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Stoichiometry

Stoichiometry. The Mole—Quantifying Equations. Atomic Mass. The mass of a single atom is far too small in grams to use conveniently. Chemists use the unit called the unified atomic mass unit ( amu ) or Dalton (Da). Definition of amu is exactly 1/12 the mass of an atom of 12 C

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Stoichiometry

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  1. Stoichiometry The Mole—Quantifying Equations

  2. Atomic Mass • The mass of a single atom is far too small in grams to use conveniently. • Chemists use the unit called the unified atomic mass unit (amu) or Dalton (Da). • Definition of amu is exactly 1/12 the mass of an atom of 12C • Amu (Da) = 1.660539 x 10-24g

  3. Atomic Mass Unit Mass of one 12C atom = 12 amu (exactly) 1 amu approximates the mass of one proton or neutron. Mass of electron is neglible in comparison.

  4. Mass Number • Elements differ in the number of protons in their atoms. • The atomic number Z • All atoms of a given element have the same number of protons. • Number of electrons equals protons. • Number of neutrons = N • Mass Number (A) = Z + N • Mass number is the total number of nucleons.

  5. Mass for Specific Elements Why do all element not have atomic mass number listed in the periodic table that is not a whole number or very close to it? Are all atoms of an element the same?

  6. Isotopes Isotopes are atoms with the same atomic number, but different mass number. The larger mass size is due to the difference in the number of neutrons that an atom contains. Although mass numbers are whole numbers, the actual masses of individual atoms are never whole numbers (except for carbon-12).  This explains how Lithium can have an atomic mass of 6.941 Da. 

  7. Calculating Average Atomic Mass The atomic masses on the periodic table take these isotopes into account, weighing them based on their abundance in nature, therefore, more weight is given to the isotopes that occur most frequently in nature. Average mass of the element E is defined as: m(E) = ∑(m(In)*p(In))  where ∑ represents a n-times summation over all isotopes In of element E, and p(I) represents the relative abundance of the isotope I.

  8. Atomic mass based on isotopes Find the average atomic mass of Boron Mass and abundance of Boron isotopes n isotope Inmass m (Da)isotopic abundance p 1 10B 10.013 0.199 2 11B 11.009 0.801 Solution: The average mass of Boron is: m(B) = (10.013 Da)(.199) + (11.009 Da)(.801) = 1.99 Da + 8.82 Da = 10.81 Da

  9. Molecular Mass • Molecular mass: sum of atomic masses of all atoms in a molecule • Formula mass: sum of atomic masses of all atoms in a formula unit of any compound, molecular or ionic.

  10. Mole Puns can you dig it? What is a mole's favorite movie? The Green Mole What do you get when you have a bunch of moles acting like idiots? Moleasses What line from Shakespeare do high school moles have to memorize? “To mole or not to mole, that is the question.” How much does Avogadro exaggerate? He makes mountains out of molehills. What element do moles love to study in chemistry? Moleybdenum

  11. Calculate the Molecular Mass • Copper (II) Nitrate Cu(NO3)2 • 63.5 + [(14 + {3 x 16}) x 2] = 187.5g • Ca3(PO4)2 3 moles of Ca 2 moles of P 2 x 4 moles of O. • 1 mole of Ca is 40.08g, so 3 moles are 120.24   g • 1 mole of P is 30.9738g, so 2 moles are 61.9476g • 1 mole of O is 15.9994g, so 8 moles are 127.9952g • 1 mole of Ca3(PO4)2 is 310.18   g

  12. Molecular Mass Ratio • From a balanced equation, the coefficients define the ratio of reactants needed for the products that result from the reaction. • Counting atoms is impractical. • Use a mass ratio: • to predict mass of products in ideal conditions. • to calculate the percentage yield in actual conditions. • to obtain the mass of each reactant needed.

  13. Example • Balanced equation: • C2H4 + HClC2H5Cl 1 : 1 yields 1 for ratio of molecules 28.0 : 35.5 yields 64.5 for mass ratio Ethylene: Atomic mass of 2C = 2 x 12.0amu = 24.0amu Atomic mass of 4H = 4 x 1.0amu = 4.0amu Molecular mass of C2H4 = 28.0amu Hydrogen chloride: at. mass of H = 1.0amu at. Mass of Cl = 35.5amu Molecular mass of HCl = 36.5amu Ethyl chloride: at. mass of 2C = 2 x 12.0amu = 24.0amu at. mass of 5H = 5 x 1.0amu = 5.0amu at. mass of Cl = 35.5amu = 35.5amu Molecular mass of C2H5Cl = 64.5amu

  14. Try these: 1. sodium fluoride 2. potassium hydroxide 3. copper (I) chloride 4. manganese (IV) oxide 5. calcium sulfate 6. magnesium phosphate

  15. Amadeo Avogadro Amadeo Avogadro was an Italian physics professor who proposed in 1811 that equal volumes of different gases at the same temperature contain equal numbers of molecules.

  16. History—or how did they find that number??? If Avogadro’s hypothesis is true, then atomic weights for gases can be derived by weighing equal volumes of different gases (Cannizzaro). Johan Loschmidt (HS teacher) took the idea and calculated the size of a molecule of air. He developed an estimate for the number of molecules in a given volume of air. These three ideas together lead to the number named for Avogadro. Loschmidt was the first to calculate this number.

  17. Avogadro’s Number 1 mole of a substance, NA = 6.02214179(30)×1023 is known as the Avogadro constant. For calculations please use 6.02 x 1023 http://www.youtube.com/watch?v=Hj83oRHdezc&safety_mode=true&persist_safety_mode=1&safe=active

  18. Mole is just a number Definition: A mole is the amount of substance that contains as many elementary particles as there are atoms in exactly 12 grams of carbon-12 (12C). 1 Mole = 6.022045 x 1023particles (atoms, molecules, ions, electrons, apples, wads of gum, elephants) = NA particles ~100 million x 100 million x 100 million

  19. Conversion factors • 6.022045 x 1023whatever kind of particles per mole • One mole of common substances: • CaCO3 :100.09g • Oxygen: 32.00g • Copper: 63.55g • Water:18.02g

  20. 2 H2 (g) + O2 (g) → 2 H2O(g) 2 dozen H2 molecules react with exactly 1 dozen O2 molecules to give exactly 2 dozen H2O molecules. 2 moles of H2 molecules react with exactly 1 mole of O2 molecules to give exactly 2 moles of H2O molecules. Why do we do this? Because these last sizes are in the gram range and easy to weigh. Conventions: 1 mole of 12C atoms weighs 12 g exactly. 1 atom of 12C weighs 12 amu exactly. (amu = atomic mass unit = ~mass of a proton or neutron)

  21. 2 H2 (g) + O2 (g) → 2 H2O(g) 2 moles of H2 molecules react with exactly 1 mole of O2 molecules to give exactly 2 moles of H2O molecules. 2 moles of H2 molecules 4 x 1.008g = 4.032g react with exactly 1 mole of O2 molecules 2 x 15.994g = 31.988g to produce exactly 2 moles of H2O molecules 2(2 x 1.008g + 15.994g) = 36.03g Sum of reactants = Sum of products Law of Conservation of Mass

  22. Number – mass relationshipsMole – mass relationships 12 red marbles @ 7g each = 84g 12 yellow marbles @4e each=48g 55.85g Fe = 6.022 x 1023 atoms Fe 32.07g S = 6.022 x 1023 atoms S

  23. Mole—Mass Relationships Element Atomic Mass Molar Mass Number of Atoms 1 atom of H = 1.008 amu 1.008 g = 6.022045 x 1023 atoms 1 atom of S = 32.07 amu 32.07 g = 6.022045 x 1023 atoms 1 atom of O = 15.994 amu 15.994 g = 6.022045 x 1023 atoms 1 molecule O2 (15.994 x 2) 32.00amu 32.00 g = 6.022045 x 1023 atoms 1 molecule S8 (32.07 x 8) 256.56amu 256.56 g = 6.022045 x 1023 atoms

  24. Calculating the Number of Moles and Atoms in a Given Mass of Element – Class Problem Convert Mass to Moles Convert Moles to Atoms Problem: Tungsten (W) is the element used as the filament in light bulbs, and has the highest melting point of any element, 3680oC. How many moles of tungsten, and atoms of the element, are contained in a 35.0 mg sample of the metal? Plan:

  25. Calculation Solution: Moles of W = 35.0x10-3g W x 1 mol W = 0.00019032 mol 183.9 g W Moles of W = 1.90 x 10 -4 mol No. of W atoms = 1.90 x 10 -4 mol W x 6.022 x 1023 atoms 1 mole of W = 1.15 x 1020 atoms of Tungsten

  26. Molecular Mass—Molar Mass (M) The molecular mass of a compound expressed in amu is numerically the same as the mass of one mole of the compound expressed in grams , called its molar mass.

  27. Calculate the Molecular Mass of Glucose: C6H12O6 Carbon—6 x 12.01 amu = 72.06 amu Hydrogen—12 x 1.008 amu = 12.096 amu Oxygen—6 x 15.994 amu = 95.964 amu 180.12 amu

  28. Calculate the Molar Mass of Glucose: C6H12O6 Carbon—6 x 12.01 g/mol = 72.06 g/mol Hydrogen—12 x 1.008 g/mol = 12.096 g/mol Oxygen—6 x 15.994 g/mol = 95.964 g/mol 180.12 g/mol

  29. Conversions Go from mass (g) to moles Go from moles to particles (NA) Go from moles to volume of gas (L) Go from moles to mass (g)

  30. Mass (g) : g/mol x g/mol Volume STP x 22.4L/mol : 22.4L/mol x NA/mol : NA/mol Number of particles (NA)

  31. Percent Composition For a compound, the percent composition for a specific element is the fraction of the compound mass that came from that element. AnBm%A = n(A g/mol) x 100 AnBmg/mol

  32. Mass Fraction and Mass % 3.0g/ball x 3 balls = 9g 9.0g/16.0g total = 0.56 0.56 x 100% = 56% red 2.0g/ball x 2 balls = 0.25 16g total 0.25 x 100 = 25% purple 1.0g/ball x 3 balls = 3.0g 16g total 0.19 x 100 = 19% yellow Mass of Red Balls = Mass Fraction Red = Mass % Red = Mass Fraction Purple = Mass Fraction Yellow = Check: 56% + 25% + 19% = 100%

  33. Calculate M and % composition of NH4C2H3O2 N __mol of N x _________ = _____g N H __mol of H x _________ = _____g H C __mol of C x _________ = _____g C H __mol of H x _________ = _____g H O __mol of O x _________ = _____g O Molar mass = M = _____g

  34. Calculate M and % composition of NH4C2H3O2 N 1 mol of N x 14.01g/mol = 14.01g N H 7 mol of H x 1.008 g/mol = 7.056 g H C 2 mol of C x 12.011g/mol = 24.022g C O 2 mol of O x 15.994g/mol = 31.988g Molar mass = M = 77.076g NH4C2H3O2 %N = 14.01g N/77.076g = 18.18% %H = 7.056g H/77.076g = 9.15% %C = 24.022g C/77.076g = 31.17% %O = 31.988g O/77.076g = 41.50% 100.00%

  35. Empirical Formula • Empirical Formula - A formula that gives the simplest whole-number ratio of atoms in a compound. • Once the empirical formula is found, the molecular formula for a compound can be determined if the molar mass of the compound is known. • Many compounds can share the same empirical formula. Alkanes are CnH2n+2

  36. Determining Empirical Formula • Start with the number of grams of each element, given in the problem.  • If percentages are given, assume that the total mass is 100 grams so that  the mass of each element = the percent given. • Convert the mass of each element to moles using the molar mass from the periodic table.  • Divide each mole value by the smallest number of moles calculated.  • Round to the nearest whole number.  This is the mole ratio of the elements and is  represented by subscripts in the empirical formula.  • If the number is too far to round (x.1 ~ x.9), then multiply each solution by the same  factor to get the lowest whole number multiple.  • e.g.  If one solution is 1.5, then multiply each solution in the problem by 2 to get 3.  • e.g.  If one solution is 1.25, then multiply each solution in the problem by 4 to get 5. 

  37. Example A compound was analyzed and found to contain 13.5 g Ca, 10.8 g O, and 0.675 g H.  What is the empirical formula of the compound?

  38. First step Find the moles for each element

  39. Find the ratio of moles Divide each by the smallest number of moles present. Round to nearest whole number.

  40. Finding the Formula This is the mole ratio of the elements and is represented by subscripts in the empirical formula. Ca—1 O—2 H—2 Therefore CaO2H2 or with the correct formula, Ca(OH)2.

  41. Nutrasweet NutraSweet is 57.14% C, 6.16% H, 9.52% N, and 27.18% O.  Calculate the empirical formula of NutraSweet and find the molecular formula.  (The molar mass of NutraSweet is 294.30 g/mol)

  42. Start with the number of grams of each element, given in the problem. If percentages are given, assume that the total mass is 100 grams so that the mass of each element = the percent given. 57.14% C, 6.16% H, 9.52% N, and 27.18% O.

  43. Convert the mass of each element to moles using the molar mass from the periodic table. Use the conversion factor: g/mol and divide or multiply by 1/g/mol.

  44. Divide each mole value by the smallest number of moles calculated.  Round to the nearest whole number. Select the smallest number of moles to divide each element (moles). Smallest one will equal 1.

  45. Mole Ratio of Elements • This is the mole ratio of the elements and is represented by subscripts in the empirical formula. • If the number is too far to round (x.1 ~ x.9), then multiply each solution by the same factor to get the lowest whole number multiple.

  46. Molecular Formula Now, we can find the molecular formula by finding the mass of the empirical formula and setting up a ratio:

  47. Empirical Formula Experiment • A sample of a pure oxide of nickel was analyzed by heating to drive off the oxygen. A team of students weighed an empty test tube, recording a mass of 32.064 g. After adding a sample to the tube, they measured a total mass of 33.076 g. The team then heated the sample in an atmosphere of natural gas reducing it to pure metal. The final mass after two heatings was 32.785 g for the tube and the metal residue. Perform calculations necessary to find results below, showing all of your work. • Mass of nickel oxide sample • Mass of nickel present in sample • Mass of oxygen present in sample • Mass percent of nickel • Mass percent of oxygen • Determine the empirical formula of the oxide of nickel, showing your work clearly. • Name the compound according to IUPAC conventions.

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