Chemical Stoichiometry Stoichiometry – The study of quantities of materials consumed and produced in chemical reactions. Copyright © Cengage Learning. All rights reserved

Copyright © Cengage Learning. All rights reserved EXERCISE! A pile of marbles weigh 394.80 g. 10 marbles weigh 37.60 g. How many marbles are in the pile? Equivalent Relationship between number of marbles and mass of marbles: 10 marbles = 37.60 g Therefore 394.80 g = ? Marbles = 394.80 g x (10 marbles/37.60 g) = 105 marbles Equivalent Relationship between number of marbles and mass of marbles: 10 marbles = 37.60 g Copyright © Cengage Learning. All rights reserved

The masses of all other atoms are given relative to this standard. 12C is the standard for atomic mass, with a mass of exactly 12 atomic mass units (amu). The masses of all other atoms are given relative to this standard. Elements occur in nature as mixtures of isotopes. Carbon = 98.89% 12C 1.11% 13C < 0.01% 14C Copyright © Cengage Learning. All rights reserved

Atomic weight of an element = Average atomic mass = a weighted average of the atomic masses of all its STABLE isotopes. For Example, the atomic weight of C = the Average Atomic Mass of C = the weighted average of the atomic masses of 12C & 13C isotopes = 98.89% of 12 amu + 1.11% of 13.0034 amu = (0.9889)(12 amu) + (0.0111)(13.0034 amu) = 12.01 amu Copyright © Cengage Learning. All rights reserved

Calculate the average atomic mass and identify the element. EXERCISE! An element consists of 62.60% of an isotope with mass 186.956 amu and 37.40% of an isotope with mass 184.953 amu. Calculate the average atomic mass and identify the element. 186.2 amu Rhenium (Re) Average Atomic Mass = (0.6260)(186.956 u) + (0.3740)(184.953 u) = 186.2 u The element is rhenium (Re). Copyright © Cengage Learning. All rights reserved

Average Atomic Mass for Carbon Even though natural carbon does not contain a single atom with mass 12.01, for stoichiometric purposes, we can consider carbon to be composed of only one type of atom with a mass of 12.01. This enables us to count atoms of natural carbon by weighing a sample of carbon. Copyright © Cengage Learning. All rights reserved

Schematic Diagram of a Mass Spectrometer Copyright © Cengage Learning. All rights reserved

The number equal to the number of carbon atoms in exactly 12 grams of pure 12C. 1 mole of something consists of 6.022 × 1023 units of that substance (Avogadro’s number). 1 mole C = 6.022 × 1023 C atoms = 12.01 g C Copyright © Cengage Learning. All rights reserved

EXERCISE! Calculate the number of iron atoms in a 4.48 mole sample of iron. 2.70×1024 Fe atoms (4.48 mol Fe) × (6.022×1023 Fe atoms / 1 mol Fe) = 2.70×1024 Fe atoms. Copyright © Cengage Learning. All rights reserved

It is Mass in grams of one mole of the substance: For any monoatomic element, molar mass in g has the same numerical value as its atomic weight in amu Example, For Na, atomic weight = 22.99 amu/atom; Its Molar Mass = 22.99 g/mol Therefore the equivalent relationships between masses in g, amount in moles and number of atoms are: 1 mol of atoms = its molar mass in g = 6.022 x 1023 atoms Example, 1mol of Na = 22.99 g of Na = 6.022 x 1023 atoms of Na Copyright © Cengage Learning. All rights reserved

For polyatomic elements, molar mass in g has the same numerical value as molecular weight in amu Molecular weight = sum of atomic weights of all atoms in molecule. Example, For N2, molecular weight = 28.03 amu/molecule; Its Molar Mass = 28.02 g/mol Therefore the equivalent relationships between masses in g, amount in moles and number of molecules are: 1 mol of molecules = its molar mass in g = 6.022 x 1023 molecules Example, 1mol of N2 = 28.02 g N2 = 6.022 x 1023 molecules of N2

For any molecular compound, molar mass in g has the same numerical value as its molecular weight in amu Example, For H2O, molecular weight = (2 × 1.008 amu) + 16.00 amu = 18.02 amu/molecule; Its Molar Mass = 18.02 g/mol Therefore the equivalent relationships between masses in g, amount in moles and number of molecules for a molecular compound are: 1 mol of molecules = its molar mass in g = 6.022 x 1023 molecules Example, 1mol of H2O = 18.02 g H2O = 6.022 x 1023 molecules of H2O Example, For Ba(NO3)2 , Formula weight = 261.35 amu/formula unit Molar Mass = 261.35 g/mol 137.33 g + (2 × 14.01 g) + (6 × 16.00 g)

For any ionic compound, molar mass in g has the same numerical value as its formula weight in amu Example, For NaNO3, formula weight = (22.99 + 14.01 + 3 x 16.00) amu = 85.00 amu/formula unit; Its Molar Mass = 85.00 g/mol Therefore the equivalent relationships between masses in g, amount in moles and number of formula units for an ionic compound are: 1 mol of formula units = its molar mass in g = 6.022 x 1023 formula units Example, 1mol of NaNO3 = 85.00 g NaNO3 = 6.022 x 1023 formula units of NaNO3 Example, For Ba(NO3)2 , Formula weight = 261.35 amu/formula unit Molar Mass = 261.35 g/mol 137.33 g + (2 × 14.01 g) + (6 × 16.00 g)

CONCEPT CHECK! Which of the following is closest to the average mass of one atom of copper? a) 63.55 g b) 52.00 g c) 58.93 g d) 65.38 g e) 1.055 x 10-22 g The correct answer is “e”. The mass of one atom of copper is going to be extremely small, so only letter “e” makes sense. The calculated solution is (1 Cu atom) × (1 mol Cu/6.022×1023 Cu atoms) × (63.55 g Cu/1 mol Cu). Copyright © Cengage Learning. All rights reserved

CONCEPT CHECK! Calculate the number of copper atoms in a 63.55 g sample of copper. 6.022×1023 Cu atoms 6.022×1023 Cu atoms; 63.55 g of Cu is 1 mole of Cu, which is equal to Avogadro’s number. Copyright © Cengage Learning. All rights reserved

CONCEPT CHECK! Which of the following 100.0 g samples contains the greatest number of atoms? Magnesium Zinc Silver The correct answer is “a”. Magnesium has the smallest average atomic mass (24.31 u). Since magnesium is lighter than zinc and silver, it will take more atoms to reach 100.0 g. Copyright © Cengage Learning. All rights reserved

Rank the following according to number of atoms (greatest to least): EXERCISE! Rank the following according to number of atoms (greatest to least): 107.9 g of silver 70.0 g of zinc 21.0 g of magnesium b) a) c) b, a, c; The greater the number of moles, the greater the number of atoms present. Zinc contains 1.07 mol, Ag contains 1.000 mol, and Mg contains 0.864 mol. Copyright © Cengage Learning. All rights reserved

Consider separate 100.0 gram samples of each of the following: EXERCISE! Consider separate 100.0 gram samples of each of the following:  H2O, N2O, C3H6O2, CO2 Rank them from greatest to least number of oxygen atoms. H2O, CO2, C3H6O2, N2O H2O, CO2, C3H6O2, N2O; The number of oxygen atoms in each is: H2O = 3.343×1024 CO2 = 2.737×1024 C3H6O2 = 1.626×1024 N2O = 1.368×1024 Copyright © Cengage Learning. All rights reserved

Mass percent of an element: For iron in iron(III) oxide, (Fe2O3): Copyright © Cengage Learning. All rights reserved

Consider separate 100.0 gram samples of each of the following: EXERCISE! Consider separate 100.0 gram samples of each of the following:  H2O, N2O, C3H6O2, CO2 Rank them from highest to lowest percent oxygen by mass. H2O, CO2, C3H6O2, N2O H2O, CO2, C3H6O2, N2O; The percent oxygen by mass in each is: H2O = 88.81% CO2 = 72.71% C3H6O2 = 43.20% N2O = 36.35% Copyright © Cengage Learning. All rights reserved

You have a 10.0 g of Ca3(PO4)2. (Molar mass = 310.18 g/mol) Exercise You have a 10.0 g of Ca3(PO4)2. (Molar mass = 310.18 g/mol) How many moles are present? 0.0322mol How many moles of Ca+2 ions are present? 0.0966 mol How many phosphate ions are present? =0.0644 mol = 3.88 x 1022 ions.

Formulas Empirical formula = CH Simplest whole-number ratio Molecular formula = (empirical formula)n [n = integer] Molecular formula = C6H6 = (CH)6 Actual formula of the compound Copyright © Cengage Learning. All rights reserved

Analyzing for Carbon and Hydrogen Device used to determine the mass percent of each element in a compound. Copyright © Cengage Learning. All rights reserved

Strategy to Determine Empirical Formula: Convert masses to moles for all elements Calculate the mole ratio by dividing by the smallest moles Change subscripts to whole numbers; (DO NOT ROUND OFF NUMBERS WITH .5; MULTIPLY BY 2 BEFORE ROUNDING OFF; DO NOT ROUND OFF NUMBERS WITH .33; MULTIPLY BY 3 BEFORE ROUNDING OFF) Copyright © Cengage Learning. All rights reserved

Strategy to Determine Molecular Formula Given Empirical Formula & Molecular Weight: Determine the ratio of the molecular weight to empirical formula weight (the molecular weight HAS to be provided): molecular weight/empirical formula weight Multiply all subscripts in the empirical formula by the above ratio to obtain the molecular formula. Copyright © Cengage Learning. All rights reserved

What is the empirical formula? C3H5O2 What is the molecular formula? EXERCISE! The composition of adipic acid is 49.3% C, 6.9% H, and 43.8% O (by mass). The molar mass of the compound is about 146 g/mol. What is the empirical formula? C3H5O2 What is the molecular formula? C6H10O4 Assume 100.0g. 49.3 g of carbon is 4.105 mol C (49.3/12.01). 6.9 g of hydrogen is 6.845 mol H (6.9/1.008). 43.8 g of oxygen is 2.7375 mol O (43.8/16.00). The ratio of C:H:O is 1.5:2.5:1 (C: 4.105/2.7375 = 1.5; H: 6.845/2.7375 = 2.5). Multiplying each by 2 to get a whole number becomes a ratio of 3:5:2. The empirical formula is therefore C3H5O2. The molar mass of the empirical formula is 73.07 g/mol, which goes into the molar mass of the molecular formula 2 times (146/73.07). The molecular formula is therefore C6H10O4. Copyright © Cengage Learning. All rights reserved

What is the empirical formula? C3H5O2 What is the molecular formula? EXERCISE! The composition of adipic acid is 49.3% C, 6.9% H, and 43.8% O (by mass). The molar mass of the compound is about 146 g/mol. What is the empirical formula? C3H5O2 What is the molecular formula? C6H10O4 Assume 100.0g. 49.3 g of carbon is 4.105 mol C (49.3/12.01). 6.9 g of hydrogen is 6.845 mol H (6.9/1.008). 43.8 g of oxygen is 2.7375 mol O (43.8/16.00). The ratio of C:H:O is 1.5:2.5:1 (C: 4.105/2.7375 = 1.5; H: 6.845/2.7375 = 2.5). Multiplying each by 2 to get a whole number becomes a ratio of 3:5:2. The empirical formula is therefore C3H5O2. The molar mass of the empirical formula is 73.07 g/mol, which goes into the molar mass of the molecular formula 2 times (146/73.07). The molecular formula is therefore C6H10O4. Copyright © Cengage Learning. All rights reserved

Exercise Determine the empirical formula of the compound whose percentage composition is 62.1% C, 5.21% H, 12.1% N, and 20.7% O Answer: C12H12N2O3

Answer: Empirical formula: C3H4O3; Molecular formula = C6H8O6 Exercise A sample of an organic compound was found to contain 9.01g C, 1.01g H, 12.0 g O. Calculate the empirical formula of this compound. If its molecular weight is 176 amu, determine its molecular formula Answer: Empirical formula: C3H4O3; Molecular formula = C6H8O6

Identify the reactants, the products of the reaction Write formulas of reactants left of arrow & those of products right of arrow Balance the equation by inspection, starting with the most complicated molecule(s). Balance equations ONLY by putting numbers IN FRONT OF formulas. Such numbers are called COEFFICIENTS NEVER CHANGE SUBSCRIPTS – doing so will change the identity of reactants, products and your equation will no longer represent the reaction that you are considering Start by balancing those elements that appear in only one reactant and one product. Finally balance those elements that appear in two or more reactants or products. Copyright © Cengage Learning. All rights reserved

Notice 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. Copyright © Cengage Learning. All rights reserved

Balance the following equations C2H6 + O2 CO2 + H2O Exercise Balance the following equations C2H6 + O2 CO2 + H2O C2H5OH + 3O2 2CO2 + 3H2O 4NH3 + 5 O2 4NO + 6 H2O

The equation is balanced. C2H5OH + 3O2 2CO2 + 3H2O The equation is balanced. All atoms present in the reactants are accounted for in the products. 1 mole of ethanol reacts with 3 moles of oxygen to produce 2 moles of carbon dioxide and 3 moles of water. Copyright © Cengage Learning. All rights reserved

The balanced equation represents an overall ratio of reactants and products, not what actually “happens” during a reaction. 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. Copyright © Cengage Learning. All rights reserved

I. CaO2 + 3C CaC2 + CO2 II. 2CaO + 5C 2CaC2 + CO2 EXERCISE! Which of the following correctly balances the chemical equation given below? There may be more than one correct balanced equation. If a balanced equation is incorrect, explain what is incorrect about it. CaO + C CaC2 + CO2 I. CaO2 + 3C CaC2 + CO2 II. 2CaO + 5C 2CaC2 + CO2 III. CaO + (2.5)C CaC2 + (0.5)CO2 IV. 4CaO + 10C 4CaC2 + 2CO2 II, III, and IV are correct. I is not correct because you cannot use subscripts to balance a chemical equation. Copyright © Cengage Learning. All rights reserved

CONCEPT CHECK! Which of the following are true concerning balanced chemical equations? There may be more than one true statement. I. The number of molecules is conserved. II. The coefficients tell you how much of each substance you have. III. Atoms are neither created nor destroyed. IV. The coefficients indicate the mass ratios of the substances used. V. The sum of the coefficients on the reactant side equals the sum of the coefficients on the product side. Only III is correct. Copyright © Cengage Learning. All rights reserved

Stoichiometric Calculations Chemical equations can be used to relate the masses of reacting chemicals. Copyright © Cengage Learning. All rights reserved

Calculating Masses of Reactants and Products in Reactions 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. Copyright © Cengage Learning. All rights reserved

Calculating Masses of Reactants and Products in Reactions Copyright © Cengage Learning. All rights reserved

EXERCISE! Consider the following reaction: If 6.25 g of phosphorus is burned, what mass of oxygen does it combine with? 8.07 g O2 (6.25 g P4)(1 mol P4 / 123.88 g P4)(5 mol O2 / 1 mol P4)(32.00 g O2 / 1 mol O2) = 8.07 g O2 Copyright © Cengage Learning. All rights reserved

Write balanced equations for each of these reactions. EXERCISE! (Part I) Methane (CH4) reacts with the oxygen in the air to produce carbon dioxide and water. Ammonia (NH3) reacts with the oxygen in the air to produce nitrogen monoxide and water. Write balanced equations for each of these reactions. CH4 + 2O2  CO2 + 2H2O 4NH3 + 5O2  4NO + 6H2O Copyright © Cengage Learning. All rights reserved

EXERCISE! (Part II) Methane (CH4) reacts with the oxygen in the air to produce carbon dioxide and water. Ammonia (NH3) reacts with the oxygen in the air to produce nitrogen monoxide and water. What mass of ammonia would produce the same amount of water as 1.00 g of methane reacting with excess oxygen? 1.42 g of ammonia is required. Use the balanced equations from the previous exercise to start this problem. Determine the amount of water produced from 1.00 g of methane (which is 0.125 moles of water). Use the moles of water and the balanced equation to determine the amount of ammonia needed. See the “Let’s Think About It” slide next to get the students going. Copyright © Cengage Learning. All rights reserved

Let’s Think About It Where are we going? To find the mass of ammonia that would produce the same amount of water as 1.00 g of methane reacting with excess oxygen. How do we get there? We need to know: How much water is produced from 1.00 g of methane and excess oxygen. How much ammonia is needed to produce the amount of water calculated above. Copyright © Cengage Learning. All rights reserved

Limiting Reactants Limiting reactant – the reactant that runs out first and thus limits the amounts of products that can be formed. Determine which reactant is limiting to calculate correctly the amounts of products that will be formed. Copyright © Cengage Learning. All rights reserved

A. The Concept of Limiting Reactants Stoichiometric mixture N2(g) + 3H2(g) 2NH3(g)

A. The Concept of Limiting Reactants Limiting reactant mixture N2(g) + 3H2(g)  2NH3(g)

A. The Concept of Limiting Reactants Limiting reactant mixture N2(g) + 3H2(g) 2NH3(g) Limiting reactant is the reactant that runs out first. H2

Limiting Reactants The amount of products that can form is limited by the methane. Methane is the limiting reactant. Water is in excess. Copyright © Cengage Learning. All rights reserved

CONCEPT CHECK! Which of the following reaction mixtures could produce the greatest amount of product? Each involves the reaction symbolized by the equation: 2H2 + O2 2H2O a) 2 moles of H2 and 2 moles of O2 2 moles of H2 and 3 moles of O2 2 moles of H2 and 1 mole of O2 d) 3 moles of H2 and 1 mole of O2 e) Each produce the same amount of product. The correct answer is “e”. All of these reaction mixtures would produce two moles of water. Copyright © Cengage Learning. All rights reserved

Notice We cannot simply add the total moles of all the reactants to decide which reactant mixture makes the most product. We must always think about how much product can be formed by using what we are given, and the ratio in the balanced equation. Copyright © Cengage Learning. All rights reserved

CONCEPT CHECK! You know that chemical A reacts with chemical B. You react 10.0 g of A with 10.0 g of B. What information do you need to know in order to determine the mass of product that will be produced? See the “Let’s Think About It” slide next to get the students going. Copyright © Cengage Learning. All rights reserved

Let’s Think About It Where are we going? To determine the mass of product that will be produced when you react 10.0 g of A with 10.0 g of B. How do we get there? We need to know: The mole ratio between A, B, and the product they form. In other words, we need to know the balanced reaction equation. The molar masses of A, B, and the product they form. Copyright © Cengage Learning. All rights reserved

EXERCISE! You react 10.0 g of A with 10.0 g of B. What mass of product will be produced given that the molar mass of A is 10.0 g/mol, B is 20.0 g/mol, and C is 25.0 g/mol? They react according to the equation: A + 3B 2C 8.33 g of C is produced. Note: Use the box animations to assist in explaining how to solve the problem. Copyright © Cengage Learning. All rights reserved

Percent Yield An important indicator of the efficiency of a particular laboratory or industrial reaction. Copyright © Cengage Learning. All rights reserved

What mass of excess reactant remains? EXERCISE! In a reaction, 124 g of Al are reacted with 650 g of Fe2O3 to form aluminum oxide and iron Calculate the thoeretical mass of Al2O3 that can form in grams. Answer = 235 g What mass of excess reactant remains? If the actual yield of Al2O3 is 220g, what is the percent yield 64.9% = (85.0 g PF3 / x)(100); x = 130.97 g PF3 (130.97 g PF3)(1 mol PF3 / 87.97 g PF3)(1 mol P4 / 4 mol PF3)(123.88 g P4 / 1 mol P4) = 46.1 g of P4 needed Copyright © Cengage Learning. All rights reserved