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Lecture Presentation Chapter 3 Stoichiometry: Calculations with Chemical Formulas and Equations LO 1.17, 1.18, 3.1, 3.5, 1.4, 3.3, 1.2, 1.3 Ashley Warren Kings High School

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Stoichiometry The area of study that examines the quantities of substances consumed and produced in chemical reactions. Stoicheion means “element” and metron means “measure.”

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**Law of Conservation of Mass**

“We may lay it down as an incontestable axiom that, in all the operations of art and nature, nothing is created; an equal amount of matter exists both before and after the experiment. Upon this principle, the whole art of performing chemical experiments depends.” --Antoine Lavoisier, 1789 Chemists came to understand the basis for this law as atoms are neither created nor destroyed during a chemical reaction. The changes that occur during any reaction merely rearrange the atoms. The same collection of atoms is present both before and after the reaction. Lavoisier observed that mass is conserved in a chemical reaction. © 2012 Pearson Education, Inc.

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**Anatomy of a Chemical Equation**

CH4(g) + 2O2(g) CO2(g) + 2H2O(g) Chemical equations give a description of a chemical reaction. We read the + sign as “reacts with” or “yields” or “produces” Remember there are two numbers in chemical equations: Reactants appear on the left side of the equation. Products appear on the right side. The states of the reactants and products are written in parentheses to the right of each compound. Coefficients are inserted to balance the equation. © 2012 Pearson Education, Inc.

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**Subscripts and Coefficients Give Different Information**

Stoichiometric coefficients give the ratio in which the reactants and products exist. Remember we can NEVER change the subscripts when we are balancing a chemical equation. Subscripts tell the number of atoms of each element in a molecule. Coefficients tell the number of molecules. © 2012 Pearson Education, Inc.

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**Interpreting and Balancing Chemical Equations**

The following diagram represents a chemical reaction in which the red spheres are oxygen atoms and the blue spheres are nitrogen atoms. Write the chemical formulas for the reactants and products. Write a balanced chemical equation for the reaction. Is the diagram consistent with the Law of conservation of mass? © 2012 Pearson Education, Inc.

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**Interpreting and Balancing Chemical Equations**

In the diagram above, the white spheres represent hydrogen atoms and the blue spheres represent nitrogen atoms. To be consistent with the law of conservation of mass, how many NH3 molecules should be shown in the right (product) box? 6 We have had so much practice last year with balancing chemical reactions that I will not focus on balancing reactions in this powerpoint. If you need additional help then just let me know and I can give you some extra problems/help. © 2012 Pearson Education, Inc.

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Reaction Types In this section we will see three types of reactions: combination, decomposition, and combustion. We will be predicting products here and finding patterns of reactivity. © 2012 Pearson Education, Inc.

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**Combination Reactions**

In combination reactions two or more substances react to form one product. Combination reactions have more reactants than products. Remember all of the types of synthesis reactions we learned last year. Those are fair game. Examples: 2Mg(s) + O2(g) 2MgO(s) N2(g) + 3H2(g) 2NH3(g) C3H6(g) + Br2(l) C3H6Br2(l) © 2012 Pearson Education, Inc.

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**Decomposition Reactions**

In a decomposition reaction one substance breaks down into two or more substances. Again refer to the decomposition reaction types that we learned last year. More products than reactants. Many decomp rxns happen when heated. Combination and Decomposition reactions are opposites. Examples: CaCO3(s) CaO(s) + CO2(g) 2KClO3(s) 2KCl(s) + O2(g) 2NaN3(s) 2Na(s) + 3N2(g) © 2012 Pearson Education, Inc.

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**© 2012 Pearson Education, Inc.**

Combustion Reactions Combustion reactions are generally rapid reactions that produce a flame. Combustion reactions most often involve hydrocarbons reacting with oxygen in the air. Examples: CH4(g) + 2O2(g) CO2(g) + 2H2O(g) C3H8(g) + 5O2(g) 3CO2(g) + 4H2O(g) © 2012 Pearson Education, Inc.

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More Reactions Addition rxns: combination rxn where a substance is added to a compound with a double bond. © 2012 Pearson Education, Inc.

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More Reactions Substitution reactions An atom or a functional group in molecule in substituted for another atom or functional group. © 2012 Pearson Education, Inc.

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Final Reactions

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Formula Weights Chemical formulas and equations have a quantitative significance. While we cannot directly count atoms or molecules we can indirectly determine their numbers if we know their masses! © 2012 Pearson Education, Inc.

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Formula Weight (FW) A formula weight is the sum of the atomic weights for the atoms in a chemical formula. So, the formula weight of calcium chloride, CaCl2, would be Ca: 1(40.08 amu) + Cl: 2( amu) amu Formula weights are generally reported for ionic compounds. What is the formula weight for sulfuric acid? Remember that atomic weights are simply what we find as the mass on the periodic table. For example, the atomic weight of oxygen is amu. © 2012 Pearson Education, Inc.

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**© 2012 Pearson Education, Inc.**

Molecular Weight (MW) A molecular weight is the sum of the atomic weights of the atoms in a molecule. For the molecule ethane, C2H6, the molecular weight would be C: 2( amu) What is the molecule weight for glucose? C6H12O6 + H: 6( amu) amu © 2012 Pearson Education, Inc.

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Percent Composition One can find the percentage of the mass of a compound that comes from each of the elements in the compound by using this equation: % Element = (number of atoms)(atomic weight) (FW of the compound) x 100 © 2012 Pearson Education, Inc.

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**© 2012 Pearson Education, Inc.**

Percent Composition So the percentage of carbon in ethane (C2H6) is %C = (2)( amu) ( amu) amu amu = x 100 = % © 2012 Pearson Education, Inc.

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Moles Even the smallest samples we deal with in the lab contain enormous numbers of atoms, ions, or molecules. We have devised a unit for describing such large numbers of atoms or molecules. © 2012 Pearson Education, Inc.

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Avogadro’s Number 1 mole = 6.02 x 1023, symbolized NA. 1 mole of 12C has a mass of g. Abbreviation = mol © 2012 Pearson Education, Inc.

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Avogadro’s Number See exercise 3.7 and 3.8 © 2012 Pearson Education, Inc.

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Molar Mass By definition, a molar mass is the mass of 1 mol of a substance (i.e., g/mol). The molar mass of an element is the mass number for the element that we find on the periodic table. The formula weight (in amu’s) will be the same number as the molar mass (in g/mol). Cl has an atomic weight of 35.5 amu 1 mol Cl has a mass of 35.5 grams *”The atomic weight of an element in atomic mass units is numerically equal to the mass in grams of 1 mol of that element.” © 2012 Pearson Education, Inc.

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**© 2012 Pearson Education, Inc.**

Mole Relationships One mole of atoms, ions, or molecules contains Avogadro’s number of those particles. One mole of molecules or formula units contains Avogadro’s number times the number of atoms or ions of each element in the compound. © 2012 Pearson Education, Inc.

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Molar Mass Which has more mass, a mole of water or a mole of glucose? Which contains more molecules, a mole of water or a mole of glucose? Calculate the molar mass of glucose. Calculate the molar mass of calcium nitrate. © 2012 Pearson Education, Inc.

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Using Moles Moles provide a bridge from the molecular scale to the real-world scale. Let’s think about units in all of these cases! If you know the units you will ALWAYS be able to figure out what to do! © 2012 Pearson Education, Inc.

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Masses and Moles Calculate the number of moles of water in grams of water. Calculate the mass, in grams, of mol of potassium chlorite. How many molecules are in grams of C3H8? © 2012 Pearson Education, Inc.

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**Finding Empirical Formulas**

Remember from chapter 2 that an empirical formula Is the smallest whole number ratio of a compound. It tells us the relative number of atoms of each element in the substance. This also applies on the molar level. 1 mol of water contains 2 mols of Hydrogen and 1 mol of oxygen. Conversely, the ratio of the numbers of moles of all elements in a compound gives the subscripts in the compound’s empirical formula. © 2012 Pearson Education, Inc.

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**Calculating Empirical Formulas**

One can calculate the empirical formula from the percent composition. © 2012 Pearson Education, Inc.

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**Calculating Empirical Formulas**

The compound para-aminobenzoic acid (you may have seen it listed as PABA on your bottle of sunscreen) is composed of carbon (61.31%), hydrogen (5.14%), nitrogen (10.21%), and oxygen (23.33%). Find the empirical formula of PABA. © 2012 Pearson Education, Inc.

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**Calculating Empirical Formulas**

Assuming g of para-aminobenzoic acid, C: g x = mol C H: g x = 5.09 mol H N: g x = mol N O: g x = mol O 1 mol 12.01 g 14.01 g 1.01 g 16.00 g © 2012 Pearson Education, Inc.

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**Calculating Empirical Formulas**

Calculate the mole ratio by dividing by the smallest number of moles: C: = 7 H: = 7 N: = 1.000 O: = 2 5.105 mol mol 5.09 mol 1.458 mol © 2012 Pearson Education, Inc.

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**Calculating Empirical Formulas**

These are the subscripts for the empirical formula: C7H7NO2 © 2012 Pearson Education, Inc.

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**Molecular Formulas from Empirical**

Mesitylene, a hydrocarbon found in crude oil, has an empirical formula of C3H4, and an experimentally determined molecular weight of 121 amu. What is its molecular formula? The formula weight of C3H4 = 40.0 amu Whole number multiple = molecular weight empirical formula weight = 121 = 3.02 40 This means the molecular formula is triple the empirical. Therefore the molecular formula is C9H12. © 2012 Pearson Education, Inc.

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**© 2012 Pearson Education, Inc.**

Combustion Analysis Compounds containing C, H, and O are routinely analyzed through combustion in a chamber like the one shown in Figure 3.14. C is determined from the mass of CO2 produced. H is determined from the mass of H2O produced. O is determined by difference after the C and H have been determined. ***See Exercise 3.15 Empirical formulas are routinely determined by combustion analysis. When a compound containing carbon and hydrogen is completely combusted the carbon is converted to CO2 and the hydrogen is converted to H2O. The amounts of CO2 and H2O produced are determined by measuring the mass increase in the CO2 and H2O absorbers. From the masses of CO2 and H2O we can calculate the number of moles and therefore the empirical formula. If a third element (like oxygen) is present then we can determine the mass by subtracting the measured masses of carbon and hydrogen from the original sample mass. © 2012 Pearson Education, Inc.

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**Stoichiometric Calculations**

It is important to realize that the stoichiometric ratios are the ideal proportions in which reactants are needed to form products. The coefficients in the balanced equation give the ratio of moles of reactants and products as well as the relative numbers of molecules. © 2012 Pearson Education, Inc.

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**Stoichiometric Calculations**

Starting with the mass of Substance A, you can use the ratio of the coefficients of A and B to calculate the mass of Substance B formed (if it’s a product) or used (if it’s a reactant). Remember grams to moles to moles to grams! © 2012 Pearson Education, Inc.

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**Stoichiometric Calculations**

C6H12O6 + 6 O2 6 CO2 + 6 H2O Starting with 1.00 g of C6H12O6… we calculate the moles of C6H12O6…(using molar mass) use the coefficients to find the moles of H2O…(using mole to mole ratio) and then turn the moles of water to grams…(using molar mass again) © 2012 Pearson Education, Inc.

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**Stoichiometric Calculations**

If given the following reaction, how many grams of O2 can be prepared from 4.50 grams of KClO3? 2KClO3 (s) 2KCl (s) + 3 O2 (g) © 2012 Pearson Education, Inc.

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Limiting Reactants Often one or more reactants is present in excess. Therefore, at the end of the reaction those reactants present in excess will still be in the reaction mixture. © 2012 Pearson Education, Inc.

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**© 2012 Pearson Education, Inc.**

Limiting Reactants 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 (in this case, the H2). In the example below, the O2 would be the excess reagent. The limiting reactant is completely consumed. The excess reagent is the reactant that’s present in excess. © 2012 Pearson Education, Inc.

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**© 2012 Pearson Education, Inc.**

Theoretical Yield The theoretical yield is the maximum amount of product that can be made. In other words, it’s the amount of product possible as calculated through the stoichiometry problem. This is different from the actual yield, which is the amount one actually produces and measures. © 2012 Pearson Education, Inc.

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**© 2012 Pearson Education, Inc.**

Percent Yield One finds the percent yield by comparing the amount actually obtained (actual yield) to the amount it was possible to make (theoretical yield): Percent yield = x 100 actual yield theoretical yield © 2012 Pearson Education, Inc.

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**Limiting Reactant/Theoretical Yield Calculations**

See Exercise 3.18, 3.19, and 3.20 © 2012 Pearson Education, Inc.

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**Chapter 3 Homework Problems**

1, 3, 5, 6, 8, 10, 12, 13, 15, 17, 19, 22, 23, 25, 27, 30, 31, 33, 35, 37, 40, 41, 43, 46, 47, 49, 52, 53, 55, 61, 63, 65, 67, 69, 71, 76, 77, 80, 81, 83, 87, 89a,b, 97, 101 © 2012 Pearson Education, Inc.

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