Formulas and Models 1

A molecular formula shows the exact number of atoms of each element in the smallest unit of a substance An empirical formula shows the simplest whole-number ratio of the atoms in a substance molecular H 2 O C 6 H 12 O 6 O 3 empirical H 2 O CH 2 O O N 2 H 4 NH 2 2

Ionic compounds consist of a combination of cations and an anions The formula is usually the same as the empirical formula The sum of the charges on the cation(s) and anion(s) in each formula unit must equal zero The ionic compound NaCl 3

The most reactive metals (green) and the most reactive nonmetals (blue) combine to form ionic compounds. 4

Formula of Ionic Compounds 2 x +3 = +6 3 x -2 = -6 Al 2 O 3 Al 3+ O 2-1 x +2 = +2 2 x -1 = -2 CaBr 2 Ca 2+ Br - 1 x +2 = +2 1 x -2 = -2 Na 2 CO 3 Na + CO 2-3 5

Chemical Nomenclature Ionic Compounds Often a metal + nonmetal Anion (nonmetal), add ide to element name BaCl 2 K 2 O Mg(OH) 2 KNO 3 barium chloride potassium oxide magnesium hydroxide potassium nitrate 6

Transition metal ionic compounds indicate charge on metal with Roman numerals FeCl 2 2 Cl - -2 so Fe is +2 iron(ii) chloride FeCl 3 3 Cl - -3 so Fe is +3 iron(iii) chloride Cr 2 S 3 3 S -2-6 so Cr is +3 (6/2) chromium(iii) sulfide 7

Molecular compounds Nonmetals or nonmetals + metalloids Common names H 2 O, NH 3, CH 4, Element furthest to the left in a period and closest to the bottom of a group on periodic table is placed first in formula If more than one compound can be formed from the same elements, use prefixes to indicate number of each kind of atom Last element name ends in ide 10

Molecular Compounds HI NF 3 SO 2 N 2 Cl 4 NO 2 N 2 O hydrogen iodide nitrogen trifluoride sulfur dioxide dinitrogen tetrachloride nitrogen dioxide dinitrogen monoxide 11

An acid can be defined as a substance that yields hydrogen ions (H + ) when dissolved in water. For example: HCl gas and HCl in water Pure substance, hydrogen chloride Dissolved in water (H 3 O + and Cl ), hydrochloric acid 13

An oxoacid is an acid that contains hydrogen, oxygen, and another element. HNO 3 nitric acid H 2 CO 3 carbonic acid H 3 PO 4 phosphoric acid 15

Naming Oxoacids and Oxoanions 16

The rules for naming oxoanions, anions of oxoacids, are as follows: 1. When all the H ions are removed from the -ic acid, the anion s name ends with -ate. 2. When all the H ions are removed from the -ous acid, the anion s name ends with -ite. 3. The names of anions in which one or more but not all the hydrogen ions have been removed must indicate the number of H ions present. For example: H 2 PO - 4 dihydrogen phosphate HPO 2-4 hydrogen phosphate PO 3-4 phosphate 17

A base can be defined as a substance that yields hydroxide ions (OH - ) when dissolved in water. NaOH KOH Ba(OH) 2 sodium hydroxide potassium hydroxide barium hydroxide 19

Hydrates are compounds that have a specific number of water molecules attached to them. BaCl 2 2H 2 O LiCl H 2 O MgSO 4 7H 2 O Sr(NO 3 ) 2 4H 2 O barium chloride dihydrate lithium chloride monohydrate magnesium sulfate heptahydrate strontium nitrate tetrahydrate CuSO 4 5H 2 O CuSO 4 20

Stoichiometry Copyright The McGraw-Hill Companies, Inc. Permission required for reproduction or display.

Extensive and Intensive Properties An extensive property of a material depends upon how much matter is is being considered. mass length volume An intensive property of a material does not depend upon how much matter is being considered. density temperature color 22

Matter - anything that occupies space and has mass. mass measure of the quantity of matter SI unit of mass is the kilogram (kg) 1 kg = 1000 g = 1 x 10 3 g weight force that gravity exerts on an object weight = c x mass on earth, c = 1.0 on moon, c ~ 0.1 A 1 kg bar will weigh 1 kg on earth 0.1 kg on moon 23

International System of Units (SI) 24

Volume SI derived unit for volume is cubic meter (m 3 ) 1 cm 3 = (1 x 10-2 m) 3 = 1 x 10-6 m 3 1 dm 3 = (1 x 10-1 m) 3 = 1 x 10-3 m 3 1 L = 1000 ml = 1000 cm 3 = 1 dm 3 1 ml = 1 cm 3 26

Density SI derived unit for density is kg/m 3 1 g/cm 3 = 1 g/ml = 1000 kg/m 3 density = mass volume d = m V A piece of platinum metal with a density of 21.5 g/ cm 3 has a volume of 4.49 cm 3. What is its mass? d = m V m = d x V = 21.5 g/cm 3 x 4.49 cm 3 = 96.5 g 27

A Comparison of Temperature Scales K = C + 273.15 273 K = 0 C 373 K = 100 C F = 9 x C + 32 5 32 0 F = 0 C 212 0 F = 100 C 29

Scientific Notation The number of atoms in 12 g of carbon: 602,200,000,000,000,000,000,000 6.022 x 10 23 The mass of a single carbon atom in grams: 0.0000000000000000000000199 1.99 x 10-23 N is a number between 1 and 10 N x 10 n n is a positive or negative integer 30

Scientific Notation 568.762 move decimal left n > 0 568.762 = 5.68762 x 10 2 0.00000772 move decimal right n < 0 0.00000772 = 7.72 x 10-6 Addition or Subtraction 1. Write each quantity with the same exponent n 2. Combine N 1 and N 2 3. The exponent, n, remains the same 4.31 x 10 4 + 3.9 x 10 3 = 4.31 x 10 4 + 0.39 x 10 4 = 4.70 x 10 4 31

Scientific Notation Multiplication 1. Multiply N 1 and N 2 2. Add exponents n 1 and n 2 (4.0 x 10-5 ) x (7.0 x 10 3 ) = (4.0 x 7.0) x (10-5+3 ) = 28 x 10-2 = 2.8 x 10-1 Division 1. Divide N 1 and N 2 2. Subtract exponents n 1 and n 2 8.5 x 10 4 5.0 x 10 9 = (8.5 5.0) x 10 4 9 = 1.7 x 10-5 32

Any digit that is not zero is significant 1.234 kg 4 significant figures Significant Figures Zeros between nonzero digits are significant 606 m 3 significant figures Zeros to the left of the first nonzero digit are not significant 0.08 L 1 significant figure If a number is greater than 1, then all zeros to the right of the decimal point are significant 2.0 mg 2 significant figures If a number is less than 1, then only the zeros that are at the end and in the middle of the number are significant 0.00420 g 3 significant figures 33

How many significant figures are in each of the following measurements? 24 ml 2 significant figures 3001 g 4 significant figures 0.0320 m 3 3 significant figures 6.4 x 10 4 molecules 2 significant figures 560 kg 2 significant figures 34

Addition or Subtraction Significant Figures The answer cannot have more digits to the right of the decimal point than any of the original numbers. 89.332 + 1.1 one significant figure after decimal point 90.432 round off to 90.4 3.70-2.9133 0.7867 two significant figures after decimal point round off to 0.79 35

Multiplication or Division Significant Figures The number of significant figures in the result is set by the original number that has the smallest number of significant figures 4.51 x 3.6666 = 16.536366 = 16.5 3 sig figs round to 3 sig figs 6.8 112.04 = 0.0606926 = 0.061 2 sig figs round to 2 sig figs 36

Exact Numbers Significant Figures Numbers from definitions or numbers of objects are considered to have an infinite number of significant figures The average of three measured lengths: 6.64, 6.68 and 6.70? 6.64 + 6.68 + 6.70 3 = 6.67333 = 6.67 = 7 Because 3 is an exact number 37

Accuracy how close a measurement is to the true value Precision how close a set of measurements are to each other accurate & precise precise but not accurate not accurate & not precise 38

Micro World atoms & molecules Macro World grams Atomic mass is the mass of an atom in atomic mass units (amu) By definition: 1 atom 12 C weighs 12 amu On this scale 1 H = 1.008 amu 16 O = 16.00 amu 39

The average atomic mass is the weighted average of all of the naturally occurring isotopes of the element. 40

Naturally occurring lithium is: 7.42% 6 Li (6.015 amu) 92.58% 7 Li (7.016 amu) Average atomic mass of lithium: (7.42 x 6.015) + (92.58 x 7.016) 100 = 6.941 amu 41

Average atomic mass (6.941) 42

Molar mass is the mass of 1 mole of eggs shoes marbles atoms in grams 1 mole 12 C atoms = 6.022 x 10 23 atoms = 12.00 g 1 12 C atom = 12.00 amu 1 mole 12 C atoms = 12.00 g 12 C 1 mole lithium atoms = 6.941 g of Li For any element atomic mass (amu) = molar mass (grams) 43

One Mole of: C S Hg Cu Fe 44

1 12 C atom 12.00 amu x 12.00 g 6.022 x 10 23 12 C atoms = 1.66 x 10-24 g 1 amu 1 amu = 1.66 x 10-24 g or 1 g = 6.022 x 10 23 amu M = molar mass in g/mol N A = Avogadro s number 45

How many atoms are in 0.551 g of potassium (K)? 1 mol K = 39.10 g K 1 mol K = 6.022 x 10 23 atoms K 0.551 g K x 1 mol K 39.10 g K x 6.022 x 1023 atoms K 1 mol K = 8.49 x 10 21 atoms K 46

Molecular mass (or molecular weight) is the sum of the atomic masses (in amu) in a molecule. SO 2 1S 2O SO 2 32.07 amu + 2 x 16.00 amu 64.07 amu For any molecule molecular mass (amu) = molar mass (grams) 1 molecule SO 2 = 64.07 amu 1 mole SO 2 = 64.07 g SO 2 47

How many H atoms are in 72.5 g of C 3 H 8 O? 1 mol C 3 H 8 O = (3 x 12) + (8 x 1) + 16 = 60 g C 3 H 8 O 1 mol C 3 H 8 O molecules = 8 mol H atoms 1 mol H = 6.022 x 10 23 atoms H 72.5 g C 3 H 8 O 1 mol C 3 H 8 O x 60 g C 3 H 8 O x 8 mol H atoms 1 mol C 3 H 8 O 6.022 x 10 23 H atoms x = 1 mol H atoms 5.82 x 10 24 atoms H 48

Formula mass is the sum of the atomic masses (in amu) in a formula unit of an ionic compound. NaCl 1Na 1Cl NaCl 22.99 amu + 35.45 amu 58.44 amu For any ionic compound formula mass (amu) = molar mass (grams) 1 formula unit NaCl = 58.44 amu 1 mole NaCl = 58.44 g NaCl 49

What is the formula mass of Ca 3 (PO 4 ) 2? 1 formula unit of Ca 3 (PO 4 ) 2 3 Ca 3 x 40.08 2 P 2 x 30.97 8 O + 8 x 16.00 310.18 amu 50

Mass Spectrometer Mass Spectrum of Ne Light Heavy Light Heavy 51

Percent composition of an element in a compound = n x molar mass of element molar mass of compound x 100% n is the number of moles of the element in 1 mole of the compound C 2 H 6 O %C = 2 x (12.01 g) 46.07 g x 100% = 52.14% %H = 6 x (1.008 g) 46.07 g x 100% = 13.13% %O = 1 x (16.00 g) 46.07 g x 100% = 34.73% 52.14% + 13.13% + 34.73% = 100.00% 52

Percent Composition and Empirical Formulas Determine the empirical formula of a compound that has the following percent composition by mass: K 24.75, Mn 34.77, O 40.51 percent. n K = 24.75 g K x n Mn = 34.77 g Mn x 1 mol K 39.10 g K 1 mol Mn 54.94 g Mn = 0.6330 mol K = 0.6329 mol Mn n O = 40.51 g O x 1 mol O 16.00 g O = 2.532 mol O 53

Percent Composition and Empirical Formulas n K = 0.6330, n Mn = 0.6329, n O = 2.532 0.6330 K : ~ 0.6329 1.0 Mn : 0.6329 0.6329 = 1.0 2.532 O : ~ 0.6329 4.0 KMnO 4 54

Combust 11.5 g ethanol Collect 22.0 g CO 2 and 13.5 g H 2 O g CO 2 mol CO 2 mol C g C 6.0 g C = 0.5 mol C g H 2 O mol H 2 O mol H g H 1.5 g H = 1.5 mol H g of O = g of sample (g of C + g of H) 4.0 g O = 0.25 mol O Empirical formula C 0.5 H 1.5 O 0.25 Divide by smallest subscript (0.25) Empirical formula C 2 H 6 O 55

A process in which one or more substances is changed into one or more new substances is a chemical reaction A chemical equation uses chemical symbols to show what happens during a chemical reaction reactants products 3 ways of representing the reaction of H 2 with O 2 to form H 2 O 56

How to Read Chemical Equations 2 Mg + O 2 2 MgO 2 atoms Mg + 1 molecule O 2 makes 2 formula units MgO 2 moles Mg + 1 mole O 2 makes 2 moles MgO 48.6 grams Mg + 32.0 grams O 2 makes 80.6 g MgO NOT 2 grams Mg + 1 gram O 2 makes 2 g MgO 57

Balancing Chemical Equations 1. Write the correct formula(s) for the reactants on the left side and the correct formula(s) for the product(s) on the right side of the equation. Ethane reacts with oxygen to form carbon dioxide and water C 2 H 6 + O 2 CO 2 + H 2 O 2. Change the numbers in front of the formulas (coefficients) to make the number of atoms of each element the same on both sides of the equation. Do not change the subscripts. 2C 2 H 6 NOT C 4 H 12 58

Balancing Chemical Equations 3. Start by balancing those elements that appear in only one reactant and one product. C 2 H 6 + O 2 CO 2 + H 2 O start with C or H but not O 2 carbon on left C 2 H 6 + O 2 1 carbon on right 2CO 2 + H 2 O multiply CO 2 by 2 6 hydrogen on left 2 hydrogen on right multiply H 2 O by 3 C 2 H 6 + O 2 2CO 2 + 3H 2 O 59

Balancing Chemical Equations 4. Balance those elements that appear in two or more reactants or products. C 2 H 6 + O 2 2CO 2 + 3H 2 O multiply O 2 by 7 2 2 oxygen on left 4 oxygen (2x2) + 3 oxygen (3x1) = 7 oxygen on right 7 C 2 H 6 + O 2 2CO 2 + 3H 2 O 2 remove fraction multiply both sides by 2 2C 2 H 6 + 7O 2 4CO 2 + 6H 2 O 60

Balancing Chemical Equations 5. Check to make sure that you have the same number of each type of atom on both sides of the equation. 2C 2 H 6 + 7O 2 4CO 2 + 6H 2 O 4 C (2 x 2) 4 C 12 H (2 x 6) 12 H (6 x 2) 14 O (7 x 2) 14 O (4 x 2 + 6) Reactants 4 C 12 H 14 O Products 4 C 12 H 14 O 61

Amounts of Reactants and Products 1. Write balanced chemical equation 2. Convert quantities of known substances into moles 3. Use coefficients in balanced equation to calculate the number of moles of the sought quantity 4. Convert moles of sought quantity into desired units 62

Methanol burns in air according to the equation 2CH 3 OH + 3O 2 2CO 2 + 4H 2 O If 209 g of methanol are used up in the combustion, what mass of water is produced? grams CH 3 OH moles CH 3 OH moles H 2 O grams H 2 O molar mass CH 3 OH coefficients chemical equation molar mass H 2 O 209 g CH 3 OH 1 mol CH 3 OH x 32.0 g CH 3 OH x 4 mol H 2 O 2 mol CH 3 OH 18.0 g H 2 O x = 1 mol H 2 O 235 g H 2 O 63

Limiting Reagent: Reactant used up first in the reaction. 2NO + O 2 2NO 2 NO is the limiting reagent O 2 is the excess reagent 64

In one process, 124 g of Al are reacted with 601 g of Fe 2 O 3 2Al + Fe 2 O 3 Al 2 O 3 + 2Fe Calculate the mass of Al 2 O 3 formed. g Al mol Al mol Fe 2 O 3 needed g Fe 2 O 3 needed g Fe 2 O 3 mol Fe 2 O 3 mol Al needed g Al needed OR 124 g Al x 1 mol Al 27.0 g Al x 1 mol Fe 2 O 3 2 mol Al 160. g Fe 2 O x 3 = 367 g Fe 1 mol Fe 2 O 2 O 3 3 Start with 124 g Al need 367 g Fe 2 O 3 Have more Fe 2 O 3 (601 g) so Al is limiting reagent 65

Use limiting reagent (Al) to calculate amount of product that can be formed. g Al mol Al mol Al 2 O 3 g Al 2 O 3 2Al + Fe 2 O 3 Al 2 O 3 + 2Fe 124 g Al x 1 mol Al 27.0 g Al x 1 mol Al 2 O 3 2 mol Al 102. g Al 2 O x 3 = 234 g Al 1 mol Al 2 O 2 O 3 3 At this point, all the Al is consumed and Fe 2 O 3 remains in excess. 66

Reaction Yield Theoretical Yield is the amount of product that would result if all the limiting reagent reacted. Actual Yield is the amount of product actually obtained from a reaction. % Yield = Actual Yield Theoretical Yield x 100% 67