Stoichiometry: Ratios of Combination Chemistry Fifth Edition Julia Burdge Lecture PowerPoints Chapter 3 Stoichiometry: Ratios of Combination ©2020 McGraw-Hill Education. All rights reserved. Authorized only for instructor use in the classroom.  No reproduction or further distribution permitted without the prior written consent of McGraw-Hill Education.

3.1 Molecular and Formula Masses 2 Molecular and Formula Masses Although an ionic compound does not have a molecular mass, we can use its empirical formula to determine its formula mass.

SAMPLE PROBLEM 3.1 Calculate the molecular mass or the formula mass, as appropriate, for each of the following compounds: propane, (C3H8). lithium hydroxide, (LiOH). barium acetate, [Ba(C2H3O2)2].

SAMPLE PROBLEM 3.1 Solution Calculate the molecular mass or the formula mass, as appropriate, for each of the following compounds: propane, (C3H8). lithium hydroxide, (LiOH). barium acetate, [Ba(C2H3O2)2]. Solution 3(12.01 amu ) + 8(1.008 amu ) = 44.09 amu. 6.941 amu + 16.00 amu + 1.008 amu = 23.95 amu. 137.3 amu + 4(12.01 amu + 6(1.008 amu ) + 4(16.00 amu ) = 255.4 amu.

3.3 Chemical Equations 2 Interpreting and Writing Chemical Equations NH3 + HCl ⟶ NH4Cl “Ammonia and hydrogen chloride react to produce ammonium chloride.” CaCO3 ⟶ CaO + CO2 “Calcium carbonate reacts to produce calcium oxide and carbon dioxide.”

3.3 Chemical Equations 7 Balancing Chemical Equations Balancing a chemical equation requires something of a trial-and-error approach. In general, it will facilitate the balancing process if you do the following: Change the coefficients of compounds (for example, CO2) before changing the coefficients of elements (for example, O2).

3.3 Chemical Equations 8 Balancing Chemical Equations Treat polyatomic ions that appear on both sides of the equation (for example, CO32− as units, rather than counting their constituent atoms individually. Count atoms and/or polyatomic ions carefully, and track their numbers each time you change a coefficient.

3.4 The Mole and Molar Masses 2 The Mole A “mole” is a counting group , defined as the number of atoms in exactly 12 g of carbon-12. This number of atoms in 12 g of carbon-12 is known as Avogadro’s Number (NA): NA = 6.0221418 × 1023 objects If you order one dozen doughnut, you are asking for 12 doughnuts. If you order one mole of doughnuts, you are asking for 6.022 × 1023 doughnuts!

3.4 The Mole and Molar Masses 3 The Mole Conversion factors for moles of objects and number of objects:

3.4 The Mole and Molar Masses 4 Determining Molar Mass Chemists determine how many moles there are of a substance by measuring its mass (usually in grams). The molar mass of the substance is then used to convert from grams to moles. The molar mass (M) of a substance is the mass in grams of 1 mole of the substance.

3.4 The Mole and Molar Masses 5 Determining Molar Mass By definition, the mass of a mole of carbon-12 is exactly 12 g. Note that the molar mass of carbon is numerically equal to its atomic mass. Likewise, the atomic mass of calcium is 40.08 amu and its molar mass is 40.08 g, the atomic mass of sodium is 22.99 amu and its molar mass is 22.99 g, etc.

3.4 The Mole and Molar Masses 6 Determining Molar Mass 1 amu = 1.661×10−24 g ⟶ 1g = 6.022×1023 amu In effect, there is 1 mole of atomic mass units in 1 gram. So, the molar mass (in grams) of any compound is numerically equal to its molecular or formula mass (in amu).

3.4 The Mole and Molar Masses 7 Interconverting Mass, Moles, and Numbers of Particles Access the text alternative for these images

SAMPLE PROBLEM 3.6 Setup Determine (a) the number of moles of C in 10.00 g of naturally occurring carbon and (b) the mass of 0.905 mole of sodium chloride. Setup The molar mass of carbon is 12.01 g/mol. The molar mass of a compound is numerically equal to its formula mass. The molar mass of sodium chloride (NaCl) is 58.44 g/mol.

SAMPLE PROBLEM 3.6 Solution

SAMPLE PROBLEM 3.7 Determine the number of water molecules and the numbers of H and O atoms in 3.26 g of water. Determine the mass of 7.92 × 1019 carbon dioxide molecules.

SAMPLE PROBLEM 3.7 Solution

SAMPLE PROBLEM 3.7 Solution 2 Solution (b)

3.4 The Mole and Molar Masses 8 Empirical Formula from Percent Composition With the concepts of the mole and molar mass, we can now use the experimentally determined percent composition to determine the empirical formula of a compound.

SAMPLE PROBLEM 3.8 Strategy Determine the empirical formula of a compound that is 30.45 percent nitrogen and 69.55 percent oxygen by mass. Strategy Assume a 100-g sample so that the mass percentages of nitrogen and oxygen given in the problem statement correspond to the masses of N and O in the compound. Then, using the appropriate molar masses, convert the grams of each element to moles.

SAMPLE PROBLEM 3.8 Setup Setup The empirical formula of a compound consisting of N and O is NxOy. The molar masses of N and O are 14.01 and 16.00 g/mol, respectively. One hundred grams of a compound that is 30.45 percent nitrogen and 69.55 percent oxygen by mass contains 30.45 g N and 69.55 g O.

SAMPLE PROBLEM 3.8 Solution

3.6 Calculations with Balanced Chemical Equations Topics Moles of Reactants and Products Mass of Reactants and Products

3.6 Calculations with Balanced Chemical Equations 4 Moles of Reactants and Products Balanced chemical equations give us the relative amounts of reactants and products in terms of moles. However, because we measure reactants and products in the laboratory by weighing them, most often such calculations start with mass rather than the number of moles.

SAMPLE PROBLEM 3.11 Nitrous oxide (N2O) is commonly used as an anesthetic in dentistry. It is manufactured by heating ammonium nitrate. The balanced equation is Calculate the mass of ammonium nitrate that must be heated in order to produce 10.0 g of nitrous oxide. Determine the corresponding mass of water produced in the reaction.

SAMPLE PROBLEM 3.11 Setup Setup The molar masses are as follows: 80.05 g/mol for NH4NO3, 44.02 g/mol for N2O, and 18.02 g/mol for H2O.

SAMPLE PROBLEM 3.11 Solution 1 Solution (a) Thus, 18.2 g of ammonium nitrate must be heated in order to produce 10.0 g of nitrous oxide.

SAMPLE PROBLEM 3.11 Solution 2 Solution (b) Therefore, 8.18 g of water will also be produced in the reaction.

3.7 Limiting Reactants Topics Determining the Limiting Reactant Reaction Yield Types of Chemical Reactions

3.7 Limiting Reactants 1 Determining the Limiting Reactant The reactant used up first in a reaction is called the limiting reactant, because the amount of this reactant limits the amount of product that can form. When all the limiting reactant has been consumed, no more product can be formed. Excess reactants are those present in quantities greater than necessary to react with the quantity of the limiting reactant.

3.7 Limiting Reactants 2 Determining the Limiting Reactant CO(g) + 2H2(g)⟶CH3OH(l) Suppose that initially we have 5 moles of CO and 8 moles of H2:

3.7 Limiting Reactants 3 Determining the Limiting Reactant Because there are only 8 moles of H2 available, there is insufficient H2 to react with all the CO. Therefore, H2 is the limiting reactant and CO is the excess reactant.

3.7 Limiting Reactants 4 Determining the Limiting Reactant To determine how much CO will be left over when the reaction is complete, we must first calculate the amount of CO that will react with all 8 moles of H2: Thus, there will be 4 moles of CO consumed and 1 mole (5 mol − 4 mol) left over.

3.7 Limiting Reactants 5 Determining the Limiting Reactant When you use stoichiometry to calculate the amount of product formed in a reaction, you are calculating the theoretical yield of the reaction. The theoretical yield is the amount of product that forms when all the limiting reactant reacts to form the desired product. It is the maximum obtainable yield, predicted by the balanced equation.

3.7 Limiting Reactants 6 Reaction Yield In practice, the actual yield—the amount of product actually obtained from a reaction—is almost always less than the theoretical yield. The percent yield tells what percentage the actual yield is of the theoretical yield.