# Chapter Three: STOICHIOMETRY.

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Chapter Three: STOICHIOMETRY

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

Counting by Weighing Need average mass of the object.
Objects behave as though they were all identical. 3.1 Copyright © Houghton Mifflin Company. All rights reserved.

Atomic Masses Elements occur in nature as mixtures of isotopes.
Carbon = % 12C 1.11% 13C <0.01% 14C Carbon atomic mass = amu 3.2 Copyright © Houghton Mifflin Company. All rights reserved.

Schematic Diagram of a Mass Spectrometer

The Mole The number equal to the number of carbon atoms in exactly 12 grams of pure 12C 1 mole of anything = x 1023 units of that thing (Avogadro’s number) 1 mole C = x 1023 C atoms = g C 3.3 Copyright © Houghton Mifflin Company. All rights reserved.

Molar Mass Mass in grams of one mole of the substance:
Molar Mass of N = g/mol Molar Mass of H2O = g/mol (2 × 1.008) Molar Mass of Ba(NO3)2 = g/mol (2 × 14.01) + (6 × 16.00) 3.4 Copyright © Houghton Mifflin Company. All rights reserved.

Concept Check 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 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 © Houghton Mifflin Company. All rights reserved.

Concept Check Calculate the number of copper atoms in a g sample of copper. 6.022×1023 Cu atoms; g of Cu is 1 mole of Cu, which is equal to Avogadro’s number. Copyright © Houghton Mifflin Company. All rights reserved.

Concept Check Which of the following 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 amu). Since magnesium is lighter than zinc and silver, it will take more atoms to reach g. Copyright © Houghton Mifflin Company. All rights reserved.

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; The greater the number of moles, the greater the number of atoms present. Zinc contains 1.07 mol, Ag contains mol, and Mg contains mol. Copyright © Houghton Mifflin Company. All rights reserved.

Exercise Consider separate 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; 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 © Houghton Mifflin Company. All rights reserved.

Percent Composition Mass percent of an element:

Exercise Consider separate 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; The percent oxygen by mass in each is: H2O = 88.81% CO2 = 72.71% C3H6O2 = 43.20% N2O = 36.35% Copyright © Houghton Mifflin Company. All rights reserved.

Formulas Molecular formula = (empirical formula)n [n = integer]
Molecular formula = C6H6 = (CH)6 Actual formula of the compound Empirical formula = CH Simplest whole-number ratio 3.6 Copyright © Houghton Mifflin Company. All rights reserved.

Analyzing for Carbon and Hydrogen
Device used to determine the mass percent of each element in a compound. 3.6 Copyright © Houghton Mifflin Company. All rights reserved.

Balancing Chemical Equations

Chemical Equation A representation of a chemical reaction:
C2H5OH + 3O2  2CO H2O reactants products Reactants are only placed on the left side of the arrow, products are only placed on the right side of the arrow. 3.7 and 3.8 Copyright © Houghton Mifflin Company. All rights reserved.

Chemical Equation 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. 3.7 and 3.8 Copyright © Houghton Mifflin Company. All rights reserved.

Chemical Equations The coefficients in the balanced equation have nothing to do with the amount of each reactant that is given in the problem. 3.7 and 3.8 Copyright © Houghton Mifflin Company. All rights reserved.

Chemical Equations The balanced equation represents a 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. 3.7 and 3.8 Copyright © Houghton Mifflin Company. All rights reserved.

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 © Houghton Mifflin Company. 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 © Houghton Mifflin Company. 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. 3.8 Copyright © Houghton Mifflin Company. All rights reserved.

Stoichiometric Calculations
Chemical equations can be used to relate the masses of reacting chemicals. 3.9 Copyright © Houghton Mifflin Company. All rights reserved.

Calculating Masses in Reactions
Balance the equation. Convert mass to moles. Set up mole ratios from the balanced equation. Calculate number of moles of desired reactant or product. Convert back to grams. 3.9 Copyright © Houghton Mifflin Company. All rights reserved.

Exercise 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. Set up balanced equations for each reaction. Determine the amount of water produced from 1.00 g of methane (which is 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 © Houghton Mifflin Company. All rights reserved.

Let’s Think About It 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 © Houghton Mifflin Company. All rights reserved.

Limiting Reactants Limiting reactant – the reactant that is consumed first and therefore 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. 3.10 Copyright © Houghton Mifflin Company. All rights reserved.

Limiting Reactants Methane and water will react to form products according to the equation: CH4 + H2O  3H2 + CO 3.10 Copyright © Houghton Mifflin Company. All rights reserved.

Mixture of CH4 and H2O Molecules Reacting

CH4 and H2O Reacting to Form H2 and CO

Limiting Reactants The amount of products that can form is limited by the methane. Methane is the limiting reactant. Water is in excess. 3.10 Copyright © Houghton Mifflin Company. 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 © Houghton Mifflin Company. 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. 3.10 Copyright © Houghton Mifflin Company. 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 © Houghton Mifflin Company. All rights reserved.

Let’s Think About It 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 © Houghton Mifflin Company. 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. Copyright © Houghton Mifflin Company. All rights reserved.

Percent Yield An important indicator of the efficiency of a particular laboratory or industrial reaction. 3.10 Copyright © Houghton Mifflin Company. All rights reserved.