Moles Balanced chemical equations Molar ratios Mass Composition Empirical and Molecular Mass Predicting Quantities 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 3.1 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 ) 3.2
Molar mass is the mass of 1 mole of Na atoms Pb atoms in grams Kr atoms Li atoms 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) 3.2 Converting Units Unit Factor Label Method I. Write Unit of Answer II. Write Starting Quantity from Problem III. Add appropriate unit factor(s) to cancel out starting quantity and put in unit of answer. Do You Understand Molar Mass? How many atoms are in 0.551 g of potassium (K)? 0.551 g K 1 mol K = 39.10 g K 1 mol K = 6.022 x 10 23 atoms K 1 mol K x 39.10 g K x 6.022 x 1023 atoms K 1 mol K = 8.49 x 10 21 atoms K 3.2
Avogadro s Number Molar Mass 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 3.3
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 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 C 2 H 6 O 52.14% + 13.13% + 34.73% = 100.0% 3.5 The balanced chemical equation represents the ratio Of one species to any others in the equation: NaOH (aq) + HCl (aq) NaCl (aq) + H 2 O (l) 1mole of NaOH produces 1mole of NaCl 1 mol NaCl 1mol NaOH 2H 2 O 2 (l) 2 H 2 O (l) + O 2 (g) 2 moles of H 2 O 2 (l) produces 1 mole of O 2 gas 1 mol O 2 2 mol H 2 O 2 (l) Mole composition gives ratio of each element In a compound 0.345 moles of Al 2 (CO 3 ) 3 has the following moles of Al in the compound: Mol ratio of Al in Al 2 (CO 3 ) 3 : 0.345 mol Al 2 (CO 3 ) 3 X 2 mol Al = 0.690 mol Al 1 mol Al 2 (CO 3 ) 3 Moles of compound X mol ratio (element/molecule) = mol of element
Mass of Aluminum is obtained from the molar mass: 0.690 mol Al X 26.98 g = 18.62 g Al mol Al Mol X grams/mol = mass The mass of any substance can be determined if The molar mass is known. The molar mass comes From the periodic table. Al 2 (CO 3 ) 3 continued: C: 0.345 mol X 3 mol C = 1.04 mol C X 12.011g/mol 1 mol Al 2 (CO 3 ) 3 = 12.43 g C O: 0.345 mol X 9 mol O = 3.11 mol O X 15.999g/mol 1 mol Al 2 (CO 3 ) 3 = 49.677 g O Formula Weight converted to mass from moles: 2 mol X 26.98g/mol (Al) =53.96g 3mol X 12.011g/mol (C) =36.033g 9 mol X 15.999g/mol (O) = 143.991 g total molar mass 233.984 g in 1 mol Al 2 (CO 3 ) 3 233.984 g/mol X 0.345 mol = 80.724 g Al 2 (CO 3 ) 3 The real usefulness of balanced equations can be Observed in the following series of predictions: I Predict moles of any products of a reaction II Predict the amount of any and all reactants needed for a given amount of product III Determine precisely how much of all reactants are used, and which are present in excess. The basis of chemical calculations for reactions Is the balanced reaction equation.
So far, we have seen mole ratio s from an equation That tells us how much to expect in the reaction: N 2 (g) + H 2 (g) NH 3 (g) N 2 (g) + 3H 2 (g) 2NH 3 (g) (Balanced equation) Think: mols reactant x moles product = mols product mols reactant What we are asked for goes on top of the ratio. What we Are given goes on bottom. How many moles of NH 3 can be produced from 5.00 mol H 2 5.00 mol H 2 (given) x 2 mol NH 3 = 3.33 mol NH 3 formed 3 mol H 2 In the laboratory, we must measure the mols by Mass. This requires an initial conversion to mols From the given mass of substance: How many moles of NH 3 are formed from 33.6 g Of N 2 : 33.6 g x 1 mol N 2 x 2 mol NH 3 = 2.40 mol NH 3 28.0 g N 2 1 mol N 2 Mass reactant x 1 x mol product = mol product F.W react. mol reactant Finally, we must determine the mass of a product In the laboratory, since we cannot evaluate moles: What mass of H 2 is needed to produce 119 g of NH 3 What is asked for: mass of H 2 What is given: mass of NH 3 Mass reactant x 1 x mol product x FW prod =mass F.W react. mol reactant product 119 g NH 3 x1 mol NH 3 x 3 mol H 2 x 2.02 g H 2 = 21.2 g H 2 17.0 g NH 3 2 mol NH 3 mol H 2
Memorize these steps: Mass react x 1 mol react x mol product x prod. g = mass grams react mol react mol prod. Mass react x 1 mol react x mol product x prod. g = mass grams react mol react mol prod. SiO 2 is etched with HF (aq): SiO 2 (s) + 4 HF (aq) SiF 4 (g) + 2 H 2 O (l) If there is 4.86 moles HF and 60 g SiO 2, will there be Any glass left over? 4.86 mol HF x 1 mol SiF 4 x 104 g SiF 4 = 126 g SiF 4 ; 4 mol HF mol SiF 4 126 g SiF 4 x 1 mol SiF 4 x SiO 2 x 60g SiO 2 = 72.7 g SiO 2 104 g SiF 4 SiF 4 mol SiO 2 Since I have only 60g (1 mol) of SiO 2, I will run out of Glass before I use up the acid! SiO 2 is LIMITING 4 Ag (s) + 2 H 2 S (g) + O 2 (g) 2 Ag 2 S (s) + 2 H 2 O (l) IF there is 0.145 mol Ag, 0.0872 mol H 2 S, and excess O 2, how much Ag 2 S is produced and what mass of Reactant is in excess? Ag: 0.145 mol x 2 mol Ag 2 S = 0.0725 mol Ag 2 S 4 mol Ag H 2 S: 0.0872 mol H 2 S x 2 mol Ag 2 S = 0.0872 mol Ag 2 S 2 mol H 2 S Ag is limiting and H 2 S is present in excess. We must use the limiting reagent quantity to predict actual products: 0.0725 mol Ag 2 S x 248 g Ag 2 S = 18.0 g Ag 2 S formed mol Ag 2 S
A reaction in the laboratory results in a product mass That is not quite what the balanced equation calculation Predicts. The difference between the theoretical amount and the Actual one is called Percent Yield: Actual Yield (mass) x 100% = percent yield Theoretical yield (mass) Thermo-chemical equations A balanced chemical equation that indicates the heat Energy transformation in the reaction. Heat of reaction: H Exothermic: a reaction in which heat energy is liberated H = -value Endothermic: a reaction in which heat energy is absorbed H= + value Heat energy is measured in Kilojoules 2 C 2 H 2 (g) + 5O 2 (g) 4CO 2 (g) + 2H 2 O (l) H=-2602 KJ This reaction gives off heat. (Acetylene Torch) If a sample of acetylene is burned, and releases 550 KJ of Energy, how many grams of CO 2 can we expect to form? 2 C 2 H 2 (g) + 5O 2 (g) 4CO 2 (g) + 2H 2 O (l) H=-2602 KJ We can relate the moles of CO 2 to the amount of heat, since It is part of the balanced equation: 550 KJ x 4 mol CO 2 x 44.0 g CO 2 = 37.2 g CO 2 2602 KJ mol CO 2
I. Formulas show chemistry at a standstill. Equations show chemistry in action. A. Equations show: 1. the reactants which enter into a reaction. 2. the products which are formed by the reaction. 3. the amounts of each substance used and each substance produced. B. Two important principles to remember: 1. Every chemical compound has a formula which cannot be altered. 2. A chemical reaction must account for every atom that is used. This is an application of the Law of Conservation of Matter which states that in a chemical reaction atoms are neither created nor destroyed. Some things to remember about writing equations: 1. The diatomic elements when they stand alone are always written H 2, N 2, O 2, F 2, Cl 2, Br 2, I 2 2. The sign, ----->, means "yields" and shows the direction of the action. 3. A small delta, ( ), above the arrow shows that heat has been added. 4. A double arrow, <----->, shows that the reaction is reversible and can go in both directions. 5. Before beginning to balance an equation, check each formula to see that it is correct. NEVER change a formula during the balancing of an equation. 6. Balancing is done by placing coefficients in front of the formulas to insure the same number of atoms of each element on both sides of the arrow. 7. Always consult the Activity Series of metals and non-metals before attempting to write equations for replacement reactions. 8. If a reactant or product is solid, place (s) after the formula 9. If the reactant or product is a liquid, place (l) after the formula 10. If the reactant or product is a gas, place (g) after the formula 11. If the reactant or product is in water, place (aq) after the formula 12. A category of reaction will produce an unstable product which decomposes: H 2 CO 3 (aq) H 2 O (l) + CO 2 (g) carbonic acid H 2 SO 3 (aq) H 2 O (l) + SO 2 (g) sulfurous acid NH 4 OH (aq) NH 3 (g) + H 2 O (l) ammonium hydroxide
Rules for Writing Chemical Equations: 1. Write down the formula(s) for the reactants and add a + between them then put a yields arrow ( ) at the end 2. Examine the formula(s) to determine which of four basic types of reactions will occur. On the basis of your decision, write down correct formula(s) for the products on the right side of the yield arrow. 3. Place a coefficient in front of each species in the reactants and the products to ensure that the conservation of matter is observed: Balance the reaction II. Four Basic Types of Chemical reactions A. Synthesis two or more elements or compounds combine to give a more complex product Examples of Synthesis reactions: 1. Metal + oxygen -----> metal oxide 2Mg (s) + O 2(g) ----> 2MgO (s) 2. Nonmetal + oxygen -----> nonmetallic oxide C (s) + O 2(g) ----> CO 2(g) 3. Metal oxide + water -----> metallic hydroxide MgO (s) + H 2 O (l) ----> Mg(OH) 2(s) 4. Nonmetallic oxide + water -----> acid CO 2(g) + H 2 O (l) ----> ; H 2 CO 3(aq) 5. Metal + nonmetal -----> salt 2 Na (s) + Cl 2(g) ----> 2NaCl (s) 6. A few nonmetals combine with each other. 2P (s) + 3Cl 2(g) ----> 2PCl 3(g) B. Decomposition: A single compound breaks down into simpler compounds. Basic form: AX -----> A + X Examples of decomposition reactions (when heated): 1. Metallic carbonates form metallic oxides and CO 2(g). CaCO 3(s) ----> CaO (s) + CO 2(g) 2. Most metallic hydroxides decompose into metal oxides and water. Ca(OH) 2(s) ----> CaO (s) + H 2 O (g) 3. Metallic chlorates decompose into metallic chlorides and oxygen. 2KClO 3(s) ----> 2KCl (s) + 3O 2(g) 4. Some acids decompose into nonmetallic oxides and water. H 2 SO 4 ----> H 2 O (l) + SO 3(g) 5. Some oxides decompose. 2HgO (s) ----> 2Hg (l) + O 2(g) 6. Some decomposition reactions are produced by electricity. 2H 2 O (l) ----> 2H 2(g) + O 2(g) 2NaCl (l) ----> 2Na (s) + Cl 2(g)
C. Replacement: a more active element takes the place of another element and frees the less active one. Basic form: A + BX -----> AX + B or AX + Y -----> AY + X Examples of replacement reactions: 1. Replacement of a metal in a compound by a more active metal. Fe (s) + CuSO 4(aq) ----> FeSO 4(aq) + Cu (s) 2. Replacement of hydrogen in water by an active metal. 2Na (s) + 2H 2 O (l) ----> 2NaOH (aq) + H 2(g) Mg (s) + H 2 O (g) ----> MgO (s) + H 2(g) 3. Replacement of hydrogen in acids by active metals. Zn (s) + 2HCl (aq) ----> ZnCl 2(aq) + H 2(g) 4. Replacement of nonmetals by more active nonmetals. Cl 2(g) + 2NaBr (aq) ----> 2NaCl (aq) + Br 2(l) NOTE: Refer to the Activity Series for metals and nonmetals to predict products of replacement reactions. If the free element is above the element to be replaced in the compound, then the reaction will occur. If it is below, then no reaction occurs. If the free element is above the element to be replaced in the compound, Then the reaction will occur. If the free element is below it, no reaction occurs. D. Ionic or Double Displacement: occurs between ions in aqueous solution. A reaction will occur when a pair of ions come together to produce at least one of the following: 1. a precipitate 2. a gas 3. water or some other non-ionized substance. Basic form: AX + BY -----> AY + BX Examples of ionic reactions: 1. Formation of precipitate. NaCl (aq) + AgNO 3(aq) ----> NaNO 3(aq) + AgCl (s) BaCl 2(aq) + Na 2 SO 4(aq) ----> 2NaCl (aq) + BaSO 4(s) 2. Formation of a gas. HCl (aq) + FeS (s) ----> FeCl 2(aq) + H 2 S (g) 3. Formation of water. (If the reaction is between an acid and a base it is called a neutralization reaction.) HCl (aq) + NaOH (aq) ----> NaCl (aq) + H 2 O (l) 4. Formation of a product which decomposes. CaCO 3(s) + HCl (aq) ----> CaCl 2(aq) + CO 2(g) + H 2 O (l)
Use the Solubility Rules to decide whether a product of an ionic reaction is insoluble in water and will thus form a precipitate. If a compound is soluble in water then it should be shown as being in aqueous solution, or left as separate ions. It is, in fact, often more desirable to show only those ions that are actually taking part in the actual reaction. Equations of this type are called Net Ionic equations. Combustion of Hydrocarbons: Another important type of reaction, in addition to the four types above, is that of the combustion of a hydrocarbon. When a hydrocarbon is burned with sufficient oxygen supply, the products are always carbon dioxide and water vapor. If the supply of oxygen is low or restricted, then carbon monoxide will be produced. This is why it is so dangerous to have an automobile engine running inside a closed garage or to use a charcoal grill indoors. Hydrocarbon (C x H y ) + O 2(g) -----> CO 2(g) + H 2 O (g) CH 4(g) + 2O 2(g) ----> CO 2(g) + 2H 2 O (g) 2C 4 H 10(g) + 13O 2(g) ----> 8CO 2(g) + 10H 2 O (g) NOTE: Complete combustion means the higher oxidation number is attained. Incomplete combustion means the lower oxidation number is attained. The phrase "To burn" means to add oxygen unless told otherwise. Conservation of matter requires identical numbers of atoms on both sides Begin by inspecting the number of each type of atom on each side. Nitrogen must separate before the reaction Hydrogen must separate before the reaction. Re-combine in a different chemical species Same atoms, different ratio!
D. Ionic or Double Displacement: occurs between ions in aqueous solution. A reaction will occur when a pair of ions come together to produce at least one of the following: 1. a precipitate 2. a gas 3. water or some other non-ionized substance. Basic form: AX + BY -----> AY + BX Examples of ionic reactions: 1. Formation of precipitate. NaCl (aq) + AgNO 3(aq) ----> NaNO 3(aq) + AgCl (s) BaCl 2(aq) + Na 2 SO 4(aq) ----> 2NaCl (aq) + BaSO 4(s) 2. Formation of a gas. HCl (aq) + FeS (s) ----> FeCl 2(aq) + H 2 S (g) 3. Formation of water. (If the reaction is between an acid and a base it is called a neutralization reaction.) HCl (aq) + NaOH (aq) ----> NaCl (aq) + H 2 O (l) 4. Formation of a product which decomposes. CaCO 3(s) + HCl (aq) ----> CaCl 2(aq) + CO 2(g) + H 2 O (l) Use the Solubility Rules to decide whether a product of an ionic reaction is insoluble in water and will thus form a precipitate. If a compound is soluble in water then it should be shown as being in aqueous solution, or left as separate ions. It is, in fact, often more desirable to show only those ions that are actually taking part in the actual reaction. Equations of this type are called Net Ionic equations.
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
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 O H (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 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