Chapter 3. Stoichiometry - PDF Free Download

Chapter 3 Stoichiometry

Chapter 3 Chemical Stoichiometry Stoichiometry The study of quantities of materials consumed and produced in chemical reactions. Since atoms are so small, we must use the average mass we will count by weighing! Copyright Cengage Learning. All rights reserved 2

Section 3.1 Counting by Weighing Ex: A pile of marbles weighs 394.80g. You randomly count out 10 marbles and determine their weight to be 37.60 g. A) What is the average mass of 1 marble? B) How many marbles are in the pile? Avg. Mass of 1 Marble = 37.60 g 10 marbles = 3.760 g / marble 394.80 g = 105.0 marbles 3.760 g Return to TOC Copyright Cengage Learning. All rights reserved 3

Section 3.3 The Counting Mole by Weighing By observing proportions in which elements combine, 19 th century chemists calculated relative atomic masses. The modern system is based on 12 C 12 C is assigned a mass of exactly 12 atomic mass units (amu), and the masses of all other atoms are given relative to this. The most accurate method for comparing masses of atoms involves the use of the mass spectrometer. Return to TOC Copyright Cengage Learning. All rights reserved 4

Section 3.2 Atomic Counting Masses by Weighing Since elements occur in nature as mixtures of isotopes, chemists use the average atomic mass (aka atomic mass or atomic weight) Atomic mass can be used to weigh out large numbers of atoms Return to TOC Copyright Cengage Learning. All rights reserved 6

Section 3.4 The Molar Mole Mass Avogadro s number The number equal to the number of carbon atoms in exactly 12 grams of pure 12 C. 1 mole = 6.022 x 10 23 units of anything! Named in honor of Avogadro (although he did not discover it!) Measuring moles Ex: 1 mole C = 6.022 x 10 23 C atoms = 12.01 g C Return to TOC Copyright Cengage Learning. All rights reserved 7

Section 3.4 Molar Mass Molar Mass is the mass in grams of one mole of a substance: Ex: Molar Mass of H = 1.008 g/mol Molar Mass of O = 16.00 g/mol (Use 4 significant figures when determining molar mass) Practice: Determine the molar mass of: 1. H 2 O 2. Ba(NO 3 ) 2 (2 1.008 g) + 16.00 g = 18.02 g/mol 137.33 g + (2 14.01 g) + (6 16.00 g) = 261.35 g/mol Return to TOC Copyright Cengage Learning. All rights reserved 8

Section 3.3 Concept The Counting Mole Check by Weighing 1. Calculate the number of iron atoms in a 4.48 mole sample of iron. (4.48 mol Fe) (6.022 10 23 Fe atoms / 1 mol Fe) = 2.70 10 24 Fe atoms 2. Calculate the number of aluminum atoms in a 10.0 g sample of aluminum. (10.0 g Al) (1 mol Al / 26.98 g Al) x (6.022 10 23 Al atoms / 1 mol Al) 2.23 10 23 Al atoms 3. Calculate the mass of 5.00 x 10 20 atoms of cobalt. (5.00 10 20 Co atoms) x (1 mol Co / 6.022 10 23 atoms Co) x (58.93 g Co / 1 mol Co) = 4.89 x 10-2 g Co Return to TOC Copyright Cengage Learning. All rights reserved 11

Section 3.3 Concept Check The Counting Mole by Weighing 4. Calculate the number of liters in 6.8g of carbon dioxide at STP (assuming ideal gas behavior). (6.8 g CO2) (1 mol CO2 / 44.01 g CO2 ) x (22.42 L CO2 / 1 mol CO2 ) = 3.5 L CO 2 5. Determine the density of carbon dioxide at STP (assuming ideal gas behavior). D = m/v *assume 1 mol = 44.01 / 22.42-1.963 g/mol CO 2 6. What is the mass of the carbonate ions present in a sample of calcium carbonate containing 4.86 mol? (4.86 mol CaCO3) (1 mol CO3 / 1 mol CaCO3) x (60.01 g CO3 / 1 mol CO3 ) = 292 g CO 3 2- Return to TOC Copyright Cengage Learning. All rights reserved 12

Section 3.6 Percent Composition of Compounds Calculating Percent Composition Calculating any percentage: "The part, divided by the whole, multiplied by 100" Percentage Composition Calculate the percent of each element in the total mass of the compound mass of element in compound mass % = 100% mass of compound Return to TOC Copyright Cengage Learning. All rights reserved 14

Section 3.6 Percent Composition of Compounds Example: Determine the percent of iron, by mass, in iron(iii) oxide, (Fe 2 O 3 ): mass of element in compound mass % = 100% mass of compound 2( 55.85 g) 111.70 g mass % Fe = = 100% = 69.94% 2( 55.85 g)+ 3( 16.00 g) 159.70 g Return to TOC Copyright Cengage Learning. All rights reserved 15

Ex: Calculate the percent composition by mass of magnesium carbonate. Molar mass of MgCO 3 (remember, 4 SF!): 24.31 g + 12.01 g + 3(16.00 g) = 84.32 g 100.00 Check!

Section 3.7 Determining the Formula of a Compound Formulas Empirical formula Formula written as the simplest whole-number ratio Molecular formula Actual formula of the compound Return to TOC Copyright Cengage Learning. All rights reserved 17

Formulas for ionic compounds are ALWAYS empirical (lowest whole number ratio). Ex: NaCl MgCl 2 CaO Al 2 (SO 4 ) 3 Formulas for molecular compounds MIGHT be empirical (lowest whole number ratio). Molecular: H 2 O C 6 H 12 O 6 C 12 H 22 O 11 Empirical: H 2 O CH 2 O C 12 H 22 O 11

Section 3.7 Determining the Formula of a Compound The Poem: Percent to mass, Mass to mole Divide by small, Multiply til whole! Example: 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 146 g/mol. a. What is the empirical formula? b. What is the molecular formula? Return to TOC Copyright Cengage Learning. All rights reserved 19

A) To determine Empirical Formula, Use the poem! Percent to mass, Mass to mole:

Divide by small... Carbon: Hydrogen: Oxygen:

Multiply til whole... (If the answer is already whole, skip this step!) Carbon: 1.50 Hydrogen: 2.50 Oxygen: 1.00 x 2 x 2 x 2 3 5 2 Empirical formula: C 3 H 5 O 2

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. b) Determine the Molecular Formula Use the empirical formula. Divide the molecular mass by the empirical formula mass. It s all about M/E Mass of Molecular Given (146 g/mol) Mass of Empirical: C 3 H 5 O 2 3(12.01 g) + 5(1.01) + 2(16.00) = 73.08 g (C 3 H 5 O 2 ) x 2 = C 6 H 10 O 4

Percent to mass In 100.0 g of the compound there are 81.10 g of B and 18.90 g of H. Mass to mole Divide by small 7.502 mol B = 1mol B 7.502 mol Multiply til whole 1 mol B 81.10 g B = 7.502 mol B 10.81 g B 1 mol H 18.90 g H = 18.75 mol H 1.008 g H 1 mol B 2 = 2 mol B 18.72 mol H = 2.499 mol H = 2.5 mol H 7.502 mol Answer : B H 2.5 mol H 2 = 5 mol H 2 5 Copyright Cengage Learning. All rights reserved 26

Section 3.8 Chemical Equations A representation of a balanced chemical equation: C 2 H 5 OH (l) + 3O 2(g) 2CO 2(g) + 3H 2 O (g) reactants products Reactants on the left, products on the right. Since atoms are neither created nor destroyed, all atoms present in the reactants must be present in the products, in the same number. Return to TOC Copyright Cengage Learning. All rights reserved 27

Section 3.8 Chemical Equations 1C 2 H 5 OH (l) + 3O 2(g) 2CO 2(g) + 3H 2 O (g) The coefficients (numbers in front) give the atomic/molecular/mole ratios Ex: 1 mole of ethanol reacts with 3 moles of oxygen to produce 2 moles of carbon dioxide and 3 moles of water. (s) = solid (g) = gas (l) = liquid (aq) = aqueous: dissolved in water Return to TOC Copyright Cengage Learning. All rights reserved 28

Section 3.9 Balancing Chemical Equations 1. Determine what reaction is occurring. What are the reactants, the products, and the physical states involved? 2. Write the unbalanced equation. 3. Balance the equation systematically, starting with the most complicated molecule(s). 4. Check - The same number of each type of atom needs to appear on both reactant and product sides. Balancing equations takes practice! Return to TOC Copyright Cengage Learning. All rights reserved 29

Section 3.9 Balancing Chemical Equations Important The number of atoms of each type of element must be the same on both sides of a balanced equation. Subscripts can NOT be changed to balance an equation. A balanced equation tells us the ratio of the number of molecules/formula units which react and are produced in a chemical reaction. Coefficients can be fractions, although they are usually given as lowest integer multiples. Return to TOC Copyright Cengage Learning. All rights reserved 31

Section 3.9 Balancing Chemical Equations 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 CaC 2 + CO 2 I. CaO 2 + 3C CaC 2 + CO 2 II. 2CaO + 5C 2CaC 2 + CO 2 III. CaO + (2.5)C CaC 2 + (0.5)CO 2 IV. 4CaO + 10C 4CaC 2 + 2CO 2 Return to TOC Copyright Cengage Learning. All rights reserved 32

Section 3.9 Balancing Chemical Equations 1. NH 3 + O 2 NO + H 2 O 2. SnO 2 + H 2 Sn + H 2 O 3. Fe + H 2 S0 4 Fe 2 (SO 4 ) 3 + H 2 4. KOH + H 3 PO 4 K 3 PO 4 + H 2 O 5. C 2 H 6 + O 2 H 2 O + CO 2 6. KNO 3 + H 2 CO 3 K 2 CO 3 + HNO 3 7. B 2 Br 6 + HNO 3 B(NO 3 ) 3 + HBr Return to TOC Copyright Cengage Learning. All rights reserved 33

Section 3.9 Balancing Chemical Equations 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. Return to TOC Copyright Cengage Learning. All rights reserved 34

Section 3.10 Stoichiometric Calculations: Amounts of Reactants and Products Stoichiometry is the calculation of chemical quantities from balanced equations. The four quantities involved are: particles (atoms, ions, unit formulas or molecules) moles mass volume - of gaseous reactants or products Remember: Atoms and mass are always conserved in chemical reactions. Return to TOC Copyright Cengage Learning. All rights reserved 35

Section 3.10 Stoichiometric Calculations: Amounts of Reactants and Products 1. Begin by balancing the equation! 2. Convert the unit given to moles of that substance. 3. Use the balanced equation to set up a mole ratios. Every stoich problem has a mole-mole ratio! 4. Convert moles of given to moles of the wanted. If needed, convert moles to unit wanted. Return to TOC Copyright Cengage Learning. All rights reserved 36

Volume 1 mol A = 22.42 L A Get x & y from the coefficients in the balanced equation Volume 1 mol B = 22.42 L B Mass 1 mol A = (molar mass) g A Mol A (given) 1 mol A x x mol B = mol B y mol A Mol B (wanted) Mass 1 mol B = (molar mass) g B Molecules 1 mol A = 6.022 x 10 23 particles A Molecules 1 mol B = 6.022 x 10 23 particles B

Stoichiometry Calculations Practice

1. How many moles of ammonia are produced from 0.35 moles of nitrogen? N 2 + 3H 2 2NH 3 0.35 mol N 2 X 2 1 mol NH 3 mol N 2 = 7.0 x 10-1 mol NH 3

2. How many moles of aluminum are needed to react completely with 49.0 moles of chlorine? 2Al + 3Cl 2 2AlCl 3 49.0 mol Cl 2 x 2 3 mol Al mol Cl 2 = 3.27 x 10 1 mol Al

3. How many moles of aluminum are needed to react completely with 49.0 g of chlorine? 2Al + 3Cl 2 2AlCl 3 49.0 g Cl 2 X 1 mol Cl 2 X 2 mol Al 70.90 g Cl 2 3 mol Cl 2 = 4.61 x 10-1 mol Al

4. How many grams of chlorine are needed to react completely with Al to produce 8.1 mol of aluminum chloride? 2Al + 3Cl 2 2AlCl 3 8.1 mol AlCl 3 X 3 2 mol Cl 2 mol AlCl 3 X 70.90 1 g Cl 2 mol Cl 2 = 8.6 x 10 2 g Cl 2

5. How many grams of aluminum chloride are produced by the reaction of 5.4 g of Al? 2Al + 3Cl 2 2AlCl 3 mol Al 5.4 g Al 1 X X 26.98 g Al 2 mol Al 2 mol AlCl 3 X 133.3 1 g AlCl 3 mol AlCl 3 = 2.7 x 10 1 g AlCl 3

6. How many liters of carbon monoxide are needed to react with 4.8 g of oxygen? 2CO + O 2 2CO 2 4.8 g O 2 1 mol O 2 2 mol CO 22.42 L CO 32.00 g O 2 1 mol O 2 1 mol CO = 6.7 L CO

7. What mass of ammonium is necessary to react with 2.1 x 10 24 molecules of oxygen? NH + 4 + 2O 2 2H 2 O + NO 2 2.1 x 10 24 molec O 2 1 mol O 2 1 mol NH 4 18.04 g NH 4 6.022 x 10 23 2 mol O 2 1 mol NH 4 molec O 2 = 3.1 x 10 1 g NH 4

8. How many liters of ammonium are necessary to produce 3.0 mol of water? NH 4 + + 2 O 2 2 H 2 O + NO 2 1 mol NH 22.42 3.0 mol H 4 2 O L NH 4 2 mol H 2 O 1 mol NH 4 = 3.4 x 10 1 L of NH 4

9. How many molecules of water are produced from 222 moles of oxygen? NH + 4 + 2 O 2 2 H 2 O + NO 2 222 mol O 2 2 mol H 2 O 6.022 x 10 23 molec H 2 O 2 mol O 2 1 mol H 2 O = 1.34 x 10 26 molec H 2 O

10. How many moles of nitrogen dioxide are produced from 46 moles of oxygen? NH 4 + + 2 O 2 2 H 2 O + NO 2 46 mol O 2 1 mol NO 2 2 mol O 2 = 2.3 x 10 1 mol NO 2

Limiting and Excess reagents The limiting reagent is consumed first and therefore limits the amount of product that can be formed. The excess reagent is the reactant that is left over or remaining The limiting and excess reagents are always reactants (located on the left side of the equation!)

Section 3.11 The Concept of Limiting Reagent CH 4 and H 2 O form H 2 and CO notice that the methane is limiting, the water was in excess. Return to TOC Copyright Cengage Learning. All rights reserved 53

How to identify a limiting problem Two amounts of reactants are given You need to solve for the amount of product. Find which reactant you will run out of first (the limiting reagent) and use this reactant to solve the problem.

Sample Problem 1 If 10.0 g of aluminum is combined with 10.0g of oxygen gas, how many grams of aluminum oxide can be made? Step One: Predict & balance the equation. 4Al+ 3O 2 2Al 2 O 3 +

Step 2: Use mol/coefficient ratio Convert reactants to moles. Aluminum : 10.0g x 1 mol = 0.371 mol Al 26.98 g Oxygen: 10.0g x 1 mol = 0.313 mol O 2 32.00 g Divide each moles by its coefficient. Aluminum: 0.371 / 4 = 0.0928 Oxygen: 0.313 / 3 = 0.104

Step 3: Identify the Limiting Reagent. The limiting reagent is the smallest number. Aluminum is the limiting reagent! Step 4: Use the limiting reagent to solve the problem. 10.0gAl x 1 mol Al x 2 mol Al 2 0 3 26.98 g Al 4 mol Al x 101.98 g Al 2 0 3 1 mol Al 2 0 3 Answer: 1.89 x 10 1 g Aluminum Oxide

If the problem asks how much excess reagent is left: Use the limiting to see how much of the excess you need (do a stoich problem from limiting reagent to excess reagent) Subtract the amount you need from the original amount given in the problem. ** Be careful of units!!

Sample problem 2: If 10.0 g of aluminum is combined with 10.0g of oxygen gas, how many grams of the excess reagent remains unreacted? 4 Al + 3 O 2 2 Al 2 O 3 (Excess Reagent given) (What you need) = Excess USE THE LIMITING REAGENT to figure out how much you need. 10.0 g Al 1 mol Al 3 mol O 2 32.00 g O 2 26.98 g Al 4 mol Al 1 mol O 2 = 8.90 g O 2 10.0g O 2-8.90 g O 2 = 1.1 g O 2 remains unreacted

6.4 mol 3.4 mol 2 H 2 + O 2 2 H 2 O Problem 3: How many grams of water are produced given 6.4 mol of hydrogen and 3.4 mol of oxygen? Step #1: Find the limiting & excess reagent. 6.4 mol H 2 / 2 = 3.2 3.4 mol O 2 / 1 = 3.4 H 2 is the limiting reagent O 2 is the excess reagent

6.4 mol 3.4 mol 2H 2 + O 2 2H 2 O Step #2 Use the limiting reagent to solve the problem! 6.4 mol H 2 X 2 mol H 2 O 2 mol H 2 X 18.02 g H 2 O 1 mol H 2 O =1.2 x 10 2 g H 2 O

6.4 mol 3.4 mol 2H 2 + O 2 2H 2 O How much of the excess reagent remains unreacted? (Excess Reagent given) (What you need) = Excess USE THE LIMITING REAGENT to figure out how much you need. 6.4 mol H 2 x 1 mol O 2 = 3.2 mol O 2 2 mol H2 3.4 mol O 2-3.2 mol O 2 = 0.2 mol O 2 remains unreacted

Section 3.11 The Concept of Limiting Reagent Concept Check 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. Return to TOC Copyright Cengage Learning. All rights reserved 64

Section 3.11 The Concept of Limiting Reagent Percent Yield An important indicator of the efficiency of a particular laboratory or industrial reaction. Return to TOC Copyright Cengage Learning. All rights reserved 65

Section 3.11 The Concept of Limiting Reagent Calculating Percent Yield 1. Actual yield - what you got by actually performing the reaction 2. Theoretical yield - what stoichiometric calculation says the reaction SHOULD have produced Αctual Yield Percent Yield = Τheoretical Yield 100 Return to TOC Copyright Cengage Learning. All rights reserved 66

Example 1 Find the percent yield when 68.0 g silver reacts with oxygen to form 50.0 g silver oxide. Step 1: Determine and balance the equation: 68.0 g 50.0 g 4 Ag + O 2 2 Ag 2 O

Step 2: Calculate the theoretical yield from the reactant given. 68.0 g 50.0 g 4 Ag + O 2 2 Ag 2 O 68.0 g Ag 1 mol Ag 2 mol Ag 2 O 231.8 g Ag 2 O 107.9 g Ag 4 mol Ag 1 mol Ag 2 O Theoretical yield = 73.0 g Ag 2 O

Step 3: Plug numbers into the equation and... solve! Actual Yield (product given) Theoretical Yield (product calculated) 50.0 g (product given) 73.0 g (product calculated) X 100 = % Yield X 100 = % Yield % Yield = 68.5%

Example 2 Find the actual product made when the incomplete reaction between 16.5 g hydrogen gas and oxygen gas produces a 70.0% yield of water. Step 1: Determine and balance the equation: 16.5 g 70.0% 2 H 2 + O 2 2 H 2 O

Step 2: Calculate the theoretical yield from the reactant given. 16.5 g 2 H 2 + O 2 2 H 2 O 16.5 g H 2 1 mol H 2 g H 2 2 mol H 2 O 18.02 g H 2 O 2.016 2 mol H 2 1 mol H 2 O = 1.47 x 10 2 g H 2 O Theoretical yield

Step 3: Plug numbers into the equation and... solve! Actual Yield (product given) Theoretical Yield (product calculated) Actual Yield (product given) 1.47 x 10 2 g (product calculated) X 100 = % Yield X 100 = 70.0 Actual Yield = 1.03 x 10 2 g H 2 O