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Lecture 3

Lecture 3. Stoichiometry. Contents. Elements Atomic number, mass number, atomic mass Mole concepts, Avogadro’s number Conversions between mole and number of atoms Relationship between mole and mass Molar mass Percent composition Empirical formula, molecular formula. Stoichiometry is:.

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Lecture 3

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  1. Lecture 3 Stoichiometry

  2. Contents • Elements • Atomic number, mass number, atomic mass • Mole concepts, Avogadro’s number • Conversions between mole and number of atoms • Relationship between mole and mass • Molar mass • Percent composition • Empirical formula, molecular formula

  3. Stoichiometry is: • Study of the quantities of materials consumed and produced in chemical reactions • Mole/mass relationships between reactants and products

  4. Atomic number Element symbol Atomic mass The Periodic Table of Elements

  5. Atomic number, Z • the number of protons in the nucleus • the number of electrons in a neutral atom • the integer on the periodic table for each element

  6. Mass Number, A • integer representing the approximate mass of an atom • equal to the sum of the number of protons and neutrons in the nucleus

  7. Masses of Atoms Carbon-12 Scale Masses of the atoms are compared to the mass of C-12 isotope having a mass of 12.0000 amu The masses of all other atoms are given relative to this standard

  8. Mass Spectrometerthe most accurate methods for comparing masses of atoms

  9. Atomic Masses andIsotopic Abundances natural atomic masses = sum[(atomic mass of isotope)  (fractional isotopic abundance)] C-12 98.89% mass – 12 amu C-13 1.11% mass – 13.003355 amu The average atomic mass for natural carbon is: 12amu x 0.9889+13.0034amu x 0.0111= 12.01amu

  10. The Mole • a unit of measurement, quantity of matter present • The number equal to the number of carbon atoms in exactly 12 grams of pure C-12 (6.022  1023) • Avogadro’s Number 6.022  1023 particles

  11. Chemical Packages—Moles • Mole = Number of things equal to the number of atoms in 12 g of C-12. • 1 atom of C-12 weighs exactly 12 amu. • 1 mole of C-12 weighs exactly 12 g. • In 12 g of C-12 there are 6.022 x1023 C-12 atoms.

  12. Molecular Mass - Molar Mass ( M ) The Molecular mass of a compound expressed in amuis numerically the same as the mass of one mole of the compound expressed in grams, called itsmolar mass. For water: H2O Molecular mass = (2 x atomic mass of H ) + atomic mass of O = 2 ( amu) + amu = amu Mass of one molecules of water = amu Molar mass = ( 2 x molar mass of H ) + (1 x molar mass of O) = 2 ( g ) + g = g g H2O = 6.022 x 1023 molecules of water = 1 mole H2O

  13. Molecular Mass - Molar Mass ( M ) The Molecular mass of a compound expressed in amuis numerically the same as the mass of one mole of the compound expressed in grams, called itsmolar mass. For water: H2O Molecular mass = (2 x atomic mass of H ) + atomic mass of O = 2 ( 1.008 amu) + 16.00 amu = 18.02 amu Mass of one molecules of water = 18.02 amu Molar mass = ( 2 x molar mass of H ) + (1 x molar mass of O) = 2 ( 1.008 g ) + 16.00 g = 18.02 g per mole H2O 18.02 g H2O = 6.022 x 1023 molecules of water = 1 mole H2O

  14. Molar Mass The mass in grams of one mole of molecules or formula units of a substance, also called molar mass or molecular weight Molar mass of: C - 12.01 g/mole O - 16.00 g/mole Fe – 55,85 g/mole

  15. Conversion factors between Moles, Number of particles, Grams

  16. atoms Ag mol Ag Example 1. A Silver Ring Contains 1.1 x 1022 Silver Atoms. How Many Moles of Silver Are in the Ring? Given: Find: 1.1 x 1022 atoms Ag moles Ag Solution Map: Relationships: 1 mol = 6.022 x 1023 atoms Solution: Check: Since the number of atoms given is less than Avogadro’s number, the answer makes sense.

  17. Write down the given quantity and its units. Given: 1.1 x 1022 Ag atoms Example:A silver ring contains 1.1 x 1022 silver atoms. How many moles of silver are in the ring?

  18. Write down the quantity to find and/or its units. Find: ? moles Information: Given: 1.1 x 1022 Ag atoms Example:A silver ring contains 1.1 x 1022 silver atoms. How many moles of silver are in the ring?

  19. Collect needed conversion factors: 1 mole Ag atoms = 6.022 x 1023 Ag atoms. Information: Given: 1.1 x 1022 Ag atoms Find: ? moles Example:A silver ring contains 1.1 x 1022 silver atoms. How many moles of silver are in the ring?

  20. Write a solution map for converting the units: Information: Given: 1.1 x 1022 Ag atoms Find: ? moles Conversion Factor: 1 mole = 6.022 x 1023 Example:A silver ring contains 1.1 x 1022 silver atoms. How many moles of silver are in the ring? atoms Ag moles Ag

  21. Apply the solution map: Information: Given: 1.1 x 1022 Ag atoms Find: ? moles Conversion Factor: 1 mole = 6.022 x 1023 Solution Map: atoms  mole Example:A silver ring contains 1.1 x 1022 silver atoms. How many moles of silver are in the ring? = 1.8266 x 1022 moles Ag • Significant figures and round: = 1.8 x 1022 moles Ag

  22. Practice 1 —Calculate the Number of Atoms in 2.45 Mol of Copper.

  23. mol Cu atoms Cu Practice 1 —Calculate the Number of Atoms in 2.45 Mol of Copper, Continued. Given: Find: 2.45 mol Cu atoms Cu Solution Map: Relationships: 1 mol = 6.022 x 1023 atoms Solution: Check: Since atoms are small, the large number of atoms makes sense.

  24. Relationship Between Moles and Mass • The mass of one mole of atoms is called the molar mass. • The molar mass of an element, in grams, is numerically equal to the element’s atomic mass, in amu. • The lighter the atom, the less a mole weighs. • The lighter the atom, the more atoms there are in 1 g.

  25. Mole and Mass Relationships 1 mole carbon 12.01 g 1 mole sulfur 32.06 g

  26. g S mol S Example 2 —Calculate the Moles of Sulfur in 57.8 G of Sulfur. Given: Find: 57.8 g S mol S Solution Map: Relationships: 1 mol S = 32.07 g Solution: Check: Since the given amount is much less than 1 mol S, the number makes sense.

  27. Example 2: Calculate the number of moles of sulfur in 57.8 g of sulfur.

  28. Write down the given quantity and its units. Given: 57.8 g S Example:Calculate the number of moles of sulfur in 57.8 g of sulfur.

  29. Write down the quantity to find and/or its units. Find: ? moles S Information: Given: 57.8 g S Example:Calculate the number of moles of sulfur in 57.8 g of sulfur.

  30. Collect needed conversion factors: 1 mole S atoms = 32.07 g Information: Given: 57.8 g S Find: ? moles S Example:Calculate the number of moles of sulfur in 57.8 g of sulfur.

  31. Write a solution map for converting the units: Information: Given: 57.8 g S Find: ? moles S Conversion Factor: 1 mole S = 32.07 g Example:Calculate the number of moles of sulfur in 57.8 g of sulfur. g S moles S

  32. Apply the solution map: Information: Given: 57.8 g S Find: ? moles S Conversion Factor: 1 mole S = 32.07 g Solution Map: g  moles Example:Calculate the number of moles of sulfur in 57.8 g of sulfur. = 1.802307 moles S • Significant figures and round: = 1.80 moles S

  33. Practice 2 —How Many Copper Atoms Are in a Penny Weighing 3.10 g?

  34. g Cu mol Cu atoms Cu Practice 2 —How Many Copper Atoms Are in a Penny Weighing 3.10 g?, Continued Given: Find: 3.10 g Cu atoms Cu Solution Map: Relationships: 1 mol Cu = 63.55 g, 1 mol = 6.022 x 1023 Solution: Check: Since the given amount is much less than 1 mol Cu, the number makes sense.

  35. Molar Mass of Compounds • The relative weights of molecules can be calculated from atomic weights. Formula mass = 1 molecule of H2O = 2(1.01 amu H) + 16.00 amu O = 18.02 amu. • Since 1 mole of H2O contains 2 moles of H and 1 mole of O. Molar mass = 1 mole H2O = 2(1.01 g H) + 16.00 g O = 18.02 g.

  36. mol H2O g H2O Example 3—Calculate the Mass of 1.75 Mol of H2O. Given: Find: 1.75 mol H2O g H2O Solution Map: Relationships: 1 mol H2O = 18.02 g Solution: Check: Since the given amount is more than 1 mol, the mass being > 18 g makes sense.

  37. g PbO2 mol PbO2 units PbO2 Practice 3 —How Many Formula Units Are in 50.0 g of PbO2? (PbO2 = 239.2), Continued Given: Find: 50.0 g PbO2 formula units PbO2 Solution Map: Relationships: 1 mol PbO2 = 239.2 g,1 mol = 6.022 x 1023 Solution: Check: Since the given amount is much less than 1 mol PbO2, the number makes sense.

  38. Wednesday, October14. • Percent composition • Empirical formula • Molecular formula • Chemical equation • Balancing chemical equation • Mass - to - mass conversion

  39. Percent Composition • Percentage of each element in a compound. • By mass. • Can be determined from: • The formula of the compound. • The experimental mass analysis of the compound.

  40. Example 4 —Find the Mass Percent of Cl in C2Cl4F2. Given: Find: C2Cl4F2 % Cl by mass Solution Map: Relationships: Solution: Check: Since the percentage is less than 100 and Cl is much heavier than the other atoms, the number makes sense.

  41. Practice 4 —Determine the Percent Composition of the Following, Continued: CaCl2

  42. 100g MMA % A mass A (g) moles A moles A moles B 100g MMB % B mass B (g) moles B Empirical Formulas • The simplest, whole-number ratio of atoms in a molecule is called the empirical formula. • Can be determined from percent composition or combining masses. • The molecular formula is a multiple of the empirical formula.

  43. Empirical Formulas, Continued Hydrogen Peroxide Molecular formula = H2O2 Empirical formula = HO Benzene Molecular formula = C6H6 Empirical formula = CH Glucose Molecular formula = C6H12O6 Empirical formula = CH2O

  44. Practice 5 —Determine the Empirical Formula of Benzopyrene, C20H12.

  45. Practice 5 —Determine the Empirical Formula of Benzopyrene, C20H12, Continued • Find the greatest common factor (GCF) of the subscripts. 20 factors = (10 x 2), (5 x 4) 12 factors = (6 x 2), (4 x 3) GCF = 4 • Divide each subscript by the GCF to get the empirical formula. C20H12 = (C5H3)4 Empirical formula = C5H3

  46. Finding an Empirical Formula 1. Convert the percentages to grams. a. Skip if already grams. 2. Convert grams to moles. a. Use molar mass of each element. 3. Write a pseudoformula using moles as subscripts. 4. Divide all by smallest number of moles. 5. Multiply all mole ratios by number to make all whole numbers, if necessary. a. If ratio ?.5, multiply all by 2; if ratio ?.33 or ?.67, multiply all by 3, etc. b. Skip if already whole numbers after Step 4.

  47. Example 6 — Finding an Empirical Formulafrom Experimental Data

  48. Example: A laboratory analysis of aspirin determined the following mass percent composition. Find the empirical formula. C = 60.00% H = 4.48% O = 35.53%

  49. Write down the given quantity and its units. Given: C = 60.00% H = 4.48% O = 35.53% Therefore, in 100 g of aspirin there are 60.00 g C, 4.48 g H, and 35.53 g O. Example:Find the empirical formula of aspirin with the given mass percent composition.

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