Formulas, Equations and Moles

Save this PDF as:

Size: px
Start display at page:

Transcription

1 Chapter 3 Formulas, Equations and Moles Interpreting Chemical Equations You can interpret a balanced chemical equation in many ways. On a microscopic level, two molecules of H 2 react with one molecule of O 2 to produce two molecules of H 2 O On a macroscopic level, these formulas and equations represent large-scale behaviors of atoms and molecules that result in observable properties! A chemical equation can mean whatever you want it to mean depending on the context. Chapter 3 2 1

2 Moving from Molecules to Mass A balanced chemical equation is the scientist s guide to the preparation or formation of product compounds. What does this equation tell us about the preparation of Ethyl Chloride? Chapter 3 3 Atomic Mass The atomic mass of any element on the periodic table is expressed in atomic mass units (amu) This value is listed on the periodic table and represents the mass of a single atom of an element. This relationship can be used to write conversion factors For example, the atomic mass of iron is amu, so: amu 1 Fe atom OR 1 Fe atom amu Chapter 3 4 2

3 Formula Mass The formula mass of a compound is the sum of the atomic masses of all the atoms in its formula. To determine the formula mass, you multiply each element s atomic mass by its formula subscript and then add them all up. Chapter 3 5 Avogadro s Number How do we keep track of atoms or molecules? They are very small, so we group them in a large bunch We use Avogadro s Number (N A ) to represent this bunch of atoms or molecules. Avogadro s Number was experimentally determined to be the number of atoms in grams of carbon. Its numerical value is Therefore, a g sample of carbon contains carbon atoms. Chapter 3 6 3

4 The Mole The mole (mol and abbreviated n) is a unit of measure that allows us to make comparisons between substances that have different masses A mole is Avogadro s number of atoms, that is atoms. 1 mol = atoms Notice how this mole relationship resembles one of our unit equations. Therefore, we can use to write conversion factors to convert between the number of atoms and the mass of a substance. Chapter 3 7 The atomic mass of any substance expressed in grams per mole (g/mol) is the molar mass (MM) of that substance. Molar Mass Molar Mass = Mass mole = m n If the atomic mass of iron is amu, then the molar mass of iron is g/mol. Chapter 3 8 4

5 Calculating Molar Mass The molar mass of a substance is the sum of the molar masses of each element present in the substance. What is the molar mass of potassium phosphate? Chapter 3 9 Mole Calculations Grams to Moles What is the mass of 2.87 mol of sulfuric acid? Step 1: Determine what you have: 2.87 mol sulfuric acid Step 2: Determine what you want:??? g sulfuric acid Step 3: Write a plan to convert from what you have to what you want. Step 4: Select conversion factor(s) that allows you to perform your plan: 1 mol SA = g SA Grams Molar Mass Moles 2.87 mole SA g SA 1 mole SA = 282 g SA Chapter

6 Mole Calculations Grams to Atoms Now we will use the molar mass of a compound to convert between grams of a substance and particles of a substance particles = 1 mol = molar mass If we want to convert particles to mass, we must first convert particles to moles and then we can convert moles to mass. Grams Molar Mass Moles Avogadro s Number Atoms Chapter 3 11 Stoichiometry: Mole - Mole Relationships We can use a balanced chemical equation to write mole ratios which can be used as unit factors: N 2 (g) + O 2 (g) 2 NO(g) Since 1 mol of N 2 reacts with 1 mol of O 2 to produce 2 mol of NO, we can write the following mole relationships: 1 mol N 2 1 mol O 2 1 mol N 2 2 mol NO 1 mol O 2 2 mol NO 1 mol O 2 2 mol NO 2 mol NO 1 mol N 2 1 mol N 2 1 mol O 2 Chapter

7 Stoichiometry Problems In a mass-mass stoichiometry problem, we will convert a given mass of a reactant or product to an unknown mass of reactant or product. There are four steps: Step 1: Check that the equation is balanced!! Step 2: Convert the given mass to moles using the molar mass as a conversion factor. Step 3: Convert the moles of given to moles of the unknown using the coefficients in the balanced equation as a conversion factor. Step 4: Convert the moles of unknown to grams using the molar mass as a conversion factor. A + B C + D Molar Mole Molar Mass of A Ratio Mass of B Grams of A Moles of A Moles of B Grams of B Chapter 3 13 Stoichiometry Problem Aqueous solutions of NaOCl (household bleach) are prepared by the reaction of NaOH with Cl 2 : NaOH (aq) + Cl 2 (g) NaOCl (aq) + NaCl (aq) + H 2 O (l) How many grams of NaOH are needed to react with 25.0 g of Cl 2? Step 1: Balance the equation!!! Step 2: Convert grams Cl 2 to moles Cl 2 using the molar mass of Cl 2. Step 3: Convert moles Cl 2 to moles NaOH using the balanced equation. Step 4: Convert moles NaOH to grams NaOH using the molar mass. Molar Mole Molar Mass of Cl 2 Ratio Mass of NaOH Grams of Cl 2 Moles of Cl 2 Mol of NaOH Grams of NaOH 25.0 g Cl 2 1 mol Cl g Cl 2 2 mol NaOH 1 mol Cl g NaOH g NaOH 1 mol NaOH Chapter

8 Percent Yield: Efficiency of Reaction When you perform a laboratory experiment, the amount of product collected is the actual yield. The amount of product calculated from the reaction is the theoretical yield. The percent yield is a measure of the efficiency of the lab reaction by comparing the amount of the actual yield to the theoretical yield. actual yield theoretical yield 100 % = percent yield Chapter 3 15 Calculations with Percent Yield Dichloromethane (CH 2 Cl 2 ) is prepared by reaction of methane (CH 4 ) with chlorine (Cl 2 ) giving hydrogen chloride as a by-product. How many grams of dichloromethane result from the reaction of 1.85 kg of methane if the yield is 43.1%? Chapter

9 Limiting Reactant Concept Say you re making grilled cheese sandwiches. You need 1 slice of cheese and 2 slices of bread to make one sandwich. 1 Cheese + 2 Bread 1 Sandwich If you have 5 slices of cheese and 8 slices of bread, how many sandwiches can you make? With 5 slices of cheese you could make 5 sandwiches (a 1:1 ratio) With 8 slices of bread you could make 4 sandwiches (a 2:1 ratio) You would run out of bread after 4 sandwiches so that s all you can make! Chapter 3 17 Limiting Reactant Concept Since you run out of bread first, bread is the ingredient that limits how many sandwiches you can make. In a chemical reaction, the limiting reactant is the reactant that controls the amount of product you can make. A limiting reactant is used up before the other reactants. The other reactants are present in excess. Chapter

10 Limiting Reactant Concept Which Reactant below is limiting? How do you know? Chapter 3 19 Limiting Reactant Problems There are three steps to a limiting reactant problem: Step 1: Calculate the moles of product that can be produced from the first reactant. mass reactant #1 mol reactant #1 mol product Step 2: Calculate the moles of product that can be produced from the second reactant. mass reactant #2 mol reactant #2 mol product Step 3: The limiting reactant is the reactant that produces the least amount of product. Chapter

11 Limiting Reactants: Table Method One way to keep your limiting reactant problems organized is to use a table. You do the exact same calculations as shown before but you organize your answers in the table as shown below. This helps to see the limiting reactant and how much product is made. g initial MM n initial g final Li 2 O(s) + H 2 O(l) 2.50 g g/mol mol 0 g 2.50 g g/mol mol g 2 LiOH (s) g/mol 0 mol n- / n n final 0 mol mol mol Limiting Reactant 2.00 g Chapter g Determining the Limiting Reactant Lithium oxide is a drying agent used on the space shuttle. If 80.0 g of water is to be removed and 65.0 g of lithium oxide is available, which reactant is limiting? Li 2 O(s) + H 2 O(l) LiOH (s) Chapter

12 Limiting Reactant Problem Cisplatin is an anti-cancer agent prepared as follows: K 2 PtCl 4 + NH 3 Pt(NH 3 ) 2 Cl 2 + KCl If 10.0 g of K 2 PtCl 4 and 10.0 g of NH 3 are allowed to react: (a) Which is the limiting reagent? (b) How many grams of the excess reagent are consumed? (c) How many grams of cisplatin are formed? Chapter 3 23 Solution Concentration The molar concentration, or molarity (M), is the number of moles of solute per liter of solution, is expressed as moles/liter. moles of solute liters of solution = M Chapter

13 Calculating Molarity What is the molarity of a solution containing 27.5 g of KOH in L of solution? Step 1: We have 27.5 g of KOH so we need to convert to moles of KOH. Step 2: We want molarity of the solution (mol/l), so check your units. Step 3: Divide the moles of KOH by the volume of solution (0.100 L). moles of solute liters of solution = M Chapter 3 25 Molar Concentration Problem How many grams of Ca(OH) 2 would you use to prepare ml of a 1.25 M calcium hydroxide solution? Step 1: We have ml of solution and molarity Ca(OH) 2 Step 2: We want grams of Ca(OH) 2 Step 3: Use the molarity formula to solve for moles of Ca(OH) 2. Step 4: Convert the moles of Ca(OH) 2 to grams of Ca(OH) 2 using its molar mass. Chapter

14 Dilution of a Solution Dilution is the process of reducing a solution s concentration by adding more solvent. Rather than prepare a dilute solution by dissolving a solid in water, we can prepare a solution by diluting a more concentrated solution. When performing a dilution, the amount of solute does not change, only the amount of solvent. M concentrated x V concentrated = M dilute x V dilute Chapter 3 27 Dilution Problem What volume of 18.0 M H 2 SO 4 is required to prepare ml of M aqueous H 2 SO 4? Step 1: Determine your unknown (V C ) Step 2: Identify M C, M D and V D Step 3: Plug these values into the dilution equation and solve for V C M C V C = M D V D (18.0 M) V C = (0.500 M) (250.0 ml) V C = (0.500 M) (250.0 ml) 18.0 M = 6.94 ml Chapter

15 Solution Stoichiometry A solution stoichiometry problem uses molarity as a conversion factor between volume and moles of a substance in solution. There are three steps: Step 1: Convert the given volume of solution to moles using the molarity (mol/l) as a unit factor. Step 2: Convert the moles of given to moles of the unknown using the coefficients in the balanced equation. Step 3: Convert the moles of unknown to molarity by dividing by the volume of the solution. Chapter 3 29 Solution Stoichiometry Problem Stomach acid, a dilute solution of HCl in water, can be neutralized by reaction with sodium hydrogen carbonate, NaHCO 3. How many milliliters of M NaHCO 3 solution are needed to neutralize 18.0 ml of M HCl? NaHCO 3 (aq) + HCl (aq) CO 2 (g) + H 2 O (l) + NaCl (aq) Chapter

16 Summary of Stoichiometry Problems Chapter 3 31 Acid-Base Titrations A titration is used to analyze an acid solution using a solution of a base. A measured volume of base is added to the acid solution. When all of the acid has been neutralized, the ph is 7. One extra drop of base solution after the endpoint increases the ph dramatically. Chapter

17 Acid Base Titration Problem What is the molarity of a sulfuric acid solution if a 25.0 ml sample is titrated to equivalence with 50.0 ml of M potassium hydroxide solution? Step 1: Write the balanced neutralization reaction. H 2 SO 4 (aq) + KOH (aq) K 2 SO 4 (aq) + H 2 O (l) Step 2: We want concentration sulfuric acid (SA), we have concentration and volume of potassium hydroxide (KOH) so we need to convert from potassium hydroxide to sulfuric acid. Volume Mole Volume of KOH Ratio of SA Conc of KOH Moles of KOH Moles of SA Conc of SA Chapter 3 33 Redox Titrations As with acids and bases, a titration can be used to analyze the concentration of oxidizing and reducing agents. Chapter

18 Percent Composition A percent, %, expresses the amount of a single portion compared to an entire sample. portion of interest ("Part") % = 100% total sample ("Whole") The percent composition of a compound lists the mass percent of each element.0 % = Mass of element Mass of compound X 100% Chapter 3 35 Calculating Percent Composition Saccharin has the molecular formula C 7 H 5 NO 3 S. What is the percentage composition of saccharin? Step 1: Determine the molar mass of the compound (this will be your whole ) Step 3: Determine the mass of each element in the compound (you should have this from your molar mass calculation). These will be your parts. Step 3: Find the percent composition of the compound by dividing the mass of the part (the element) by the mass of the whole (saccharin) then multiplying by 100 Chapter

19 Empirical Formulas The empirical formula of a compound is the simplest whole number ratio of elements in a formula unit The molecular formula of a compound is some multiple of the empirical formula However, the molecular formula and empirical formula can be the same for a compound too! Compound Formula Empirical Formula Hydrogen peroxide H 2 O 2 Benzene C 6 H 6 Ethylene C 2 H 4 Propane C 3 H 8 Chapter 3 37 Determining Empirical Formula: Mass Percents A compound s empirical formula can be determined from its percent composition. Rhyme to Remember: Percent to Mass Mass to mole Divide by smallest Multiply til whole Chapter

20 Determining Empirical Formula: Component Mass Combustion analysis is one of the most common methods for determining empirical formulas. A weighed compound is burned in oxygen and its products analyzed by a gas chromatogram. It is particularly useful for analysis of hydrocarbons (Combustion Reactions). Chapter 3 39 Converting Decimals to Whole Numbers When calculating empirical formulas, you don t always get a nice whole number. Sometimes the result of dividing by the smallest number of moles gives a decimal instead. Decimal values that are close to a whole number can be rounded to that number: 2.04 becomes 2.00 and 6.98 becomes 7.00 However, a decimal that is greater than 0.1 or less than 0.9 has to be multiplied by a small integer: Chapter

21 Determining Empirical Formulas A compound was analyzed to be 82.67% carbon and 17.33% hydrogen by mass. What is the empirical formula of the compound? Chapter 3 41 Determining Empirical Formulas Combustion analysis of mg of toluene gives mg of H 2 O and mg of CO 2. What is the empirical formula of toluene? Chapter

22 Determining Empirical Formulas Menthol, a flavoring agent obtained from peppermint oil, contains carbon, hydrogen and oxygen. Combustion analysis of 1.00 g of menthol yields g of H 2 O and g of CO 2. What is the empirical formula of menthol? Chapter 3 43 Molecular Formulas The empirical formula for benzene is CH. This represents the ratio of C to H atoms in a benzene molecule. The actual molecular formula is some multiple (f) of the empirical formula, (CH) f. To determine f, we divide the mass of the molecular formula by the mass of the empirical formula: Mass of Molecular Formula Mass of Empirical Formula = f Chapter

23 Finding Molecular Formulas There are 4 steps to determining the molecular formula of a compound: Step 1: Determine the empirical formula of the compound. May need to calculate yourself or may be given in the problem Step 2: Calculate the mass of the empirical formula. Step 3: Divide the mass of the molecular formula (usually given in the problem!) by the mass of the empirical formula to determine the multiplier factor (f). Step 4: Multiply all the subscripts in the empirical formula by the factor f to get the molecular formula. Chapter 3 45 Molecular Formulas A compound was analyzed to be 82.67% carbon and 17.33% hydrogen by mass. An osmotic pressure experiment determined that its molar mass is g/mol. What is the molecular formula of the compound? Step 1: Determine the empirical formula. We did so in earlier problem: C 2 H 5 Step 2: Determine the mass of the empirical formula C 2 H 5 : g/mol Step 3: Divide the mass of the molecular formula (given above: g/mol) by the mass of the empirical formula to determine f Step 4: Multiply all the subscripts in the empirical formula by f (C H ) = g/mol 2 5 f C 2 H g/mol f = 2 and the molecular formula is C 4 H 10 Chapter

Chapter 3 Mass Relationships in Chemical Reactions 國防醫學院生化學科王明芳老師

Chapter 3 Mass Relationships in Chemical Reactions 國防醫學院生化學科王明芳老師 2011-9-20 1 Balancing Chemical Equations A balanced chemical equation shows that the law of conservation of mass is adhered to. In a balanced

Balance the following equation: KClO 3 + C 12 H 22 O 11 KCl + CO 2 + H 2 O

Balance the following equation: KClO 3 + C 12 H 22 O 11 KCl + CO 2 + H 2 O Ans: 8 KClO 3 + C 12 H 22 O 11 8 KCl + 12 CO 2 + 11 H 2 O 3.2 Chemical Symbols at Different levels Chemical symbols represent

Calculation of Molar Masses. Molar Mass. Solutions. Solutions

Molar Mass Molar mass = Mass in grams of one mole of any element, numerically equal to its atomic weight Molar mass of molecules can be determined from the chemical formula and molar masses of elements

Part One: Mass and Moles of Substance. Molecular Mass = sum of the Atomic Masses in a molecule

CHAPTER THREE: CALCULATIONS WITH CHEMICAL FORMULAS AND EQUATIONS Part One: Mass and Moles of Substance A. Molecular Mass and Formula Mass. (Section 3.1) 1. Just as we can talk about mass of one atom of

Element of same atomic number, but different atomic mass o Example: Hydrogen

Atomic mass: p + = protons; e - = electrons; n 0 = neutrons p + + n 0 = atomic mass o For carbon-12, 6p + + 6n 0 = atomic mass of 12.0 o For chlorine-35, 17p + + 18n 0 = atomic mass of 35.0 atomic mass

Mole - Mass Relationships in Chemical Systems

Chapter 3: Stoichiometry Mole - Mass Relationships in Chemical Systems 3.1 The Mole 3.2 Determining the Formula of an Unknown Compound 3.3 Writing and Balancing Chemical Equations 3.4 Calculating the Amounts

2 Stoichiometry: Chemical Arithmetic Formula Conventions (1 of 24) 2 Stoichiometry: Chemical Arithmetic Stoichiometry Terms (2 of 24)

Formula Conventions (1 of 24) Superscripts used to show the charges on ions Mg 2+ the 2 means a 2+ charge (lost 2 electrons) Subscripts used to show numbers of atoms in a formula unit H 2 SO 4 two H s,

Chapter 3: Stoichiometry

Chapter 3: Stoichiometry Key Skills: Balance chemical equations Predict the products of simple combination, decomposition, and combustion reactions. Calculate formula weights Convert grams to moles and

Formulas, Equations, and Moles. + "reacts with" "to produce" Equations must be balanced. Equal amounts of each element on each side of the equation.

Chapter 3 Formulas, Equations, and Moles Chemical Equations 2 2 + 2 2 2 reactants products + "reacts with" "to produce" coefficients - indicate amount of substance Equations must be balanced. Equal amounts

Chemical calculations

Chemical calculations Stoichiometry refers to the quantities of material which react according to a balanced chemical equation. Compounds are formed when atoms combine in fixed proportions. E.g. 2Mg +

Chapter 3. Stoichiometry: Ratios of Combination. Insert picture from First page of chapter. Copyright McGraw-Hill 2009 1

Chapter 3 Insert picture from First page of chapter Stoichiometry: Ratios of Combination Copyright McGraw-Hill 2009 1 3.1 Molecular and Formula Masses Molecular mass - (molecular weight) The mass in amu

Unit 2: Quantities in Chemistry

Mass, Moles, & Molar Mass Relative quantities of isotopes in a natural occurring element (%) E.g. Carbon has 2 isotopes C-12 and C-13. Of Carbon s two isotopes, there is 98.9% C-12 and 11.1% C-13. Find

Mass and Moles of a Substance

Chapter Three Calculations with Chemical Formulas and Equations Mass and Moles of a Substance Chemistry requires a method for determining the numbers of molecules in a given mass of a substance. This allows

The Mole Concept. The Mole. Masses of molecules

The Mole Concept Ron Robertson r2 c:\files\courses\1110-20\2010 final slides for web\mole concept.docx The Mole The mole is a unit of measurement equal to 6.022 x 10 23 things (to 4 sf) just like there

Calculating Atoms, Ions, or Molecules Using Moles

TEKS REVIEW 8B Calculating Atoms, Ions, or Molecules Using Moles TEKS 8B READINESS Use the mole concept to calculate the number of atoms, ions, or molecules in a sample TEKS_TXT of material. Vocabulary

Molecular Formula: Example

Molecular Formula: Example A compound is found to contain 85.63% C and 14.37% H by mass. In another experiment its molar mass is found to be 56.1 g/mol. What is its molecular formula? 1 CHAPTER 3 Chemical

Chapter 3. Stoichiometry of Formulas and Equations

Chapter 3 Stoichiometry of Formulas and Equations Chapter 3 Outline: Mole - Mass Relationships in Chemical Systems 3.1 The Mole 3.2 Determining the Formula of an Unknown Compound 3.3 Writing and Balancing

Chapter 6 Chemical Calculations

Chapter 6 Chemical Calculations 1 Submicroscopic Macroscopic 2 Chapter Outline 1. Formula Masses (Ch 6.1) 2. Percent Composition (supplemental material) 3. The Mole & Avogadro s Number (Ch 6.2) 4. Molar

Stoichiometry. What is the atomic mass for carbon? For zinc?

Stoichiometry Atomic Mass (atomic weight) Atoms are so small, it is difficult to discuss how much they weigh in grams We use atomic mass units an atomic mass unit (AMU) is one twelfth the mass of the catbon-12

Chapter 3 Calculation with Chemical Formulas and Equations

Chapter 3 Calculation with Chemical Formulas and Equations Practical Applications of Chemistry Determining chemical formula of a substance Predicting the amount of substances consumed during a reaction

Other Stoich Calculations A. mole mass (mass mole) calculations. GIVEN mol A x CE mol B. PT g A CE mol A MOLE MASS :

Chem. I Notes Ch. 12, part 2 Using Moles NOTE: Vocabulary terms are in boldface and underlined. Supporting details are in italics. 1 MOLE = 6.02 x 10 23 representative particles (representative particles

Chapter 3. Chemical Reactions and Reaction Stoichiometry. Lecture Presentation. James F. Kirby Quinnipiac University Hamden, CT

Lecture Presentation Chapter 3 Chemical Reactions and Reaction James F. Kirby Quinnipiac University Hamden, CT The study of the mass relationships in chemistry Based on the Law of Conservation of Mass

Tutorial 4 SOLUTION STOICHIOMETRY. Solution stoichiometry calculations involve chemical reactions taking place in solution.

T-27 Tutorial 4 SOLUTION STOICHIOMETRY Solution stoichiometry calculations involve chemical reactions taking place in solution. Of the various methods of expressing solution concentration the most convenient

4. Aluminum chloride is 20.2% aluminum by mass. Calculate the mass of aluminum in a 35.0 gram sample of aluminum chloride.

1. Calculate the molecular mass of table sugar sucrose (C 12 H 22 O 11 ). A. 342.30 amu C. 320.05 amu B. 160.03 amu D. 171.15 amu 2. How many oxygen atoms are in 34.5 g of NaNO 3? A. 2.34 10 23 atoms C.

Chapter 4. Chemical Composition. Chapter 4 Topics H 2 S. 4.1 Mole Quantities. The Mole Scale. Molar Mass The Mass of 1 Mole

Chapter 4 Chemical Composition Chapter 4 Topics 1. Mole Quantities 2. Moles, Masses, and Particles 3. Determining Empirical Formulas 4. Chemical Composition of Solutions Copyright The McGraw-Hill Companies,

How much does a single atom weigh? Different elements weigh different amounts related to what makes them unique.

How much does a single atom weigh? Different elements weigh different amounts related to what makes them unique. What units do we use to define the weight of an atom? amu units of atomic weight. (atomic

STOICHIOMETRY STOICHIOMETRY. Measurements in Chemical Reactions. Mole-Mole Relationships. Mass-Mass Problem. Mole-Mole Relationships

STOICHIOMETRY STOICHIOMETRY The analysis of the quantities of substances in a chemical reaction. Stoichiometric calculations depend on the MOLE- MOLE relationships of substances. Measurements in Chemical

CHAPTER 3 Calculations with Chemical Formulas and Equations. atoms in a FORMULA UNIT

CHAPTER 3 Calculations with Chemical Formulas and Equations MOLECULAR WEIGHT (M. W.) Sum of the Atomic Weights of all atoms in a MOLECULE of a substance. FORMULA WEIGHT (F. W.) Sum of the atomic Weights

We know from the information given that we have an equal mass of each compound, but no real numbers to plug in and find moles. So what can we do?

How do we figure this out? We know that: 1) the number of oxygen atoms can be found by using Avogadro s number, if we know the moles of oxygen atoms; 2) the number of moles of oxygen atoms can be found

Chemistry 65 Chapter 6 THE MOLE CONCEPT

THE MOLE CONCEPT Chemists find it more convenient to use mass relationships in the laboratory, while chemical reactions depend on the number of atoms present. In order to relate the mass and number of

Concept 1. The meaning and usefulness of the mole. The mole (or mol) represents a certain number of objects.

Chapter 3. Stoichiometry: Mole-Mass Relationships in Chemical Reactions Concept 1. The meaning and usefulness of the mole The mole (or mol) represents a certain number of objects. SI def.: the amount of

Chapter 3! Stoichiometry: Calculations with Chemical Formulas and Equations. Stoichiometry

Chapter 3! : Calculations with Chemical Formulas and Equations Anatomy of a Chemical Equation CH 4 (g) + 2O 2 (g) CO 2 (g) + 2 H 2 O (g) Anatomy of a Chemical Equation CH 4 (g) + 2 O 2 (g) CO 2 (g) + 2

MASS RELATIONS IN CHEMISTRY; STOICHIOMETRY

MASS RELATIONS IN CHEMISTRY; STOICHIOMETRY [MH5; Ch. 3] Each element has its own unique mass. The mass of each element is found on the Periodic Table under the chemical symbol for the element (it is usually

Stoichiometry Lecture Examples Answer Key Ex. 1 Balance the following chemical equations: 3 NaBr + 1 H 3 PO 4 3 HBr + 1 Na 3 PO 4 2 C 3 H 5 N 3 O 9 6 CO 2 + 3 N 2 + 5 H 2 O + 9 O 2 2 Ca(OH) 2 + 2 SO 2

CHAPTER 3 STOICHIOMETRY OF FORMULAS AND EQUATIONS

CHAPTER 3 STOICHIOMETRY OF FORMULAS AND EQUATIONS FOLLOW UP PROBLEMS 3.1A Plan: The mass of carbon must be changed from mg to g. The molar mass of carbon can then be used to determine the number of moles.

Chemistry Stoichiometry. 45 points. Do not turn page until told to do so.

Chemistry 2010 Stoichiometry 45 points Do not turn page until told to do so. Stoichiometry Multiple Choice (Each worth 2 points) Identify the letter of the choice that best completes the statement or

Practice questions for Ch. 3

Name: Class: Date: ID: A Practice questions for Ch. 3 1. A hypothetical element consists of two isotopes of masses 69.95 amu and 71.95 amu with abundances of 25.7% and 74.3%, respectively. What is the

Formula Stoichiometry. Text pages

Formula Stoichiometry Text pages 237-250 Formula Mass Review Write a chemical formula for the compound. H 2 CO 3 Look up the average atomic mass for each of the elements. H = 1.008 C= 12.01 O = 16.00 Multiply

Ch. 10 The Mole I. Molar Conversions

Ch. 10 The Mole I. Molar Conversions I II III IV A. What is the Mole? A counting number (like a dozen) Avogadro s number (N A ) 1 mole = 6.022 10 23 representative particles B. Mole/Particle Conversions

Chemical Reactions. Chemical Equations. Mole as Conversion Factor: To convert between number of particles and an equivalent number of moles:

Quantities of Reactants and Products CHAPTER 3 Chemical Reactions Stoichiometry Application of The Law of Conservation of Matter Chemical book-keeping Chemical Equations Chemical equations: Describe proportions

Moles. Balanced chemical equations Molar ratios Mass Composition Empirical and Molecular Mass Predicting Quantities Equations

Moles Balanced chemical equations Molar ratios Mass Composition Empirical and Molecular Mass Predicting Quantities Equations Micro World atoms & molecules Macro World grams Atomic mass is the mass of an

Stoichiometry. Unit Outline

3 Stoichiometry Unit Outline 3.1 The Mole and Molar Mass 3.2 Stoichiometry and Compound Formulas 3.3 Stoichiometry and Chemical Reactions 3.4 Stoichiometry and Limiting Reactants 3.5 Chemical Analysis

Atomic Masses. Chapter 3. Stoichiometry. Chemical Stoichiometry. Mass and Moles of a Substance. Average Atomic Mass

Atomic Masses Chapter 3 Stoichiometry 1 atomic mass unit (amu) = 1/12 of the mass of a 12 C atom so one 12 C atom has a mass of 12 amu (exact number). From mass spectrometry: 13 C/ 12 C = 1.0836129 amu

Chapter 4 Chemical Composition. Moles of Various Elements and Compounds Figure 4.8

Chapter 4 Chemical Composition Mole Quantities Moles, Masses, and Particles Determining Empirical and Molecular Formulas Chemical Composition of Solutions 4-1 Copyright The McGraw-Hill Companies, Inc.

Moles and Chemical Reactions. Moles and Chemical Reactions. Molar mass = 2 x 12.011 + 6 x 1.008 + 1 x15.999 = 46.069 g/mol

We have used the mole concept to calculate mass relationships in chemical formulas Molar mass of ethanol (C 2 H 5 OH)? Molar mass = 2 x 12.011 + 6 x 1.008 + 1 x15.999 = 46.069 g/mol Mass percentage of

Chapter 3 Chemical Reactions and Reaction Stoichiometry. 許富銀 ( Hsu Fu-Yin)

Chapter 3 Chemical Reactions and Reaction Stoichiometry 許富銀 ( Hsu Fu-Yin) 1 Stoichiometry The study of the numerical relationship between chemical quantities in a chemical reaction is called stoichiometry.

Unit 10A Stoichiometry Notes

Unit 10A Stoichiometry Notes Stoichiometry is a big word for a process that chemist s use to calculate amounts in reactions. It makes use of the coefficient ratio set up by balanced reaction equations

The mass of the formula unit is called the formula mass Formula masses are calculated the same way as molecular masses

Chapter 4: The Mole Atomic mass provides a means to count atoms by measuring the mass of a sample The periodic table on the inside cover of the text gives atomic masses of the elements The mass of an atom

stoichiometry = the numerical relationships between chemical amounts in a reaction.

1 REACTIONS AND YIELD ANSWERS stoichiometry = the numerical relationships between chemical amounts in a reaction. 2C 8 H 18 (l) + 25O 2 16CO 2 (g) + 18H 2 O(g) From the equation, 16 moles of CO 2 (a greenhouse

Mole Notes.notebook. October 29, 2014

1 2 How do chemists count atoms/formula units/molecules? How do we go from the atomic scale to the scale of everyday measurements (macroscopic scale)? The gateway is the mole! But before we get to the

Unit 9 Stoichiometry Notes (The Mole Continues)

Unit 9 Stoichiometry Notes (The Mole Continues) is a big word for a process that chemist s use to calculate amounts in reactions. It makes use of the coefficient ratio set up by balanced reaction equations

Lecture 5 Outline. Derived from the Greek stoicheion ( element ) and metron ( measure )

Lecture 5 Outline 5.1 Stoichiometry,, the mole etc. 5.2 Chemical Equations 5.3 Molarity 5.4 Limiting reagents and yields 5.5 Reaction enthalpies and Gibbs free energy 5.6 Catalyst Lecture 5 Stoichiometry

Sample Exercise 3.1 Interpreting and Balancing Chemical Equations

Sample Exercise 3.1 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.

Atomic mass is the mass of an atom in atomic mass units (amu)

Micro World atoms & molecules Laboratory scale measurements Atomic mass is the mass of an atom in atomic mass units (amu) By definition: 1 atom 12 C weighs 12 amu On this scale 1 H = 1.008 amu 16 O = 16.00

Chemical Equations & Stoichiometry

Chemical Equations & Stoichiometry Chapter Goals Balance equations for simple chemical reactions. Perform stoichiometry calculations using balanced chemical equations. Understand the meaning of the term

Daily Practice Review 2/28-29/08

Daily Practice Review 2/28-29/08 1. Why is it not correct to balance an equation by changing the subscripts in one or more of the formulas? If you change the subscripts in a formula you change the chemical

CHEMISTRY. (i) It failed to explain how atoms of different elements differ from each other.

CHEMISTRY MOLE CONCEPT DALTON S ATOMIC THEORY By observing the laws of chemical combination, John Dalton proposed an atomic theory of matter. The main points of Dalton s atomic theory are as follows: (i)

Stoichiometry. Stoichiometry Which of the following forms a compound having the formula KXO 4? (A) F (B) S (C) Mg (D) Ar (E) Mn

The Advanced Placement Examination in Chemistry Part I Multiple Choice Questions Part II - Free Response Questions Selected Questions from 1970 to 2010 Stoichiometry Part I 1984 2. Which of the following

Calculations and Chemical Equations. Example: Hydrogen atomic weight = 1.008 amu Carbon atomic weight = 12.001 amu

Calculations and Chemical Equations Atomic mass: Mass of an atom of an element, expressed in atomic mass units Atomic mass unit (amu): 1.661 x 10-24 g Atomic weight: Average mass of all isotopes of a given

Moles. Moles. Moles. Moles. Balancing Eqns. Balancing. Balancing Eqns. Symbols Yields or Produces. Like a recipe:

Like a recipe: Balancing Eqns Reactants Products 2H 2 (g) + O 2 (g) 2H 2 O(l) coefficients subscripts Balancing Eqns Balancing Symbols (s) (l) (aq) (g) or Yields or Produces solid liquid (pure liquid)

Chapter 3 Mass Relations in Chemistry; Stoichiometry

Chapter 3 Mass Relations in Chemistry; Stoichiometry MULTIPLE CHOICE 1. An atomic mass unit (amu) is defined as a. 1.60 10-19 C. b. the mass of 1 mole of hydrogen-s. c. the mass of 1 hydrogen-. d. the

EXPERIMENT 12: Empirical Formula of a Compound

EXPERIMENT 12: Empirical Formula of a Compound INTRODUCTION Chemical formulas indicate the composition of compounds. A formula that gives only the simplest ratio of the relative number of atoms in a compound

IB Chemistry. DP Chemistry Review

DP Chemistry Review Topic 1: Quantitative chemistry 1.1 The mole concept and Avogadro s constant Assessment statement Apply the mole concept to substances. Determine the number of particles and the amount

1. How many hydrogen atoms are in 1.00 g of hydrogen?

MOLES AND CALCULATIONS USING THE MOLE CONCEPT INTRODUCTORY TERMS A. What is an amu? 1.66 x 10-24 g B. We need a conversion to the macroscopic world. 1. How many hydrogen atoms are in 1.00 g of hydrogen?

Stoichiometry. Types of Problems. Stoichiometry. Chemistry 1010 Review Tutorial 4/9/2013. Stoichiometry and Lewis Structures

Stoichiometry Chemistry 1010 Review Tutorial Stoichiometry and Lewis Structures April 9 th, 2013 Stoichiometry Stoichiometry involves MOLES Elements/compounds can only be compared side by side using moles

AP Chemistry. Unit #3. Chapter 3 Zumdahl

AP Chemistry Unit #3 Chapter 3 Zumdahl Stoichiometry C6H12O6 + 6 O2 6 CO2 + 6 H2O Students should be able to: Calculate the atomic weight (average atomic mass) of an element from the relative abundances

Chapter 3 Stoichiometry

Chapter 3 Stoichiometry 3-1 Chapter 3 Stoichiometry In This Chapter As you have learned in previous chapters, much of chemistry involves using macroscopic measurements to deduce what happens between atoms

0.786 mol carbon dioxide to grams g lithium carbonate to mol

1 2 Convert: 2.54 x 10 22 atoms of Cr to mol 4.32 mol NaCl to grams 0.786 mol carbon dioxide to grams 2.67 g lithium carbonate to mol 1.000 atom of C 12 to grams 3 Convert: 2.54 x 10 22 atoms of Cr to

Percent Yield = Actual yield of product x 100% Theoretical yield of product

Yields of Chemical Reactions In the stoichiometry examples so far we have made the unstated assumption that the reaction goes to completion, that all reactants are converted to products. In actual fact,

IB Chemistry 1 Mole. One atom of C-12 has a mass of 12 amu. One mole of C-12 has a mass of 12 g. Grams we can use more easily.

The Mole Atomic mass units and atoms are not convenient units to work with. The concept of the mole was invented. This was the number of atoms of carbon-12 that were needed to make 12 g of carbon. 1 mole

10/21/2013. Chemical REACTIONS you should know. Chemical Reactions. 1 st Write Reaction

Chapter 3: Chemical Stoichiometry Chemical Equations (Write, Balance, Interpret) Reactions You Should Know Formula Weights (Must know chemical formula Avogadro s number and the Mole Limiting Reactants

Chemical Calculations: The Mole Concept and Chemical Formulas. AW Atomic weight (mass of the atom of an element) was determined by relative weights.

1 Introduction to Chemistry Atomic Weights (Definitions) Chemical Calculations: The Mole Concept and Chemical Formulas AW Atomic weight (mass of the atom of an element) was determined by relative weights.

2 NO + O 2 2 NO 2 3/18/2014. iclicker Participation Question: A B C

Today: Stoichiometric Analysis: Gram to Gram Conversions: Use MOLAR MASS to get to moles Limiting Reagents: Method 1 Method 2 Actual Yield & Percent Yield Combustion Analysis Titrations Next Meeting Reading

Stoichiometry Dr. M. E. Bridge

Preliminary Chemistry Course Stoichiometry Dr. M. E. Bridge What is stoichiometry? The meaning of the word: The word stoichiometry comes from two Greek words: (meaning element ) and (meaning measure )

Formulae, stoichiometry and the mole concept

3 Formulae, stoichiometry and the mole concept Content 3.1 Symbols, Formulae and Chemical equations 3.2 Concept of Relative Mass 3.3 Mole Concept and Stoichiometry Learning Outcomes Candidates should be

Name Date Class CHEMICAL QUANTITIES. SECTION 10.1 THE MOLE: A MEASUREMENT OF MATTER (pages 287 296)

10 CHEMICAL QUANTITIES SECTION 10.1 THE MOLE: A MEASUREMENT OF MATTER (pages 287 296) This section defines the mole and explains how the mole is used to measure matter. It also teaches you how to calculate

Chemistry I: Using Chemical Formulas. Formula Mass The sum of the average atomic masses of all elements in the compound. Units are amu.

Chemistry I: Using Chemical Formulas Formula Mass The sum of the average atomic masses of all elements in the compound. Units are amu. Molar Mass - The mass in grams of 1 mole of a substance. Substance

Chem 1100 Chapter Three Study Guide Answers Outline I. Molar Mass and Moles A. Calculations of Molar Masses

Chem 1100 Chapter Three Study Guide Answers Outline I. Molar Mass and Moles A. Calculations of Molar Masses B. Calculations of moles C. Calculations of number of atoms from moles/molar masses 1. Avagadro

3. How many moles of KCl and O 2 are formed from the decomposition of 6 moles of KClO 3?

What coefficients mean: 2 Na + Cl 2 2NaCl 2 Na 1 Cl 2 2NaCl 4 Na 2 Na + Cl 2 4Cl 2 6 moles Na 2NaCl 10 atoms Na ONLY WORKS FOR MOLES, MOLECULES, ATOMS 1. How many moles of H 2 and O 2 must react to form

Chapter 4 Notes - Types of Chemical Reactions and Solution Chemistry

AP Chemistry A. Allan Chapter 4 Notes - Types of Chemical Reactions and Solution Chemistry 4.1 Water, the Common Solvent A. Structure of water 1. Oxygen's electronegativity is high (3.5) and hydrogen's

Liquid phase. Balance equation Moles A Stoic. coefficient. Aqueous phase

STOICHIOMETRY Objective The purpose of this exercise is to give you some practice on some Stoichiometry calculations. Discussion The molecular mass of a compound is the sum of the atomic masses of all

Molecular Masses Recall: periodic table gives us the average mass (u) of an atom of a specified element

Chapter 3 (Hill/Petrucci/McCreary/Perry Stoichiometry: Chemical Calculations This chapter deals with quantitative relationships in compounds and between compounds in chemical reactions. These quantitative

CHEM J-2 June /01(a) What is the molarity of the solution formed when 0.50 g of aluminium fluoride is dissolved in ml of water?

CHEM1001 2014-J-2 June 2014 22/01(a) What is the molarity of the solution formed when 0.50 g of aluminium fluoride is dissolved in 800.0 ml of water? 2 The molar mass of AlF 3 is: molar mass = (26.98 (Al)

Stoichiometry Chapter 9 Assignment & Problem Set

Stoichiometry Name Warm-Ups (Show your work for credit) Date 1. Date 2. Date 3. Date 4. Date 5. Date 6. Date 7. Date 8. Stoichiometry 2 Study Guide: Things You Must Know Vocabulary (know the definition

Calculations involving concentrations, stoichiometry

Calculations involving concentrations, stoichiometry MUDr. Jan Pláteník, PhD Mole Unit of amount of substance the amount of substance containing as many particles (atoms, ions, molecules, etc.) as present

Chapter 1: Moles and equations. Learning outcomes. you should be able to:

Chapter 1: Moles and equations 1 Learning outcomes you should be able to: define and use the terms: relative atomic mass, isotopic mass and formula mass based on the 12 C scale perform calculations, including

One element, usually a metal, replaces another element in a compound. This forms a new compound and leaves behind a new free element!

124 SINGLE REPLACEMENT REACTIONS One element, usually a metal, replaces another element in a compound. This forms a new compound and leaves behind a new free element! example: Copper loses electrons, goes

Chapter 3 Mass Relationships in Chemical Reactions

Chapter 3 Mass Relationships in Chemical Reactions Student: 1. An atom of bromine has a mass about four times greater than that of an atom of neon. Which choice makes the correct comparison of the relative

Honors Chemistry: Unit 6 Test Stoichiometry PRACTICE TEST ANSWER KEY Page 1. A chemical equation. (C-4.4)

Honors Chemistry: Unit 6 Test Stoichiometry PRACTICE TEST ANSWER KEY Page 1 1. 2. 3. 4. 5. 6. Question What is a symbolic representation of a chemical reaction? What 3 things (values) is a mole of a chemical

Chemical Calculations: Formula Masses, Moles, and Chemical Equations

Chemical Calculations: Formula Masses, Moles, and Chemical Equations Atomic Mass & Formula Mass Recall from Chapter Three that the average mass of an atom of a given element can be found on the periodic

Chapter 3: ex. P2O5 molecular mass = 2(30.97 amu) + 5(16.00 amu) = amu

Molecular Mass and Formula Mass for molecular compounds: the molecular mass is the mass (in amu) of one molecule of the compound molecular mass = atomic masses of elements present Chapter 3: ex. P2O5 molecular

Subscripts and Coefficients Give Different Information

Chapter 3: Stoichiometry Goal is to understand and become proficient at working with: 1. Chemical equations (Balancing REVIEW) 2. Some simple patterns of reactivity 3. Formula weights (REVIEW) 4. Avogadro's

6/27/2014. Periodic Table of the ELEMENTS. Chemical REACTIONS you should know. Brief Review for 1311 Honors Exam 2

Brief Review for 3 Honors Exam 2 Chapter 2: Periodic Table I. Metals. Representative Metals Alkali Metals Group Alkaline Earth Metals. Group 2 2. Transition Metals II. Metalloids Chapter 3: All Chapter

EMPIRICAL AND MOLECULAR FORMULA

EMPIRICAL AND MOLECULAR FORMULA Percent Composition: law of constant composition states that any sample of a pure compound always consists of the same elements combined in the same proportions by mass

Name Date Class STOICHIOMETRY. SECTION 12.1 THE ARITHMETIC OF EQUATIONS (pages 353 358)

Name Date Class 1 STOICHIOMETRY SECTION 1.1 THE ARITHMETIC OF EQUATIONS (pages 353 358) This section explains how to calculate the amount of reactants required or product formed in a nonchemical process.

CHEMISTRY CP Name: KEY Period: UNIT 5: TEST REVIEW SHEET MOLE CONVERSIONS, EMPIRICAL/MOLECULAR FORMULA, STOICHIOMETRY

CHEMISTRY CP Name: KEY Period: UNIT 5: TEST REVIEW SHEET MOLE CONVERSIONS, EMPIRICAL/MOLECULAR FORMULA, STOICHIOMETRY MOLES AND MOLE CONVERSIONS 1. Things to Know: a. What is a mole? Avogadro s number

CHEMICAL FORMULA COEFFICIENTS AND SUBSCRIPTS. Chapter 3: Molecular analysis 3O 2 2O 3

Chapter 3: Molecular analysis Read: BLB 3.3 3.5 H W : BLB 3:21a, c, e, f, 25, 29, 37,49, 51, 53 Supplemental 3:1 8 CHEMICAL FORMULA Formula that gives the TOTAL number of elements in a molecule or formula

2. In a double displacement reaction, 3. In the chemical equation, H 2 O 2 H 2 + O 2, the H 2 O 2 is a

What are the missing coefficients for the skeleton equation below? Al 2 (SO 4 ) 3 + KOH Al(OH) 3 + K 2 SO 4 2. In a double displacement reaction, A. 1,6,2,3 B. 2,12,4,6 C. 1,3,2,3 D. 4,6,2,3 E. 2,3,1,1