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Chapter 6 Chemical Composition 2006, Prentice Hall

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2 CHAPTER OUTLINE The Mole Concept The Mole Concept Molecular & Molar Mass Calculations Using the Mole Percent Composition Chemical Formulas Calculating Empirical Formulas Molecular Formulas

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Tro's Introductory Chemistry, Chapter 6 Why is Knowledge of Composition Important? everything in nature is either chemically or physically combined with other substances to know the amount of a material in a sample, you need to know what fraction of the sample it is Some Applications: –the amount of sodium in sodium chloride for diet –the amount of iron in iron ore for steel production –the amount of hydrogen in water for hydrogen fuel –the amount of chlorine in freon to estimate ozone depletion

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4 THE MOLE CONCEPT Chemists find it more convenient to use mass relationships in the laboratory, while chemical reactions depend on the number of atoms present. In order to relate the mass and number of atoms, chemists use the SI unit mole (abbreviated mol).

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5 THE MOLE CONCEPT The number of particles in a mole is called Avogadro’s number and is 6.02x10 23. 1 mole equals to 6.02 x 10 23 Avogadro’s number (N A )

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6 THE MOLE CONCEPT A mole is a very large quantity 6.02x10 23 If 10,000 people started to count Avogadro’s number and counted at the rate of 100 numbers per minute each minute of the day, it would take over 1 trillion years to count the total number. MOLE

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7 THE MOLE CONCEPT 1 mole H 2 molecules= 6.02x10 23 H 2 molecules = 2 x (6.02x10 23 ) H atoms 1 mole Na + ions 1 mole H 2 O molecules= 6.02x10 23 H 2 O molecules = 2 x (6.02x10 23 ) H atoms = 6.02x10 23 O atoms = 6.02x10 23 Na + ions 1 mole H atoms= 6.02x10 23 H atoms

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8 THE MOLE CONCEPT The atomic mass of one atom expressed in amu is numerically the same as the mass of 1 mole of atoms of the element expressed in grams. Mass of 1 H atom = 1.008 amu Mass of 1 mol H atoms = 1.008 grams Mass of 1 Mg atom = 24.31 amu Mass of 1 mol Mg atoms = 24.31 g Mass of 1 Cl atom = 35.45 amu Mass of 1 mol Cl atoms = 35.45 g

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9 MOLE DAY Chemists and chemistry students celebrate two days in the year in honor of the Mole and call them Mole Days. October 23 rd Jun 2 nd 6:02 a.m. 10:23 a.m.

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10 MOLECULAR MASS The sum of atomic masses of all the atoms in one molecule of a substance is called molecular mass, and is measured in amu. 1 O atom = 2 H atoms = Mass of one molecule of H 2 O 2 (1.008 amu) = 2.016 amu 1 (16.00 amu) = 16.00 amu 18.02 amu Molecular mass

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Relationship Between Moles and Mass The mass of one mole of atoms is called the molar mass The molar mass of an element, in grams, is numerically equal to the element’s atomic mass, in amu

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12 MOLAR MASS 1 mol O atom = 2 mol H atoms = Mass of one mole of H 2 O 2 (1.008 g) = 2.016 g 1 (16.00 g) = 16.00 g 18.02 g Molar mass The mass of one mole of a substance is called molar mass, and is measured in grams.

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13 CALCULATIONS USING THE MOLE Conversions between mass, mole and particles can be done using molar mass and Avogadro’s number. Mass of a substance Moles of a substance MM Particles of a substance NANA Molar mass Avogadro’s number

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14 What is the mass of 5.00 mol of water? Example 1: 3 significant figures 1 18.02 90.1 Molar mass

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15 How many Mg atoms are present in 5.00 g of Mg? Example 2: 1 24.3 mass mol atoms Avogadro’s number 1 6.02 x 10 23 1.24x10 23 atoms Mg Molar mass 3 significant figures

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16 How many molecules of HCl are present in 25.0 g of HCl? Example 3: 1 36.45 mass mol molecules 1 6.02 x 10 23 4.13 x 10 23 molecules HCl 3 significant figures

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17 PERCENT COMPOSITION The percent composition of a compound is the mass percent of each element in the compound. Total mass of element Total mass of compound

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18 Calculate the percent composition of sodium chloride (NaCl). Example 1: Step 1:determine molar mass of NaCl 1 mol Na atoms = 1 (22.99 g) = 22.99 g 1 mol Cl atom = 1 (35.45 g) = 35.45 g 58.44 g/mol Molar mass

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19 Example 1: Step 2:calculate the mass % of each element % Na = % Cl = Sum = 100%

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20 1.63 g of zinc combines with 0.40 g of oxygen to form zinc oxide. Determine the % composition of the compound formed. Example 2: Step 1:determine total mass of sample 1.63 g + 0.40 g = 2.03 g mass of sample =

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21 Example 2: Step 2:calculate the mass % of each element % Zn = % O = Sum = 100%

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22 Calculate the percent composition of sodium hydroxide (NaOH). Example 3: Step 1:determine molar mass of NaOH 1 mol Na atoms =1 (23.0 g) = 23.0 g 1 mol O atom = 1 (16.0 g) = 16.0 g 1 mol H atoms = 1 (1.01 g) = 1.01 g 40.0 g/mol Molar mass

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23 Calculate the percent composition of sodium hydroxide (NaOH). Example 3: Step 1:determine molar mass of NaOH 1 mol Na atoms =1 (23.0 g) = 23.0 g 1 mol O atom = 1 (16.0 g) = 16.0 g 1 mol H atoms = 1 (1.01 g) = 1.01 g 40.0 g/mol Molar mass

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24 CHEMICAL FORMULAS Molecular formula Empirical formula Shows the actual number of atoms in a compound Shows the simplest ratio of atoms in a compound Can be written for molecular compounds only Can be written for molecular and ionic compounds

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25 CHEMICAL FORMULAS MolecularEmpiricalMultiplier H2O2H2O2 H2OH2O C 6 H 12 O 6 C6H6C6H6 C2H2C2H2 C 6 H 12 HO2 H2OH2O1 CH 2 O6 CH 6 2 CH 2 6

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26 CHEMICAL FORMULAS Several compounds may possess the same percent composition and empirical formula, but different molecular formulas. Formula Composition % C % H Molar mass (g/mol) CH92.37.7 C2H2C2H2 92.37.7 C6H6C6H6 92.37.7 13.02 26.04 (2x13.02) 78.12 (6x13.02)

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Finding an Empirical Formula 1)convert the percentages to grams a)skip if already grams 2)convert grams to moles a)use molar mass of each element 3)write a pseudoformula using moles as subscripts 4)divide all by smallest number of moles 5)multiply all mole ratios by number to make all whole numbers a)if ratio ?.5, multiply all by 2; if ratio ?.33 or ?.67, multiply all by 3, etc. b)skip if already whole numbers

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If, after dividing by the smallest number of moles, the subscripts are not whole numbers, multiply all the subscripts by a small whole number to arrive at whole- number subscripts.

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29 CALCULATING EMPIRICAL FORMULAS Arsenic (As) reacts with oxygen (O) to form a compound that is 75.7% As and 24.3% oxygen by mass. What is the empirical formula for this compound? Step 1:Percent to mass 75.7% As75.7 g As 24.3% O24.3 g O Assume 100 g

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30 CALCULATING EMPIRICAL FORMULAS Step 2:Mass to mole 75.7 g As 24.3 g O Use atomic mass of oxygen

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31 CALCULATING EMPIRICAL FORMULAS Step 3:Divide by small As = O =

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32 CALCULATING EMPIRICAL FORMULAS Step 4:Multiply till Whole As 1.00 O 1.50 x 2 = 2 x 2 = 3 As 2 O 3

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33 Determine the empirical formula for a compound containing 11.2% H and 88.8% O. Example 1: mol H = mol O = H2OH2O

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34 Determine the empirical formula for a compound with the following percent composition: 52.14% C, 13.12% H, 34.73% O. Example 2: mol C = mol H = C2H6OC2H6O mol O =

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35 MOLECULAR FORMULAS Molecular formula can be calculated from empirical formula if molar mass is known. Molecular formula = (empirical formula) n

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36 A compound of N and O with a molar mass of 92.0 g, has the empirical formula of NO 2. What is its molecular formula? Example 1: Mass of empirical formula =14.0 + 2(16.0) = 46.0 Molecular formula =2 x (NO 2 ) = N 2 O 4

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37 Calculate the empirical and molecular formulas of a compound that contains 80.0% C and 20.0% H, and has a molar mass of 30.00 g. Example 2: mol C = mol H = CH 3 Empirical formula

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38 Example 2: Mass of empirical formula =12.0 + 3(1.01) = 15.0 Molecular formula =2 x (CH 3 ) = C 2 H 6 Empirical formula =CH 3

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39 THE END

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