Turn in last week’s catalyst and grab calculator and worksheet. Sit in lab seats. Objective: Convert between moles of substances in an equation. Convert.

Slides:



Advertisements
Similar presentations
Mole relationships in chemical equations
Advertisements

Mole Review 1.) Calculate the number of moles in 60.4L of O2. 2.) How many moles are there in 63.2g of Cl2? 1 mol O2 60.4L O2 = 2.7 mol O2 22.4L O2 1mol.
LIMITING REACTANT The reactant that gives the least number of product moles “limits” the reaction. To understand this concept, let’s suppose you were an.
Chapter 11 “Stoichiometry”
Chapter 12 “Stoichiometry”
Chapter 12 “Stoichiometry” Chemistry Tutorial Stoichiometry Mr. Mole.
Stoichiometry Chapter 12.
HONORS CHEMISTRY Feb 27, Brain Teaser Cu + 2 AgNO 3  2 Ag + Cu(NO 3 ) 2 – How many moles of silver are produced when 25 grams of silver nitrate.
Stoichiometry Chapter 12.
Unit 3 Test (Part 2) Review Stoichiometry. Balancing Equations Balance the following chemical equation: ____ Ag 2 O  ____ Ag + ____O 2.
Limiting Reagent What happens in a chemical reaction, if there is an insufficient amount of one reactant?
Stoichiometry #3 October 22 & Ms. Boon Chemistry.
Stoichiometry Chapter 11 & Chapter 13.3.
Limiting reagent, Excess reactant, Theoretical or Percent yield
Chapter 12 Stoichiometry
Chapter 9 Stoichiometry
Unit 3: Chemical Equations and Stoichiometry
“Stoichiometry” Original slides by Stephen L. Cotton Mr. Mole.
Information given by chemical equations
Ch 100: Fundamentals for Chemistry Ch 9: Calculations from Chemical Reactions Lecture Notes (Sections 9.1 to 9.5)
Information given by chemical equations
Stoichiometry The Math of Chemical Reactions Unit 9.
Stoichiometry: Calculations with Chemical Formulas and Equations
1 STOICHIOMETRY 2 General Approach For Problem Solving 1. Clearly identify the Goal or Goals and the UNITS involved. (starting and ending unit) 2. Determine.
Chapter 12 “Stoichiometry” Chemistry Chemistry Pioneer High School Mr. David Norton.
Chapter 9 Calculations from Chemical Equations (Stoichiometry)
Chapter 3 Stoichiometry. Atomic Mass Carbon-12 is assigned a mass of exactly atomic mass units (amu) Masses of other elements are compared to Carbon-12.
Chapter 12 Review “Stoichiometry”
Irrelevant Information to Take Your Mind Off of Stoichiometry The first person to use the word “stoichiometry” was Nicephorus I, the ecumenical Patriarch.
Review. Stoichiometry u Greek for “measuring elements” u The calculations of quantities in chemical reactions based on a balanced equation. u We can interpret.
Stoichiometric Calculations Stoichiometry – Ch. 9.
Stoichiometric Calculations Stoichiometry – Ch. 8.
Solution types of stoichiometry problems are no harder than any other stoichiometry problem. You must use the concentration given (molarity) to convert.
Stoichiometry. Information Given by the Chemical Equation  The coefficients in the balanced chemical equation show the molecules and mole ratio of the.
Stoichiometry. The study of chemical changes is at the heart of chemistry. Stoichiometry is the area of study that examines the quantities of substances.
Do Now: 9/8 AgNO 3(aq) + CaCl 2(aq) --> Ca(NO 3 ) 2(aq) + AgCl (s) 1.What is the molar mass of Silver Nitrate? (AgNO 3 ) 2.What does it mean to have a.
 Objective: Understand molecular formulas and balancing equations.  Before: Introduction to molecular formulas  During: Discuss molecular formulas.
Stoichiometry. What Is It? Branch of chemistry that shows the relationships among reactants and products in a chemical reaction Equations must be balanced.
$100 $200 $300 $400 $500 $100 $200 $300 $400 $500 $100 $200 $300 $400 $500 $100 $200 $300 $400 $500 $100 $200 $300 $400 $500 $100 $200 $300.
Chapter 12 Notes. Information given by chemical equations 2 C 6 H 6 (l) + 15 O 2 (g)  12 CO 2 (g) + 6 H 2 O (g)  In this equation there are 2 molecules.
Unit 3 Test (Part 2) Review Stoichiometry. Balancing Equations Balance the following chemical equation: ____ Ag 2 O  ____ Ag + ____O 2.
The Mole, part II Glencoe, Chapter 11 Sections 11.3 & 11.4.
1 Chapter 9 Mole Factors Calculations with Equations Limiting Reactions Percent Yield.
Unit 8 Review Stoichiometry Complete on Markerboard or in your notes.
The Math of Chemical Reactions
Chapter 12 “Stoichiometry” Pre-AP Chemistry Charles Page High School Stephen L. Cotton Mr. Mole.
Can’t directly measure moles Measure units related to moles: –Mass (molar mass) –Number of particles (6.02 x ) –Liters of gas (22.4 Liters at STP)
I. I.Stoichiometric Calculations Stoichiometry – Ch. 10.
Good Morning! Today is Tuesday, January 5, 2016 HW Due: Balancing Review Please put in the Inbox Do-Now: What do you think the term stoichiometry means?
Stoichiometry. Stoichiometry is the study of the mass relationships of the reactants and the products in a chemical reaction. How much stuff it takes.
Stoichiometry Chapter 9 Limiting Reagents Stoich ppt _5 Limiting Reagents Practice.
Stoichiometry Introduction to Chemistry. Stoichiometry Example: 2H 2 + O 2 → 2H 2 O Equivalencies: 2 mol H 2 for every 1 mol O 2 2 mol H 2 for every 2.
Limiting Reagent What happens in a chemical reaction, if there is an insufficient amount of one reactant?
7.4 Calculations Involving Limiting Reagents
Unit 8: Stoichiometry: Part 1
Chapter 12 Stoichiometry.
Stoichiometric Calculations (p )
“Stoichiometry” Mr. Mole.
HONORS CHEMISTRY Feb 27, 2012.
Stoichiometric Calculations (p. 352 – 367)
Chapter 11 “Stoichiometry”
CHAPTER 11 Stoichiometry 11.2 Percent Yield and Concentration.
4/27/16 Today I will define stoichiometry and calculate mole-mole stoichiometry problems Warm Up Write a balanced equation for the reaction of magnesium.
Unit 5 “Stoichiometry” Mr. Mole.
Information given by chemical equations
Chapter 12 Stoichiometry
Chapter 12 Stoichiometry
Mole ratios Mole to mole conversions
Chapter 9 “Stoichiometry”
Presentation transcript:

Turn in last week’s catalyst and grab calculator and worksheet. Sit in lab seats. Objective: Convert between moles of substances in an equation. Convert between grams of substances in an equation. Catalyst: (on scratch paper) How many moles are there in the following masses of substances? A g of N 2 B g of NaOH C g of Cu(SO 4 ) 2 Homework: Complete lab write-up (not full report) Complete C.7 supplement

Group assignments Green: 1 g baking soda, 1 mL acetic acid Blue: 1 g baking soda, 2 mL acetic acid Yellow: 1 g baking soda, 3 mL acetic acid White: 1 g baking soda,4 mL acetic acid Pink: 1 g baking soda, 5 mL acetic acid Purple 1 g baking soda, 6 mL acetic acid Goldenrod: 1 g baking soda, 7 mL acetic acid

Some considerations NaHCO 3 : Baking Soda CH 3 COOH : Acetic Acid – Molarity of acetic acid:.003 mol/mL H 2 0: Water – 1 mL of water = 1 gram of water

Modified grading Group Grades – 20 points: Pre-Lab & lab Individual Grades: – 40 points: Graph and Data Analysis

Grading Group Grades – 10 points: Pre-Lab – 30 points: Presentation Data Quality (group) Individual Grades: – 15 points: Safe and Productive experimentation – 10 points: Presentation Skills (individual) – 15 points: Peer Grade

Questions to consider: What products are being formed? What evidence do you have that these products are being formed? Besides water displacement into the graduated cylinder, what other indications do you notice that a chemical reaction is taking place? In which flask is a chemical reaction taking place? How do you know? What variables contribute to the amount of water displaced? What is the best way to represent your data?

Your Presentation/Paper… What was your procedure (you may exclude the set- up as it will be the same for everyone) – Independent, dependent and control variables! What were your results – Graph and describe trend Analysis – What do you think is happening at the molecular level? – Describe what happens as you increase one variable? Why do you think this is happening? – Compute the moles of baking soda and vinegar at this point: What is the molar ratio? – Determine a theoretical weight/volume of your independent variable that would produce the most gas with the least of the substance.

Information given by chemical equations 2 C 6 H 6 (l) + 15 O 2 (g)  12 CO 2 (g) + 6 H 2 O (g)  This equation could be read as 2 moles of benzene reacts with 15 moles of oxygen to produce 12 moles of carbon dioxide and 6 moles of water. MOLE RATIO Since the number of molecules in any mole is 6.02 x 10 23, a common factor between all species involved in the equation, a MOLE RATIO relationship can be discussed.

Information given by chemical equations 2 C 6 H 6 (l) + 15 O 2 (g)  12 CO 2 (g) + 6 H 2 O (g) The MOLE RATIO for benzene and oxygen is 2 : 15. It can be written as: 2 moles C 6 H 6 or as 15 moles of O 2 15 moles O 2 2 moles of C 6 H 6 The MOLE RATIO for oxygen and carbon dioxide is 15 : 12. It can be written as: 12 moles CO 2 or as 15 moles of O 2 15 moles O 2 12 moles of CO 2 The MOLE RATIO is used for converting moles of one substance into moles of another substance. Without the balanced equation there is no other relationship between two different compounds.

Using the mole ratio to relate the moles of one compound to the moles of another compound is the part of chemistry called STOICHIOMETRY !!!!! 2 H 2 (g) + O 2 (g) 2 H 2 O (g) 2 H 2 (g) + O 2 (g)  2 H 2 O (g) Q. How many mole of hydrogen are necessary to react with 2 moles of oxygen in order to produce exactly 4 moles of water? A. 2 mol O 2 (2 moles H 2 / 1 mole O 2 ) = 4 mole H 2

STOICHIOMETRY The Stoichiometry Flow Chart Use Molar mass (A) Use mole ratio from equation Use Molar mass (B)

STOICHIOMETRY 2 H 2 (g) + O 2 (g) 2 H 2 O (g) 2 H 2 (g) + O 2 (g)  2 H 2 O (g) Q1. How many moles of hydrogen are necessary to react with 15.0 g of oxygen? A. 15.0g O 2 ( 1 mole O 2 ) ( 2 mole H 2 ) = moles H g 1 mole O 2 Q2. How many grams of hydrogen are necessary to react with 15.0 g of oxygen? A. 15.0g O 2 ( 1 mole O 2 ) ( 2 mole H 2 ) ( g H 2 ) = 1.89 g H g 1 mole O 2 1 mole H 2

STOICHIOMETRY 2 H 2 (g) + O 2 (g)  2 H 2 O (g) Q3. How many grams of water are produced from 15.0 g of oxygen? A. 15.0g O 2 ( 1 mole O 2 ) ( 2 mole H 2 O ) ( 18.0 g H 2 O ) =16.9 g H 2 O 32.0 g 1 mole O 2 1 mole H 2 O Q4. How much hydrogen and oxygen is needed to produce 25.0 grams of water? A. 25.0g H 2 O ( 1 mole H 2 O ) ( 2 mole H 2 ) ( g H 2 ) = 2.80 g H g 2 mole H 2 O 1 mole H 2 A. 25.0g H 2 O ( 1 mole H 2 O ) ( 1 mole O 2 ) ( 32.0 g O 2 ) = 22.2 g O g 2 mole H 2 O 1 mole O 2 Notice that the Law of Conservation of Mass still applies.

How many grams of solid are formed when 10.0 g of lead reacts with excess phosphoric acid? 3. Make a list under the appropriate substance 3 Pb+2 H 3 PO 4  Pb 3 (PO 4 ) 2 (s) + 3 H 2 (g) 10.0gm=? Start with what is given: 13.1 g Pb 3 (PO 4 ) gPb ( 1 mole Pb )( 1 mole Pb 3 (PO 4 ) 2 )( 811 g Pb 3 (PO 4 ) 2 ) = 13.1 g Pb 3 (PO 4 ) g Pb 3 mole Pb 1 mole Pb 3 (PO 4 ) 2

GROUP STUDY PROBLEMS 1.How many grams of O2 liquid product can be produced from 3.55 moles of HgO? 2 HgO  2 Hg + O 2 2.How many moles of fluorine are required to produce 3.0 grams of KrF 6 ? Kr + 3 F 2  KrF 6 3.How many grams of Na 2 CO 3 will be produced from 20.0g of NaHCO 3 ? 2 NaHCO 3 ---> Na2CO 3 + H 2 O + CO 2 4.How many grams of O 2 are needed to combust 55.0 grams of C 2 H 4 ? C 2 H 4 + 3O 2 -> 2CO 2 + 2H 2 O 5.How many grams of silver is produced when 50.0 g of copper is reacted with excess silver nitrate solution? Cu + 2 AgNO 3  Cu(NO 3 ) Ag.

PRACTICE PROBLEMS 1. How many grams of gas can be produced from moles of HgO?2 HgO  2 Hg + O 2 2. How many moles of fluorine are required to produce 12.0 grams of KrF 6 ? Given the equation: Kr + 3 F 2  KrF 6 3.How many grams of Na 2 CO 3 will be produced from the thermal decomposition of g of NaHCO 3 ? 2 NaHCO 3 ---> Na2CO 3 + H 2 O + CO 2 5. How many grams of silver is produced when g of copper is reacted with excess silver nitrate solution? Cu + 2 AgNO 3  Cu(NO 3 ) Ag g mol g g