The Atom, The Mole & Stoichiometry. Chapter 2 I. The Atomic Theory A. proposed the modern atomic model to explain the laws of chemical combination.
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1 Unit 2: The Atom, The Mole & Stoichiometry Chapter 2 I. The Atomic Theory A. proposed the modern atomic model to explain the laws of chemical combination. Postulates of the atomic theory: 1. All matter is composed of extremely small, indivisible particles called. 2. All atoms of a given element are the. 3. Compounds form when atoms of different elements combine in proportions. 4. A chemical reaction only involves a rearrangement of atoms. No atoms are created or. Modifications of the atomic theory: 1. Atoms are. 2. All atoms of a given element are NOT the same; there are different of atoms, which differ in the number of. B. Dalton s Atomic Theory was based on scientific laws and experimentation: 1. Law of Conservation of Mass: The total mass remains constant during a reaction. This was determined quantitatively by during the late 18 th century in France. 2 HgO! 2 Hg + O 2 2. Law of Definite Proportions (Law of Composition): Joseph concluded that compounds always had a fixed composition. For example, water is 11.2% hydrogen and 88.8% oxygen by mass. 3. Law of Multiple Proportions: concluded that two elements can often combine in multiple ways. For example, carbon and oxygen can combine to form or. II. Subatomic Particles A. By the end of the 19 th century it was determined that the atom was divisible. 1. discovered the negatively-charged electron using a. 2. discovered the positively-charged nucleus with the experiment. 3. discovered the neutron with a nuclear rxn. B. The Nucleus 1. Contains the and the. It is extremely. 2. The relative mass of each of these particles is amu. 3. The proton is charged; the neutron is. C. The Electrons 1. The electron is charged, and it has a relative mass of amu. 2. In a neutral atom, the number of equals the number of. 1
2 D. Atomic Number ( ): The number of in an atom. This the element. H has proton; U has protons. (Written at the of the symbol). E. Mass Number ( ): The sum of the and in an atom. (Written at the of the symbol). F. Isotopes: Atoms that have the same number of but a different number of. 1. There are three isotopes of hydrogen: 2. Chlorine has two main isotopes: G. Number of neutrons = -. H. Example: Determine the number of neutrons, protons, and electrons in each of the following atoms: 1. Oxygen Uranium Carbon-14 III. Atomic Masses A. John Dalton arbitrarily assigned the mass of the hydrogen atom as, then determined the relative masses of all other atoms and called these atomic weights (we now use the term ). B. IUPAC has now agreed to use the pure isotope of as the standard mass. Therefore we state that 1 atom of is exactly 12 amu. 1 amu = 1/12 the mass of Carbon-12 = x g. C. Atomic mass is defined as the average of the masses of the occurring isotopes of that element. The atomic mass of oxygen is. The atomic mass of iron is. D. Calculations of atomic mass: atomic mass = Σ (fractional abundance x mass of isotope) E. Example: There are two naturally occurring isotopes of carbon. Their percent abundances and atomic masses are given: Carbon % amu Carbon % amu Calculate the atomic mass of carbon. 2
3 IV. The Periodic Table A. The first periodic table was organized by in It was organized by increasing. When organized in this manner, many elements within the columns showed similar chemical characteristics, though some elements seemed out of order. For example, he correctly switched and, due to their properties. He was able to predict the properties of yet undiscovered elements: B. The modern periodic table is organized by increasing. The table is divided into groups ( ) and periods ( ). ========================================================================================= Chapter 3 I. Molecular Mass and Formula Mass A. Molecular mass 1. Sum of the masses of atoms in a formula Example: Molecular mass of CO 2 2. The mass of ONE, therefore it is measured in 3. Molecular mass only refers to molecules, which are compounds comprised of. B. Formula mass 1. Sum of the masses of atoms in an compound. 2. Ionic compounds do not form, so a different term must be used. 3. The ionic compound of cupric bromite has a formula unit of. Example: Formula mass of cupric bromite: II. The Mole A. The mole ( ) is the number of atoms in exactly g of. B. This amount is known as Number. C. The mol allows us scale up chemical formulas and reactions from individual atoms and molecules to a measurable amount. Example: C + O 2! CO 2 3
4 D. Molar Mass 1. The mass of 1 of a substance. This term can be used for ANY substance: 1 mol of carbon = = atoms of carbon 1 mol of carbon dioxide = = molecules of carbon dioxide 1 mol of cupric bromite = = formula units of cupric bromite 2. Molar mass is more useful to chemists than atomic mass, molecular mass, or formula mass because it is measured in the unit of! 3. Determine the molar mass of: (1) ammonium sulfate, (2) 1-propanol E. Calculations: 1. Determine the mass of mol Na in grams. How many atoms is this? 2. Determine the mass of 3.77 x molecules of ethane. 3. Calculate the number of formula units in 85.6 g of calcium iodide. 4. Determine the number of ammonium ions in a 145 g sample of ammonium sulfate. 5. Determine the mass of 1-propanol that contains 1.5 mol of carbon. 4
5 6. How many atoms of gold would be found in a sphere with a diameter of 25 cm? (D = 19.3 g/ml) III. Percent Composition A. Percent composition: The percent by mass of each element in a compound. (Calculate to the hundredth s place). 1. Determine the percent composition of carbon dioxide. 2. Determine the percent composition of benzene. 3. Determine the percent water in cupric sulfate pentahydrate. IV. Chemical Formulas: Ratio of atoms or ratio of moles of atoms (e.g. H 2 O) A. Empirical Formula: The chemical formula in the LOWEST possible ratio. All compounds have only empirical formulas. 1. Determination of the empirical formula (simplest formula) from % composition: a. Assume 100 g of sample (% of the element becomes mass in grams). b. Convert the mass of each element into moles. c. Write the moles as a subscript in a preliminary formula. d. Divide by the smallest number of moles (to try to reach a whole number). e. If subscripts are still not whole number (or within +/ of a whole number), then multiply all subscripts by a multiple to achieve a whole number. 2. Examples: a. Determine the empirical formula of a compound with 82.66% C and 17.34% H. 5
6 b. Determine the empirical formula of a compound with 76.57%C, 6.43% H, and 17.00% O. B. Molecular Formula: The true chemical formula of a molecule. It is a multiple of the integral multiple of the empirical formula. 1. Determination of the empirical formula from the empirical formula: a. Determine the empirical formula. b. Calculate the empirical mass and the integral multiple: Multiple = molecular mass empirical mass c. Multiply the empirical formula subscripts by the calculated multiple. 2. Example: Determine the empirical and molecular formulas of a compound that contains 65.44% C, 5.50% H, and 29.06% O. It has a molecular mass of 110 amu. V. Elemental Analysis A. Technique used to experimentally determine the % composition of a compound. 1. Combustion analysis is used for organic molecules that contain C, H, and O. 2. Apparatus for combustion analysis: a. A sample of known mass (organic molecule) is placed in the boat inside the high T furnace. b. A stream of oxygen (excess) is passed over the sample in the furnace. c. The products in this combustion reaction ( and ) move to stage 2. d. A substance (such as magnesium perchlorate, ) is placed in stage 2 to remove water from the gas stream. e. The carbon dioxide continues to stage 3 where it is absorbed by sodium hydroxide ( ) which is a basic substance. f. The mass differences are determined for the substances in stage 2 and 3 to determine the amount of CO 2 and H 2 O absorbed. 6
7 3. Example: Burning a g sample of a compound in excess oxygen yields g CO 2 and g H 2 O. A separate experiment shows that the molecular mass is 90 amu. (a) Determine the % composition; (b) Determine the empirical formula; (c) Determine the molecular formula. (a) Step 1: Calculate % C (g CO 2! mol CO 2! mol C! g C! % C) Step 2: Calculate % H (g H 2 O! mol H 2 O! mol H! g H! %H) Step 3: Calculate % O (100% - %C - %H) (b) Empirical formula (c) Molecular formula VI. Chemical Reactions A. A chemical equation is the shorthand description of a chemical reaction (chemical change). B. If heat or catalysts are used in a chemical reaction, they are often indicated above the arrow. C. Coefficients in front of a chemical substance can represent the number of atoms, molecules or formula units OR they can be scaled up to represent the number of. 7
8 D. All chemical equations must be balanced to follow the Law of Conservation of Mass! VII. Balancing Chemical Reactions by Inspection (Trial and Error) A. Atoms on the reactants and products side of the reaction must be equal. B. Helpful strategies: 1. If a group (polyatomic ion) remains unchanged from reactants to products, keep it as a group. 2. If an element is present in just one compound on each side, try starting with that element. 3. Generally balance free elements last. 4. Sometimes it is easier to use fractions to balance, then clear the fractions at the end. C. Examples: Balance the following equations by inspection. 1. Fe + O2! Fe2O3 2. H 3 PO 4 + NaCN! HCN + Na 3 PO 4 3. C 2 H 6 + O 2! CO 2 + H 2 O 4. CaO + P 4 O 10! Ca 3 (PO 4 ) 2 VIII. Balancing Chemical Reactions Using the Algebraic Technique A. Balance the following equation: Ca 3 (PO 4 ) 2 + H 2 SO 4 CaSO 4 + H 3 PO 4 STEP 1: Assign letter to unknown coefficients: a Ca 3 (PO 4 ) 2 + b H 2 SO 4 c CaSO 4 + d H 3 PO 4 STEP 2: Make a grid indicating the appearance of element or ion in each species of the equation. Use whole number and the coefficient to indicate the appearance. Ca 3a + 0 = c + 0 PO 4 2a + 0 = 0 + d H 0 + 2b = 0 + 3d SO b = c + 0 STEP 2: Reduce the equations: 3a = c 2a = d 2b = 3d b = c STEP 3: Assume a = 1; solve for coefficients (NOTE: Any variable can be assumed to be 1) a = 1 b = 3 c = 3 d = 2 STEP 4: Write equation with coefficients (A coefficient of 1 is implied, not written) B. Examples: Ca 3 (PO 4 ) H 2 SO 4 3 CaSO H 3 PO 4 1. Na 2 CO 3 + C + Sb 2 S 3 Sb + Na 2 S + CO 2 8
9 2. C 6 H 6 + O 2 CO 2 + H 2 O 3. MnO 4 + CaC 2 O 4 + H 2 SO 4 CaSO 4 + Mn + CO 2 + H 2 O IX. Stoichiometry A. Coefficients in a chemical equation can represent the number of or. Example Write the balanced reaction: Carbon monoxide reacts with hydrogen to form methanol. B. In stoichiometry, the coefficients are used as the mol ratio. 1. (a) When mol propane is burned in an excess of oxygen, how many mol oxygen are consumed? (b) How many mol of carbon dioxide are formed? C. In the lab, most reactants and products are measured by mass (grams), so it is necessary to convert mol of reactants and products to. D. Stoichiometric road map (Use the balanced chemical equation) 9
10 Examples: (1) Aluminum reacts with oxygen to form aluminum oxide. If you are given 80.0 g of aluminum, (a) how many grams of aluminum oxide can be made? (b) How many moles of oxygen gas are required? (2) Pentane reacts with oxygen to form carbon dioxide and water. If 125 g of pentane is reacted with an excess of oxygen, (a) how many grams of carbon dioxide will form? (b) How many grams of water will form? (3) Titanium chloride (TiCl 4 ) reacts with magnesium to form titanium metal and magnesium chloride. How many gram of magnesium are required to react with 83.6 g of titanium chloride? (4) Ammonium sulfate is produced commercially by passing gaseous ammonia into an aqueous solution that is 65% sulfuric acid by mass and has a density of 1.55 g/ml. X. Limiting Reactants (Reagents) A. Stoichiometric amounts: The proportions indicated in the rxn. Most reactions do not have stoichiometric amounts. Generally one reactant is depleted before the other. The substance that runs out first is known as the limiting reagent. B. Analogy: How to make a cheese sandwich. 2 slices of bread + 1 slice of cheese 1 cheese sandwich If you have 9 slices of bread and 3 slices of cheese, how many sandwiches can you make? (theoretical yield) What is the limiting reagent? What is the excess reagent? How much of the excess reagent is left at the end of the rxn? 10
11 C. Limiting reagent problems (1) Magnesium nitride can be formed by the reaction of magnesium metal with nitrogen gas. (a) How many grams of magnesium nitride can be made in the reaction of g of magnesium and g of nitrogen? (b) How many grams of excess reagent remain after the reaction? (2) One way to produce hydrogen sulfide gas is by the reaction of ferrous sulfide with hydrochloric acid (ferrous chloride is also produced). If 10.2 g of hydrochloric acid is added to 13.2 g of ferrous sulfide how many grams of hydrogen sulfide can be formed? What is the mass of the excess reagent remaining? (3) A laboratory source of hydrogen gas is the reaction between aqueous hydrochloric acid and aluminum metal. An aqueous solution of aluminum chloride is the other product in the reaction. How many grams of hydrogen are produced in the reaction of 12.5 g of aluminum and ml of an aqueous hydrochloric acid solution that is 25.6% HCl by mass and has a density of 1.13 g/ml? XI. Yields of Chemical Reactions A. Theoretical yield: The amount of product that forms if all of the reagent has reacted. (This number is CALCULATED!) B. Actual yield: The amount of product that is actually made (Experimental). C. Percent yield: The comparison of the actual yield to the theoretical yield. % yield = actual yield x 100 theoretical yield 11
12 D. Examples: (1) Solid ferric oxide reacts with carbon monoxide to form iron metal and carbon dioxide. (a) If you start with g of ferric oxide as the limiting reagent, what is the theoretical yield of Fe? (b) If the actual yield of Fe in your reaction was 87.9 g, what was the percent yield? (2) Ethyl acetate (CH 3 COOCH 2 CH 3 ) is a solvent used as fingernail polish remover. What mass of acetic acid is needed to prepare 252 g ethyl acetate if the expected percent yield is 85.0%? Assume that the other reactant, ethanol, is present in excess and water is the other product. (3) Isopentyl acetate (C 7 H 14 O 2 ) is the main component of banana flavoring. (a) Calculate the theoretical yield of isopentyl acetate that can be made from 20.0 g isopentyl alcohol (C 5 H 12 O) and acetic acid (water is also produced). (b) If the percent yield is 90.0%, what is the actual yield of isopentyl acetate? XII. Solution Stoichiometry A. Solutions are mixtures in which the is dissolved into the. The solute is generally present in amount. B. Example solvents: water (aqueous solutions), ethanol, isopropanol, benzene, acetone, hexane C. Concentration: Quantity of solute dissolved in a given amount of solvent. 1. % by mass: = mass solute / mass solution x Molarity M = mol solute / L solution D. To prepare an aqueous solution, a flask should be used. The solute is weighed out and added to the flask; distilled water is added to bring the solution to the mark on the glassware. Examples: 1. What is the molarity of a solution in which 333 g potassium bicarbonate is dissolved in enough water to make a 10.0 L solution? 12
13 2. What is the molarity of the acid solution when g hydrochloric acid is dissolved in L of solution? 3. Calculate the molarity of ethanol in a solution containing 10.5 ml of ethanol (D = g/ml) in 25.0 ml of solution. 4. What mass of sodium chloride is needed to make 250 ml of a 5.0 M solution? E. Dilutions of Solutions: Mol remains constant as a solution is diluted with solvent. Therefore: M 1 V 1 = M 2 V 2 Examples: 1. How many ml of concentrated sulfuric acid (18 M) would be needed to make 1 L of a 0.1 M solution? How would the diluted solution be prepared? 2. If 50.0 ml of 2.00 M CuSO 4 is added to a 250 ml volumetric flask and diluted with deionized water, what is the new concentration? F. Solution Stoichiometry Problems 1. To test commercial antacids, you use 0.10 M HCl to stimulate the acid concentration of the stomach. How many liters of this acid would react with a table containing 0.10 g of Mg(OH) 2 (Products are MgCl 2 and water). 13
14 2. One way to remove mercury from wastewater is with the following reaction: Hg(NO 3 ) 2 (aq) + Na 2 S(aq)! HgS(s) + 2 NaNO 3 In a lab, 50.0 ml of M mercury (II) nitrate reacts with 20.0 ml of 0.10 M sodium sulfide. (a) How many grams of mercury (II) sulfide form? (b) What is the concentration of the excess reagent at the end of the reaction? 14
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