Mass. Unit 1: Stoichiometry (Review) Adapted from Stoichiometry (NMSI, Rene McCormick) and Chemistry (Brown/LeMay)

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1 Unit 1: Stoichiometry (Review) Adapted from Stoichiometry (NMSI, Rene McCormick) and Chemistry (Brown/LeMay) 1.18 Atomic Masses (pp ) 12 C Carbon 12 In 1961 it was agreed that this would serve as the standard and would be defined to have a mass of EXACTLY 12 atomic mass units (amu). All other atomic masses are measured relative to this. mass spectrometer a device for measuring the mass of atoms or molecules atoms or molecules are passed into a beam of high speed electrons this knocks electrons OFF the atoms or molecules transforming them into cations apply an electric field this accelerates the cations since they are repelled from the (+) pole and attracted toward the ( ) pole send the accelerated cations into a magnetic field an accelerated cation creates it s OWN magnetic field which perturbs the original magnetic field this perturbation changes the path of the cation the amount of deflection is proportional to the mass; heavy cations deflect little ions hit a detector plate where measurements can be obtained. 13 Mass C 12 Mass C Mass 13 C ( )(12amu) amu Exact by definition average atomic masses atoms have masses of whole numbers, HOWEVER samples of quadrillions of atoms have a few that are heavier or lighter [isotopes] due to different numbers of neutrons present percent abundance percentage of atoms in a natural sample of the pure element represented by a particular isotope percent abundance = number of atoms of a given isotope 100 Total number of atoms of all isotopes of that element counting by mass when particles are small this is a matter of convenience. Just as you buy 5 lbs of sugar rather than a number of sugar crystals, or a pound of peanuts rather than counting the individual peanuts.this concept works very well if your know an average mass. mass spectrometer to determine isotopic composition load in a pure sample of natural neon or other substance. The areas of the peaks or heights of the bars indicate the relative abundances of 10 Ne, 10 Ne, and 10 Ne Exercise 1.24 Average Atomic Mass When a sample of natural copper is vaporized and injected into a mass spectrometer, the results shown in the figure are obtained. Use these data to compute the average mass of natural copper. (The mass values for 63 Cu and 65 Cu are amu and amu, respectively.) amu

2 1.19 The Mole (pp ) mole the number of C atoms in exactly 12.0 grams of 12 C; also a number, just as the word dozen means 12 and couple means 2. Avogadro s number , the number of particles in a mole of anything Calculated to help chemists estimate very large numbers of very small particles Mass (g) Molar Mass g/mol Moles 22.4 L = 1 mol (gas at STP) Volume (L) 6.02x10 23 part = 1 mol Mol/L # of Particles Molarity (mol/l) Exercise 1.25 Mole Conversions Mass/Particles Americium is an element that does not occur naturally. It can be made in very small amounts in a device known as a particle accelerator. Compute the mass in grams of a sample of americium containing six atoms. 2x10-21 g Exercise 1.26 Mole Conversions Mass/Particles Aluminum (A1) is a metal with a high strength to mass ratio and a high resistance to corrosion; thus it is often used for structural purposes. Compute both the number of moles of atoms and the number of atoms in a 10.0 g sample of aluminum. 2.23x10 23 atoms Exercise 1.27 Mole Conversions Moles/Mass Cobalt (Co) is a metal that is added to steel to improve its resistance to corrosion. Calculate both the number of moles in a sample of cobalt containing atoms and the mass of the sample. 8.31x10-4 mol; g Co 1.20 Molar mass, Molecular weight, and Formula weight (pp , 92 93) molar mass, MM the mass in grams of Avogadro s number of molecules; i.e. the mass of a mole! molecular weight, MW sum of all the atomic weights of all the atoms in the formula (must have a correct formula!) empirical formula that ratio in the network for an ionic substance. formula weight same as molecular weight, just a language problem [ molecular implies covalent bonding while A formula implies ionic bonding {just consider this to be a giant conspiracy designed to keep the uneducated from ever understanding chemistry kind of like the scoring scheme in tennis}. We ll use MM for all formula masses. Units: grams per mole (g/mol)

3 The mass of 1 atom in amu = the mass of 1 mole of atoms of that element in g Ex: 1 formula unit of NaCl has a mass of 58.5 amu but 1 mole of NaCl (6.02x10 23 units of NaCl) has a mass of 58.5 g Significant Figures: It is correct to use as many significant figures for the molar mass as you are given in your problem. However, most of the time you ll be fine if you round the molar masses to 2 places after the decimal. Exercise 1.28 Calculating Molar Mass Juglone, a dye known for centuries, is produced from the husks of black walnuts. It is also a natural herbicide (weed killer) that kills off competitive plants around the black walnut tree but does not affect grass and other noncompetitive plants [a concept called allelopathy]. The formula for juglone is C 10 H 6 O 3. a. Calculate the molar mass of juglone g/mol b. A sample of 1.56 x 10 2 g of pure juglone was extracted from black walnut husks. How many moles of juglone does this sample represent? 8.96x10-5 mol Exercise 1.29 Calculating Molar Mass Calcium carbonate (CaCO 3 ), also called calcite, is the principal mineral found in limestone, marble, chalk, pearls, and the shells of marine animals such as clams. a. Calculate the molar mass of calcium carbonate g/mol b. A certain sample of calcium carbonate contains 4.86 moles. What is the mass in grams of this sample? What is the mass of the CO 3 2 ions present? 486 g; 292 g Exercise 1.30 Molar Mass and Molecules Isopentyl acetate (C 7 H 14 O 2 ), the compound responsible for the scent of bananas, can be produced commercially. Interestingly, bees release about 1µg ( g) of this compound when they sting. The resulting scent attracts other bees to join the attack. How many molecules of isopentyl acetate are released in a typical bee sting? How many atoms of carbon are present? 5x10 15 molecules 3x10 16 atoms ELEMENTS THAT EXIST AS MOLECULES Pure hydrogen, nitrogen, oxygen and the halogens exist as DIATOMIC molecules under normal conditions. MEMORIZE!!! Be sure you compute their molar masses as diatomics. Others to be aware of, but not memorize: P 4 tetratomic form of elemental phosphorous; an allotrope S 8 sulfur s elemental form; also an allotrope Carbon diamond and graphite covalent networks of atoms

4 1.21 Percent Composition (pp ) Percent Composition the percent by mass of each element in the compound Two common ways of describing the composition of a compound: in terms of the number of its constituent atoms and in terms of the percentages (by mass) of its elements. Percent (by mass) Composition: law of constant composition states that any sample of a pure compound always consists of the same elements combined in the same proportions by mass. % comp = mass of desired element 100 Total mass of compound Consider ethanol, C 2 H 5 OH g Mass % of C = 2 mol = g mol g Mass % of H = 6 mol 1.01 = 6.06 g mol g Mass % of O = 1 mol = 16.00g mol Mass of 1 mol of C 2 H 5 OH = g NEXT THE MASS PERCENT CAN BE CALCULATED: Mass percent of C = g C 100% = 52.14% g Repeat for the H and O present. Exercise 1.31 Calculating Mass Percent Carvone is a substance that occurs in two forms having different arrangements of the atoms but the same molecular formula (C 10 H 14 O) and mass. One type of carvone gives caraway seeds their characteristic smell, and the other type is responsible for the smell of spearmint oil. Compute the mass percent of each element in carvone % C; 9.41% N; 10.65% O Exercise 1.32 Calculating Mass Percent Penicillin, the first of a now large number of antibiotics (antibacterial agents), was discovered accidentally by the Scottish bacteriologist Alexander Fleming in 1928, but he was never able to isolate it as a pure compound. This and similar antibiotics have saved millions of lives that might have been lost to infections. Penicillin F has the formula C 14 H 20 N 2 SO 4. Compute the mass percent of each element % C; 6.47% H; 8.97% N; 10.26% S; 20.48% O

5 1.22 Determining the Formula of a Compound (pp ) Empirical formula gives the relative number of atoms of each element present in a compound hydrates dot waters used to cement crystal structures. anhydrous without water To calculate E.F. from % composition data: 1. Assume 100 g of the sample (so the % = mass) (if given grams go straight to step 2) 2. Convert grams of each element in the sample to moles 3. Calculate the mole ratio (divide all moles by the lowest number of moles) 4. Ratio = subscripts in the E.F. **If in #4 you don t get whole numbers, multiply all by a number to get a whole number **E.F. must have the lowest whole number subscripts possible (lowest ratio possible) Calculating molecular formula from empirical formula The subscripts in the M.F. are always a whole number multiple of the subscripts in the E.F. [(empirical formula) n, where n is an integer] To find M.F., you have to know the molar mass of the molecular formula Whole Number Multiple = molar mass/empirical mass Combustion Analysis Way of analyzing unknown compounds to find composition When faced with a compound of unknown formula, one of the most common techniques is to combust it with oxygen to produce CO 2, H 2 O, and N 2 which are then collected and weighed. When an organic compound (a hydrocarbon) undergoes combustion, all of the C from the compound is converted into CO 2 and all of the hydrogen is converted into H 2 O. You can find out how much of each was in the original sample using stoichiometry. Example: A compound is composed of carbon, nitrogen and hydrogen. When g of this compound is reacted with oxygen [burned, combusted], g or carbon dioxide and g of water are collected. What is the empirical formula of the compound? Compound + O 2 CO 2 + H 2 O + N 2 but NOT balanced!! (You can see that all of the carbon ended up in CO 2 so when in doubt, FIND THE NUMBER OF MOLES!!) g CO g/mol = moles of CO 2 x 1 mol C/1 mol CO 2 = moles of C (Next, you can see that all of the hydrogen ended up in H 2 O, so.find THE NUMBER OF MOLES!!) g H 2 O g/mol = moles of H 2 O (BUT there are 2 moles of H for each mole of water [ think organ bank one heart per body, one C per molecule of carbon dioxide 2 lungs per body, 2 atoms H in water and so on ] so DOUBLE THE NUMBER OF MOLES TO GET THE NUMBER OF MOLES OF HYDROGEN!!) mol H 2 O = moles of H The rest must be nitrogen, BUT we only have mass data for the sample so convert your moles of C and H to grams: SUBTRACT! g C = moles C = grams C + g H = moles H 1.01 = grams H grams C + H g sample g thus far = grams N left = g N so g N = moles N Chemical formulas represent mole to mole ratios, so divide the number of moles of each by the smallest # of moles of any one of them to get a guaranteed ONE in your ratios multiply by 2, then 3, etc to get to a ratio of small whole numbers!!

6 Element # moles ALL Divided by C H N Therefore the correct EMPIRICAL formula is CH 5 N. Next, if we are told that the MM is g/mol, then simply use this relationship: (Empirical mass) n = MM ( ) n = Solve for n n = or essentially one, so the empirical formula and the molecular formula are the same. Exercise 1.33 Determining Empirical & Molecular Formulas Determine the empirical and molecular formulas for a compound that gives the following analysis (in mass percents): The molar mass is known to be g/mol % C % C 4.07% H CH 2 Cl; C 2 H 4 Cl 2 Exercise 1.34 Determining Empirical & Molecular Formulas A white powder is analyzed and found to contain 43.64% phosphorus and 56.36% oxygen by mass. The compound has a molar mass of g/mol. What are the compound s empirical and molecular formulas? P 2 O 5 ; P 4 O 10 Exercise 1.35 Determining a Molecular Formula Caffeine, a stimulant found in coffee, tea, and chocolate, contains 49.48% carbon, 5.15% hydrogen, 28.87% nitrogen, and 16.49% oxygen by mass and has a molar mass of g/mol. Determine the molecular formula of caffeine. C 8 H 10 N 4 O 2

7 1.23 Chemical Reactions and Equations (pp ) Chemical reactions are the result of a chemical change where atoms are reorganized into one or more new arrangements. Bonds are broken [requires energy] and new ones are formed [releases energy]. Chemical equation a representation of a chemical reaction Reactants are listed on the left side of the arrow, product on the right Coefficients represent the relative number of molecules and/or moles required for the reaction Energy is included SOMETIMES (called a thermochemical equation) to show that the rxn is either endothermic or exothermic (this isn t info you have to figure out, you ll be told if you re supposed to include it in the equation) The time required for a rxn to occur is not included Law of Conservation of Mass (Lavoisier) atoms (matter) can t be created or destroyed, so the number of atoms on the left of the arrow has to be the same as the atoms on the right To satisfy the law of conservation of mass, balance the equation Write correct formulas for all compounds in the reaction Never change a subscript in an equation Use coefficients to make the number of atoms equal on each side of the arrow States of matter for each compound in the reaction are represented by: (s) = solid (l) = liquid Information given by a chemical equation (g) = gas (aq) = aqueous Writing & Balancing Equations Begin with the most complicated looking thing (save the elemental thing for last). If you get stuck, double the most complicated looking thing. MEMORIZE THE FOLLOWING: metals + halogens M a X b CH (and/or O) + O 2 CO 2 (g) + H 2 O(g) H 2 CO 3 [any time formed!] CO 2 + H 2 O; in other words, never write carbonic acid as a product, it spontaneously decomposes [in an open container] to become carbon dioxide and water. metal carbonates metal OXIDES + CO 2 Types of Chemical Reactions (general) Combination (Synthesis) Reactions (A + B AB) Reactants can be single elements or compounds or both! Product must be one compound Decomposition Reactions (AB A + B) Generally require energy (endothermic) Reactant must be one compound Products can be single elements or compounds, or both! Combustion in Air Rapid reactions in which a flame is produced Most use O 2 in air as a reactant Organic Combustion C x H y + O 2 CO 2 + H 2 O (complete combustion) Sometimes incomplete combustion occurs C x H y + O 2 CO + H 2 O (it depends on how much oxygen is there as to whether combustion is complete or incomplete) Unless you are told otherwise, assume complete combustion! The water product can be liquid or gaseous depending on the reaction conditions

8 Exercise 1.36 Balancing Equations Chromium compounds exhibit a variety of bright colors. When solid ammonium dichromate, (NH 4 ) 2 Cr 2 O 7, a vivid orange compound, is ignited, a spectacular reaction occurs, as shown in the two photographs on page 105. Although the reaction is actually somewhat more complex, let s assume here that the products are solid chromium(iii) oxide, nitrogen gas (consisting of N 2 molecules), and water vapor. Balance the equation for this reaction. (NH 4 ) 2 Cr 2 O 7 Cr 2 O 3 + N 2 + 4H 2 O Exercise 1.37 Balancing Equations At 1000ºC, ammonia gas, NH 3 (g), reacts with oxygen gas to form gaseous nitric oxide, NO(g), and water vapor. This reaction is the first step in the commercial production of nitric acid by the Ostwald process. Balance the equation for this reaction. 4NH 3 + 5O 2 4NO + 6H 2 O 1.24 Stoichiometric Calculations: Amounts of Reactants and Products (pp ) Coefficients in a balanced equation represent the relative numbers of molecules (or moles) in a reaction Mole ratio allows me to convert between compounds present in a reaction SOOOOOO important to be good at this it s not going away! If you see a reaction, think stoich! Can solve using dimensional analysis or a new way either way is fine, whatever is easiest for you! you have to be proficient at the following no matter which method you choose!: Writing CORRECT formulas this requires knowledge of your polyatomic ions and being able to use the periodic table to deduce what you have not had to memorize. Review section 2.8 in your Chapter 2 notes or your text. Calculate CORRECT molar masses from a correctly written formula Balance a chemical equation Use the mole map to calculate the number of moles or anything else! Example: What mass of oxygen will react with 96.1 grams of propane? Option 1: Solve using dimensional analysis: 1. Write the balanced equation: C 3 H 8 + 5O 2 3CO 2 + 4H 2 O 2. Use D.A. to go from grams of propane to grams of oxygen: Option 2: Solve using the table method: 1. Make a table Molar Mass: mole:mole # moles amount

9 2. Write a chemical equation paying special attention to writing correct chemical formulas! Molar Mass: mole:mole # moles amount 3. Calculate the molar masses and put in parentheses above the formulas soon you ll figure out you don t have to do this for every reactant and product, just those you re interested in. mole:mole # moles amount 4. Look at the coefficients on the balanced equation, they ARE the mole:mole ratios! # moles amount 5. Next, re read the problem and put in an amount in this example it s 96.1 g of propane. # moles amount 96.1 g 6. Find the number of moles of something anything! Use the mole map start at 96.1 g, divide by molar mass to get the # of moles of propane. # moles 2.18 amount 96.1 g

10 7. Use the mole:mole ratio to find moles of EVERYTHING! If 1 = 2.18, then oxygen is 5(2.18) etc (if the first you find is not a 1 just devide to make it 1 ) Leave all digits in your calculator, I only rounded to save space! # moles amount 96.1 g 8. Re read the problem to determine which amount was asked for here s the payoff AP problems ask for several amounts! First we ll find mass of oxygen required since that s the problem asked mol x 32 g/mol = 349 g O 2 # moles amount 96.1 g 349 g 9. What if another part of the question asked for liters of CO2 at STP (1 atm, 273 K)? Use the mole map. Start in the middle with 6.53 moles x 22.4 L/mol = 146 L # moles amount 96.1 g 349 g 146 L 10. What if another part asked how many water molecules are produced? Use the mole map. Start in the middle with 8.71 moles x 6.02x10 23 molecules/1 mol = 5.24x10 24 molecules water # moles amount 96.1 g 349 g 146 L 5.24x10 24 molec. Either way table method or Dimensional Analysis whatever is easier and makes more sense to you! Exercise 1.38 Stoichiometry Solid lithium hydroxide is used in space vehicles to remove exhaled carbon dioxide from the living environment by forming solid lithium carbonate and liquid water. What mass of gaseous carbon dioxide can be absorbed by 1.00 kg of lithium hydroxide? 919 g

11 Exercise 1.39 Stoichiometry Baking soda (NaHCO 3 ) is often used as an antacid. It neutralizes excess hydrochloric acid secreted by the stomach: NaHCO 3 (s) + HCl(aq) NaCl(aq) + H 2 O(l) + CO 2 (aq) Milk of magnesia, which is an aqueous suspension of magnesium hydroxide, is also used as an antacid: Mg(OH) 2 (s) + 2HCl(aq) 2H 2 O(l) + MgCl 2 (aq) Which is the more effective antacid per gram, NaHCO 3 or Mg(OH) 2? Mg(OH) Calculations involving a Limiting Reactant (pp ) How to Recognize a L.R. problem: you ll be given 2 amounts for the reactants in the problem! One of the reactants will be the limiting reactant and the other will be in excess Limiting reactant reactant that is completely used up in a reaction (also called the limiting reagent) Excess reactant reactant that is not used up in the reaction (you ll have some left over when the reaction is done) Example: Suppose 25.0 kg of nitrogen reacts with 5.00 kg of hydrogen to form ammonia. What mass of ammonia can be produced? Which reactant is the limiting reactant? What is the mass of the reactant that is in excess? To Solve: (use D.A. or the table method I ll show table method since that s new!) 1. Set up your table like before only now you ll have 2 amounts to start with: Molar Mass: (28.04) (2.02) (17.04) N H 2 2 NH 3 mole:mole # moles amount 25,000 g 5,000 g 2. Find the # of moles of both reactants you re given info about Molar Mass: (28.02) (2.02) (17.04) N H 2 2 NH 3 mole:mole # moles 892 moles 2,475 moles amount 25,000 g 5,000 g 3. To find which reactant is limiting, pick one either N 2 or H 2 it doesn t matter. Let s pick H 2. Calculate how much N 2 is used if all of the H 2 is used up. WHAT IF I used up all the moles of hydrogen? I d need 1/3 2,475 moles = 825 moles of nitrogen. Clearly I have EXCESS moles of nitrogen!! Therefore, hydrogen limits me.

12 OR WHAT IF I used up all the moles of nitrogen? I d need moles = 2,676 moles of hydrogen. Clearly I don t have enough hydrogen, so it limits me!! Therefore nitrogen is in excess. Either way, I ve established that hydrogen is the limiting reactant so I modify the table: Molar Mass: (28.02) (2.02) (17.04) N H 2 2 NH 3 mole:mole # moles 825 mol used 1650 mol produced 892 moles amount 825 mol (28.02) = 23,116 g used 25,000 g 1,884 g excess!! 2,475 moles 5,000 g 1650 mol (17.04) = 28,116 g produced Here s the question again, let s clean up any sig.fig issues: Suppose 25.0 kg of nitrogen reacts with 5.00 kg of hydrogen to form ammonia. (3 sig. fig. limit) What mass of ammonia can be produced? 23,100 g produced = 23.1 kg (always polite to respond in the unit given. Which reactant is the limiting reactant? hydrogen once that s established, N 2 doesn t matter anymore! What is the mass of the reactant that is in excess? 1,884 g = 1.88 kg excess nitrogen!! Exercise 1.40 Stoichiometry: Limiting Reactant Nitrogen gas can be prepared by passing gaseous ammonia over solid copper(ii) oxide at high temperatures. The other products of the reaction are solid copper and water vapor. If a sample containing 18.1 g of NH 3 is reacted with 90.4 g of CuO, which is the limiting reactant? How many grams of N 2 will be formed? 10.7 g Theoretical yield how much product you ll make when all of the limiting reactant has been used up (this assumes perfect conditions and gives a maximum amount not likely!) Actual yield amount of product actually made 100 % Exercise 1.41 Calculating Percent Yield Methanol (CH 3 OH), also called methyl alcohol, is the simplest alcohol. It is used as a fuel in race cars and is a potential replacement for gasoline. Methanol can be manufactured by combination of gaseous carbon monoxide and hydrogen. Suppose 68.5 kg CO(g) is reacted with 8.60 kg H 2 (g). Calculate the theoretical yield of methanol. If g CH 3 OH is actually produced, what is the percent yield of methanol? 52.3 %