Lecture 2: Stoichiometry PDF

Lecture 2: Stoichiometry  Law of Conservation of Mass Neither created nor destroyed in a chemical reaction. Total mass of the reactants before the reaction is equal to the Antoine Lavoisier total mass of the products after the reaction  Law of Conservation of Energy...

Lecture 2: Stoichiometry  Law of Conservation of Mass Neither created nor destroyed in a chemical reaction. Total mass of the reactants before the reaction is equal to the Antoine Lavoisier total mass of the products after the reaction  Law of Conservation of Energy Sates that energy cannot be created or destroyed; it can only be transformed from one form to another or transferred between objects. The total amount of energy in a closed system remains constant over time.  Law of Definite Proportion Also known as the Law of Constant Composition. States that a given chemical compound always contains the same elements in the exact same proportion by mass, regardless of the amount of the compound or its source. For example, water (H2O). The mass ratio is always 1:8.  Law of Multiple Proportion When two elements combine to form more than one compound, the masses of one element that combine with a fixed mass of the other elements are in simple whole- number ratios. John Dalton Example: carbon and oxygen can combine to form both CO and CO2.  Questions to Consider Can atoms be counted? Mole concept Does the number of atoms in a reaction affect the result? Stoichiometry and balanced reactions Can the mass of a sample be used to determine the number of moles it comprises? Mass to mole using Molecular Weight or Atomic Weight.  Stoichiometry Describes quantitative relationships in substances and their reactions – Chemical equations – The mole and molar mass – Chemical formulas – Mass relationships in equations – Limiting reactant Average Mass The term “average mass” often refers to the atomic mass or molecular mass of an element or compound, respectively. Can refer to the average mass of particles in a sample when there are variations due to different isotopes or molecular forms. Atomic Mass (Atomic Weight) For individual elements, the average atomic mass is the weighted average mass of the isotopes of that element. For example, the atomic mass of carbon is 12.01 amu.  Molecular Mass (Molecular Weight) For compounds, the average mass refers to the molecular mass, which is the sum of the atomic masses of the atoms in the molecule. Example: Water (H2O) Molecular Mass = (2x 1.008 amu) + (16.00 amu) = 18.016 amu Example When a sample of natural copper is vaporized and injected into a mass spectrometer, the results shown in Fig. 5-3 are obtained. Use these data to compute the average mass of natural copper. (The mass values for 63Cu and 65Cu are 62.93 u and 64.93 u, respectively.) Solution: Mole ✔ The mole is defined as the number equal to the number of carbon atoms in exactly 12 grams of pure 12C. 12 g of 12C = 6.022 × 1023 atoms of 12C (Also called Avogadro’s number) ✔ One mole of a substance contains 6.022 × 1023 units of that substance. ✔ A sample of a natural element with a mass equal to the element’s atomic mass expressed in grams contains 1 mole of atoms. Element Number of Atoms Present Mass of Sample (g) Aluminum 6.022 × 1023 26.98 Copper 6.022 × 1023 63.55 Iron 6.022 × 1023 55.85 Sulfur 6.022 × 1023 32.07 Iodine 6.022 × 1023 126.9 Mercury 6.022 × 1023 200.6 Example Americium is an element that does not occur naturally. It can be made in very small amounts in a device known as a particle accelerator. Compute the mass in grams of a sample of americium containing six atoms. Solution Given: 1 mole Americium = 6.022x1023 atoms Americium = 243 g Americium Molar Mass ✔ It is the mass of one mole of the compound measured in grams. Traditionally called molecular weight. ✔ Formula unit - Used for compounds that do not contain molecules. Example: NaCl - Sodium chloride CaCO3 - Calcium carbonate Example Juglone, a dye known for centuries, is produced from the husks of black walnuts. It is also a natural herbicide (weed killer) that kills off competitive plants around the black walnut tree but does not affect grass and other noncompetitive plants. The formula for juglone is C10H6O3. a) Calculate the molar mass of juglone. b) A sample of 1.56 × 10−2 g of pure juglone was extracted from black walnut husks. How many moles of juglone does this sample represent? Solution Percent Composition of a Compound ⮚ In terms of the numbers of its constituent atoms ⮚ In terms of the percentages (mass) of its elements ⮚ Consider ethanol (C2H5OH) Mass Percent ⮚ It is calculated by comparing the mass percent of the element in one mole of the compound with the total mass of one mole of the compound and multiplying the result by 100%. Example: Calculating the mass of carbon in ethanol Example Carvone is a substance that occurs in two forms having different arrangements of atoms but the same molecular formula (C10H14O) and mass. One type of carvone gives caraway seeds their characteristic smell, and the other type is responsible for the smell of spearmint oil. Compute the mass percent of each element in carvone. Carvone is a substance that occurs in two forms having different arrangements of atoms but the same molecular formula (C10H14O) and mass. One type of carvone gives caraway seeds their characteristic smell, and the other type is responsible for the smell of spearmint oil. Compute the mass percent of each element in carvone. Solution  Determining the Formula of a Compounds ⮚ Determined by using a weighed sample and one of the following techniques ✔ Decomposing it into its component elements ✔ Introducing oxygen to produce substances such as CO2, H2O, etc., which are collected and weighed Analyzing for Carbon and Hydrogen Empirical and Molecular Formula ⮚ Empirical formula (EF) - the simplest whole-number ratio of the various types of atoms in a compound. Can be obtained from the mass percent of elements in a compound ⮚ The molecular formula (MF) varies for molecules and ions. ✔ For molecular substances, it is the formula of the constituent molecules. Always an integer multiple of the empirical formula. ✔ For ionic substances, it is the same as the empirical formula  Steps in Empirical Formula Determination 1. Since mass percentage gives the number of grams of a particular element per 100 grams of compound, base the calculation on 100 grams of compound. Each percent will then represent the mass in grams of that element. 2. Determine the number of moles of each element present in 100 grams of compound using the atomic masses of the elements present 3. Divide each value of the number of moles by the smallest of the values. If each resulting number is a whole number (after appropriate rounding), these numbers represent the subscripts of the elements in the empirical formula 4. If the numbers obtained in the previous step are not whole numbers, multiply each number by an integer so that the results are all whole numbers Example Determine the empirical and molecular formulas for a compound that gives the following percentages on analysis (in mass percents): 71.65% Cl, 24.27% C, 4.07% H. The molar mass is known to be 98.96 g/mol Solution Empirical Formula CH2Cl (48.47 g/mol) Molecular Formula C2H4Cl2 (48.47 g/mol) Chemical Reactions ⮚ Involve the reorganization of atoms in one or more substances. ⮚ In a chemical equation, the reactants are situated on the left side of an arrow, and the products are located on the right Bonds are broken and new bonds are formed Both sides possess the same number of atoms  Meaning of a Chemical Equation State Symbol ❖ It contains information on: Solid (s) ✔ The nature of the reactants and products Liquid (l) Gas (g) ✔ The relative numbers of each Dissolved in (aq) ❖ Equations also provide the physical state of water (in aqueous the reactants and products solution) Reactants Products CH4 (g) + 2 O2(g) CO2 (g) + 2 H2O(g) 1 molecule + 2 molecules 1 molecule + 2 molecules 1 mol + 2 mol 1 mol + 2 mol 6.022 × 1023 molecules + 2 (6.022 6.022 × 1023 molecules + 2 (6.022 × × 1023 molecules) 1023 molecules) 16 g + 2 (32 g) 44 g + 2 (18 g) 80 g reactants 80 g products Balancing Chemical Equation Atoms are neither created nor destroyed in a chemical reaction. The number of atoms on each side of the equation must be the same. Consider the reaction between methane and oxygen  Steps in Balancing a Chemical Equation Step 1. Start with the most complicated molecules. Consider the following unbalanced equation Step 2. The most complicated molecule is C2H5OH. Balancing carbon Balancing hydrogen Balancing oxygen  Steps in Balancing a Chemical Equation Step 3. Verifying the results Example Chromium compounds exhibit a variety of bright colors. When solid ammonium dichromate, (NH4)2Cr2O7, a vivid orange compound, is ignited, a spectacular reaction occurs. Although the reaction is actually somewhat more complex, let’s assume here that the products are solid chromium (III) oxide, nitrogen gas (consisting of N2 molecules), and water vapor. Balance the equation for this reaction. The unbalanced equation is:  The unbalanced equation is: Note that nitrogen and chromium are balanced (two nitrogen atoms and two chromium atoms on each side), but hydrogen and oxygen are not. A coefficient of 4 for H2O balances the hydrogen atoms. Obtaining Ratios from a Balanced Chemical Equation MOLE RATIOS: Problem-Solving Strategy - Calculating Masses of Reactants and Products  Sample Problem 1 In the combustion of methane, how many moles of O2 are required if 6.75 mol of CH4 is to be consumed? CH4(g) + 2O2(g) CO2(g) + 2 H2O(l)  Sample Problem 2 How many grams of water can be produced if sufficient hydrogen reacts with 26.0 g of oxygen? 2H2(g) + O2(g) 2 H2O(g)  Sample Problem 3 Tetraphosphorus trisulfide, P4S3, is used in the manufacture of matches. Elemental phosphorus and sulfur react directly to form P4S3: 8 P4 + 3 S8 8 P4S3 If we have 153 g of S8 and an excess of phosphorus, what mass of P4S3 can be produced by this reaction?  Sample Problem 3 Tetraphosphorus trisulfide, P4S3, is used in the manufacture of matches. Elemental phosphorus and sulfur react directly to form P4S3: 8 P4 + 3 S8 8 P4S3 If we have 153 g of S8 and an excess of phosphorus, what mass of P4S3 can be produced by this reaction?  Limiting Reactants LIMITING REACTANT: - reactant that is completely consumed - when used up, the reaction stops - limits the quantities of products formed EXCESS REACTANT: - other reactants present - not completely consumed  Sample Problem 4 A solution of hydrochloric acid contains 5.22 g of HCl. When it is allowed to react with 3.25 g of solid K2CO3, the products are KCl, CO2, and H2O. Which is the limiting reactant?  Sample Problem 5 If 28.2 g of P4 is allowed to react with 18.3 g of S8, which is the limiting reactant?  Sample Problem 6 MTBE (methyl tert-butyl ether) has been used as an additive in gasoline? The compound is produced by reacting methanol and isobutene, according to the following equation: If 45.0 kg of methanol is allowed to react with 70.0 kg of isobutene, what is the maximum mass of MTBE that can be obtained?  Sample Problem 6 If 45.0 kg of methanol is allowed to react with 70.0 kg of isobutene, what is the maximum mass of MTBE that can be obtained? Isobutene is the limiting reactant, use it to calculate mass of MTBE produced.  Sample Problem 6 If 45.0 kg of methanol is allowed to react with 70.0 kg of isobutene, what is the maximum mass of MTBE that can be obtained? Use 70.0 kg of isobutene to calculate mass of MTBE.  Reaction Yields Actual Yield - quantity of product measured through experiments Theoretical Yield - quantity of product calculated from reaction stoichiometry  Sample Problem 7 Solvay process is important in the commercial production of sodium carbonate (Na2CO3), which is used in the manufacture of most glass. The last step in Solvay process is the conversion of NaHCO3 (sodium bicarbonate) to Na2CO3 by heating. When 42.0 g of NaHCO3 is heated, 22.3 g of Na2CO3 is formed. What is the percentage yield of this reaction?  Sample Problem 7 When 42.0 g of NaHCO3 is heated, 22.3 g of Na2CO3 is formed. What is the percentage yield of this reaction?  Solution Stoichiometry Determining number of moles of substances when volumes are measured rather than masses.  Sample Problem 8 The fuel hydrazine can be produced by the reaction of solutions of sodium hypochlorite (NaClO) and ammonia (NH3) as shown in the equation below: If 750.0 mL of 0.806 M NaClO is mixed with excess ammonia, how many moles of hydrazine can be formed? If the final volume of the resulting solution is 1.25 L, what will be the molarity of hydrazine?  Sample Problem 8 If 750.0 mL of 0.806 M NaClO is mixed with excess ammonia, how many moles of hydrazine can be formed? If the final volume of the resulting solution is 1.25 L, what will be the molarity of hydrazine? Use 0.605 mol NaClO to calculate moles of hydrazine: Molarity of hydrazine can be calculated as:  Sample Problem 9 Many common titrations involve the reaction of an acid with a base. If 24.75 mL of 0.503 M NaOH solution is used to titrate a 15.00-mL sample of sulfuric acid, H2SO4, what is the concentration of the acid? Calculate moles of NaOH from given molarity and volume: Use balance equation to calculate moles of H2SO4 :  Sample Problem 9 Many common titrations involve the reaction of an acid with a base. If 24.75 mL of 0.503 M NaOH solution is used to titrate a 15.00-mL sample of sulfuric acid, H2SO4, what is the concentration of the acid? Use balance equation to calculate moles of H2SO4 : Calculate molarity of H2SO4 :