Objective/Warm-Up SWBAT review topics for the final. If you have 30 g of NaCl or 100 g of NaCl, which would have the greater % of Cl? Why? Calculate the.

Slides:

Advertisements

Similar presentations

Calculations What you need to know: Relative formula mass

Advertisements

Chapter 3: Calculations with Chemical Formulas and Equations MASS AND MOLES OF SUBSTANCE 3.1 MOLECULAR WEIGHT AND FORMULA WEIGHT -Molecular weight: (MW)

STOICHIOMETRY Study of the amount of substances consumed and produced in a chemical reaction.

Stoichiometry A measure of the quantities consumed and produced in chemical reactions.

Chemistry 101 Chapter 9 Chemical Quantities.

Chemical Reactions Unit

What Is a Mole?.

 CHEM.B Apply the mole concept to representative particles (e.g., counting, determining mass of atoms, ions, molecules, and/or formula units). 

Chapter 3 Stoichiometry.

Percentage Composition

Ch. 9: Calculations from Chemical Equations

Solutions Solute Solvent Solutions are homogenous mixtures. When a substance is mixed with a liquid and it disintegrates into sub microscopic particles.

Stoichiometry: the mass relationships between reactants and products. We will use the molar masses ( amount of grams in one mole of a element or compound)

1 © 2006 Brooks/Cole - Thomson Chapter 3 Stoichiometry of Formulas and Equations Chemical Reactions.

Chapter 3: Calculations with Chemical Formulas

Ch 11 & 12 DefinitionsSolubilityRandomCalculations $100 $200 $300 $400 $500.

Atomic Mass l Atoms are so small, it is difficult to discuss how much they weigh in grams. l Use atomic mass units. l an atomic mass unit (amu) is one.

Chemical Stoichiometry

Reaction Stoichiometry.   Deals with the mass relationships that exist between reactants and product  In this type of chemistry, a quantity is given,

Jeopardy Molar Mass Avogadro’s Number Stoichiometry Limiting Reactant Emp/Molec Formula Q $100 Q $200 Q $300 Q $400 Q $500 Q $100 Q $200 Q $300 Q $400.

The Mole and Chemical Composition

The S-word Stoichiometry Chapter 3. By definition: 1 atom 12 C “weighs” 12 amu On this scale 1 H = amu 16 O = amu Atomic mass is the mass.

Things you must KNOW and what to expect  Things you must KNOW  You must KNOW your polyatomics  You must KNOW how to write a balanced formula  You have.

Conversion Factors Molar mass Molar mass atomic mass in g = 1 mole atomic mass in g = 1 mole Volume of gas at STP Volume of gas at STP 1 mole gas = 22.4L.

Stoichiometry Chapters 7 and 9.

Stoichiometry Balancing Equations Molecular and Empirical Formulas Percent Composition Mole Conversions

Unit 5: The Mole.

Mole Jeopardy Moles and Grams Avagadro!Percent Composition Empirical Formula Molecular Formula

Stoichiometry Warm-Up Question(s). Warm-up How many seconds are there in 72 hours? Show your work.

By: Hilary Megeney & Kristen Jorgensen Chemistry 11 adv.

Calculations with Chemical Formulas and Equations.

The Mole & Stoichiometry!

Stoichiometry – Chemical Quantities Notes. Stoichiometry Stoichiometry – Study of quantitative relationships that can be derived from chemical formulas.

Mass Relationships in Chemical Reactions. By definition: 1 atom 12 C “weighs” 12 amu On this scale 1 H = amu 16 O = amu Atomic mass is the.

Solving a Stoichiometry Problem 1.Balance the equation. 2.Convert given to moles. 3.Determine which reactant is limiting. 4.Use moles of limiting reactant.

Calculate the mass of Cu produced? Mass of beaker and Cu – mass of beaker.

Define mole ratio (What is it? How is it determined?)

What is stoichiometry? A problem solving method used to calculate the amount of product made or amount of reactant needed in a chemical reaction What is.

Unit 4: Formula Stoichiometry. What is stoichiometry? Deals with the quantitative information in chemical formula or chemical reaction. Deals with the.

Stoichiometry Chapter 3 Chemical Formulas and Equations.

Unit 17: Review All the Things. Part I. LPChem1415 Groups of similar elements have special names:

Stoichiometry is… Greek for “measuring elements” Defined as: calculations of the quantities in chemical reactions, based on a balanced equation. There.

GENERAL CHEMISTRY CHE 101 Lecture 3: Mass Relationship in Chemical Reactions Course Instructor: HbR.

Unit 9: Stoichiometry Chapter 12 Exam: February 11, 2009.

Chapter 9 Rev Chemical Change Changes the chemical composition of a compound Burns Odors Color Change Release or Absorbs Energy.

Solving a Stoichiometry Problem

Stoichiometry II.

Stoichiometry Chapter 3

MOLE AND STOIKIOMETRI We measure mass in grams. g

Wednesday Do Now Agenda Welcome back! What element am I?

Empirical Formula Molecular Formula

Chemical Reactions Unit

Ch. 9: Calculations from Chemical Equations

Chapter 12 Review.

Bell Ringer 4/4/16 When 6.58 g SO3 and 1.64 g H2O react, what is the expected yield of sulfuric acid? If the actual yield is 7.99 g sulfuric acid,

Stoichiometry Review.

Ch. 9 Stoichiometry Stoichiometry is a branch of chemistry that deals with the mass relationships of elements in compounds and the mass relationships between.

Stoichiometric Calculations

Formation of Ammonia.

mole (symbolized mol) = 6.02 x particles

Bell Ringer 4/3/13 When 6.58 g SO3 and 1.64 g H2O react, what is the expected yield of sulfuric acid? If the actual yield is 7.99 g sulfuric acid,

Moles & Stoichiometry Whiteboard Review

Chapter 9 Key Terms Mole Molar Mass Avogadro's Number Percent Composition Stoichiometry Limiting Reactant Excess Reactant Actual Yield Theoretical Yield.

Presentation transcript:

Objective/Warm-Up SWBAT review topics for the final. If you have 30 g of NaCl or 100 g of NaCl, which would have the greater % of Cl? Why? Calculate the % of Cl in 100 g of NaCl. 5:004:304:003:303:002:302:001:301:000:300:00

Objective/Warm-Up SWBAT review topics for the final. Perform the following conversions: g Pb(CO 3 ) 2 = ______________ moles 33.2 g KCl = _____________ formula units 7.2 x atoms Cu = _____________ grams Cu 5:004:304:003:303:002:302:001:301:000:300:00

Percent Composition- mass of element x100% total mass Empirical Formula- simplest ratio of elements Molecular Formula- “unreduced”, real, actual formula of elements Examples: Empirical Formula CH 2 ONO 2 HO Molecular Formula C 6 H 12 O 6 N2O4N2O4 H2O2H2O2

Change from Empirical to Molecular A compound has a molecular mass of 90 g/mol and an empirical formula of HCO 2, what is the molecular formula?

Percent Hydrate If you have a hydrate that weighs a total of 150 grams and you heat it so that only 70 grams are left, what percent was water?

C. Types of Yields of Products 1. Theoretical Yield the calculated amount that should be produced 2. Actual Yield the measured amount of a product 3. Percent Yield the efficiency of the reaction found from the ratio of the actual yield to the theoretical yield x 100 ( A/T x 100)

1. When the proper amount of NH 3 reacts with enough O 2, g of NO was produced. Tom performed the experiment, but actually produced 98.0 g NO. What is the % yield of NO? D. % Yield Stoichiometry Problems 4 NH O 2  4 NO + 6 H 2 O 98.0 g NO g NO X 100= 81.7%

2. When 36.8 g of NH 3 react with excess of oxygen, the actual yield of water is 56.2 g. What is the % yield of water? 36.8 g NH g NH 3 X108.1 g H 2 O= 58.4 g H 2 O 56.2 g H 2 O 58.4 g H 2 O X 100= 96.2%

Warm-Up: 2AlCl 3  2Al + 3Cl 2 If 35.0 g of aluminum chloride are decomposed, how many grams of chlorine are produced? What is the percent yield if only 8.50 g Cl 2 are produced? 5:004:594:584:574:564:554:544:534:524:514:504:494:484:474:464:454:444:434:424:414:404:394:384:374:364:354:344:334:324:314:304:294:284:274:264:254:244:234:224:214:204:194:184:174:164:154:144:134:124:114:104:094:084:074:064:054:044:034:024:014:003:593:583:573:563:553:543:533:523:513:503:493:483:473:463:453:443:433:423:413:403:393:383:373:363:353:343:333:323:313:303:293:283:273:263:253:243:233:223:213:203:193:183:173:163:153:143:133:123:113:103:093:083:073:063:053:043:033:023:013:002:592:582:572:562:552:542:532:522:512:502:492:482:472:462:452:442:432:422:412:402:392:382:372:362:352:342:332:322:312:302:292:282:272:262:252:242:232:222:212:202:192:182:172:162:152:142:132:122:112:102:092:082:072:062:052:042:032:022:012:001:591:581:571:561:551:541:531:521:511:501:491:481:471:461:451:441:431:421:411:401:391:381:371:361:351:341:331:321:311:301:291:281:271:261:251:241:231:221:211:201:191:181:171:161:151:141:131:121:111:101:091:081:071:061:051:041:031:021:011:00 YOU HAVE ONLY ONE MINUTE LEFT ON THIS QUESTION 0:590:580:570:560:550:540:530:520:510:500:490:480:470:460:450:440:430:420:410:400:390:380:370:360:350:340:330:320:310:300:290:280:270:260:250:240:230:220:210:200:190:180:170:160:150:140:130:120:110:100:090:080:070:060:050:040:030:020:010:00 Objective: SWBAT use stoichiometry to solve limiting reactant problems using a balanced chemical equation g AlCl 3 X 1 mol AlCl g AlCl 3 X 3 mol Cl 2 2 mol AlCl 3 X 70.9 g Cl 2 1 mol Cl 2 =27.9 g Cl 2 % yield=8.50 g/27.9 g Cl 2 x100% = 30.5%

Concentration Concentration- amount of solute in a given amount of solvent or solution Molarity- moles solute L solution molality- moles solute kg solvent

Acids and Bases AcidBase ArrheniusReleases H+ ions (HCl) Releases OH- ions (NaOH) Bronsted- Lowery H+ (proton or hydronium ion) donor (HCl) H+ (proton or hydronium ion) acceptor (NH 3 ) LewisElectron pair acceptor (BH 3 ) Electron pair donor (NH 3 )

Types of Mixtures Solution- a homogeneous mixture, particles in a solution are about the size of molecules, approximately 1 nanometer Suspension- Those that make up suspensions are larger than 1,000 nanometers. Colloid- mixtures whose particles are larger than the size of a molecule but smaller than particles that can be seen with the naked eye. Finally, colloidal particles range in size between 1 and 1,000 nanometers. Colloids are also called colloidal dispersions because the particles of which they are made are dispersed, or spread out, through the mixture.