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Stoichiometry Calculations with Chemical Formulas and Equations

Stoichiometry Calculations with Chemical Formulas and Equations. THE MOLE. Mole -The number of C atoms in exactly 12.0 g of carbon., also a number 6.022 x 10 23 Avogadro’s number 6.022 x 10 23 , the number of particles in a mole of anything More on the mole later……. ATOMIC MASSES.

allistair-guthrie
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Stoichiometry Calculations with Chemical Formulas and Equations

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  1. Stoichiometry Calculations with Chemical Formulas and Equations

  2. THE MOLE • Mole-The number of C atoms in exactly 12.0 g of carbon., also a number 6.022 x 1023 • Avogadro’s number 6.022 x 1023, the number of particles in a mole of anything • More on the mole later……

  3. ATOMIC MASSES • 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.

  4. AVERAGE ATOMIC MASSES • The average atomic mass of an element is seen on the periodic table. It is calculated using the percent abundance of eachisotope (same # protons, differ in # neutrons and atomic mass) of an element and the mass of that particular isotope .

  5. Measuring atomic masses and isotope abundance • 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

  6. Measuring atomic masses and isotope abundance • 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.

  7. Measuring atomic masses and isotope abundance

  8. Obtain data from mass spec that looks like this:

  9. Converting between atoms and grams • Americium is an element that does not occur naturally. It can be produced in minute amounts in a device known as a particle accelerator. Compute the mass in grams of a sample of Americium containing 6 atoms: 2.42 x 10-21 grams

  10. Problem Type 2: Determining Moles of Atoms • Aluminum 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 • 2.23 x 1023 atoms 0.307moles

  11. Molar mass, molecular weight and Formula weight • Molar Mass MM- the mass in grams of one mole (or Avogadros number of particles) of a substance (unit g/mol or g mol-1) • Molecular weight MW the sum of all of the atomic weights of all atoms in a formula, “molecular” implies compounds in which atoms are covalently bonded • Side note: atoms that exist as molecules :Mr “BrINClHOF” • Formula weight- same as molecular weight, but it implies ionic bonding • ***The AP Exam uses the term Molar Mass for all formula masses.

  12. Problem Type 1 Calculating Molar Mass Calcium carbonate, also known as calcite, is the principle mineral found in limestone, marble, pearls, and chalk. • Calculate the molar mass of calcium carbonate • CaCO3 Ca = 1 x 40.08= 40.08 • C = 1 x 12.01= 12.01 • O = 3 x 16.00= 48.00 • 100. 09 g/ mol (g mol-1 ) • A certain sample of calcium carbonate contains 4.86 moles. What is the mass in grams of the sample? What is the mass of the carbonate ions present? • X g CaCO3 = 4.86 mole x I00.09 g/mol • X g CO32- = 4.86 mole x 60.01 g/mol = 292 g CO32- 486 g CaCO3 292 g CO32-

  13. Isopentyl acetate (C7H14O2) , the compound responsible for the scent of bananas, can be produced commercially. Interestingly, bees release about 1 mg of this compound when they sting. The resulting scent attracts other bees to join in the attack. How many molecules of isopentyl acetate are released in a typical bee sting (Remember Avogadro’s number???) 5 x 1015 molecules   C= 7 x 12.01 = 84.07 H= 14 x 1.01 = 14.14 O= 2 x 16.00 = 32.00 130.21 g/mol x molecules = 1 mg C7H14O2 x 1 g x I mol x 6.022 x 1023 molecules 1x106mg 130.21 g 1 mol • How many atoms of carbons are present ?   x atoms C = 5 x 1015 C7H14O2 molecules x 7 atoms C_______ 1 molecule C7H14O2 5 x 1015 molecules 4 x 1016 atoms C

  14. AP Style Multiple Choice Question In the multiple choice section of the AP Exam, you are not permitted to use a calculator!!!—You have 90 minutes to answer 75 questions, which gives you about 1 minute 20 seconds per question)---NOT MUCH TIME!

  15. 1. Which pair of samples contains the same number of atoms of oxygen in each member? (A) 0.10 mol, Al2O3 and 0.50 mol BaO (B) 0.20 mol Cl2O and 0.10 HClO (C) 0.20 mol SnO and 0.20 mol SnO2 (D) 0.10 mol Na2O and 0.10 mol Na2SO4 (E) 0.20 mol Ca(OH)2 and 0.10 mol H2C2O4

  16. AP Exam Free Response Question (There are 6 free response questions, broken into two Sections, each section consisting of 3 multi-part questions. The Free response section of the test lasts a total of 90 minutes. Each of the two sections have a time allotment of 45 minutes. The Section I is done with a calculator, and in Section 2 a calculator is not permitted). This gives you 15 minutes per question

  17. AP Style Molar Mass Question • a) One molecule of the antibiotic known as penicillin G has a mass of 5.342x10-21g. What is the molar mass of penicillin G? 3217 g/mol

  18. AP Style Molar Mass Question Hemoglobin, the oxygen carrying protein in red blood cells, has four iron atoms per molecule and contains 0.340 percent iron by mass. Calculate the molar mass of hemoglobin. 6.57 x 104 g/mol

  19. Percent Composition H2O H = 2 x 1.0 g = 2.02 g / 18.0 g = 0.111 x 100 = 11.1% O = 1 x 16.0 g = 16.00 g / 18.0 g = 0.889 x 100 = 88.9% 18.02 g 100.0 %

  20. 1. Which oxides of manganese, Mn, have a percent by mass of manganese that is greater than 50%? I . MnO II. MnO2 III. Mn2O3 (A) II only (B) III only (C) I and III only (D) II and III only (E) I, II, and III

  21. Don’t forget: You can copy-paste this slide into other presentations, and move or resize the poll.

  22. MASS PERCENT OF A HYDRATE-Gravimetric Analysis

  23. Hydrates • are salts (ionic compounds) who have water particles physically bound to their crystal lattice. Barium chloride dihydrate is an example BaCl2 . 2H2O. When heated, their water evaporates leaving just the ionic compound (called an anhydrous salt).

  24. Calculatethepercentage of wáter in bariumchloridedihydrate. 14.75 % water • Ba 1 x 137.34 = 137.34 • Cl 2 x 35.45= 70.90 • H2O 2 x 18.02= 36.04 243.28 %H2O= 36.04/243.28 = 14.75 If 15.0 grams of bariumchloridedihydrateisheated in a crucibleto a constantmass, howmanygrams of anhydroussaltshouldremain100-14.75= 85.25% anhydroussalt12.8 g BaCl2 15.0 x .8525= 12.8 g

  25. c. Thepercentbymass of carbón in oxalicacid, H2C2O4 . 2H2O isclosestto (A)2/14 x 100 (B)12/90 x 100 (C)24/66 x 100 (D)24/90 x 100 (E)24/126 x 100

  26. AP levelPercentCompositionproblems • 2)A mixture of BaCl2. 2H2O and NaCl with a mass of 1.6992g is placed in a vial weighing 3.3531g. The mixture is heated at 105oC to drive off the water. The residue of the mixture now has a mass of 1.4804g. Use the preceding data to calculate the % NaCl in the mixture.

  27. Empirical Formulas • Empirical formula- • the most reduced form of a true formula. • Represent the mole to mole ratio of elements in a compound H2 O • Can be calculated from the percentage composition of a compound 1 mole O AP 1 mole O yleLeve 1 Mass Question • 2 m • 2 moles H • 2 ol2 moles H • es H

  28. Empirical Formulas • Assume that you have 100 g of the compound and change your percentages to g. • Convert the g to moles. • Divide these answers by the smallest number of moles. • Use these whole numbers as the subscripts for the formula. • If they aren’t whole numbers, you may need to multiply each answer by a 2, 3, or 4 to make it a whole number. Use 2 if it ends close to .5, use 3 if it ends close to .33 or .67, and use 4 if it ends close to .25 or .75.

  29. Example A compound is 63% Mn and 37% O. What is the empirical formula? Step 1 Convert to Moles 63 g Mn x 1 mol Mn = 1.15 mol Mn 54.9 g Mn 37 g O x 1 mol O = 2.31 mol O 16.0 g O Step 2 Divide by smallest The lowest number of moles is 1.15 Mn 1.15 mol Mn = 1 mol Mn 1.15 mol  O 2.31 mol O = 2 mol O 1.15 mol The formula is MnO2­

  30. A compound is reported to contain 26.6% potassium, 35.3 % Chromium, and 38.1 % oxygen by mass. Which set of values, when substituted for x, y and z below, give the best representation of the empirical formula for the unknown compound? KxCryOz x y z (A) 39.152.016.0 26.6 35.3 38.1 (B) 26.635.338.1 16.0 16.0 16.0 (C) 26.635.338.1 39.1 52.0 16.0 (D) 39.152.016.0 26.6 26.6 26.6 (E) 26.635.338.1 52.0 52.0 52.0

  31. Practice Problem Determine empirical formula and Molecular formula • A White powderisanalyzed and foundtocontain 43.64% phosphorus and 56.36% oxygenbymass. Thecompound has a molar mass of 283.88 g/mol. What are thecompound’sempirical and molecular formula? EF P2O5 • MF P4O10

  32. AP levelEmpirical Formula problems 1. An oxybromate compound, KBrOx, where x is unknown, is analyzed and found to contain 52.92 percent Br. What is the value of x?

  33. AP levelEmpirical Formula problemsusingcombustionanalysis In a combustionreaction, a reactantisburned (reactedwithoxygen). Frequently, thisreacatantis a hydrocarbon. Theproducts of the complete combustión of a hydrocanbon are CO2 and H2O CxHy (Z) + O2 CO2 + H2O + (Z) (where Z is sometimes oxygen or a halogen)

  34. AP levelEmpirical Formula problemsusingcombustionanalysis • CxHy (Z) + O2 CO2 + H2O + (Z) • ***AccordingtotheLaw of Conservation of Mass: • themass of the carbón in thehydrocarbon, equalsthemass of the carbón in CO2. • Themass of hydrogen in wáter= mass of thehydrogen in thehydrocarbon. • Ifthemass of the CO2 and H2O isgiven, thenyou can use stoichiometryto determine themass of Carbon and hydrogen in yourhydrocarbon, and then determine theempirical formula

  35. AP levelEmpirical Formula problems 1. An organic compound was found to contain only three elements, C, H, and Cl. When a 1.50 g sample of the compound was completely combusted in air, 3.52g of CO2 was formed. In a separate experiment the chlorine in a 1.00g sample of the compound was converted to 1.27g of AgCl. Determine the empirical formula of the compound. C6H5Cl

  36. Chemical Reactions 5 signs of a chemicalChange: 1. Precipitate (new solid) isformed 2. Color Change 3. Bubbles/new odor—gas isformed 4. Heatreleasedortaken in 5. Explosion

  37. Chemical Reactions • Chemicalreactions shows amounts of reactants and productsbymoleculeorby mole. Shownbychemicalequation • Symbols denote states: (s), (l), (g), (aq)

  38. Balancing Chemical EquationsAcid base • H2SO4 + LiOH → H2O + + Li2SO4

  39. Balancing Chemical EquationsCombustion 2. C3H8 + O2→ H2O + CO2

  40. Balancing Chemical EquationsDouble displacement 3. Fe(OH)2 + NaCl→ FeCl2 + NaOH

  41. Balancing Chemical EquationsSynthesis 4. Li + O2→ Li2O

  42. Balancing Chemical Equationsdecomposision 5. SO2→ S8 + O2

  43. Balancing Chemical Equationssingle displacement (water HOH) 6. Li + H2O →LiOH + H2

  44. AP Level Question Write a plausible chemical equation for the combustion of liquid triethylene glycol in an abundant supply of oxygen gas. Triethylene glycol is 47.99% C, 9.40% H, and 42.62% O by mass and has a molecular mass of 150.2u.

  45. Stoichiometry! Stoichiometry is the most important skill that you can learn as you embark upon AP Chemistry. Get good at this and you will do well all year—this never goes away. Stoichiometry measures the quantitative relationship between reactants and products in a chemical reaction

  46. THE MOLE • All stoichiometric calculations utilize the mole • There are a variety of ways to “get to the mole”

  47. THE MOLE • 1 mole= Molar mass (grams) • = Avogadro’s number of particles (6.022 x 1024) • = 22.4 L gas at STP (Standard temperature and pressure) • Ideal gas law (when dealing with gases): n= PV/RT ( of a gas ) from ideal gas law PV=nRT • Concentration in Molarity (use with solutions) moles= Molarity (moles solute/liters solution) x volume(L)

  48. Problem 1 Mole-mole problem One way to change the iron mineral, Fe2O3, into metallic iron is to heat it with hydrogen. Fe2O3 + 3 H2 --> 2 Fe + 3H2O How many moles of iron are made from 20.0 mol of Fe2O3?   x moles Fe = 20.0 mol Fe2O3 x 2 mol Fe 1 mol Fe2O3 = 40.0 moles

  49. Problem 2 Mass-mass problem What mass of oxygen will react with 96.1 grams of propane C3H8 C3H8 + 5 O2  3 CO2 + 4 H2 O X grams O2 = 96.1 g C3H8 x 1 mol C3H8 x5 mol O2 x 32.00 g O2 44.11 g C3H8 1 mol C3H8 1 mol O2 = 349 grams O2

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