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Stoichiometry: Chemical Calculations

Chapter Three. Stoichiometry: Chemical Calculations. Molecular Masses and Formula Masses. Molecular mass : sum of the masses of the atoms represented in a molecular formula. Simply put: it is the mass of a molecule. Molecular mass is specifically for molecules .

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Stoichiometry: Chemical Calculations

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  1. Chapter Three Stoichiometry: Chemical Calculations

  2. Molecular Masses andFormula Masses • Molecular mass: sum of the masses of the atoms represented in a molecular formula. • Simply put: it is the mass of a molecule. • Molecular mass is specifically for molecules. • Ionic compounds don’t exist as molecules; for them we use … • Formula mass: sum of the masses of the atoms or ions present in a formula unit.

  3. Example 3.1Calculate the molecular mass of glycerol (1,2,3-propanetriol). In glycerol, one –OH group replaces one H atom on each of the three C atoms in propane, leading to the condensed structural formula CH2OH – CHOH - CH2OH which translates to the molecular formula C3H8O3. To obtain the molecular mass, we must add together three times the atomic mass of carbon, eighttimes the atomic mass of hydrogen, and three times the atomic mass of oxygen: 3 x atomic mass of C = 3 x 12.011 u = 36.033 u 8 x atomic mass of H = 8 x 1.00794 u = 8.06352 u 3 x atomic mass of O = 3 x 15.9994 u = 47.9982 u Molecular mass of C3H8O3 = 92.095 u Solution

  4. Example 3.2Calculate the formula mass of ammonium sulfate, a fertilizer commonly used by homegardeners. Solution Ammonium sulfate is an ionic compound consisting of ammonium ions (NH4+) and sulfate ions (SO42–); its formula is therefore (NH4)2SO4. To derive a formula mass from a complex formula like this one, we must make certain that all the atoms in the formula unit are accounted for, which means paying particular attention to all the subscripts and parentheses in the formula. Let’s first note the relevant atomic masses and the way in which they must be combined:

  5. Assessment As long as every atom in the formula unit is accounted for, we can check our answer by using a different summation, for example, by considering each element separately: Formula mass = (2 x atomic mass N) + (8 x atomic mass H) + (1 x atomic mass of S) + (4 x atomic mass O) = (2 x 14.0067 u) + (8 x 1.00794 u) + (1 x 32.066 u) + (4 x 15.9994 u) = 132.141 u Calculate the formula mass of ammonium sulfate, a fertilizer commonly used by homegardeners. • Summing the atomic masses • (NH4)2 : [14.0067 u + 4 x 1.00794 u)] x 2 = 36.0769 u • SO4 : 32.066 u + (4 x 15.9994 u) = 96.064 u • Formula mass of (NH4)2SO4 = 132.141 u

  6. 3.2 The Mole & Avogadro’s Number • Mole (mol): amount of substance that contains as many elementary entities, as there are atoms in exactly 12 g of the carbon-12 isotope. • Atoms are small, so this is a BIG number … • Avogadro’s number (NA) = 6.022 × 1023/mol • 1 mol = 6.022 × 1023“things” (atoms, molecules, ions, formula units, oranges, etc.) • A mole of oranges would weigh about as much as the earth! • Mole is NOT abbreviated as either M or m (but mol).

  7. 12.011 amu 31.9988 amu 44.010 amu Two views of the combination of carbon and oxygen to form carbon dioxide At the microscopic (molecular) level, chemical reactions occur between atoms and molecules (top) but we can show only a few atoms and molecules to represent the enormous numbers that actually make up the samples. In reality, we usually observe substances at the macroscopic level.

  8. One Mole of Four Elements One mole each of helium, sulfur, copper, and mercury. 6.022 × 1023atoms of each 4,002g He, 32.066g S, 63,55g Cu, 200,59 Hg

  9. Example 3.3 Determine (a) the mass of a 0.0750-mol sample of Na, (b) the number of moles of Na in a 62.5-g sample, (c) the mass of a sample of Na containing 1.00 x 1025 Na atoms, and (d) the mass of a single Na atom. Solution (a) To convert from moles to grams, we need to the conversion factor: 22.99 g Na/1 mol Na: (b) This time we write the conversion factor as the inverse- 1 mol Na/22.99 g Na- because we are to convert from grams to moles:

  10. (c) the mass of a sample of Na containing 1.00 x 1025 Na atoms (S1) Here we can use conversion factors, to convert first from number of atoms to number of moles, and then from moles to the mass in grams: (S2) Alternatively, we can use the conversion factor 22.99 g Na/6.022 x 1023 Na atoms, to convert directly from number of atoms to mass in grams:

  11. (d) the mass of a single Na atom. (d) The answer must have the unit grams per sodium atom (g/Na atom). We know these quantities from the molar mass (22.99 g Na/mol Na) and Avogadro’s number (1 mol Na/6.022 x 1023 atoms). Our answer is the product of these two factors: ? g Na atom = 1 Na atom

  12. (d) the mass of a single Na atom. (d) The answer must have the unit grams per sodium atom (g/Na atom). We know these quantities from the molar mass (22.99 g Na/mol Na) and Avogadro’s number (1 mol Na/6.022 x 1023 atoms). Our answer is the product of these two factors: ? g Na atom = 1 Na atom x 1 mol Na atom 6.022 x 1023 Na atom

  13. (d) the mass of a single Na atom. (d) The answer must have the unit grams per sodium atom (g/Na atom). We know these quantities from the molar mass (22.99 g Na/mol Na) and Avogadro’s number (1 mol Na/6.022 x 1023 atoms). Our answer is the product of these two factors: ? g Na atom = 1 Na atom x 22.99 g Na atom X 1 mol Na atom 1 mol Na atom 6.022 x 1023 Na atom

  14. (d) the mass of a single Na atom. (d) The answer must have the unit grams per sodium atom (g/Na atom). We know these quantities from the molar mass (22.99 g Na/mol Na) and Avogadro’s number (1 mol Na/6.022 x 1023 atoms). Our answer is the product of these two factors: ? g Na atom = 1 Na atom x 22.99 g Na atom X 1 mol Na atom 1 mol Na atom 6.022 x 1023 Na atom

  15. (d) the mass of a single Na atom. (d) The answer must have the unit grams per sodium atom (g/Na atom). We know these quantities from the molar mass (22.99 g Na/mol Na) and Avogadro’s number (1 mol Na/6.022 x 1023 atoms). Our answer is the product of these two factors: ? g Na atom = 1 Na atom x 22.99 g Na atom X 1 mol Na atom 1 mol Na atom 6.022 x 1023 Na atom = 3.818 x 10 -23 g / Na atom

  16. The Mole and Molar Mass • Molar massis the mass of one mole of a substance. • Molar mass is numerically equal to atomic mass, molecular mass, or formula mass. However … • … the units of molar mass are grams (g/mol). • Examples: 1 atom Na = 22.99 u 1 molNa = 22.99 g 1 molecule CO2 = 44.01 u 1 mol CO2 = 44.01 g 1 formula unit KCl = 74.56 u 1 mol KCl = 74.56 g

  17. Conversions involving Mass, Moles, and Number of Atoms/Molecules 1 mol Na = 6.022 × 1023 Na atoms = 22.99 g Na We can use these equalities to construct conversion factors,such as: 1 mol Na ––––––––– 22.99 g Na 1 mol Na –––––––––––––––––– 6.022 × 1023 Na atoms 22.99 g Na ––––––––– 1 mol Na 22.99 g Na –––––––––––––––––– 6.022 × 1023 Na atoms

  18. We can read formulas in terms of moles of atoms or ions.

  19. Example 3.4 Determine • the number of NH4+ ions in a 145-g sample of (NH4)2SO4 and • the volume of 1,2,3-propanetriol (glycerol)(d= 1.261 g/mL) that contains 1.00 mol O atoms.

  20. Determine the number of NH4+ ions in • a 145-g sample of (NH4)2SO4 Solution • In this calculation, we must use the relationship between the NH4+ ion and the (NH4)2SO4 formula unit; (132.14 g / mol) that is, • 2 mol NH4+/1 mol (NH4)2SO4. • The path to the desired answer is

  21. b) Determine the volume of 1,2,3-propanetriol (glycerol, d = 1.261 g/mL) that contains 1.00 mol O atoms. • (b) The conversion factors needed for this calculation are • the relationship between moles of O atoms and C3H8O3 molecules, • the molar mass of C3H8O3, and • the inverse of the density of C3H8O3:

  22. Example 3.5 An Estimation Example Which of the following is a reasonable value for the number of atoms in 1.00 g of helium? • 4.1 ×10–25 • 4.0 • 1.5 ×1023 (d) 1.5 ×1024

  23. number of atoms in 1.00 g of helium? • a) 4.1 × 10–25b)4.0 c) 1.5 × 1023 (d) 1.5 × 1024 (a) is what we get if, in error, we divide the number of moles of He, 0.25, by Avogadro’s number. Clearly, we can’t have anything less than one atom. (b) expressed as 4.0 u is the atomic mass of He; expressed as 4.0 g/mol He, it is the molar mass. In either case, it is far too small or big to be the number of atoms for any macroscopic sample. (c) is the correct order of magnitude, and the correct answer. (d) Order of magnetude is reasonable, however, it is larger than Avogadro’s number.We would expect a fraction of a mole of helium to have fewer than Avogadro’s number of atoms. Response (c) is the correct answer, obtained from the calculation: 0.25 xNA = 1.5 X 10 23

  24. 3.4 Mass Percent Compositionfrom Chemical Formulas The mass percent composition of a compound refers to the proportion of the constituent elements, expressed as the number of grams of each element per 100 grams of the compound. In other words … X g element X % element = –––––––––––––– OR … 100 g compound g element % element = ––––––––––– × 100 g compound

  25. Percentage Composition of Butane

  26. Example 3.6 Calculate, to four significant figures, the mass percent of each element in ammonium nitrate(NH4 NO3).

  27. Calculate, to four significant figures, the mass percent of each element in ammonium nitrate NH4NO3. Solution

  28. Example 3.7 How many grams of nitrogen are present in 46.34 g ammonium nitrate?

  29. Solution The central factor (shown in red) in the conversion is based on the chemical formula NH4NO3. The other factors in the following setup are based on molar masses: • How many grams of nitrogen are present in 46.34 g ammonium nitrate?

  30. Example 3.8 An Estimation Example Without doing detailed calculations, determine which of these compounds contains the greatest mass of sulfur per gram of compound: barium sulfate, (BaSO4) lithium sulfate, (Li2SO4) sodium sulfate, (Na2SO4) lead sulfate. (PbSO4)

  31. The Periodic table

  32. the greatest mass of sulfur per gram of compound: • a) barium sulfate, b) lithium sulfate, c) sodium sulfate, d) lead sulfate. Analysis and Conclusions To make this comparison, we need formulas of the compounds, which we can get from their names: BaSO4 Li2SO4 Na2SO4 PbSO4 The compound with the greatest mass of sulfur per gram of compound also has the greatest mass of sulfur per 100 g of compound—in other words, the greatest % S by mass. From the formulas, we see that in one mole of each compound there is one mole of sulfur, which means 32.066 g S. Thus, the compound with the greatest % S is the one with the smallest formula mass. Because each formula unit has oneSO42– ion, all we have to do is compare some atomic masses: that of barium to twice that of lithium, and so on. With just a glance at an atomic mass table:we see the answer must be lithium sulfate, Li2SO4.

  33. 3.5 Chemical Formulas from Mass Percent Composition • We can “reverse” the process of finding percentage composition. • First we use the percentage or mass of each element to find moles of each element. • Then we can obtain the empirical formula by finding the smallest whole-numberratio of moles. • Find the whole-number ratio by dividing each number of moles by the smallest number of moles.

  34. Example 3.9 Phenol, a general disinfectant, has the composition 76.57% C, 6.43% H, and 17.00% O by mass. Determine its empirical formula.

  35. the composition 76.57% C, 6.43% H, and 17.00% O by mass. Determine its empirical formula. Solution Step 1:A 100.00-g sample of phenol contains 76.57 g C, 6.43 g H, and 17.00 g O. Step 2: We convert the masses of C, H, and O to amounts in moles:

  36. Step 3: Now we use the numbers of moles in Step 2 as subscripts in a tentative formula: C6.375H6.38O1.063 Step 4:Next we divide each subscript by the smallest (1.063) to try to get integralsubscripts: Step 5: The subscripts in Step 4 are all integers. We need do nothing further. The empirical formula of phenol is C6H6O.

  37. Example 3.10 Diethylene glycol, used in antifreeze, as a softening agent for textile fibers and some leathers, and as a moistening agent for glues and paper, has the composition 45.27% C, 9.50% H, and 45.23% O by mass. Determine its empirical formula.

  38. the composition 45.27% C, 9.50% H, and 45.23% O by mass. • Determine its empirical formula. Solution Step 1:A 100.00-g sample of diethylene glycol contains 45.27 g C, 9.50 g H, and 45.23 g O. Step 2: We convert the masses of C, H, and O to amounts in moles:

  39. the composition 45.27% C, 9.50% H, and 45.23% O by mass. • Determine its empirical formula. Step 3: Now we use the numbers of moles in Step 2 as subscripts in a tentative formula: C3.769H9.42O2.827 Step 4: Next we must divide each subscript by the smallest (2.827) in an attempt to get integral subscripts: Step 5: Finally we multiply all the subscripts from Step 4 by a common factor to convert them all to integers. By recognizing that 1.333 = 4/3 and 3.33 = 10/3, we can see that the common factor we need is 3:

  40. Diethylene glycol m.w. = 106.12g f.p.dep. for Water 1.86oC/mol (Na+)(Cl-) m.w. = 58.8 Which is better to use as Antifreeze = ?

  41. molecular formula mass = integer (nearly) empirical formula mass Relating Empirical FormulastoMolecular Formulas • A molecular formula is a simple integer multiple of the empirical formula. • That is, an empirical formula of CH2 means that the molecular formula is CH2, or C2H4, or C3H6, or C4H8, etc. • So: we find the molecular formula by: We then multiply each subscript in the empirical formula by the integer.

  42. Example 3.11 The empirical formula of hydroquinone, a chemical used in photography, is C3H3O, and its molecular mass is 110 u. What is its molecular formula?

  43. The empirical formula of hydroquinone, is C3H3O, and its molecular mass is 110 u. • What is its molecular formula? Solution The empirical formula mass is (3 x 12.0 u) + (3 x 1.0 u) + 16.0 u = 55.0 u. The multiplier we need to convert the subscripts in the empirical formula to those in the molecular formula is the integral factor The molecular formula is (C3H3O) x 2, or C6H6O2.

  44. 3.6 Elemental Analysis … • … is one method of determining empirical formulas in the laboratory. • This method is used primarily for simple organic compounds (that contain carbon, hydrogen, oxygen). • The organic compound is burned in oxygen. • The products of combustion (usually CO2 and H2O) are weighed. • The amount of each element is determined from the mass of products.

  45. Elemental Analysis … H2O formed, is absorbed by MgClO4, and … The sample is burned in a stream of oxygen gas, producing … … CO2formed, is absorbed by NaOH.

  46. Example 3.12 Burning a 0.1000-g sample of a carbon–hydrogen–oxygen compound in oxygen yields 0.1953 g CO2 and 0.1000 g H2O. A separate experiment shows that the molecular mass of the compound is 90 u. Determine • the mass percent composition, • the empirical formula, and • the molecular formula of the compound.

  47. Burning a 0.1000-g sample of a carbon–hydrogen–oxygen compound in oxygenyields 0.1953 g CO2 and 0.1000 g H2O. • A separate experiment shows that the molecular mass of the compound is 90 u. • Determine a) the mass percent composition, Solution (a)First, we do the conversions outlined to calculate the mass of carbon in the CO2produced: This mass of carbon originated from the 0.1000-g sample. Thus, the mass percent carbon in the compound is

  48. Burning a 0.1000-g sample of a carbon–hydrogen–oxygen compound in oxygenyields 0.1953 g CO2 and 0.1000 g H2O. • A separate experiment shows that the molecular mass of the compound is 90 u. • Determine a) the mass percent composition, We can use similar calculations to determine first the mass of hydrogen and then the mass percent of hydrogen in the compound: Finally, we can calculate the mass percent oxygen by subtracting the mass percents of C and H from 100.00%: % O = 100.00% – 53.30% – 11.19% = 35.51% Thereforethemasspercentcompositionis 53.30% C, 11.19% H and35.51% O

  49. Burning a 0.1000-g sample of a carbon–hydrogen–oxygen compound in • oxygen yields 0.1953 g CO2 and 0.1000 g H2O. A separate experiment • shows that the molecular mass of the compound is 90 u. Determine • b) the empirical formula, (b)Here we apply the method of Examples 3.9 and 3.10, but we need only the firstfour steps. Step 1: A 100.00-g sample of the compound contains 53.30 g C, 11.19 g H, and 35.51 g O. Step 2: We convert the masses of C, H, and O to numbers of moles: C4.438H11.10O2.220

  50. Burning a 0.1000-g sample of a carbon–hydrogen–oxygen compound in • oxygen yields 0.1953 g CO2 and 0.1000 g H2O. A separate experiment • shows that the molecular mass of the compound is 90 u. Determine • b) the empirical formula, • . Step 3: Next, we use the numbers of moles in Step 2 as subscripts in a tentative formula: C4.438H11.10O2.220 Step 4: We divide all subscripts by the smallest subscript (2.220) to get integral subscripts:

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