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STOICHIOMETRY (The Mathematics of Chemical Equations!) Rice Atomic Mass Atoms are so small, it is difficult to discuss how much they weigh in grams. So……. We use atomic mass units. an atomic mass unit (amu) is 1/12 the mass of a carbon-12 atom. This gives us a basis for comparison. The decimal numbers on the periodic table are average atomic masses in amu. They are not whole numbers Why???? Because they are based on averages of atoms and of isotopes. We figure out the average atomic mass of each element from the mass of the isotopes and their relative abundance. Massaverage = Σ (% abundance x Massisotope) Examples There are two isotopes of carbon C-12 with a mass of 12.00000 amu (98.892%), and C-13 with a mass of 13.00335 amu (1.108%). What is the average atomic mass. C-12 = 0.98892 x 12.00000 = 11.86704 C-13 = 0.01108 x 13.00335 = 0.14408 C = 12.01 Wow! it matches the periodic table Your Turn There are two isotopes of nitrogen, one with an atomic mass of 14.0031 amu and one with a mass of 15.0001 amu. What is the percent abundance of each isotope? Solution 14.0067 • 100 = Σ [(14.0034 • x) + (15.0001 • (100-x)] 99.668907 = x So % abundance of N-14 = 99.668907% % abundance of N-15 = 0.3310926% CHECK: N-14 = 14.0031 x 0.9968907 = 13.95956 N-15 = 15.0001 x 0.003310926 = + 0.04966422 close enough 14.0092 amu The Mole The Mole The mole is a number. Amounts in chemistry are measured in moles. The Mole The average mass of each element on the periodic table is = 1 mole. How we measure how much? We can measure mass, or volume, or we can count pieces. We measure mass in grams. We measure volume in liters. We count pieces in MOLES. 1 mole is equal to: 1. The GMW of any compound 2. 22.4 L volume 3. 6.02 x 1023 molecules, particles, or atoms (6.02 x 1023 is called Avagadro’s number) Treat it like a very large dozen! Molar Mass Some Confusing Vocabulary We call the smallest pieces of a substance … particles. We refer to molecular compounds as …molecules. We refer to ionic compound as … formula units. We call an element …an atom. Molar Mass: The mass of one mole of anything. Also called: 1. 2. 3. 4. 5. 6. 7. gram molecular mass (MM) gram formula mass (GFM) gram atomic mass (GAM) molecular weight (MW) formula weight (FW) atomic mass (AM) formula mass (FM) Molar mass The molar mass of an element is the mass written on the periodic table. molar mass of C is 12.011g/mol FOR DIATOMCS: molar mass of O2 is 32.00g/mol What about compounds? In 1 mole of H2O molecules there are two moles of H atoms and 1 mole of O atoms To find the mass of one mole of a compound 1. Determine the number of moles of each element in the compound 2. Find out the mass of 1 mole then multiply it by the # of moles 3. Add them all up Example What is the mass of one mole of CH4? 1 mole of C = 12.01 g 4 moles of H so (4 x 1.01g) = 4.04g Add them together = 16.05g So the GMW of CH4 is 16.05g Note: We call this the Gram Molecular mass of CH4 since it is the mass of a molecular compound. Gram Formula Mass The mass of one mole of an ionic compound. Calculated the same way. What is the GFM of Fe2O3? 2 moles of Fe (2 x 55.85g) = 111.70 g 3 moles of O (3 x 16.00g) = 48.00 g The GFM = 111.70 g + 48.00 g = 159.70g Your Turn Find the molar mass of: 1) CH4 2) Mg3P2 3) Ca(NO3)3 4) Al2(Cr2O7)3 5) CaSO4 · 2H2O Solution 1) CH4 = 16.04 g/mole 2) Mg3P2 = 134.85 g/mole 3) Ca(NO3)3 = 226.10 g/mole 4) Al2(Cr2O7)3 = 701.96 g/mole 5) CaSO4 · 2H2O = 172.14 g/mole More Examples Calculate the molar mass of the following: 1. Na2S 2. N2O4 3. C 4. Ca(NO3)2 5. C6H12O6 6. (NH4)3PO4 All the things we can change Mass Moles MW Moles Mass Volume MW Moles Mass Volume 22.4 L Moles MW Mass Volume 22.4 L MW Moles Representative Particles Mass Volume 22.4 L MW Moles 6.02 x 23 10 Representative Particles Mass Volume 22.4 L MW Moles 6.02 x 23 10 Representative Particles Atoms Mass Volume 22.4 L MW Mass Moles 6.02 x 23 10 Representative Particles Atoms Ions Can You answer these Questions? How many oxygen atoms in the following? CaCO3 Al2(SO4)3 How many ions in the following? CaCl2 NaOH Al2(SO4)3 Using Molar Mass Conversions using Molar Mass We make conversion factors using molar mass. How many moles in 5.69 g of NaOH? Need to change grams to moles for NaOH: 1 mole Na = 22.99g 1 mol O = 16.00 g 1 mole of H = 1.01 g So 1 mole NaOH = 40.00 g Use Factor-Label: 5.69g NaOH 1 1 mol NaOH 40g NaOH = 0.142 mol NaOH Your Turn How many moles is 4.56 g of CO2 ? How many grams is 9.87 moles of H2O? How many molecules in 6.8 g of CH4? 49 molecules of C6H12O6 has what mass? How many molecules of CO2 are the in 7.61 moles of CO2 ? How many moles of water is 5.87 x 1022 molecules? How many atoms of carbon are there in 1.23 moles of C6H12O6 ? How many moles is 7.78 x 1024 formula units of MgCl2? Other Types of Questions How many oxygen atoms in the following? CaCO3 Al2(SO4)3 How many ions in the following? CaCl2 NaOH Al2(SO4)3 Gases and the Mole + THE Gases Many of the chemicals we deal with are gases. Two things effect the volume of a gas: 1) Temperature 2) pressure Always compare gases at the same temperature and pressure. Standard Temperature and Pressure (abbreviated STP) Standard Temperature = 0ºC Standard Pressure = 1 atm pressure At STP 1 mole of gas occupies 22.4 L 22.4L/mole is called the molar volume Avagadro’s Hypothesis - at the same temperature and pressure equal volumes of gas have the same number of particles. Example 1. What is the volume of 4.59 mole of CO2 gas at STP? 4.59 mol CO2 1 22.4L CO2 = 102.82 L CO2 1 mol CO2 Your Turn 1) What is the volume of 4.59 mole of CO2 gas at STP? 2) How many moles is 5.67 L of O2 at STP? 3) What is the volume of 8.8g of CH4 gas at STP? More Problems 4) What is the mass of 2.34 moles of carbon? 5) How many moles of magnesium in 24.31g of Mg? 6) How many atoms of lithium in 1.00g of Li? 7) What is the mass of 3.45 x 1022 atoms of U? MASS DENSITY VOLUME D=m/V 3 (g/cm Density of a Solid or g/ml) is easy, but how do you get the density of a gas? Density of a gas D=m/V For a gas the units will be g / L We can determine the density of any gas at STP if we know its chemical formula (CO2, H2, Cl2) To find the density we need the mass and the volume. Examples 1. Find the density of CO2 at STP. 1 mole = 22.4L and 44.011g D = 44.011g = 2.0g/L 22.4L Your Turn 1. Find the density of CH4 at STP. The other way Given the density, we can find the molar mass of any gas. Ex: What is the molar mass of a gas with a density of 1.964 g/L? 1. Pretend you have a mole @ STP, V= 22.4L 2. D = m / V 3. 1.964g/L = x / 22.4L x = 43.994g Try This!!! What is the molar mass of a gas with a density of 2.86 g/L? We have learned how to change moles to grams moles to atoms moles to formula units moles to molecules moles to liters molecules to atoms formula units to atoms formula units to ions Percent Composition Percent Composition Percent of each element (by mass) in a compound: Part x 100 = % Whole (must equal 100%) Example: Find the % composition of CH4 C = 12.011 x 100 = 74.8722 = 74.87% 16.042 H = (1.0079 x 4) x 100 = 25.131 = 25.13% 16.042 100% Your Turn Find the % composition of each element in: 1) Al2(Cr2O7)3 2) CaSO4 · 2H2O Solution 1) Al2(Cr2O7)3 Al = 7.68 % Cr = 44.45 % O = 47.87 % 2) CaSO4 · 2H2O Ca = 23.28 % S = 18.62 % O = 55.76 % H = 2.34 % Empirical & Molecular Formulas Empirical Formula The lowest ratio of atoms in a molecule. From percent composition, you can determine the empirical formula. Empirical Formulas are based on mole ratios. Steps: 1) Find the moles 2) Divide by the smallest # of moles 3) Use answer from #2 as subscripts of each element Example A sample is 52.11% C, 13.14% H and 34.75% O. What is its empirical formula? C = 52.11g C 1 mole = 4.339 moles = 2 12.011g C 2.17 H = 13.14g H O = 34.75g O 1 mole = 13.037 moles = 6 1.0079g H 2.17 1 mole = 2.17 moles = 1 16.00g O 2.17 So the empirical formula is: C2H6O Your Turn A 1.334 gram sample of a compound contains 0.365g Na, 0.758g O and Nitrogen. What is the empirical formula of this compound? Answer NaNO3 Molecular Formula – The True Formula Empirical To Molecular Formulas Empirical is lowest ratio. Molecular is actual formula of the molecule. Need Molar mass of the empirical molecule. Ratio of empirical to molar mass will tell you the molecular formula. Must be a whole number because subscripts are never decimals. Empirical to molecular Since the empirical formula is the lowest ratio, the actual molecule could have a greater mass…. By a whole number multiple or the empirical and the molecular formula could be the same. To Find Molecular Formula from the Empirical Formula Divide the actual molar mass of the molecule given by the mass of one mole of the empirical formula molecule. Molecular Mass of Molecular Formula Molar Mass of the Empirical Formula = # to multiply the subscripts of the empirical formula by to give you the molecular formula. Example Determine the molecular formula of the compound with an empirical formula of C4H5N2O. Formula mass of the compound is 194.19 g. Formula mass of Molecular Formula = 194.19g = Molar mass of Empirical Formula = 97.09g 2 Multiply each subscript by 2 to get molecular formula: Molecular Formula = C8H10N4O2 HEY… THAT’S CAFFIENE !! Example A compound is made of only sulfur and oxygen. It is 60.9% S by mass. Its molar mass is 192.24 g/mol. What is its molecular formula? S = 60.9g 1mol = 1.9 mol = 1 32.06g 1.9 O = 30.4g 1mol = 1.9 mol = 1 16.00g 1.9 SO = 48.06g so 192.24 = 4 48.06 so Molecular Formula is: S4O4 Your Turn The molar mass of a compound is 92g/mol. Analysis of the compound indicates that it contains 0.606gN and 1.390g O. Find its molecular formula. Solution N = 0.606g N 1mol = 0.043 = 1 14.007g 0.043 O = 1.390g O 1mol = 0.087 = 2 16.00g 0.043 NO2 = 46.007g so 92 =2 46.007 so Molecular Formula is: N2O4 Your Turn Again A compound is known to be composed of 71.65 % Cl, 24.27% C and 4.07% H. Its molar mass is known to be 98.96g/mol. What is its molecular formula? Chemical Equations Types of Chemical Reactions Key: A, B = cations X, Y = anions Combination or Synthesis: A + X AX Decomposition: AX A + X Single Replacement / Displacement: AX + B BX + A or AX + Y AY + X 4) Double Replacement / Displacement: AX + BY AY + BX 5) Combustion: ____ + O2 ____ + ____ Balancing Equations Balancing equations CH4 + O2 → CO2 + H2O Products Reactants 1 C 1 4 H 2 2 O 3 Balancing equations CH4 + O2 → CO2 + 2 H2O Reactants Products 1 C 1 4 H 2 4 2 O 3 4 Balancing equations CH4 + 2O2 → CO2 + 2 H2O Reactants Products 1 C 1 4 H 2 4 4 2 O 3 4 And We’re Done!!! Abbreviations (s) ↓ = solid (l) = liquid (g) ↑ = gas (aq) = aqueous Δ or heat or Catalyst Ex: ~ MnO2 = yields = reversible rxn ↔ = resonance structure 20 C = heat = catalyst = electricity Example Solid iron (III) sulfide reacts with gaseous hydrogen chloride to form solid iron (III) chloride and hydrogen sulfide gas. Fe2S3(s) + 6HCl(aq) 2FeCl3(s) + 3H2S(g) Your Turn Write the word equations: KClO3 (s) Cl2 (g) + O2 (g) Fe2O3 (s) + Al (s) Fe (s) + Al2O3 (s) What does a balanced equation mean? 1) A balanced equation can be used to describe a reaction in molecules, atoms and mole ratios. NOT grams. 2) Chemical reactions happen molecules at a time. or dozens of molecules at a time. or moles of molecules. LET’S DO STOICHIOMETRY (The Mathematics of Chemical Equations!) Stoichiometry Given an amount of either a reactant or product STOICHIOMETRY allows us to determine the other quantities. We use conversion factors from: molar mass (g/mole, L/mole, atoms/mole) mole ratios in the balanced equation Examples One way of producing O2(g) involves the decomposition of potassium chlorate into potassium chloride and oxygen gas. A 25.5 g sample of Potassium chlorate is decomposed. 1) How many moles of O2(g) are produced? 2) How many grams of potassium chloride are produced? 3) How many grams of oxygen? Solution 25.5g ? mol K+1ClO3-1(s) K+1Cl-1(aq) + O2(g) KClO3(s) KCl(aq) + O2(g) Balance: 2KClO3(s) 2KCl(aq) + 3O2(g) Answers to questions: 1) 2) 3) 25.5 g KClO3 1 mol KClO3 3 O2 = 0.3 molO2 1 122.553g KClO3 2KClO3 0.312 mol O2 2KCl 74.553g KCl = 15.5 g KCl 1 3O2 1 mol KCl 0.312 mol O2 1 32.00g O2 1 mol O2 = 9.98 g O2 10. g O2 Your Turn A piece of aluminum foil 5.11in x 3.23 in x 0.0381in is dissolved in excess HCl(aq). How many grams of H2(g) are produced? (Hint: the density of Al is 2.7g/cm3) How many grams of each reactant are needed to produce 15g of iron from the following reaction? Fe2O3 (s) + Al (s) Fe (s) + Al2O3 (s) Examples Given: K2PtCl4 (aq) + NH3 (aq) Pt(NH3)2Cl2 (s) + KCl (aq) Find: 1. The mass of Pt(NH3)2Cl2 that can be produced from 65g of K2PtCl4 ? 2. How much KCl will be produced? 3. How much from 65 grams of NH3? Limiting Reactant Limiting Reagent Reactant that determines the amount of product formed. The one you run out of first. Makes the least product. You can use a ratio method or… Limiting reagent To determine the limiting reagent requires that you do two stoichiometry problems. Figure out how much product each reactant makes. The one that makes the least is the limiting reagent. Excess Reactant The reactant you don’t run out of. The amount of stuff you make is the yield. The theoretical yield is the amount you would make if everything went perfect. The actual yield is what you make in the lab. Example Ammonia is produced by the following reaction: N2 + H2 NH3 1. What mass of ammonia can be produced from a mixture of 100. g N2 and 500. g H2 ? 2. How much unreacted material remains? 3.57mol 247.52mol 100.g 500.g N2 + 3H2 100. g N2 1 2NH3 1 mol N2 2NH3 17.03g NH3 = 121.58g 28.014g N2 N2 1 mol NH3 NH3 500. G H2 1 mol H2 2NH3 17.03g NH3 = 2810.23g 1 2.02g H2 3H2 1 mol NH3 NH3 Can only produce 121.58g NH3 ! Moles 3.57mol 247.52mol Given: 100.g 500.g N2 + 3H2 LIMITING 2NH3 EXCESS To determine the unreacted material start with limiting: 3.57 mol N2 1 HAVE So 3H2 N2 = 10.71 mol H2 (needed) NEED 247.52 mol – 10.71 mol = 236.81mol H2 (excess) Your Turn Hg + Br2 HgBr2 1) If 10.0g of Hg and 9.00g of Br2 are reacted, how much HgBr2 will be produced? 2) If the reaction did go to completion, how much excess reagent would be left? % Percent Yield How much you get from a chemical reaction! Percent Yield % yield = Actual (Experimental) Theoretical (calculated using stoichiometry) % yield = Experiment results Calculated results x 100 x 100 % Yield vs % Error % error = (experimental – accepted) accepted % yield = Actual Theoretical x 100 x 100 What are the three types of yield? Answer 1. % Yield 2. Actual Yield 3. Theoretical Yield Your Turn Aluminum burns in bromine producing aluminum bromide. In a laboratory 6.0g of aluminum reacts with excess bromine and 50.3g of aluminum bromide are produced. What is the % yield?