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Understanding Atomic Mass and Stoichiometry

Learn about atomic mass, isotopes, Avogadro's number, and molar mass, and how they are used in stoichiometry calculations.

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Understanding Atomic Mass and Stoichiometry

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  1. Stoichiometry

  2. Atomic Mass • It is impossible to weigh a single atom, but one can easily establish the weight of an atom relative to another • We must establish a standard: • An atom of carbon-12 (six protons, six neutrons, six electrons) has a mass of exactly 12 atomic mass units (u) • Therefore, 1 atomic mass unit (u) is equivalent to exactly a twelfth of the mass of an atom of carbon 12 • If it is found that a hydrogen atom has only 8.400% of the mass of a carbon-12 atom, the atomic mass of hydrogen must be: 0.08400 x 12 u = 1.008 u

  3. Average Atomic Mass • Most natural elements have more than one isotope • e.g.; there are two isotopes of carbon; carbon-12 (98.89%) and carbon-13 (1.11%, atomic mass = 13.00335 u), and the average atomic mass is: (0.9889)(12.00000u) + (0.0111)(13.00335) = 12.01 u • Since carbon-12 is the dominant isotope, the average atomic mass is approximately 12 • N.B. 12.01 u is an average value, and no single carbon atom has this mass, i.e., it’s either 12.00000 u or 13.00335 u

  4. Average Atomic Mass • The atomic masses of the two stable isotopes of boron, boron-10 (19.78 %) and boron-11 (80.22 %), are 10.0129 u and 11.0093 u, respectively. Calculate the average atomic mass of boron. • Solution: atomic mass of B = (0.1978)(10.0129u) + (0.8022)(11.0093 u) = 10.81 u

  5. Avogadro’s Number • One mole describes a particular number of objects, just like a pair (2), a dozen (12), or a gross (144) • A mole is the number of carbon-12 atoms in exactly 12 g of carbon-12 • The current accepted value is 1 mole = 6.022045 x 1023 particles

  6. Molar Mass • The molar mass of an atom or molecule is the mass of one mole of these atoms or molecules • The molar mass of carbon-12 is thus exactly 12 g, by definition • The numerical value of the molar mass (in grams) for an atom is the same as the numerical value of the average atomic mass (in u) • e.g.; For Na, the average atomic mass = 22.99 u and the molar mass = 22.99 g

  7. The Mass of an Atom • We know that, by definition, 12 g de carbon-12 contains 6.022045 x 1023 atoms of carbon-12 • The mass of a single atom of carbon-12 is therefore • A single atomic mass unit is 12 times smaller, and therefore 1 u = 1.660565 x 10-24 g, and 1 g = 6.022045 x 1023 u

  8. The Molar Mass of an Element and Avogadro’s Number • Example: What is the mass (in grams) of an atom of iodine? • Solution: • The molar mass of iodine = 126.9 g/mol, therefore

  9. Molecular Mass • The molecular mass is the sum of the atomic masses (in atomic mass units) of the atoms that make up a molecule • ex.; the molecular mass of water is (2)(1.008 u) + (1)(16.00 u) = 18.02 u • The molar mass of a molecule (in grams) has a numerical value equal to its average molecular mass (in atomic mass units) • e.g.; the molar mass of water is 18.02 g, and thus 18.02 g of water contains one mole (or 6.022045 x 1023 molecules) of water

  10. Molecular Mass • Example: What is the molecular mass of methanol (CH3OH)? • Solution: molecular mass of CH3OH = (4)(1.008 u) +(1)(12.01 u) +(1)(16.00 u) = 32.04 u

  11. Molar Mass • Example: Calculate the number of moles in 198 g of chloroform (CHCl3). • Solution:

  12. Mass Spectrometry • An electron beam ionizes the molecules, and a magnet affects the trajectory of the ions produced according to their charge : mass ratio • The signal produced by a particular ion is proportional to its abundance

  13. Mass Spectrometry • In this example, we can determine the relative abundance of each Ne isotope • Mass spectrometry can also be used to identify the molecular mass of a compound

  14. Percent Composition • The percent composition is the mass percentage of each element contained within a compound • The percent composition allows us to establish the empirical formula of a compound • With the molecular mass, obtained from mass spectrometry, or the ideal gas law in the case of a gas, we can then determine the molecular formula

  15. Percent Composition • Example: The percent composition of hydrogen peroxide is 5.94% H and 94.06% O. What is its empirical formula? If the molecular mass is 34.02 u, what is the molecular formula? • Solution: • If we have 100 g of hydrogen peroxide, we’d have 94.06 g of O and 5.94 g of H, and therefore (94.06 g)/(16.00 g/mol) = 5.88 mol of O and (5.94 g)/(1.008 g/mol) = 5.89 mol of H • This gives a H:O ratio of about 1:1, and therefore, the empirical formula is HO • If the molecular formula was HO, the molecular mass would be 17.01 u (twice as small) • Therefore, the molecular formula is H2O2

  16. Percent Composition • Example: Determine the empirical formula of a compound with the following percent composition: 24.75% K, 34.77% Mn, and 40.51% O. • Solution: • If we have 100 g of this compound, we’d have 24.75 g of K, 34.77 g of Mn, and 40.51 g of O, or (24.75 g)/(39.10 g/mol) = 0.6330 mol of K, (34.77 g)/(54.94 g/mol) = 0.6329 mol of Mn, and (40.51 g)/(16.00 g/mol) = 2.532 mol of O • This gives a K:Mn:O ratio of about 1:1:4 • Therefore, the empirical formula is KMnO4

  17. The percent composition of a compound is 38.08% C, 31.71% O, 25.42% S, and 4.79% H. The molecular mass of the compound is approximately 504.6 g/mol. What is the empirical formula of this compound? What is the molecular formula? What is the mass (in g) of a molecule of this compound?

  18. Balancing Chemical Equations • Once all of the reactants and products are known (this is not always trivial), we can balance the chemical equation so as to respect the law of conservation of mass • We can balance the equation by playing with the coefficients in front of each reactant or product (we can not play with the molecular formula!!!) • It is often easier to start balancing an element that appears in only one place on each side of the reaction

  19. Balancing Chemical Equations • e.g.; KClO3 KCl + O2 • K and Cl are balanced (for now) • We balance O 2 KClO3 KCl + 3 O2 • O is now balanced • K et Cl are not balanced anymore 2 KClO3 2 KCl + 3 O2 • The equation is now balanced

  20. Balancing Chemical Equations • N.B. the chemical equations KClO3 KCl + (3/2) O2 4 KClO3 4KCl + 6 O2 are both acceptable and balanced • However, it is common to balance equations using the smallest possible integers as coefficients

  21. Balancing Chemical Equations • e.g.; C2H6 + O2 CO2 + H2O • None of the elements are balanced • We balance the C C2H6 + O2 2 CO2 + H2O • We balance the H C2H6 + O2 2 CO2 + 3 H2O • We balance the O C2H6 + (7/2) O2 2 CO2 + 3 H2O • The equation is now balanced • We double the coefficients in order to obtain integers 2 C2H6 + 7 O2 4 CO2 + 6 H2O

  22. Balancing Chemical Equations • e.g.; Fe2O3 + CO Fe + CO2

  23. Calculating Masses of Reactants and Products • Stoichiometry is the study of the relationships between the masses of the reactants and products in a chemical reaction • If we know the amount of reactants (products), we can calculate the amount of products (reactants) that are formed (required) • To simplify the process, we use the moles method, i.e., we work directly with the moles (rather than grams or liters)

  24. Calculating Masses of Reactants and Products • For the reaction 2 CO(g) + O2(g) 2 CO2(g) at least two moles of CO (g) are needed to produce two moles of CO2 (g) and at least one mole of O2 (g ) is also necessary • In this example, we say that 2 mol of CO are stoichiometrically equivalent to 1 mol of O2(g) as well as 2 mol of CO2(g) • In the same way, we say that 1 mol of O2(g) is stoichiometrically equivalent to 2 mol of CO2(g)

  25. The Mole Method • Step (1): Identify all of the reactants and products • Step (2): Balance the chemical equation • Step (3): Convert all the known quantities into moles • Step (4): Use the coefficients of the balanced chemical equation to determine the number of moles of the quantities sought • Step (5): If necessary, convert the number of the quantities sought into grams, litres, etc.

  26. The Mole Method

  27. The Mole Method • Example: The reaction between NO and O2is a key step in the formation of photochemical smog: 2 NO(g) + O2(g) 2 NO2(g) (a) How many moles of NO2(g) are formed by the complete reaction of 0.254 mol of O2(g)? (b) What mass, in grams, of NO2(g) is obtained by the complete reaction of 1.44 g of NO(g)?

  28. The Mole Method • Solution: (a) The equation is already balanced and we know that et 0.254 mol of O2(g) reacts. Thus, we produce twice as much NO2. We produce 0.508 mol. (b) 1.44 g of NO is (1.44 g)/(14.01 g/mol + 16.00 g/mol) = 0.0480 mol of NO This will produce 0.0480 mol of NO2. This has a mass of (0.0480 mol) x [14.01 g/mol + (2)(16.00 g/mol)] = 2.21 g

  29. Limiting Reagents and the Yield of Reactions • The reagent which is first exhausted is called the limiting reagent • The maximum amount of product depends on the initial amount of the limiting reagent • The other reactants are called the excess reagents

  30. Limiting Reagents and the Yield of Reactions • e.g.; For the reaction S(l) + 3 F2(g) SF6(g) • We have 4 moles of S and 9 moles of F2 • 4 moles of S will react with 12 moles of F2 • But we only have 9 moles of F2 • The F2 will be used up before all of the S can react • The F2 is the limiting reagent • N.B. We arrive at the same conclusion when saying that the 9 moles of F2 will react with only 3 moles of S, thus leaving 1 mole of S in excess • 3 moles of SF6 is produced

  31. What mass of POCl3 is produced when 790.0 g of PCl3(l), 345.0 g of Cl2(g), and 265.0 g of P4O10(s) reacts together in the following fashion? 6 PCl3(l) + 6 Cl2(g) + P4O10(s) →10 POCl3(l)

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