particles of a substance

14 Specialized units: The correct name for a particle of a substance based on the type of matter Atom – the representative particle for an element example – Fe, S, etc. ion – the representative particle for a charged particle example – Na +1, Cl -1, NH 4 +1,etc. Molecule – the representative particle for a molecular compound (made up of non-metals) example – CO 2, CH 4, etc. Formula unit – the representative particle for an ionic compound (metal and non-metal or polyatomic ion) example – KCl, MgSO 4, etc.

15 Molar mass: Molar mass – (MM) – the mass of 1 mole of a substance; equal to the atomic mass on the periodic table

16 Calculating molar mass: 1. find the mass of the element on the periodic table 2. multiply by the number of atoms of that element in the formula 3. sum the relative masses of the individual elements

17 Molar Masses of Some Substances

18 Practice: Example: Determine the molar masses of the following substances. AluminumCaSO 4 (NH 4 ) 3 P

19 The Significance of the Mole (Not in Notes) In the laboratory, substances are weighed on balances, in units of grams The mole allows us to relate the number of grams of a substance to the number of atoms or molecules of a substance

20 Mole-Gram Conversions Molar mass can be used like any other conversion factor: Molar mass (g) = 1 mole

21 Example 3.3

22 3.3 Mass Relations in Chemical Formulas Percent composition from formula percent composition - the number of grams of each element in 100 g of the compound Part x 100 = % composition whole

23 2 types of % Comp. problems: 1.Given data as masses of elements, etc. a.Use the masses of each element and the total mass of the compound

24 2 types of % Comp. problems: 2. Given the chemical formula of the compound a. use the relative molar masses of each element and the molar mass of the compound

25 Example 3.4

26 Subscripts: 1.Represent the atom ratio in a compound 2.Represent the mole ratio in a compound

27 Diatomic Elements: Elements that due to there chemical reactivity exist only as molecules of 2 atoms in nature. 7 diatomic elements: Br I N Cl H O F H O N Cl Br I F

28 Simplest Formula from Chemical Analysis (Empirical Formula): Simplest formula – (empirical formula) – the simplest whole number ratio of the atoms in a compound

29 Calculating the empirical formula: 1. can be determined from masses of the individual elements or the % composition of the elements in a compound 2. if %’s are given consider the sample to be of 100 grams and so the %’s become the masses in grams: 25.6% = 25.6 g 3. convert the mass of each element to moles 4. divide each number of moles by the smallest number of moles of all of the answers to #3 5. *If the answers to #4 are whole numbers, these are the subscripts in the empirical formula. * If any of the answers to #4 is not a whole number, convert all answers to a common fraction. Multiply each fraction by the denominator resulting in a whole number and these are the subscripts in the empirical formula.

30 Common Fractions: 0.250.330.500.660.75 ¼1/3½2/3¾

31 Example 3.5 – Simplest Formula from Masses of Elements

32 Example 3.6 – Simplest Formula from Mass Percents The compound that gives vinegar its sour taste is acetic acid, which contains the elements carbon, hydrogen, and oxygen. When 5.00g of acetic acid is analyzed it is found to contain 2.00g of carbon, 0.336g of hydrogen, and 2.66g of oxygen. What is the empirical formula of acetic acid?

33 Molecular Formula: Molecular formula – the true ratio of the atoms in a compound as it exists naturally; may be the same as the empirical formula; will relate to the empirical formula in whole number ratios

34 Molecular Formula: Calculating the molecular formula: 1. find the molar mass of the empirical formula 2. divide the molecular molar mass (given in the question) by the empirical molar mass MMM EMM 3. multiply each subscript in the empirical formula by the answer to #2, these are the subscripts for the molecular formula

35 Example 3.7

36 3.4 Mass Relations in Reactions: Reactants – the starting substances in a chemical reaction; found on the left-side Products – the ending substances in a chemical reaction; found on the right-side Yield – the arrow written to separate the reactants and products

37 3.4 Mass Relations in Reactions: Subscripts – indicate the number of atoms of each element or ion in a chemical formula Coefficients – the numbers written in front of substances in a chemical equation to indicate the number of “particles” of that substance

38 States of matter Symbols indicating state of condition of a substance in the reaction Aqueous – (aq) a substance dissolved in water Precipitate – (ppt. or ) a solid substance formed from the reaction of aqueous substances Solid – (s) Liquid – (l) Gas – (g)

39 2. Types of equations a. Synthesis or combination – when 2 or more reactants react to form 1 product

40 2. Types of equations b. Decomposition – when 1 reactant reacts to form 2 or more products

41 2. Types of equations c. Single replacement or single displacement – when one element and one compound react and the element replaces one of the ions in the compound to form a new compound and different single element

42 2. Types of equations d. Double replacement or double displacement – when 2 compounds react and switch ions to form 2 new compounds

43 2. Types of equations e. Combustion – “burning”; when a hydrocarbon reacts with oxygen to form carbon dioxide and water

44 Mole ratios or stoichiometric factors: Equations are balanced in order to obey the Law of Conservation of Mass. As a result, there are mathematical relationships between the substances in a balanced chemical equation which are the foundation of stoichiometry.

45 Example: CS 2 + 3O 2  CO 2 + 2 SO 2

46 a. Interpretation in terms of moles 1 mole of carbon disulfide to 3 moles of oxygen to 1 moles of carbon dioxide to 2 moles of sulfur dioxide

47 b. Conversion factors extracted from balanced equation 1 mole CS 2 3 moles O 2 3 moles of O 2 1 mole CO 2 2 moles SO 2

48 c. For an equation to be balanced… a. sum all atoms of each type on a side, even if an element is in more than one substance b. work from left to right to stay organized c. if a polyatomic ion is present in the same form on both sides it can be counted as a unit rather than as individual elements d. look to balance H’s and O’s last in more difficult equations e. combustion reactions should always be balanced in the order of carbon, hydrogen, oxygens on the right and then the left – and use the doubling rule when you have an odd number of oxygens

49 Example 3.8


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