Stoichiometry. Chapter 3

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1 Stoichiometry Chapter 3

2 Chemical Stoichiometry Stoichiometry: The study of quantities of materials consumed and produced in chemical reactions. In macroworld, we can count objects by weighing assuming that Objects behave as though they were all identical We know the average mass of the objects In microworld, atoms are too small to count directly. But If we know the average mass of atoms, we can count them. 12 C is the standard for atomic mass, with a mass of exactly 12 atomic mass units (u). The masses of all other atoms are given relative to this standard. 2

3 Atomic Mass Atomic mass is the mass of an atom in atomic mass units (amu or u) One atomic mass unit is defined as a mass exactly equal to one-twelfth the mass of one carbon-12 atom 1amu mass of onecarbon 12atom 12 By definition: 1 12 C (6 1 p n) atom weighs 12 u On this scale and on average 1 H = u 16 O = u 56 Fe = u 3

Average Atomic Mass Elements occur in nature as mixtures of isotopes The atomic mass of an element is the average mass of a representative sample of atoms of the element (also call the weighted

4 Average Atomic Mass Elements occur in nature as mixtures of isotopes The atomic mass of an element is the average mass of a representative sample of atoms of the element (also call the weighted average atomic mass) Isotope 12 C 13 C 14 C Natural abundance 98.89% 1.11% <0.01% 6 C atomic number atomic mass The weighted average is the average calculated by considering the relative amounts! average atomic mass of natural carbon 4

5 Calculation of Average Atomic Mass The weighted average is the average calculated by considering the relative amounts Calculate the average atomic mass of chlorine: Isotope Mass relative to C-12 Percentage natural abundance Chlorine amu Chlorine amu

6 The Mole (mol) A unit to count numbers of particles. Defined as the amount of a substance that contains as many elementary entities (atoms, molecules, or other particles) as there are atoms in exactly 12 g of the carbon-12 isotope. The actual number of atoms in 12 g of carbon-12 is called Avogadro s Number (N A ), in honor of the Italian chemist Amadeo Avogadro ( ). The currently accepted value is N A = N A = (4 significant figures) 6

7 The Molar Mass (M) 1 mole of anything = units of that thing Molar mass Is the mass (g or kg) of 1 mole of units of a substance Has units of g/mol Is numerically equal to the atomic mass in amu Thus, for any element, atomic mass (amu) = molar mass (grams) 1 carbon-12 atom = amu 1 mole carbon-12 atoms = g = atoms 1 mole carbon atoms = g of C 1 mole lithium atoms = g of Li 7

8 One Mole of 8

9 Q1. Which of the following is closest to the average mass of one atom of copper? a g b g c g d g e x g. Q2. Which of the following g samples contains the greatest number of atoms? a. Magnesium b. Silicon c. Zinc d. Silver e. Cesium 9

10 Q3. Rank the following according to number of atoms (greatest to least): a g of silver b g of zinc c g of magnesium d g of potassium e g of chromium. Q4. Calculate the number of copper atoms in a g sample of copper. a Cu atoms b Cu atoms c Cu atoms d Cu atoms e Cu atoms 10

11 Molecular Mass A molecular mass (or molecular weight, MW) is the sum of the atomic weights of the atoms in a molecule. For the molecule sulfur dioxide, SO 2, the molecular weight would be S: O: 11

12 The Molar Mass Is the mass (g or kg) of 1 mole of units of a substance Has units of g/mol Is numerically equal to the molecular mass Thus, for a molecule, molecular mass (u) = molar mass (grams) Cl has an AW of u 1 mol Cl has a mass of g H 2 O has an MW of u 1 mol H 2 O has a mass of g 12

13 Formula Mass A formula mass (or formula weight, FW) is the sum of the atomic weights for the atoms in a formula unit of an ionic compound. So, the formula weight of calcium chloride, CaCl 2, would be The molar mass is also numerically equal to the molecular mass. NaCl has an FW of u 1 mol NaCl has a mass of g 13

14 EX. 1 Consider separate gram samples of each of the following: H 2 O, CO 2, C 3 H 6 O 2, N 2 O 4 Rank them from greatest to least number of oxygen atoms. You need to calculate the number of oxygen atoms by first dividing the mass of each sample by its molar mass to obtain the moles, then multiplying the moles by the number of oxygen in the formula, and finally multiplying the Avogadro s number 14

15 Percent Composition of Compounds The percent composition by mass is the percent by mass of each element in a compound. You can use the following formula to find the percent by mass of each element in a compound: no.of atomes atomic weight % composition of element 100% formulaweight of substance Example: C 2 H 5 OH %C = 100% = 52.14% %H = 100% = 13.13% %O = 100% = 34.73% Check: 15

16 EX. 2 Consider separate gram samples of each of the following: H 2 O, CO 2, C 3 H 6 O 2, N 2 O 4 Rank them from highest to lowest percent oxygen by mass. Using the molecular weight obtained in EX. 1, we can calculate the percent oxygen by mass as follows: 16

17 Analyzing for Carbon and Hydrogen A schematic diagram of the combustion device used to analyze substances for carbon and hydrogen. The sample is burned in the presence of excess oxygen, which converts all its carbon to carbon dioxide and all its hydrogen to water. These products are collected by absorption using appropriate materials, and their amounts are determined by measuring the increase in masses of the absorbents. 17

18 Review: Formulas Empirical formula: the simplest whole-number ratio of atoms in a compound. Molecular formula: the exact number of atoms of each element in the formula of a compound. Molecular formula = (empirical formula) n [n = integer] Empirical formula of dinitrogen tetroxide: molecular formula: N 2 O 4 = Empirical formula of benzene: Molecular formula: C 6 H 6 = 18

19 Empirical Formula Determination 1. Convert the percentages to grams a. assume you start with 100 g of the compound b. skip if already grams 2. Convert grams to moles a. use molar mass of each element 3. Divide all by smallest number of moles a. if result is within 0.1 of whole number, round to whole number 4. Multiply all mole ratios by number to make all whole numbers a. if ratio?.5, multiply all by 2; if ratio?.33 or?.67, multiply all by 3; if ratio 0.25 or 0.75, multiply all by 4, b. skip if already whole numbers 19

20 Molecular Formula Determination 1. Obtain the empirical formula (see previous slide) 2. Compute the empirical formula mass or molar mass 3. Calculate the ratio: Molarmass Empiricalformulamass 4. The integer from Step 3 represents the no. of empirical formula units in one molecule. When the empirical formula subscripts are multiplied by this integer, the molecular formula results. Molecularformula empiricalformula molarmass empiricalformulamass 20

21 Example 3.10 Determine the empirical and molecular formulas for a compound that gives the following percentages on analysis (in mass percent): 71.65% Cl 24.27%C 4.07% H The molar mass is known to be g/mol. 21

22 Example 3.10 Determine the empirical and molecular formulas for a compound that gives the following percentages on analysis (in mass percent): 71.65% Cl 24.27%C 4.07% H The molar mass is known to be g/mol. 22

23 Chemical Rxn and Chemical Eqn A process in which one or more substances is changed into one or more new substances is a chemical reaction A chemical equation uses chemical symbols to show what happens during a chemical reaction When the methane (CH 4 ) in natural gas combines with oxygen (O 2 ) in the air and burns, carbon dioxide (CO 2 ) and water (H 2 O) are formed. This process is represented by an unbalanced chemical reaction: CH 4 + O 2 CO 2 + H 2 O reactants products Reactants are only placed on the left side of the arrow, products are only placed on the right side of the arrow. 23

24 How to Read Chemical Equations? An unbalanced chemical equation gives qualitative information: CH 4 + O 2 CO 2 + H 2 O A balanced chemical equation gives both qualitative and quantitative information. CH 4 + 2O 2 CO 2 + 2H 2 O Use the coefficients in the balanced equation to decide the amount of each reactant that is used, and the amount of each product that is formed. 1 CH 4 molecule + 2 O 2 molecules makes 1 CO 2 molecule & 2 H 2 O molecules 1 mole CH moles O 2 makes 1 mole CO 2 and 2 moles H 2 O 16 g CH g O 2 makes 44 g CO 2 and 36 g H 2 O But NOT 1 g CH g O 2 makes 1 g CO 2 and 2 g H 2 O 24

25 Writing and Balancing the Equation for a Chemical Reaction 1. Determine what reaction is occurring. What are the reactants, the products, and the physical states involved? 2. Write the unbalanced equation that summarizes the reaction described in Step Balance the equation by inspection, starting with the most complicated molecule(s). The same number of each type of atom needs to appear on both reactant and product sides. Do NOT change the formulas of any of the reactants or products. 25

26 Notice for Balancing Chem Eqns The number of atoms of each type of element must be the same on both sides of a balanced equation. Subscripts must not be changed to balance an equation. A balanced equation tells us the ratio of the number of molecules which react and are produced in a chemical reaction. Coefficients can be fractions, although they are usually given as lowest integer multiples. 26

27 Balancing Chemical Equations 27

28 Balancing Chemical Equations 1. Write the correct formula(s) for the reactants on the left side and the correct formula(s) for the product(s) on the right side of the equation. Ethane reacts with oxygen to form carbon dioxide & water: C 2 H 6 + O 2 CO 2 + H 2 O (unbalanced) 2. Change the numbers in front of the formulas (stoichiometric coefficients) to make the number of atoms of each element the same on both sides of the equation. Do not change the subscripts. 2C 2 H 4 NOT C 4 H 8 28

29 Balancing Chem Eqns (cont d) 3. Start by balancing those elements that occur in the fewest chemical formulas. C 2 H 6 + O 2 CO 2 + H 2 O start with C or H but not O carbon on left C 2 H 6 + O 2 carbon on right CO 2 + H 2 O hydrogen on left hydrogen on right C 2 H 6 + O 2 CO 2 + H 2 O 29

30 Balancing Chem Eqns (cont d) 4. Balance those elements that appear in two or more reactants or products. C 2 H 6 + O 2 CO 2 + H 2 O oxygen on left C 2 H 6 + O 2 CO 2 + H 2 O 30

31 Balancing Chem Eqns (cont d) 5. Check to make sure that you have the same number of each type of atom on both sides of the equation. C 2 H 6 + O 2 CO 2 + H 2 O Reactants 4 C 12 H 14 O Products 4 C 12 H 14 O 31

32 Stoichiometric Calculations Chemical equations can be used to relate the masses of reacting chemicals. Mole method: treat the stoichiometric coefficients in a balanced chemical equation as the number of moles of each substance. C 2 H 5 OH + 3O 2 2CO 2 + 3H 2 O 1 molecule 3 molecules 2 molecules 3 molecules 1 mol 3 mol 2 mol 3 mol 1 mole of ethanol reacts with 3 moles of oxygen to form 2 moles of carbon dioxide and 3 moles of water. 1 mol C 2 H 5 OH = 3 mol O 2 3 mol O 2 = 3 mol H 2 O 1 mol C 2 H 5 OH = 2 mol CO 2 1 mol C 2 H 5 OH = 3 mol H 2 O The equal sign (=) here means stoichiometrically equivalent to or simply equivalent to 32

33 Calculating Masses of Rxts & Prdts in Rxns 1. Balance the equation for the reaction. 2. Convert the known mass of the reactant or product to moles of that substance. 3. Use the balanced equation to set up the appropriate mole ratios. 4. Use the appropriate mole ratios to calculate the number of moles of the desired reactant or product. 5. Convert from moles back to grams if required by the problem. 33

34 Calculating Masses of Rxts & Prdts in Rxns 34

35 EX. 3 Methanol burns in air according to the equation 2CH 3 OH + 3O 2 2CO 2 + 4H 2 O If 209 g of methanol are used up in the combustion, (a) what mass of water is produced? grams CH 3 OH moles CH 3 OH moles H 2 O grams H 2 O (b) what mass of oxygen is required? = 235 g H 2 O = 314 g O 2 35

Stoichiometric Mixture Consider the reaction: N 2 (g) + 3H 2 (g) 2NH 3 (g) If we start with a stoichiometric mixture, i.e. the molar ratio of N 2 :H 2 is 1:3, all reactants will be consumed to form products.

36 Stoichiometric Mixture Consider the reaction: N 2 (g) + 3H 2 (g) 2NH 3 (g) If we start with a stoichiometric mixture, i.e. the molar ratio of N 2 :H 2 is 1:3, all reactants will be consumed to form products. 5 N 2 molecules + 15 H 2 molecules 10 NH 3 molecules with no reactants left over. 36

37 Non-Stoichiometric Mixture Consider the reaction: N 2 (g) + 3H 2 (g) 2NH 3 (g) If we start with 5 N 2 molecules and 9 H 2 molecules, after the reaction is complete, not all reactants will be consumed to form products. 5 N 2 molecules + 9 H 2 molecules 6 NH 3 molecules with 2 N 2 molecules left over. 37

38 Limiting Reactants Consider the reaction: N 2 (g) + 3H 2 (g) 2NH 3 (g) The limiting reactant is the reactant present in the smallest stoichiometric amount. In other words, it s the reactant you ll run out of first (H 2 ). In the example (start with 5 N 2 and 9 H 2 ), the N 2 would be the excess reactant. 38

39 Limiting Reactants 39

40 Q5. Which of the following reaction mixtures could produce the greatest amount of product? Each involves the reaction symbolized by the equation: 2H 2 + O 2 2H 2 O a. 2 moles of H 2 and 2 moles of O 2 b. 2 moles of H 2 and 3 moles of O 2 c. 2 moles of H 2 and 1 mole of O 2 d. 3 moles of H 2 and 1 mole of O 2 e. Each produce the same amount of product. 40

41 Reaction Yield Reaction yield is an important indicator of the efficiency of a particular laboratory or industrial reaction. Theoretical Yield is the amount of product that would result if all the limiting reactant reacted. Actual Yield is the amount of product actually obtained from a reaction. Percent Yield is the proportion of the actual yield to the theoretical yield. Percent Yield Actual Yield Theoretical Yield 100% 41

42 EX. 4 Consider the reaction, 2NH 3 + 5F 2 N 2 F 4 + 6HF. If 25.0 g of NH 3 are reacted with 150. g of F 2, (a) What is the limiting reactant? (b) Calculate the theoretical yield of N 2 F 4 in grams. (c) Calculate the percent yield if 56.8 g of N 2 F 4 are actually obtained. (d) Calculate the actual yield of N 2 F 4 in grams if the percent yield is 90%. (a) What is the limiting reactant? Start with 25.0 g NH 3, calculate g of F 2 needed. = 140. g F 2 42

43 EX. 4 Consider the reaction, 2NH 3 + 5F 2 N 2 F 4 + 6HF. If 25.0 g of NH 3 are reacted with 150. g of F 2, (a) What is the limiting reactant? (b) Calculate the theoretical yield of N 2 F 4 in grams. (c) Calculate the percent yield if 56.8 g of N 2 F 4 are actually obtained. (d) Calculate the actual yield of N 2 F 4 in grams if the percent yield is 90%. (b) Calculate the theoretical yield of N 2 F 4 in grams. Start with 25.0 g NH 3, calculate g of N 2 F 4 produced. = 76.5 g N 2 F 4 43

44 EX. 4 Consider the reaction, 2NH 3 + 5F 2 N 2 F 4 + 6HF. If 25.0 g of NH 3 are reacted with 150. g of F 2, (a) What is the limiting reactant? (b) Calculate the theoretical yield of N 2 F 4 in grams. (c) Calculate the percent yield if 56.8 g of N 2 F 4 are actually obtained. (d) Calculate the actual yield of N 2 F 4 in grams if the percent yield is 90%. (c) Calculate the percent yield if 56.8 g of N 2 F 4 are actually obtained. = 74.2% (d) Calculate the actual yield of N 2 F 4 in grams if the percent yield is 90%. = 68.9 g 44

Chapter 3. Mass Relationships in Chemical Reactions

Chapter 3. Mass Relationships in Chemical Reactions Chapter 3 Mass Relationships in Chemical Reactions In this chapter, Chemical structure and formulas in studying the mass relationships of atoms and molecules. To explain the composition of compounds and