Modern Chemistry Chapter 9 Stoichiometry

composition stoichiometry deals with the mass relationships of elements in compounds. composition stoichiometry deals with the mass relationships of elements in compounds. reaction stoichiometry involves the mass relationships between reactants and products in a chemical reaction. reaction stoichiometry involves the mass relationships between reactants and products in a chemical reaction.

Types of Stoichiometry Problems 1) mole to mole (Both the given and the unknown quantities are amounts in moles.) 2) mole to mass (The given amount is in moles and the unknown amount is in grams.) 3) mass to mole (The given amount is in grams and the unknown amount is in moles.) 4) mass to mass (Both the given and the unknown amount is in grams.)

Mole Ratio & Molar Mass mole ratio- A conversion factor that relates the amounts in moles of any two substances involved in a chemical reaction. mole ratio- A conversion factor that relates the amounts in moles of any two substances involved in a chemical reaction. Found by using the coefficients in the balanced formula equation of the reaction. Found by using the coefficients in the balanced formula equation of the reaction. molar mass- Equal to the mass in grams of one mole of an element or a compound. molar mass- Equal to the mass in grams of one mole of an element or a compound. Found by adding the individual element atomic masses from the formula of the compound. Found by adding the individual element atomic masses from the formula of the compound.

Section Review Problems Do section review problems Do section review problems #1 through #4 on page 301 of the textbook.

Section Review page The branch of chemistry that deals with mass relationships in compounds and in chemical reactions. 2 HgO  2 Hg + O 2 2-a) 2 mol HgO &2 mol HgO 2 mol Hg 1 mol O 2 2 mol Hg &2 mol Hg 2 mol Hg &2 mol Hg 2 mol HgO1 mol O 2 2 mol HgO1 mol O 2 1 mol O 2 &1 mol O 2 2 mol HgO2 mol Hg

Section Review page It is used to convert moles of one substance into moles of another substance. 4- The formula equation MUST be BALANCED so mole ratios can be determined.

Using Conversion Factors Amount coefficientamount of of given x of unknown = unknown substance coefficientsubstance in moles of knownin moles # moles x mole ratio = #moles of given unknown given unknown Do practice problems #1 & #2 on page 306 of text.

Page 306 #1 3 H 2 + N 2  2 NH 3 6 mol H 2 x 2 mol NH 3 = 4mol NH 3 3 mol H 2 3 mol H 2

Page 306 #2 2 KClO 3  2 KCl + 3 O 2 15 mol O 2 x 2 mol KClO 3 = 10 mol KClO 3 3 mol O 2

Chapter 9 quiz #1- mole to mole calculations 6 NaBr + Mg 3 (PO 4 ) 2  2 Na 3 PO MgBr 2 6 NaBr + Mg 3 (PO 4 ) 2  2 Na 3 PO MgBr 2 Use the above balanced formula equation to solve the following: moles of NaBr will produce ? moles Na 3 PO moles of Mg 3 (PO 4 ) 2 will yield ? moles MgBr moles of NaBr will react with ? moles Mg 3 (PO 4 ) moles of NaBr will yield ? moles Na 3 PO moles NaBr will produce ? moles MgBr 2

6 NaBr + Mg 3 (PO 4 ) 2  2 Na 3 PO MgBr 2 6 NaBr + Mg 3 (PO 4 ) 2  2 Na 3 PO MgBr mol NaBr x 2 mol Na 3 PO 4 = 2.3 mol Na 3 PO 4 6 mol NaBr 6 mol NaBr 2-3 mol Mg 3 (PO 4 ) 2 x 3.0 mol MgBr 2 = 9 mol MgBr 2 1 mol Mg 3 (PO 4 ) 2 1 mol Mg 3 (PO 4 ) mol NaBr x 1 mol Mg 3 (PO 4 ) 2 = mol 6 mol NaBr mol NaBr x 2 mol Na 3 PO 4 = 0.83 mol Na 3 PO 4 6 mol NaBr mol NaBr x 3 mol MgBr 2 = 1.25 mol MgBr 2 6 mol NaBr

Using Conversion Factors amount of mass in givenx moles unknown x molar = grams of substance moles known mass of unknown in moles unknown substance # moles x mole x molar = mass of unknown givenratio mass(in grams) givenratio mass(in grams) unknown unknown Do practice problems #1 & #2 on page 308 of the textbook.

Page 308 #1 & 2 2 Mg + O 2  2 MgO 2 Mg + O 2  2 MgO 2.00 mol Mg x 2 mol MgO x 40.3 g MgO = 80.6 g MgO 2.00 mol Mg x 2 mol MgO x 40.3 g MgO = 80.6 g MgO 2 mol Mg1 mol MgO 2 mol Mg1 mol MgO 6 CO H 2 O  C 6 H 12 O O 2 10 mol CO 2 x 1 mol C 6 H 12 O 6 x 180 g C 6 H 12 O 6 = 300 g C 6 H 12 O 6 6 mol CO 2 1 mol C 6 H 12 O 6 6 mol CO 2 1 mol C 6 H 12 O 6

Chapter 9 Quiz #2- mole-mass problems 3 MgF 2 + Al 2 O 3  3 MgO + 2 AlF 3 Use the above balanced formula equation to answer the following questions mol MgF 2 will yield ? grams of MgO mol of Al 2 O 3  ? grams of AlF 3 3- If 6.0 mol of MgO is produced, ? grams of AlF mol MgF 2  ? grams of AlF mol Al 2 O 3  ? grams of MgO

3 MgF 2 + Al 2 O 3  3 MgO + 2 AlF mol MgF 2 x 3 mol MgO x 40.3 g MgO = 80.6 g 3 mol MgF 2 1 mol MgOMgO 3 mol MgF 2 1 mol MgOMgO mol Al 2 O 3 x 2 mol AlF 3 x 84.0 g AlF 3 = 672.0g 1 mol Al 2 O 3 1 mol AlF 3 AlF 3 1 mol Al 2 O 3 1 mol AlF 3 AlF mol MgO x 2 mol AlF 3 x 84.0 g AlF 3 = 336 g AlF 3 3 mol MgO 1 mol AlF 3 3 mol MgO 1 mol AlF mol MgF 2 x 2 mol AlF 3 x 84.0 g AlF 3 = 33.6 g AlF 3 3 mol MgF 2 1 mol AlF 3 3 mol MgF 2 1 mol AlF mol Al 2 O 3 x 3 mol MgO x 40.3 g MgO = 332gMgO 1 mol Al 2 O 3 1 mol MgO 1 mol Al 2 O 3 1 mol MgO

Using Conversion Factors mass (g) x 1 mol given x mol unknown = moles of of given molar mass mol given unknown substance of given substance grams x 1 x mole ratio = moles unknown molar mass molar mass Do practice problems #1 & #2 on page 309 of the textbook.

Practice problems page HgO  2 Hg + O g O 2 x 1 mol O 2 x 2 mol HgO = 7.81 mol HgO 32 g O 2 1 mol O 2 32 g O 2 1 mol O g O 2 x 1 mol O 2 x 2 mol Hg = 7.81 mol Hg 32 g O 2 1 mol O 2 32 g O 2 1 mol O 2

chapter 9 quiz #3- mass-mole problems Na 2 O + CaF 2  2 NaF + CaO Use the above equation to solve the problems grams of CaF 2  ? mol CaO g Na 2 O  ? mol NaF g Na 2 O  ? mol CaO g Na 2 O  ? mol NaF 5- A yield of 84 g NaF  ? mol CaO

Na 2 O + CaF 2  2 NaF + CaO g CaF 2 x 1 mol CaF 2 x 1 mol CaO = 2.00 mol CaO 78.1 g CaF 2 1 mol CaF g CaF 2 1 mol CaF g Na 2 O x 1 mol Na 2 O x 2 mol NaF = 6.0 mol NaF 62 g Na 2 O 1 mol Na 2 O 62 g Na 2 O 1 mol Na 2 O g Na 2 O x 1 mol Na 2 O x 1 mol CaO = 0.5 mol CaO 62.0 g Na 2 O 1 mol Na 2 O 62.0 g Na 2 O 1 mol Na 2 O g Na 2 O x 1 mol Na 2 O x 2 mol NaF = 1.0 mol NaF 62.0 g Na 2 O 1 mol Na 2 O 62.0 g Na 2 O 1 mol Na 2 O g NaF x 1 mol NaF x 1 mol CaO = 1.0 mol CaO 42.0 g NaF 2 mol NaF 42.0 g NaF 2 mol NaF

Chemistry Chapter 9- Stoichiometry Practice Problems 2 NaF + CaO  Na 2 O + CaF moles of NaF will produce -?- moles of Na 2 O ? 4.5 mol NaF x 1 mol Na 2 O/2 mol NaF = 2.25 moles Na 2 O moles of CaO will produce -?- grams of CaF 2 ? 3.2 mol CaO x 1 mol CaF 2 /1 mol CaO x 78.1 g CaF 2 /mol CaF 2 = g CaF 2

2 NaF + CaO  Na 2 O + CaF grams of NaF will produce -?- moles of Na 2 O ? /42.0 x 1 /2 = 2.0 mol Na 2 O grams of CaO will produce -?- moles of CaF 2 ? 112.2/56.1 x 1/1 = 2.0 mol CaF 2

5-Calculate the molar mass of each of the reactants & products of the above balanced formula equation. Use the molar masses in the following problems. a-AlN = (1 x 27.0) + (1 x 14.0) = 41.0 g/mol b-Na 2 O = (2 x 23.0) + (1 x 16.0) = 62.0 g/mol c-Al 2 O 3 = (2 x 27.0) + (3 x 16.0) = g/mol d-Na 3 N = (3 x 23.0) + (1 x 14.0) = 83.0 g/mol

2 AlN + 3 Na 2 O  Al 2 O Na 3 N grams of AlN will produce -?- moles of Al 2 O 3 ? 82.0/41.0 x 1/2 = 1.0 mol Al 2 O grams AlN will produce -?- moles of Na 3 N ? 164.0/41.0 x 2/2 = 4.0 mol Na 3 N

2 AlN + 3 Na 2 O  Al 2 O Na 3 N moles of Na 2 O will produce -?- grams of Na 3 N ? 2.5 x 2/3 x 83.0 = g Na 3 N moles of Na 2 O will produce -?- moles of Al 2 O 3 ? 0.75 x 1/3 = 0.25 mol Al 2 O moles of Na 2 O will produce -?- grams of Na 3 N ? 11.0 x 2/3 x 83.0 = g Na 3 N

2 H 2 + O 2  2 H 2 O grams of H 2 will react with -?- moles of O 2 ? 8.0/2.0 x 1/2 = 2.0 mol O grams of O 2 will produce -?- moles of H 2 O ? 64.0/32.0 x 2/1 = 4.0 mol H 2 O moles of H 2 will produce -?- moles of H 2 O ? 0.25 x 2/2 = 0.25 mol H 2 O

2 H 2 + O 2  2 H 2 O moles of H 2 will produce -?- grams of H 2 O ? 1.5 x 2/2 x 18.0 = 27.0 g H 2 O moles of O 2 will produce -?- moles of H 2 O ? 14 x 2/1 = 28.0 mol H 2 O

Using Conversion Factors mass (g) x 1 mol given x mol unknown x molar mass unknown = mass of given molar mass mol given 1 mol unknown of un- substance of given known grams x 1 x mole ratio x molar mass = mass of unknown given molar mass given molar mass Do practice problems #1, #2, & #3 on page 311 of textbook. Do section review problems #1 - #5 on page 311 of textbook.

Practice problems page 311 NH 4 NO 3  N 2 O + 2 H 2 O NH 4 NO 3  N 2 O + 2 H 2 O 32 g N 2 O x 1 mol N 2 O x 1 mol NH 4 NO 3 x 80 g NH 4 NO 3 = 60 g NH 4 NO 3 44 g N 2 O 1 mol N 2 O 1 mol NH 4 NO 3 44 g N 2 O 1 mol N 2 O 1 mol NH 4 NO 3

Chapter 9 quiz #4- mass-mass problems Na 2 O + CaF 2  2 NaF + CaO g Na 2 O  ? grams NaF g Na 2 O  ? grams CaO g CaF 2  ? g NaF g CaF 2  ? g CaO g NaF  ? g CaO

Na 2 O + CaF 2  2 NaF + CaO g Na 2 O x 1 mol Na 2 O x 2 mol NaF x 42.0 g NaF = 168 g 62 g Na 2 O 1 mol Na 2 O 1 mol NaF 62 g Na 2 O 1 mol Na 2 O 1 mol NaF g Na 2 O x 1 mol Na 2 O x 1 mol CaO x 56.1 g CaO = g 62 g Na 2 O 1 mol Na 2 O 1 mol CaO 62 g Na 2 O 1 mol Na 2 O 1 mol CaO g CaF 2 x 1 mol CaF 2 x 2 mol NaF x 42.0 g NaF = 252 g 78.1 gCaF 2 1 mol CaF 2 1 mol NaF 78.1 gCaF 2 1 mol CaF 2 1 mol NaF g CaF 2 x 1 mol CaF 2 x 1 mol CaO x 56.1 g CaO = g 78.1 gCaF 2 1 mol CaF 2 1 mol CaO 78.1 gCaF 2 1 mol CaF 2 1 mol CaO g NaF x 1 mol NaF x 1 mol CaO x 56.1 g CaO = 56.1 g 42.0 g NaF 2 mol NaF 1 mol CaO 42.0 g NaF 2 mol NaF 1 mol CaO

The “MOLE HILL” x mole ratio x mole ratio # moles known# moles unknown ÷ molar massx molar mass of knownof unknown mass of knownmass of unknown

Stoichiometry Practice Problems 2 H 2 + O 2  2 H 2 O 1)2.5 mol H 2 x 1 mol O 2 = 1.25 mol O 2 2 mol H 2 2 mol H 2 2)2.5 mol H 2 x 2 mol H 2 O = 2.5 mol H 2 O 2 mol H 2 2 mol H 2 3)2.5 mol H 2 x 1 mol O 2 x 32 g O 2 = 40 g O 2 2 mol H 2 1 mol O 2 2 mol H 2 1 mol O 2 4)2.5 mol H 2 x 2 mol H 2 O x 18 g H 2 O = 45 g H 2 O 2 mol H 2 1 mol H 2 O 2 mol H 2 1 mol H 2 O

Stoichiometry Practice Problems 2 H 2 + O 2  2 H 2 O 5)16 g H 2 x 1 mol H 2 x 1 mol O 2 = 4.0 mol O 2 2 g H 2 2 mol H 2 2 g H 2 2 mol H 2 6)16 g H 2 x 1 mol H 2 x 2 mol H 2 O = 8.0 mol H 2 O 2.0 g H 2 2 mol H g H 2 2 mol H 2 7) 16 g H 2 x 1 mol H 2 x 1 mol O 2 x 32 g O 2 = 128 g O g H 2 2 mol H 2 1 mol O g H 2 2 mol H 2 1 mol O 2 8) 16 g H 2 x 1 mol H 2 x 2 mol H 2 O x 18 g H 2 O = 144 g H 2 O 2.0 g H 2 2 mol H 2 1 mol H 2 O 2.0 g H 2 2 mol H 2 1 mol H 2 O

Limiting Reactant & Percentage Yield limiting reactant is the reactant that limits the amount of the other reactant that can combine and the amount of product that can be formed in a chemical reaction. limiting reactant is the reactant that limits the amount of the other reactant that can combine and the amount of product that can be formed in a chemical reaction. excess reactant is the substance that is NOT completely used up in a chemical reaction. excess reactant is the substance that is NOT completely used up in a chemical reaction.

Sample & Practice Problems See sample problem F on page 313 of textbook. Do practice problems #1 on page 313. See sample problem G on pages Do practice problems #1 & #2 on page 315.

QuickLAB Do the QuickLAB titled “Limiting Reactants in a Recipe” on page 316 of the textbook. Do the QuickLAB titled “Limiting Reactants in a Recipe” on page 316 of the textbook. Yes, cooking IS chemistry!

Percentage Yield  theoretical yield is the maximum amount of product that can be produced from a given amount of reactant  actual yield of a product is the measured amount of a product obtained from a reaction  percentage yield is the ratio of the actual yield to the theoretical yield multiplied by 100  percentage yield = actual yield x 100 theoretical yield

Problems see sample problem H on pages of textbook see sample problem H on pages of textbook Do practice problems #1 & #2 on page 318. Do practice problems #1 & #2 on page 318. Do section review problems #1 - #4 on page 318. Do section review problems #1 - #4 on page 318. Do critical thinking problems #37, #38, #39, & #40 on pages 322 & 323. Do critical thinking problems #37, #38, #39, & #40 on pages 322 & 323.

Chapter 9 test review 20 multiple choice questions 20 multiple choice questions definitions of composition & reaction stoichiometrydefinitions of composition & reaction stoichiometry mole ratios: their definition & usemole ratios: their definition & use SI units of molar massSI units of molar mass identify mole ratio from balanced formula equationidentify mole ratio from balanced formula equation 5 mole-mole problems5 mole-mole problems Definitions & practical applications of excess reactant & limiting reactantDefinitions & practical applications of excess reactant & limiting reactant definitions & practical applications of theoretical yield, actual yield, & % yielddefinitions & practical applications of theoretical yield, actual yield, & % yield

Honors Chemistry Chapter 9 Test Review 25 multiple choice 25 multiple choice Know the definitions of reaction & composition stoichiometry, mole ratio, and units of molar mass. Know the definitions of reaction & composition stoichiometry, mole ratio, and units of molar mass. Know what mole ratio means and how it is used in stoichiometry. Know what mole ratio means and how it is used in stoichiometry. Determine mole ratio using balanced formula equation. (2) Determine mole ratio using balanced formula equation. (2) Perform mole to mole stoichiometric calculations (4). Perform mole to mole stoichiometric calculations (4). Perform mole to mass stoichiometric calculations (1). Perform mole to mass stoichiometric calculations (1). Perform mass to mole stoichiometric calculations (1). Perform mass to mole stoichiometric calculations (1). Perform mass to mass stoichiometric calculations (1). Perform mass to mass stoichiometric calculations (1). Know definitions and applications of limiting and excess reactants. Know definitions and applications of limiting and excess reactants. Know the definitions & applications of actual yield, theoretical yield, and percent yield. Know the definitions & applications of actual yield, theoretical yield, and percent yield.

Practice #2 2 Na 3 PO CaSO 4  3 Na 2 SO 4 + Ca 3 (PO 4 ) 2 Known = 3 moles CaSO 4 unknown = ? Moles Na 3 PO 4 3 x 2/3 = 2 moles Na 3 PO 4 Known = 2.0 moles Na 3 PO 4 unknown = ? Moles Na 2 SO 4 2 x 3/2 = 3 moles Na 2 SO 4 Known = 1.5 moles Ca 3 (PO 4 ) 2 unknown = ? Moles CaSO x 3/1 = 4.5 moles CaSO 4 Known = 4.4 moles CaSO 4 unknown = ? Moles Na 3 PO x 2/3 = 2.9 moles Na 3 PO 4

2 Na 3 PO CaSO 4  3 Na 2 SO 4 + Ca 3 (PO 4 ) 2 Known = 6.0 moles Na 3 PO 4 unknown = ? Moles Na 2 SO x 3/2 = 9.0 moles Na 2 SO 4 Known = 5.4 moles CaSO 4 unknown = ? Moles Na 3 PO x 2/3 = 3.6 moles Na 3 PO 4 Known = 0.6 moles Na 3 PO 4 unknown = ? Moles CaSO x 3/2 = 0.9 moles CaSO 4

Stoichiometry Practice #3 4 Na 3 N + 3 O 2  6 Na 2 O + 2 N 2 Assume you have 12.0 moles of Na 3 N. 1) How many moles of O 2 do you need? 2) How many moles of Na 2 O will you get? 3) How many moles of N 2 will you get?

Stoichiometry Practice #3 4 Na 3 N + 3 O 2  6 Na 2 O + 2 N 2 Assume you have 12.0 moles of Na 3 N. How many moles of O 2 do you need? 12 x 3/4 = 9 moles O 2 How many moles of Na 2 O will you get? 12 x 6/4 = 18 moles Na 2 O How many moles of N 2 will you get? 12 x 2/4 = 6 moles N 2

Practice #3 4 Na + O 2  2 Na 2 O If you have 4 moles of Na, how many grams of O 2 will you need? If you have 64 grams of O 2, how many moles of Na 2 O will you produce? If you have 46 grams of Na, how many grams of O 2 will you need?

Practice #3 4 Na + O 2  2 Na 2 O If you have 4 moles of Na, how many grams of O 2 will you need? 4 x 1/4 x 32 = 32 grams O 2 If you have 64 grams of O 2, how many moles of Na 2 O will you produce? 64/32 x 2/1 = 4 moles Na 2 O If you have 46 grams of Na, how many grams of O 2 will you need? 46/23 x 1/4 x 32 = 16 grams O 2