Chapter 3 Stoichiometry

Similar documents
Chapter 3 Stoichiometry

Chapter 3: Phenomena. Chapter 3: Stoichiometry. Mass of A. Mass of C. Mass of A. Mass of D. Mass of B. Mass of B. Mass of C

AP Chemistry Chapter 3. Stoichiometry

9/14/ Chemistry Second Edition Julia Burdge. Stoichiometry: Ratios of Combination. Molecular and Formula Masses

Balancing Chemical Reactions. CHAPTER 3: Quantitative Relationships in Chemical Reactions. Zn + HCl ZnCl 2 + H 2. reactant atoms product atoms

Chapter 3. Mass Relationships in Chemical Reactions

The Mole. Relative Atomic Mass Ar

General Chemistry. Chapter 3. Mass Relationships in Chemical Reactions CHEM 101 (3+1+0) Dr. Mohamed El-Newehy 10/12/2017

Chapter 9 STOICHIOMETRY

Mass Relationships in Chemical Reactions

Chapter 12 Stoichiometry. Mr. Mole

CHAPTER 3: PART 2 8/9/2015. A chemical change (a chemical reaction) converts one substance into another.

L = 6.02 x mol Determine the number of particles and the amount of substance (in moles)

Chapter 3. Stoichiometry: Calculations with Chemical Formulas and Equations. Lecture Presentation

Stoichiometry. Chapter 3

Chapter 2: Mass Relations in Formulas, Chemical Reactions, and Stoichiometry

Chapter 3. Stoichiometry

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.

Unit III: Quantitative Composition of Compounds

Chapter 3 Stoichiometry. Ratios of combination

Chapter 9. Stoichiometry. Mr. Mole. NB page 189

Chemistry B11 Chapter 5 Chemical reactions

Formulas and Models 1

Chapter 3: Stoichiometry

Chapter 3. Stoichiometry

Stoichiometry is the relationship between the amount of reactants used and the amount of products produced in a chemical reaction.

6 atomic # C symbol Carbon name of element atomic mass. o Examples: # 1 mol C = g # 1 mol O = g # 1 mol H = 1.

Stoichiometry. Mr. Mole

Mass Relationships in Chemical Reactions

Chapter 3. Chemical Equations & Reaction Stoichiometry. Symbolic representation of a chemical reaction

Ch. 3 The Mole: Relating the Microscopic World of Atoms to Laboratory Measurements. Brady & Senese, 5th Ed.

EXAM 1 Review Session

Stoichiometry Ratios of Combination


Matter is anything that has mass and occupies space. Three physical states of matter

Calculations with Chemical Formulas and Equations

PowerPoint to accompany. Chapter 2. Stoichiometry: Calculations with Chemical Formulae and Equations. Dr V Paideya

NOTES Mole Concept Chapter 3

Unit III: Quantitative Composition of Compounds

3 Stoichiometry: Calculations with Chemical Formulas and Equations

Chemical Quantities: Stoichiometry and the Mole

Chapter 3 The Mole and Stoichiometry

Chemistry 101 Chapter 8 Chemical Composition

Name AP CHEM / / Chapter 3 Outline Stoichiometry

Atoms, Ions and Molecules Calculations

Unit 4: Reactions and Stoichiometry

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

Unit 3. Stoichiometry

Chapter 3 Chemical Reactions and Reaction Stoichiometry

Stoichiometry is the relationship between the amount of reactants used and the amount of products produced in a chemical reaction.

Counting by mass: The Mole. Unit 8: Quantification of Chemical Reactions. Calculating molar mass. Particles. moles and mass. moles and particles

Mass Relationships in Chemical Reactions

CHEM 120: Introduction to Inorganic Chemistry. Chapters Covered and Test dates. The mole concept and atoms. Avogadro s number

Stoichiometry is the relationship between the amount of reactants used and/or the amount of products produced in a chemical reaction.

SIC CONCEPTS TS OF CHEMISTRY. Unit. I. Multiple Choice Questions (Type-I)

Mole: base unit for an amount of substance A mole contains Avogadro s number (N A ) of particles (atoms, molecules, ions, formula units )

Chapter 3. Stoichiometry: Calculations with Chemical Formulas and Equations

Chemical Equations. Law of Conservation of Mass. Anatomy of a Chemical Equation CH4(g) + 2O2(g) Chapter 3

Importance and Scope of Chemistry

Chapter 3. Stoichiometry: Calculations with Chemical Formulas and Equations

Chapter 10 Chemical Quantities

Stoichiometry is the relationship between the amount of reactants used and/or the amount of products produced in a chemical reaction.

Atoms, Molecules, and the Mole

Solutions to the Extra Problems for Chapter 8

Chapter 1 IB Chemistry Warm Ups Stoichiometry. Mrs. Hilliard

Equation Writing for a Neutralization Reaction

Mass Relationships in Chemical Reactions

TOPIC 4: THE MOLE CONCEPTS

Chapter 5. Chemistry for Changing Times, Chemical Accounting. Lecture Outlines. John Singer, Jackson Community College. Thirteenth Edition

INTRO AND BACKGROUND: Reactions, Moles, Stoichiometry, and Solutions. Chemical Reaction Atoms are REARRANGED to form a different substance

UNIT 6 STOICHIOMETRY 1

Chapter 3. Stoichiometry: Calculations with Chemical Formulas and Equations. Lecture Presentation

Vijaykumar N. Nazare

Introductory Chemistry: A Foundation, 6 th Ed. Introductory Chemistry, 6 th Ed. Basic Chemistry, 6 th Ed.

Review Package #2 Measurement and Communication The Mole Chemical Reactions and Equations Stoichiometry

REACTANTS - materials that are needed fot a reaction. PRODUCTS - materials that are formed in a reaction

Notes: Unit 7 Moles & Stoichiometry

Notes: Unit 7 Moles & Stoichiometry

UNIT 3 Chemical Quantities Chapter 5 Counting Atoms and Molecules The Mole

Chapter 3: Stoichiometry

Stoichiometry. Introduction. Rx between Hydrogen and Oxygen can be described as: Balanced equation: Or Avogadros Number: (number of Molecules)

CHEMISTRY- I PUC. Properties of matter and their measurements in chemistry:

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

Q: How long would it take to spend a mole of $1 coins if they were being spent at a rate of 1 billion per second? A:

Chapter 3. Stoichiometry:

Stoichiometry is the relationship between the amount of reactants used and the amount of products produced in a chemical reaction.

REACTANTS - materials that are needed fot a reaction. PRODUCTS - materials that are formed in a reaction

Chapter 9: Stoichiometry The Arithmetic ti Of Equations

Class XI Chapter 1 Some Basic Concepts of Chemistry Chemistry

Molar Mass. The total of the atomic masses of all the atoms in a molecule:

REVIEW of Grade 11 Chemistry

Chapter 3: Molecules, Compounds and Chemical Equations: (continue and finish chapter 3: 8-11)

Chapter 3 Molecules, Compounds, and Chemical Equations

IGCSE (9-1) Edexcel - Chemistry

CHEMISTRY Matter and Change

Lecture 5. Percent Composition. etc. Professor Hicks General Chemistry II (CHE132) Percent Composition. (aka percent by mass) 100 g.

The Atom, The Mole & Stoichiometry. Chapter 2 I. The Atomic Theory A. proposed the modern atomic model to explain the laws of chemical combination.

Stoichiometry. Please take out your notebooks

Unit 8. The Mathematics Of Chemical Equations

Name Date Class CHEMICAL REACTIONS. SECTION 11.1 DESCRIBING CHEMICAL REACTIONS (pages )

Transcription:

Chapter 3: Phenomena Phenomena: When some substances are mixed together other substances form. Below is data for the reaction A(s) + 2B(aq) C(aq) + D(aq). Look at the data below and identify any patterns that are present. Can you predict what would happen if you had 15.00 g A, 6.00 g B, 4.00 g C, and 5.00 g D? A B Initial C D A B Final C D 10.00 g 10.00 g 0.00 g 0.00 g 0.00 g 5.70 g 2.01 g 12.29 g 20.00 g 0.00 g 0.00 g 0.00 g 20.00 g 0.00 g 0.00 g 0.00 g 0.00 g 20.00 g 0.00 g 0.00 g 0.00 g 20.00 g 0.00 g 0.00 g 0.00 g 0.00 g 10.00 g 10.00 g 0.00 g 0.00 g 10.00 g 10.00 g 5.00 g 5.00 g 5.00 g 5.00 g 0.00 g 2.86 g 6.00 g 11.14 g 13.97 g 6.03 g 0.00 g 0.00 g 0.00 g 0.00 g 2.81 g 17.18 g 15.00 g 6.00 g 4.00 g 5.00 g Big Idea: Depending on the situation, it is convenient to talk about substances in terms of number of particles, mass, or moles. Given one of the quantities, the other two can be calculated A mole is just a specific number of objects similar to a dozen. Chemical equations allow us to find the relationship between reactants and products. Chapter 3 Stoichiometry Moles Molar Mass Limiting Reagents and Theoretical Yields Percent Yield Determining Composition of Compounds Determining Chemical Formulas 2 Determining the Number of All non-zeros are significant Interior zeros are significant Leading zeros are not significant Trailing zeros, after the decimal place, are significant Note: If a decimal place is put onto a whole number, and there are no numbers to the right of the decimal, all trailing zeros are considered interior zeros. 0.00004520090 Number of Calculations with Addition/Subtraction: The answer is as precise as the least accurate number. Multiplication/Division: The number of significant figures in the answer is the same as number with the least amount of significant figures. Note: If multiple operations are performed, follow the order of operations. Perform addition and subtraction first (with correct sig figs) then do multiplication and division (with correct sig figs). What Calculator would Report What You Should Report (Correct ) 23,098,000 200. 2.34+4.678 = 7.018 0.99 3.457= 3.42243 2.93 2.910 13 0.02 0.7220769231 13 3 4 Moles How many significant figures should appear in the answer for the following mathematical operation?.... a) 1 b) 2 c) 3 d) 4 e) None of the above 5 6 1

Molar Mass Molar Mass (M): The mass of a mole of objects. m=mass n=moles 1 mole of C weighs 12.011 grams 12.011 12.011 amu Note: amu (atomic mass unit) 1 amu = 1 Average Atomic Mass: The sum of the masses of an atoms isotopes, each multiplied by its natural abundance. Molar Mass Molar Compounds Calculate the molar mass of C 2 H 5 OH ( ): 2 6 2 12.01 46.08 6 1.01 16.00 Note: Plug in fraction of isotopes in decimals. Not 20% but 0.20. 7 8 Procedure Step 1: Determine where you are starting and finishing Step 2: Start calculation by writing down what you know Step 3: Use dimensional analysis Set up problem so you can cancel out units Tools Available Avogadro s Constant 6.022 10 Periodic Table (M) Chemical Formulas Later Chemical Reactions Moles Atoms/Molecules/etc. Use: Avogadro s Constant N A = 6.0221 1023 How many moles in 5.23 10 23 atoms? Moles Grams Use: Periodic Table (M) How many moles of copper in 3.20 g? 63.55 9 10 Conversion Flow Chart Moles Moles Use: Chemical Formula/Chemical Equation How many moles of O in 3.4 mol of CuSO 4? Grams Molar Mass Moles Chemical Formula or Equation N A Moles Atoms/Molecules/etc 12 2

Chemical Reaction: A process by which one or more substances are converted into other substances Reactants: Starting material. Products: Substances formed. Note: A chemical reaction is symbolized by an arrow. Skeleton Equation (not balanced) Na + H 2 O NaOH + H 2 Reactants Products Na H O Law of Conservation of Mass: Mass is constant during a chemical reaction. Note: The same number and type of atoms that are on the reactants side of the equation must be on the product side of the equation. Na(s) + H 2 O(l) NaOH(aq) + H 2 (g) Reactants Products Na H O 13 14 Hints for Balancing Writing Chemical Reactions If an element occurs in only one compound on both sides of the equation, balance it first. Step 1: Write skeleton equation. Step 2: Balance. Note: If there are more than one such element, balance metals before nonmetals. Note: Never change subscripts to balance equations. If an element occurs as a free element on either side of the chemical equation, balance it last. Step 3: Add state symbols Examples: Solid (s), liquid (l), gas (g), aqueous (aq) compound that is dissolved in water. When possible think of polyatomic ions as a group and balance them as 1 unit instead of individual atoms. Step 4: Write other variables needed for the reaction to take place over the arrow. Examples: Heat, catalyst V 2O 5 15 16 How many of the following statements are true concerning balanced chemical equations? i. The number of molecules is conserved. ii. Coefficients indicate mass ratios of the substances involved. iii. Atoms are neither created nor destroyed. iv. The sum of the coefficients on the left side equals the sum of the coefficients on the right side. a)0 b) 1 c) 2 d) 3 e) 4 Reaction Stoichiometry: The quantitative relationship between the amounts of reactants consumed and those of products formed in chemical reactions as expressed by the balanced chemical equation for the reaction. Example: N 2 (g) +3H 2 (g) 2NH 3 (g) 1 mole of N 2 and 3 moles of H 2 can make 2 moles of NH 3 Mole Ratios: The stoichiometric relationship between two species in a chemical reaction written as a conversion factor. Note: Equation must be balanced before constructing mole ratios. 17 18 3

Limiting Reagents and Theoretical Yields Limiting Reagents and Theoretical Yields 19 2 bread + 1 bologna 1 sandwich What you have in your cupboard: 30 pieces of bread 7 pieces of bologna Initial/Change/Final ICF Tables Bread Bologna Sandwich Initial Change Final 20 Limiting Reagent: The reactant that governs the maximum amount of product that can form. Identifying The Limiting Reagent Step 1: Calculate the amount of each reactant in moles. Step 2: Choose one of the reactants and use the mole ratio to calculate the theoretical amount of the second reactant needed. If n cal < n given then the reactant used for the calculation is the limiting reagent. If n cal > n given then the reactant not used for the calculation is the limiting reagent. Step 3: Use an ICF table to determine the final number of moles of substances after the reaction has gone to completion. Limiting Reagents and Theoretical Yields Limiting Reagents and Theoretical Yields Theoretical Yield: The maximum quantity (amount, mass, or volume) of product that can be obtained from a given quantity of a specified reactant. Finding the Theoretical Yield Step 1: Identify the limiting reagent. Step 2: Calculate theoretical yield based on the limiting reagent. When Zn reacts with hydrochloric acid zinc chloride and hydrogen gas form. If 25.0 g of zinc are react with 17.5 g of hydrochloric acid. How many grams of hydrogen gas will be produced? Helpful Information: 65.41, 36.46, 100.86, and 2.02 a) 0.240 g b) 0.385 g c) 0. 772 g d) 0.970 g e) None of the above 21 22 Percent Yield Determining Composition of Compounds Percent Yield: The fraction of the theoretical yield actually produced. 100% If a student s theoretical yield is 1.44 g and they only get 1.23 g. What is their percent yield? Mass Percentage Composition: The mass of each element expressed as a percentage of the total mass. % 100% Note: Mass percentage is an intrinsic property (independent of size). Example: A 3.16 g sample is made from 2.46 g of carbon, 0.373 g of hydrogen, and 0.329 g of oxygen. Determine the mass percentages of carbon. 23 24 4

Determining Chemical Formulas Determining Chemical Formulas Empirical Formula: Tells the relative number of atoms of each element present in the compound. Example: CH 2 O (glucose) tells us that C, H, and O are present in a ratio of 1:2:1 Molecular Formula: Tells the actual numbers of atoms of each element in a molecule. Example: C 6 H 12 O 6 (glucose) tells us that exactly 6 C, 12 H, and 6 O atoms are in the compound. Note: Different molecular formulas can have same empirical formulas. Example: CH 2 O (formaldehyde) and C 2 H 4 O 2 (acetic acid) both have the empirical formula CH 2 O. Determining Empirical Formulas Step 1: Find number of grams of each atom. It is sometimes useful to assume that you have a 100 g sample. Step 2: Calculate moles of each type of atom (use molar mass). Step 3: Divide through by smallest number of moles calculated in step 2. Step 4: If necessary multiply all numbers in step 3 with a coefficient so that only whole numbers remain. 25 26 Determining Chemical Formulas Determining Chemical Formulas A 1.40 g sample of silicon is allowed to react with fluorine to produce 5.20 g of a product. What is the empirical formula of the compound? Helpful Information: 28.08, and 19.00 Combustion of 250.0 g of a compound containing solely C, H, and O results in the production of 366.4 g of CO 2 and150.0 g water. What is the empirical formula? a) SiF 2 b) SiF 3 c) SiF d) Si 2 F 3 e) None of the above 27 28 Take Away From Chapter 3 Take Away From Chapter 3 Big Idea: Depending on the situation it is convenient to talk about substances in terms of number of particles, mass, or moles. Given one of the quantities the other two can be calculated A mole is just a specific number of objects similar to a dozen. Chemical equations allow us to find the relationship between reactants and products. Be able to determine the number of significant figures in a number. Be able to preform operations (+,-,,and ) and express answers with correct significant figures. Moles Know that a mole is just a specific number of objects (1 mol = 6.0221 10 23 objects) and can be use to classify anything. Molar Mass m M n Know how to calculate the molar mass of an atom from relative abundances and atomic masses of its isotopes. (23,26,27,28) average atomic mass=(fraction isotope 1 )( )+ Be able to calculate the molar mass of a compound (34,35,36,38) Be able to convert between the following Moles molecules/atoms (Avogadro's Number) Grams moles (molar mass) Moles moles (chemical formula or equation) Be able to write and balance chemical equations.(63,65,68,69) Be able to determine the amount of products formed when given reactants.(71,74,75) 29 30 5

Take Away From Chapter 3 Limiting Reagents and Theoretical Yields Be able to identify the limiting reagent.(81,82,84,137) Be able to determine the theoretical yield. Be able to determine the amount of excess reactants. Percent Yield(86) 100% Determining Composition of Compounds Be able to determine the mass percent of an element in a compound.(41,44,47) Usually easiest if assume sample size of 1 mol Take Away From Chapter 3 Determining Chemical Formulas (Continued) Be able to determine empirical and molecular formulas from mass percentages of elements in compound. (53,54,59) Be able to determine empirical and molecular formulas when given combustion data. (60,61,132) C x H y O z + O 2 CO 2 + H 2 O general combustion equation 31 % 100% Determining Chemical Formulas Know the difference between empirical and molecular formulas. Empirical: Relative number of atoms in a compound (usually for ionic compounds) Molecular: The exact number of atoms in a molecule (molecular compounds) 32 6