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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: 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: 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 3.1 Calculate the molecular mass of glycerol (1,2,3-propanetriol). Example 3.2 Calculate the formula mass of ammonium sulfate, a fertilizer commonly used by home gardeners.

  4. Determining the molecular mass of Glycerol (1,2,3-propanetriol). • Determine molecular formula • CH2OHCHOHCH2OH = C3H8O3 • Multiply # of atoms of each element by the atomic mass of that element • C3 = 3 x 12.011u = 36.033u • H8 = 8 x 1.008u = 8.064u • O3 = 3 x 15.999u 47.997u • Add up the contributions of each element to the molecular mass • Molecular mass = 36.033 + 8.064 + 47.997 = 92.094u

  5. Determining the Formula Mass of Ammonium Sulfate

  6. 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 • 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.

  7. One Mole of Four Elements One mole each of helium, sulfur, copper, and mercury. How many atoms of helium are present? Of sulfur? Of copper? Of mercury?

  8. 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 ×1025 Na atoms, and (d) the mass of a single Na atom.

  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. Strategy To perform these calculations, we will use a conversion factor derived from molar mass to relate moles and grams, and a conversion factor derived from Avogadro’s number to relate moles and number of atoms. These are the relationships embodied in Equation (3.1). Solution (a) To convert from moles to grams, we need to use the first and third terms in Equation (3.1), written as a conversion factor: 22.99 g Na/1 mol Na: (b) Again, we need the first and third terms in Equation (3.1), but 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. Example 3.3 continued Solution continued (c) Here we can use all three terms in Equation (3.1), written as two conversion factors, to convert first from number of atoms to number of moles, and then from moles to the mass in grams: Alternatively, we can use the second and third terms in Equation (3.1), written as the conversion factor 22.99 g Na/6.022 x 1023 Na atoms, to convert directly from number of atoms to mass in grams: (d) The answer must have the unit grams per sodium atom (g/Na atom). Thus, if we know the mass of a certain number of Na atoms, our answer is simply that mass divided by the corresponding number of atoms. And we know these quantities from the molar mass (22.99 g Na/mol Na) and Avogadro’s number (1 mol Na/6.022 x 1023atoms). Our answer is the product of these two factors:

  11. Example 3.3 continued Assessment In these examples, note the common practice of expressing molar mass and Avogadro’s number with at least one significant figure more than the number of significant figures in the least precisely known quantity. Doing this ensures that the precision of the calculated results is limited only by the least precisely known quantity. In part (d), this simple fact should deter you from mistakenly multiplying instead of dividing by Avogadro’s number: Individual atoms are exceedingly small and possess masses that are many orders of magnitude less than one gram. It is also worth noting that the calculated mass is the true mass of a sodium atom—23Na has no isotopes. For elements with two or more isotopes, the mass calculated for an atom is a weighted average. Exercise 3.3A Exercise 3.3B Calculate (a) the mass in milligrams of 1.34 x 10–4 mol Ag and (b) the number of oxygen atoms in 20.5 mol O2. Calculate (a) the number of moles of Al in a cube of aluminum metal 5.5 cm on an edge (d = 2.70 g/cm3) and (b) the volume occupied by 4.06 x 1024 Br atoms present as Br2 molecules in liquid bromine (d = 3.12 g/mL).

  12. Exercise 3.3A Exercise 3.3B • Calculate (a) the number of moles of Al in a cube of aluminum metal 5.5 cm on an edge • (d = 2.70 g/cm3) and (b) the volume occupied by 4.06 x 1024 Br atoms present as Br2 molecules in liquid bromine (d = 3.12 g/mL). • Moles Al = 166 cm3 of Al x 2.70g x 1 mole Al = • cm3 26.982g • Calculate (a) the mass in milligrams of 1.34 x 10–4 mol Ag and (b) the number of oxygen atoms in 20.5 mol O2. • (a) Mass in mg = 1.34 x 10-4 mol Ag x 107.868g x 1000mg= 14.5mg Ag • 1 mol 1g • (b) # O atoms = 20.5 mol x 6.022 x 1023 atoms x _2 atoms = 2.48 x 1025 atoms of oxygen • 1 mol O2 molecule Solutions to Examples 3.3A,B

  13. 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 mol Na = 22.99 g 1 mol CO2 = 44.01 g 1 molecule CO2 = 44.01 u 1 formula unit KCl = 74.56 u 1 mol KCl = 74.56 g

  14. 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 22.99 g Na ––––––––– 1 mol Na 1 mol Na –––––––––––––––––– 6.022 × 1023 Na atoms Note: preliminary and follow-up calculations may be needed.

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

  16. Example 3.4 Determine (a) the number of NH4+ ions in a 145-g sample of (NH4)2SO4 and (b) the volume of 1,2,3-propanetriol (glycerol, d= 1.261 g/mL) that contains 1.00 mol O atoms. 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? (a) 4.1 ×10–23 (c) 1.5×1023 (b) 4.0 (d) 1.5×1024

  17. 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

  18. Percentage Composition of Butane

  19. Example 3.6 Calculate, to four significant figures, the mass percent of each element in ammonium nitrate. Example 3.7 How many grams of nitrogen are present in 46.34 g ammonium nitrate? 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, lithium sulfate, sodium sulfate, or lead sulfate.

  20. 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 molesof 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.

  21. 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. 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.

  22. molecular formula mass = integer (nearly) empirical formula mass Relating Molecular Formulasto Empirical 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.

  23. 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?

  24. 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.

  25. Elemental Analysis (cont’d) … H2O, which is absorbed by MgClO4, and … The sample is burned in a stream of oxygen gas, producing … … CO2, which is absorbed by NaOH.

  26. Elemental Analysis (cont’d) If our sample were CH3OH, every two molecules of CH3OH … … would give two molecules of CO2 … … and four molecules of H2O.

  27. 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 (a) the mass percent composition, (b) the empirical formula, and (c) the molecular formula of the compound.

  28. Writing Chemical Equations • A chemical equation is a shorthand description of a chemical reaction, using symbols and formulas to represent the elements and compounds involved.

  29. Writing Chemical Equations • Sometimes additional information about the reaction is conveyed in the equation.

  30. Balancing Equations Illustrated How can we tell that the equation is not balanced? … not by changing the equation … … and not by changing the formulas. The equation is balanced by changing the coefficients …

  31. Guidelines for BalancingChemical Equations • If an element is present in just one compound on each side of the equation, try balancing that element first. • Balance any reactants or products that exist as the free element last. • In some reactions, certain groupings of atoms (such as polyatomic ions) remain unchanged. In such cases, treat these groupings as a unit. • At times, an equation can be balanced by first using a fractional coefficient(s). The fraction is then cleared by multiplying each coefficient by a common factor.

  32. Example 3.13 Balance the equation Fe + O2 Fe2O3 (not balanced) Example 3.14 Balance the equation C2H6 + O2  CO2+ H2O Example 3.15 Balance the equation H3PO4 + NaCN  HCN + Na3PO4 Example 3.16 A Conceptual Example Write a plausible chemical equation for the reaction between water and a liquid molecular chloride of phosphorus to form an aqueous solution of hydrochloric acid and phosphorus acid. The phosphorus-chlorine compound is 77.45% Cl by mass.

  33. Stoichiometric Equivalenceand Reaction Stoichiometry • A stoichiometric factor or mole ratio is a conversion factor obtained from the stoichiometric coefficients in a chemical equation. • In the equation: CO(g) + 2 H2(g) CH3OH(l) • 1 mol CO is chemically equivalent to 2 mol H2 • 1 mol CO is chemically equivalent to 1 mol CH3OH • 2 mol H2 is chemically equivalent to 1 mol CH3OH 1 mol CO ––––––––– 2 mol H2 1 mol CO ––––––––––––– 1 mol CH3OH 2 mol H2 ––––––––––––– 1 mol CH3OH

  34. Concept of Stoichiometric Equivalence One car may be equivalent to either 25 feet or 10 feet, depending on the method of parking. One mole of CO may be equivalent to one mole of CH3OH, or to one mole of CO2, or to two moles of CH3OH, depending on the reaction(s).

  35. Outline of Simple Reaction Stoichiometry Note: preliminary and/or follow-up calculations may be needed.

  36. Example 3.17 When 0.105 mol propane is burned in an excess of oxygen, how many moles of oxygen are consumed? The reaction is C3H8+ 5 O2 3 CO2+ 4 H2O

  37. Outline of Stoichiometry Involving Mass … we’ve added a conversion from mass at the beginning … To our simple stoichiometry scheme … Substances A and B may be two reactants, two products, or reactant and product. … and a conversion to mass at the end. Think: If we are given moles of substance A initially, do we need to convert A to grams?

  38. Example 3.18 The final step in the production of nitric acid involves the reaction of nitrogen dioxide with water; nitrogen monoxide is also produced. How many grams of nitric acid are produced for every 100.0 g of nitrogen dioxide that reacts? Example 3.19 Ammonium sulfate, a common fertilizer used by gardeners, is produced commercially by passing gaseous ammonia into an aqueous solution that is 65% H2SO4 by mass and has a density of 1.55 g/mL. How many milliliters of this sulfuric acid solution are required to convert 1.00 kg NH3 to (NH4)2SO4?

  39. Limiting Reactants • Many reactions are carried out with a limited amount of one reactant and a plentiful amount of the other(s). • The reactant that is completely consumed in the reaction limits the amounts of products and is called the limiting reactant, or limiting reagent. • The limiting reactant is not necessarily the one present in smallest amount.

  40. Limiting Reactant Analogy If we have 10 sandwiches, 18 cookies, and 12 oranges … … how many packaged meals can we make?

  41. Molecular View of the Limiting Reactant Concept • Why is ethylene left over, when we started with more bromine than ethylene? (Hint: count the molecules.) • What mass of ethylene is left over after reaction is complete? (Hint: it’s an easy calculation; why?) When 28 g (1.0 mol) ethylene reacts with … … 150 g of 1,2-dibromoethane, and leftover ethylene! … 128 g (0.80 mol) bromine, we get …

  42. Recognizing and Solving Limiting Reactant Problems • We recognize limiting reactant problems by the fact that amounts of two (or more) reactants are given. • One way to solve them is to perform a normal stoichiometric calculation of the amount of product obtained, starting with each reactant. • The reactant that produces the smallest amount of product is the limiting reactant.

  43. Example 3.20 Magnesium nitride can be formed by the reaction of magnesium metal with nitrogen gas. (a) How many grams of magnesium nitride can be made in the reaction of 35.00 g of magnesium and 15.00 g of nitrogen? (b) How many grams of the excess reactant remain after the reaction?

  44. Yields of Chemical Reactions • The theoretical yield of a chemical reaction is the calculated quantity of product in the reaction. • The actual yieldis the amount you actually get when you carry out the reaction. • Actual yield will be less than the theoretical yield, for many reasons … can you name some? actual yield Percent yield = ––––––––––––– × 100 theoretical yield

  45. Actual Yield of ZnS Is Less than the Theoretical Yield

  46. Example 3.21 Ethyl acetate is a solvent used as fingernail polish remover. What mass of acetic acid is needed to prepare 252 g ethyl acetate if the expected percent yield is 85.0%? Assume that the other reactant, ethanol, is present in excess. The equation for the reaction, carried out in the presence of H2SO4, is CH3COOH + HOCH2CH3 CH3COOCH2CH3+ H2O Acetic acid Ethanol Ethyl acetate Example 3.22 A Conceptual Example What is the maximum yield of CO(g) obtainable from 725 g of C6H14(l), regardless of the reaction(s) used, assuming no other carbon-containing reactant or product?

  47. Solutions andSolution Stoichiometry • Solute: the substance being dissolved. • Solvent: the substance doing the dissolving. • Concentration of a solution: the quantity of a solute in a given quantity of solution (or solvent). • A concentrated solution contains a relatively large amount of solute vs. the solvent (or solution). • A dilute solution contains a relatively small concentration of solute vs. the solvent (or solution). • “Concentrated” and “dilute” aren’t very quantitative …

  48. Molar Concentration Molarity (M), or molar concentration, is the amount of solute, in moles, per liter of solution: moles of solute Molarity = –––––––––––––– liters of solution • A solution that is 0.35 M sucrose contains 0.35 moles of sucrose in each liter of solution. • Keep in mind that molarity signifies moles of solute per liter of solution, not liters of solvent.

  49. Preparing 0.01000 M KMnO4 Weigh 0.01000 mol (1.580 g) KMnO4. Add more water to reach the 1.000 liter mark. Dissolve in water. How much water? Doesn’t matter, as long as we don’t go over a liter.

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