Atomic mass is the mass of an atom in atomic mass units (amu)

Micro World atoms & molecules Laboratory scale measurements 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 Law of Definite (or Constant) Composition: No matter what its source, a particular chemical compound is composed of the same elements in the same parts (fractions) by mass. CaCO 3 1 atom of Ca 40.08 amu 40.08 amu 100.08 amu 12.00 amu 100.08 amu 48.00 amu 100.08 amu 1 atom of C 12.00 amu 3 atoms of O 3 x 16.00 amu 100.08 amu = 0.401 parts Ca = 0.120 parts C = 0.480 parts O 1

The Atomic Basis of the Law of Multiple Proportions 2

Dalton s s Atomic Theory 1. All matter consists of atoms. 2. Atoms of one element cannot be converted into atoms of another element. 3. Atoms of an element are identical in mass and other properties and are different from atoms of any other element. 4. Compounds result from the chemical combination of a specific ratio of atoms of different elements. The mole (mol) is the amount of a substance that contains as many elementary entities as there are atoms in exactly 12.00 grams of 12 C 1 mol = N A = 6.0221367 x 10 23 Avogadro s number (N A ) 3

One Mole of Common Substances Oxygen 32.00 g Figure 3.2 CaCO 3 100.09 g Water 18.02 g Copper 63.55 g Molar mass is the mass of 1 mole of Na atoms Pb atoms Kr atoms Li atoms in grams 1 mole 12 C atoms = 6.022 x 10 23 atoms = 12.00 g 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) 4

Interconversions of Moles, Mass, and Number of Chemical Entities Mass (g) = no. of moles x no. of grams 1 mol No. of moles = mass (g) x 1 mol no. of grams No. of entities = no. of moles x 6.022x1023 entities 1 mol 1 mol No. of moles = no. of entities x 6.022x10 23 entities Avogadro s Number Molar Mass 5

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 6

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 Calculating the Mass and the Number of Atoms in a Given Number of Moles of an Element PROBLEM: (a) Silver (Ag) is used in jewelry and tableware but no longer in U.S. coins. How many grams of Ag are in 0.0342mol of Ag? (b) Iron (Fe), the main component of steel, is the most important metal in industrial society. How many Fe atoms are in 95.8g of Fe? (a) To convert mol of Ag to g we have to use the #g Ag/mol Ag, the molar mass M. 107.9g Ag 0.0342mol Ag x = 3.69g Ag mol Ag (b) We first have to find the #mols of Fe and then convert mols to atoms. mol Fe 95.8g Fe x 55.85g Fe 6.022x10 23 atoms Fe mol Fe = 1.04x10 24 atoms Fe amount(mol) of Ag multiply by M of Ag (107.9g/mol) mass(g) of Ag mass(g) of Fe divide by M of Fe (55.85g/mol) amount(mol) of Fe multiply by 6.022x10 23 atoms/mol atoms of Fe 7

Determination of Empirical Formula from Mass % 1. Assume 100 grams of material 2. Determine moles of each element in compound 3. Divide all subscripts in empirical formula by lowest number because the number of atoms must be an integer 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 2 x (12.01 g) %C = x 100% = 52.14% 46.07 g 6 x (1.008 g) %H = x 100% = 13.13% 46.07 g 1 x (16.00 g) %O = x 100% = 34.73% 46.07 g 52.14% + 13.13% + 34.73% = 100.0% 3.5 8

Determining the Empirical Formula from Masses of Elements PROBLEM: Elemental analysis of a sample of an ionic compound gave the following results: 2.82g of Na, 4.35g of Cl, and 7.83g of O. What are the empirical formula and name of the compound? PLAN: Once we find the relative number of moles of each element, we can divide by the lowest mol amount to find the relative mol ratios (empirical formula). SOLUTION: mol Na 2.82g Na = 0.123 mol Na 22.99g Na 0.123 mol Na mol Cl = 1.00 4.35g Cl 0.123 35.45g Cl = 0.123 mol Cl 0.123 mol Cl = 1.00 7.83g O mol O 16.00 O = 0.489 mol O 0.123 Na 1 Cl 1 O 3.98 NaClO 4 0.489 mol O = 3.98 NaClO 4 is sodium perchlorate. 0.123 Structural Formulas vs. Molecular Formulas Two Compounds with Molecular Formula C 2 H 6 O Property Ethanol Dimethyl Ether M(g/mol) Color Melting Point Boiling Point Density at 20 0 C 46.07 Colorless -117 0 C 78.5 0 C 0.789g/mL(liquid) 46.07 Colorless -138.5 0 C -25 0 C 0.00195g/mL(gas) Use Structural formulas and space-filling model H H C H intoxicant in alcoholic beverages H C H O H H C O H H in refrigeration H C H H 9

Determination of Molecular Formula from Empirical Formula Empirical formula only can be determined by mass percent Multiplier = MoleculeMolarMass EmpiricalMolarMass Chemical Equations reactants products 3 ways of representing the reaction of H 2 with O 2 to form H 2 O 10

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 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 C 2 H 6 + O 2 2 hydrogen on right 2CO 2 + 3H 2 O multiply H 2 O by 3 11

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 C 2 H 6 + 7 O 2 2CO 2 + 3H 2 O 2 2C 2 H 6 + 7O 2 4CO 2 + 6H 2 O remove fraction multiply both sides by 2 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 12 14 4 C H O (2 (7 x 2) 6) 2) 1412 O H (44 (6 Cx 2 x + 2) 6) Reactants 4 C 12 H 14 O Products 4 C 12 H 14 O 12

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 IS NOT 2 grams Mg + 1 gram O 2 makes 2 g MgO 2S (s) + 3O 2 (g) ->2 SO 3 (g) S O Reactant Mixture Product Mixture 13

Mass Changes in Chemical Reactions 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 Memorize these steps: mol ratio from balanced equation molar mass of product Mass react x 1 mol react x mol product x product g = mass grams react mol react mol prod. molar mass of reactant 14

Stoichiometric Relationships Molarity of reactant a,b,c or d From balanced equation Molarity of product Volume reactant Density (g/ml) Grams reactant (mol/g) Moles reactant Volume (L) * moles product moles reactant Volume -1 (L -1 ) * Moles product (g/mol) Grams of product 1/density (ml/g) Volume of product 1/N a Atoms or molecules Of reactant N a Atoms or molecules Of products a Reactant A + b Reactant B c Product C + d Product D or a Product A + b Product B c Reactant C + d Reactant D 2004 Randal Hallford 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 15

I have found the pot at the end of the rainbow!! Limiting Reagents 6 green red left used overup 16

2S (s) + 3O 2 (g) ->2 SO 3 (g) Limiting Reagent S O Reactant Mixture Product Mixture Excess Reagent Limiting Reagent Problems 1. Balance chemical equation 2. Determine limiting reagent Do two separate calculations for the amount of product each reactant would produce if they were the limiting reagent The reactant that gives the lower number is the limiting reagent. 3. The amount of product produced is the number calculated by limiting reagent 17

Limiting Reagent Problems 4. Determination of the amount of excess reagent left over Calculate the amount of excess reagent (ER) used in chemical reaction Subtract the ER used from original amount of ER. Calculating Amounts of Reactant and Product in Reactions Involving a Limiting Reactant PROBLEM: A fuel mixture used in the early days of rocketry is composed of two liquids, hydrazine(n 2 H 4 ) and dinitrogen tetraoxide(n 2 O 4 ), which ignite on contact to form nitrogen gas and water vapor. How many grams of nitrogen gas form when 1.00x10 2 g of N 2 H 4 and 2.00x10 2 g of N 2 O 4 are mixed? PLAN: We always start with a balanced chemical equation and find the number of mols of reactants and products which have been given. In this case one of the reactants is in molar excess and the other will limit the extent of the reaction. mass of N 2 H 4 divide by M mass of N 2 O 4 limiting mol N 2 multiply by M mol of N 2 H 4 molar ratio mol of N 2 O 4 g N 2 mol of N 2 mol of N 2 18

Calculating Amounts of Reactant and Product in Reactions Involving a Limiting Reactant SOLUTION: 2 N 2 H 4 (l) + N 2 O 4 (l) 3 N 2 (g) + 4H 2 O(l) 1.00x10 2 g N 2 H mol N 2 H 4 4 = 3.12mol N2 H 4 32.05g N 2 H 4 3.12mol N 2 H 4 3 mol N 2 = 4.68mol N 2 2mol N 2 H 4 mol N 2.00x10 2 2 O 4 g N 2 O 4 = 2.17mol N 2 O 4 92.02g N 2 O 4 N 2 H 4 is the limiting reactant because it produces less product, N 2, than does N 2 O 4. 28.02g N 4.68mol N 2 2 = 131g N 2 mol N 2 2.17mol N 2 O 4 3 mol N 2 = 6.51mol N 2 mol N 2 O 4 Limiting Reagents 124 g of Al are reacts 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 OR g Fe 2 O 3 mol Fe 2 O 3 mol Al needed g Al needed 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); Al is limiting reagent 19

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 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 20

Actual Yields < 100% Calculating Percent Yield PROBLEM: PLAN: Silicon carbide (SiC) is an important ceramic material that is made by allowing sand(silicon dioxide, SiO2) to react with powdered carbon at high temperature. Carbon monoxide is also formed. When 100.0kg of sand are processed, 51.4kg of SiC are recovered. What is the percent yield of SiC in this process? SOLUTION: write balanced equation SiO 2 (s) + 3C(s) SiC(s) + 2CO(g) find mol reactant & product 10 mol SiO 100.0kg SiO 3 g SiO 2 2 2 = 1664 mol SiO 2 kg SiO 2 60.09g SiO 2 find g product predicted actual yield/theoretical yield x 100 percent yield mol SiO 2 = mol SiC = 1664 40.10g SiC kg 1664mol SiC mol SiC 10 3 g = 66.73kg 51.4kg x100 =77.0% 66.73kg 21

Calculating the Molarity of a Solution PROBLEM: Hydrobromic acid(hbr) is a solution of hydrogen bromide gas in water. Calculate the molarity of hydrobromic acid solution if 455mL contains 1.80mol of hydrogen bromide. mol of HBr divide by volume concentration(mol/ml) HBr 1.80mol HBr 455 ml soln 1000mL 1 L = 3.96M 10 3 ml = 1L molarity(mol/l) HBr Calculating Mass of Solute in a Given Volume of Solution How many grams of solute are in 1.75L of 0.460M sodium monohydrogen phosphate? Molarity is the number of moles of solute per liter of solution. Knowing the molarity and volume leaves us to find the # moles and then the # of grams of solute. The formula for the solute is Na 2 HPO 4. volume of soln multiply by M moles of solute multiply by M grams of solute SOLUTION: 1.75L 0.460moles = 0.805mol Na 2 HPO 4 1 L 0.805mol Na 2 HPO 4 141.96g Na 2 HPO 4 mol Na 2 HPO 4 = 114g Na 2 HPO 4 22

Preparing a Dilute Solution from a Concentrated Solution Isotonic saline is a 0.15M aqueous solution of NaCl that simulates the total concentration of ions found in many cellular fluids. Its uses range from a cleaning rinse for contact lenses to a washing medium for red blood cells. How would you prepare 0.80L of isotomic saline from a 6.0M stock solution? The number of moles of solute does not change during the dilution but the volume does. The new volume will be the sum of the two volumes, that is, the total final volume. M dil xv dil = #mol solute = M conc xv conc volume of dilute soln multiply by M of dilute solution moles of NaCl in dilute soln = mol NaCl in concentrated soln divide by M of concentrated soln L of concentrated soln SOLUTION: 0.80L soln 0.15mol NaCl L soln = 0.12mol NaCl 0.12mol NaCl L soln conc 6mol = 0.020L soln 23

Calculating Amounts of Reactants and Products for a Reaction in Solution A common antacid contains magnesium hydroxide, which reacts with stomach acid to form water and magnesium chloride solution. As a government chemist testing commercial antacids, you use 0.10M HCl to simulate the acid concentration in the stomach. How many liters of stomach acid react with a tablet containing 0.10g of magnesium hydroxide? Write a balanced equation for the reaction; find the grams of Mg(OH) 2 ; determine the mol ratio of reactants and products; use mols to convert to molarity. mass Mg(OH) 2 divide by M mol Mg(OH) 2 L HCl mol HCl divide by M mol ratio 24

Calculating Amounts of Reactants and Products for a Reaction in Solution Mg(OH) 2 (s) + 2HCl(aq) MgCl 2 (aq) + 2H 2 O(l) 0.10g Mg(OH) 2 mol Mg(OH) 2 58.33g Mg(OH) 2 2 mol HCl 1 mol Mg(OH) 2 1L 0.10mol HCl = 3.4x10-2 L HCl Solving Limiting-Reactant Problems for Reactions in Solution 0.050L of 0.010M mercury(ii) nitrate reacts with 0.020L of 0.10M sodium sulfide. How many grams of mercury(ii) sulfide form? As usual, write a balanced chemical reaction. Since this is a problem concerning a limiting reactant, we proceed as before and find the amount of product which would be made from each reactant. We then chose the reactant which gives the lesser amount of product. 25

Solving Limiting-Reactant Problems for Reactions in Solution SOLUTION: Hg(NO 3 ) 2 (aq) + Na 2 S(aq) HgS(s) + 2NaNO 3 (aq) L of Hg(NO 3 ) 2 0.050L Hg(NO 3 ) 2 0.020L Hg(NO 3 ) 2 L of Na 2 S multiply by M x 0.010 mol/l x 0. 10 mol/l multiply by M mol Hg(NO 3 ) 2 x 1mol HgS x 1mol HgS mol Na 2 S mol ratio 1mol Hg(NO 3 ) 2 1mol Na 2 S mol ratio mol HgS = 5.0x10-4 mol HgS = 2.0x10-3 mol HgS mol HgS Hg(NO 3 ) 2 is the limiting reagent. 5.0x10-4 mol HgS 232.7g HgS 232.7mol HgS = 0.12g HgS Preparation of Stock Solutions 26

Stoichiometric Relationships Molarity of reactant a,b,c or d From balanced equation Molarity of product Volume reactant Density (g/ml) Grams reactant (mol/g) Moles reactant Volume (L) * moles product moles reactant Volume -1 (L -1 ) * Moles product (g/mol) Grams of product 1/density (ml/g) Volume of product 1/N a Atoms or molecules Of reactant N a Atoms or molecules Of products a Reactant A + b Reactant B c Product C + d Product D or a Product A + b Product B c Reactant C + d Reactant D 2004 Randal Hallford 27