REACTION STOICHIOMETRY 1792 JEREMIAS RICHTER The amount of substances produced or consumed in chemical reactions can be quantified 4F-1 (of 14)

Slides:



Advertisements
Similar presentations
Solutions Solute – what is dissolved
Advertisements

CHEMICAL REACTIONS CHAPTER 4
Stoichiometry Chemistry 6.0. The Mathematics of Chemical Reactions: STOICHIOMETRY I. Balanced Chemical Equations A. Provide qualitative and quantitative.
A.P. Chemistry Chapter 4: Reactions in Aqueous Solutions Part
Stoichiometry of Precipitation Reactions
The following problems refer to FeSO4.
CHAPTER 4 Stoichiometry. 2 Calculations Based on Chemical Equations How many CO molecules are required to react with 25 molecules of Fe 2 O 3 ?
Chemical Stoichiometry
Stoichiometry Chapter 5. Stoichiometry Quantitative relationships between reactants and products The balanced chemical equation gives us the relationships.
Chemical Equations & Reaction Stoichiometry
1 CHAPTER THREE CHEMICAL EQUATIONS & REACTION STOICHIOMETRY CHEMICAL EQUATIONS & REACTION STOICHIOMETRY.
Stoichiometry: Quantitative Information about chemical reactions.
1 Stoichiometry Limiting Reagents: The extent to which a reaction takes place depends on the reactant that is present in limiting amounts—the limiting.
Ch 4. Chemical Quantities and Aqueous Reactions. CH 4 (g) + 2O 2 (g) → CO 2 (g) + 2H 2 O (g) 1 mol2 mol1 mol2 mol Stoichiometry of the reaction FIXED.
1 Chapter 11 Reactions in Aqueous Solutions II: Calculations Aqueous Acid-Base Reactions 1. Calculations Involving Molarity 2. Titrations 3. The Mole Method.
Chemistry Chapter 10, 11, and 12 Jeopardy
Types of Chemical Reactions and Solution Stoichiometry
1 Solution Stoichiometry The concentration of a solution is the amount of solute present in a given quantity of solvent or solution. M = molarity = moles.
Topic E conservation of atoms and mass
1 11 Reactions in Aqueous Solutions II: Calculations.
2Al (s) + 6HCl (g) 2AlCl 3(s) + 3H 2(g) Consider the reaction above. If we react 30.0 g Al and 20.0 g HCl, how many moles of aluminum chloride will be.
Solution Stoichiometry
Reactions in Aqueous Solutions II: Calculations Chapter 11.
Solutions Solute Solvent Solutions are homogenous mixtures. When a substance is mixed with a liquid and it disintegrates into sub microscopic particles.
Practice Problems Quantitative Aspects. How to keep things straight when solving quantitative problems: First identify what you are being asked to find.
Quantitative Relationships (Stoichiometry). Lets take a moment… sit back… relax… and review some previously learned concepts… Lets take a moment… sit.
THE MOLE “n”  The mole term is similar to the “dozen” term.  Just as a dozen represents “12”; the mole represents x A very large amount.
Chapter 3 - Stoichiometry It is important to be able to quantify the amount of reagent(s) that will be needed to produce a given amount of product(s).
Chapter 3 Stoichiometry
Chapter Four: Stoichiometry “ Stoichiometry is a branch of chemistry that deals with the quantitative relationships that exist between the reactants and.
Types of Chemical Reactions and Solution Stoichiometry.
Chemical Stoichiometry
1 CHAPTER THREE CHEMICAL EQUATIONS & REACTION STOICHIOMETRY.
Stoichiometry: Quantitative Information About Chemical Reactions Chapter 4.
Cookies and Chemistry…Huh!?!? Just like chocolate chip cookies have recipes, chemists have recipes as well Just like chocolate chip cookies have recipes,
Chapter 4 Chemical Quantities and Aqueous Reactions.
AP Chemistry Dickson County High School Mrs. Laura Peck 1.
If you are traveling at 65 mi/h how long will it take to travel 112 km? If your car gets 28 miles per gallon how many liters of gas will it take to travel.
Stoichiometry. What is stoichiometry? Composition stoich – deals with mass relationships of elements in compounds (review Ch 3) Reaction stoich – deals.
Chemical Equations and Reaction Stoichiometry
Stoichiometry Stoichiometry CDO High School. Stoichiometry Consider the chemical equation: 4NH 3 + 5O 2  6H 2 O + 4NO There are several numbers involved.
AP/IB Chemistry Chapter 4: Aqueous Solutions and Solution Stoichiometry.
Stoichiometry Calculations based on chemical reactions.
PROPERTIES OF SOLUTIONS
Ch 4. Chemical Quantities and Aqueous Reactions. CH 4 (g) + 2O 2 (g)  CO 2 (g) + 2H 2 O (g) 1 mol2 mol1 mol2 mol Stoichiometry of the reaction FIXED.
THE MOLE CONCEPT.
Chapter 3.  symbolic representation of a chemical reaction that shows: ◦ reactants on left side of reaction ◦ products on right side of equation ◦ relative.
SOLUTION STOICHIOMETRY LECTURE 3 ACIDS AND BASES.
Stoichiometry It’s Finally Here!. What in the world is Stoichiometry? Stoichiometry is how we figure out how the amounts of substances we need for a.
Solution Stoichiometry In solution stoichiometry, we are given a concentration and a volume which we use to determine moles. n = C x V Then we use molar.
Chapter 3.  symbolic representation of a chemical reaction that shows: ◦ reactants on left side of reaction ◦ products on right side of equation ◦ relative.
Stoichiometry: Quantitative Information About Chemical Reactions Chapter 4.
© 2009, Prentice-Hall, Inc. Molarity Two solutions can contain the same compounds but be quite different because the proportions of those compounds are.
Topic 3 (Part B): Chemical Equations. Moles A Pair is 2 A dozen is 12 A gross is 144 A ream is 500 A mole is a number, that measures the amount of a substance.
K. Cumsille, 2010 AP CHEMISTRY UNIT 2: REACTIONS IN AQUEOUS SOLUTIONS Concentration, Dilutions, Gravimetric Analysis & Titrations.
Chapter 9 Rev Chemical Change Changes the chemical composition of a compound Burns Odors Color Change Release or Absorbs Energy.
General properties of Aqueous Solutions “A solution is a homogeneous mixture of two or more substances” Solute: The substance in a smaller amount Solvent:
Solutions - Quantitatively. Solutions Mixture of at least two components Mixture of at least two components Solute Solute Solvent Solvent Components can.
Mass Relationships in Chemical Reactions Chapter 3.
Chapter Four: Stoichiometry Stoichiometry is a branch of chemistry that deals with the quantitative relationships that exist between the reactants and.
1 Chapter 4 Aqueous solutions Types of reactions.
 A process by which one or more substances are changed into one or more different substances. Reactants  Products.
THE MOLE CONCEPT.
Stoichiometry Chapter 3
KNOW, calculations based on…..
Ch. 15 & 16 - Acids & Bases III. Titration (p )
Stoichiometry: Quantitative Information About Chemical Reactions
Chemical Stoichiometry
Stoichiometry Review.
Aqueous Solutions Pt. 2.
Presentation transcript:

REACTION STOICHIOMETRY 1792 JEREMIAS RICHTER The amount of substances produced or consumed in chemical reactions can be quantified 4F-1 (of 14)

INFORMATION FROM CHEMICAL EQUATIONS 2H 2 + O 2 → 2H 2 O 2 molecules 2 moles 0.84 moles moles 1 molecule 1 moles 0.42 moles moles 2 molecules 2 moles 0.84 moles moles 4F-2 (of 14) The moles that react and form do so in the ratio of the balanced equation

INFORMATION FROM CHEMICAL EQUATIONS 2H 2 + O 2 → 2H 2 O 0.60 moles moles 0.00 moles 0.40 moles moles 0.10 moles 0.00 moles moles 0.60 moles 4F-3 (of 14) starting reacting ending 0.50 moles moles 0.10 moles 0.20 moles moles 0.00 moles moles 0.40 moles starting reacting ending 2H 2 + O 2 → 2H 2 O The moles that react and form do so in the ratio of the balanced equation

MASS CALCULATIONS Calculate the mass of oxygen needed to burn 5.00 g of propanol, C 3 H 8 O. C 3 H 8 O + O 2 →CO 2 + H 2 O 34 4½ 4F-4 (of 14)

MASS CALCULATIONS Calculate the mass of oxygen needed to burn 5.00 g of propanol, C 3 H 8 O. C 3 H 8 O + O 2 →CO 2 + H 2 O g x g 2 mol9 mol mol C 3 H 8 O = 9 mol O g C 3 H 8 O x 9 mol O 2 _________________ 2 mol C 3 H 8 O = 12.0 g O 2 x 1 mol C 3 H 8 O ____________________ g C 3 H 8 O x g O 2 ______________ 1 mol O 2 4F-5 (of 14)

Calculate the mass of carbon dioxide produced from the 5.00 g of propanol. C 3 H 8 O + O 2 →CO 2 + H 2 O gx g 2 mol6 mol mol C 3 H 8 O = 6 mol CO g C 3 H 8 O x 6 mol CO 2 _________________ 2 mol C 3 H 8 O = 11.0 g CO 2 x 1 mol C 3 H 8 O ____________________ g C 3 H 8 O x g CO 2 ________________ 1 mol CO 2 4F-6 (of 14)

C 3 H 8 O + O 2 →CO 2 + H 2 O g6.0 g g11.0 g 4F-7 (of 14)

LIMITING REACTANT CALCULATIONS LIMITING REACTANT – The reactant that is completely used up in a reaction 4F-8 (of 14)

x 12 piecesx 1 sandwich _______________ 1 piece = 12 sandwiches 22 slicesx 1 sandwich _______________ 2 slices = 11 sandwichesactual amount produced limiting reactant 12 pieces22 slices 4F-9 (of 14)

Calculate the mass of tungsten metal produced when 25.0 g of barium reacts with 26.0 g of tungsten (III) fluoride. Ba + WF 3 →BaF 2 + W 25.0 g 26.0 g 3 mol 2 mol 323 x g 2 mol 2 4F-10 (of 14)

25.0 g Ba x 2 mol W ____________ 3 mol Ba = 22.3 g W x 1 mol Ba _______________ g Ba x g W ______________ 1 mol W Ba + WF 3 → 25.0 g 26.0 g 3 mol 2 mol 23BaF 2 + W3 x g 2 mol 2 Calculate the mass of tungsten metal produced when 25.0 g of barium reacts with 26.0 g of tungsten (III) fluoride. 4F-11 (of 14)

25.0 g Ba x 2 mol W ____________ 3 mol Ba = 22.3 g W x 1 mol Ba _______________ g Ba x g W _______________ 1 mol W Ba + WF 3 → 25.0 g 26.0 g 3 mol 2 mol 23BaF 2 + W3 x g 2 mol 2 Calculate the mass of tungsten metal produced when 25.0 g of barium reacts with 26.0 g of tungsten (III) fluoride g WF 3 x 2 mol W _____________ 2 mol WF 3 = 19.8 g W x 1 mol WF 3 _________________ g WF 3 x g W _______________ 1 mol W WF 3 is the limiting reactant 19.8 g W are produced 4F-12 (of 14)

Determine the percentage of magnesium and silver in an alloy of the two metals. A 6.50 gram sample of the alloy reacts with 14.5 grams of hydrogen chloride. Mg + HCl →MgCl 2 + H 2 x g14.5 g 1 mol2 mol 2 1 mol Mg = 2 mol HCl 14.5 g HCl x 1 mol Mg ______________ 2 mol HCl = g Mg x 1 mol HCl ________________ g HCl x g Mg ______________ 1 mol Mg 4F-13 (of 14) g Mg x 100 _______________ 6.50 g alloy = 74.4% Mg100% % = 25.6% Ag

A 1.17 gram sample of the unknown is dissolved in water, treated with lead (II) ions, and 1.42 grams of precipitate are collected. Pb 2+ + I - →PbI 2 x g 2 mol g 1 mol 1.42 g PbI 2 x 2 mol I - _____________ 1 mol PbI 2 = g I - x 1 mol PbI 2 _________________ g PbI 2 x g I - _____________ 1 mol I g I - ___________________ 1.17 g sample = 66.8% I - in the sample x 100 4F-14 (of 14) Determine the percentage by mass of iodide in a solid unknown.

MOLES FROM SOLUTION DATA Find the moles of potassium carbonate contained in 275 mL of a M potassium carbonate solution. M = n ___ V MV= n x L solution = mol K 2 CO mol K 2 CO 3 ______________________ L solution 4G-1 (of 12)

Find the moles of each ion in the M potassium carbonate solution mol K 2 CO 3 = mol K + x mol K 2 CO 3 = mol CO 3 2- x 1 4G-2 (of 12)

10.0 mL of M BaCl 2 are mixed with 20.0 mL of M K 2 SO 4. (a)Find the moles of each ion in the solution. x L solution = mol BaCl mol BaCl 2 ______________________ L solution  mol Ba 2+ and mol Cl - x L solution = mol K 2 SO mol K 2 SO 4 ______________________ L solution  mol K + and mol SO G-3 (of 12)

10.0 mL of M BaCl 2 are mixed with 20.0 mL of M K 2 SO 4. (b)Find the moles of each ion after any reaction. Ba 2+ + SO 4 2- →BaSO 4 Initial moles Reacting moles Final moles  0 mol Ba mol SO mol Cl mol K + 4G-4 (of 12)

10.0 mL of M BaCl 2 are mixed with 20.0 mL of M K 2 SO 4. (c)Find the final molarities of each ion in the solution. = 0 M Ba 2+ 0 mol Ba 2+ ______________________ L solution = M SO mol SO 4 2- _______________________ L solution = M Cl mol Cl - _______________________ L solution = M K mol K + _______________________ L solution 4G-5 (of 12)

20.0 mL of M HCl are mixed with 30.0 mL of M NaOH. (a)Find the moles of each ion in the solution. x L solution = mol HCl mol HCl ___________________ L solution  mol H + and mol Cl - x L solution = mol NaOH mol NaOH ______________________ L solution  mol Na + and mol OH - 4G-6 (of 12)

20.0 mL of M HCl are mixed with 30.0 mL of M NaOH. (b)Find the moles of each ion after any reaction. H + + OH - → H 2 O Initial moles Reacting moles Final moles  0 mol H mol OH mol Cl mol Na + 4G-7 (of 12)

20.0 mL of M HCl are mixed with 30.0 mL of M NaOH. (c)Find the final molarities of each ion in the solution. = 0 M H + 0 mol H + ______________________ L solution = M OH mol OH - _______________________ L solution = M Cl mol Cl - _______________________ L solution = M Na mol Na + _______________________ L solution 4G-8 (of 12)

20.0 mL of M HCl are mixed with 30.0 mL of M NaOH. (d)Find the mass of water produced by the reaction. = g H 2 O mol H 2 O x g H 2 O __________________ 1 mol H 2 O 4G-9 (of 12)

DILUTION CALCULATIONS When a solution is diluted, only solvent is added, so the moles of solute are unchanged mol solute (concentrated) = mol solute (diluted) M C V C = M D V D 4G-10 (of 12)

Calculate the volume of 6.00 M ammonia needed to prepare 250. mL of a M ammonia solution. M C V C = M D V D V C = M D V D _______ M C = (0.100 M)(250. mL) ________________________ (6.00 M) = 4.17 mL M C = V C = 6.00 M ? M D = V D = M 250. mL 4G-11 (of 12)

Calculate the volume of water that must be added to 5.00 mL of concentrated hydrochloric acid (12.1 M) to make the acid 3.00 M. M C V C = M D V D M C V C = V D _______ M D = (12.1 M)(5.00 mL) _______________________ (3.00 M) = 20.2 mL M C = V C = 12.1 M 5.00 mL M D = V D = 3.00 M ? 20.2 mL mL ____________ Volume of dilute solution Volume of concentrated solution 15.2 mLWater that must be added 4G-12 (of 12)

TITRATION – A technique in which one solution is used to analysis another Buret: a solution of 1 reactant of known concentration REACTIONS IN SOLUTION Flask: another reactant of unknown concentration, mass, etc. STANDARD SOLUTION – A solution of known concentration 4H-1 (of 13)

The mass of sodium bicarbonate in an antacid tablet is to be determined. ACID-BASE INDICATOR – A weak organic acid or base that changes color in acidic or basic solutions The tablet is dissolved in water, an acid-base indicator added, and 21.5 mL of a M hydrochloric acid solution produces a color change. NaHCO 3 + HCl →NaCl + H 2 O + CO 2 x g 21.5 mL M 1 mol x 1 mol NaHCO 3 _________________ 1 mol HCl mol HCl x L solution __________________ L solution = g NaHCO 3 x g NaHCO 3 _______________________ mol NaHCO 3 4H-2 (of 13)

x 1 mol O.A.D. ____________________ g O.A.D. A sodium hydroxide solution is to be standardized mL of the sodium hydroxide solution are required to neutralize a solution made with grams of solid H 2 C 2 O 4. 2H 2 O (m = g/mol). NaOH + H 2 C 2 O 4 →. 2H 2 ONa 2 C 2 O 4 + H 2 O 34.2 mL x M g 2 mol 1 mol x 2 mol NaOH _________________ 1 mol O.A.D g O.A.D. = M NaOH x 1 _______________________ L solution 2 24 _ 4H-3 (of 13)

A 10.0 mL aliquot of a solution containing V 2+ ions is acidified, and 32.7 mL of a M MnO 4 - solution produces a light purple color. If the V 2+ was oxidized to V 5+, determine the molarity of the V 2+ in the original solution. 4H-4 (of 13) ( ) x 3 ( ) x 5 MnO 4 - (aq) + V 2+ (aq) → MnO 4 - → Mn V 2+ → V H 2 O 8H + +5e e - Mn 2+ + V

A 10.0 mL aliquot of a solution containing V 2+ ions is acidified, and 32.7 mL of a M MnO 4 - solution produces a light purple color. If the V 2+ was oxidized to V 5+, determine the molarity of the V 2+ in the original solution. 4H-5 (of 13) 3MnO 4 - → 3Mn 2+ 5V 2+ → 5V H 2 O 24H + +15e e - 15e H + + 3MnO V 2+ →3Mn H 2 O + 5V e - MnO 4 - (aq) + V 2+ (aq) → Mn 2+ + V 5+

A 10.0 mL aliquot of a solution containing V 2+ ions is acidified, and 32.7 mL of a M MnO 4 - solution produces a light purple color. If the V 2+ was oxidized to V 5+, determine the molarity of the V 2+ in the original solution mL x M 32.7 mL M 5 mol3 mol x 5 mol V 2+ ________________ 3 mol MnO mol MnO 4 - x L solution _____________________ L solution = M V 2+ x 1 _______________________ L solution 4H-6 (of 13) 3MnO 4 - (aq) + 5V 2+ (aq) → 3Mn 2+ (aq) + 12H 2 O (l) + 5V 5+ (aq)24H + (aq) +

molar mass H 2 X = grams H 2 X _______________ moles H 2 X molar mass H 2 X = grams H 2 X ______________________ ? moles H 2 X Calculate the molar mass of a diprotic acid if grams of it are neutralized by mL of a M sodium hydroxide solution. 4H-7 (of 13)

H 2 X + NaOH →H(OH) + Na 2 X g x mol mL M 1 mol 2 mol 2 x 1 mol H 2 X _______________ 2 mol NaOH mol NaOH x L sol’n _____________________ L sol’n = mol H 2 X 2 Calculate the molar mass of a diprotic acid if grams of it are neutralized by mL of a M sodium hydroxide solution g H 2 X ________________________ mol H 2 X = 146 g/mol 4H-8 (of 13)

The molarity of an aluminum hydroxide solution is to be determined. An acid-base indicator is added to 10.0 mL of the aluminum hydroxide solution, and 12.5 mL of M hydrochloric acid produces a color change. Al(OH) 3 + HCl →AlCl 3 + H(OH) 10.0 mL x M 12.5 mL M 1 mol3 mol 3 x 1 mol Al(OH) 3 _________________ 3 mol HCl mol HCl x L solution __________________ L solution = M Al(OH) 3 x 1 ____________ L 3 4H-9 (of 13)

( ) x 8 Cinnabar ore contains S 2- ions. A g sample of the ore is dissolved in acid, then all the S 2- is oxidized by 20.4 mL of a M Cr 2 O 7 2- solution. Determine the percentage of S 2- in cinnabar ore. 4H-10 (of 13) ( ) x 3 K 2 Cr 2 O 7 (aq) + S 2- (aq) → K + (aq) + Cr 2 O 7 2- (aq) + S 2- (aq) → Cr 2 O 7 2- → Cr S 2- → S H 2 O14H + +6e e - Cr 3+ + S 8 +30

Cinnabar ore contains S 2- ions. A g sample of the ore is dissolved in acid, then all the S 2- is oxidized by 20.4 mL of a M Cr 2 O 7 2- solution. Determine the percentage of S 2- in cinnabar ore. 4H-11 (of 13) K 2 Cr 2 O 7 (aq) + S 2- (aq) → K + (aq) + Cr 2 O 7 2- (aq) + S 2- (aq) → 8Cr 2 O 7 2- → 16Cr S 2- → 3S H 2 O112H + +48e e - 48e H + + 8Cr 2 O S 2- →16Cr H 2 O + 3S e - Cr 3+ + S 8 +30

Cinnabar ore contains S 2- ions. A g sample of the ore is dissolved in acid, then all the S 2- is oxidized by 20.4 mL of a M Cr 2 O 7 2- solution. Determine the percentage of S 2- in cinnabar ore. x g20.4 mL M x 24 mol S 2- _________________ 8 mol Cr 2 O mol Cr 2 O 7 2- x L solution _______________________ L solution = g S 2- x g S 2- _____________ mol S g S 2- x 100 _______________ g ore = 14.1% S 2- in the ore 4H-12 (of 13) 8Cr 2 O 7 2- (aq) + 24S 2- (aq) → 16Cr 3+ (aq) + 56H 2 O (l) + 3S 8 (s)112H + (aq) + 8 mol 24 mol

Calculate the molarity of a sulfuric acid solution if 25.0 mL of it are neutralized by 33.5 mL of a M potassium hydroxide solution. H 2 SO 4 + KOH →H(OH) + K 2 SO mL x M 33.5 mL M 1 mol2 mol 2 x 1 mol H 2 SO 4 _________________ 2 mol KOH mol KOH x L solution ___________________ L solution = M H 2 SO 4 x 1 ____________ L 2 4H-13 (of 13)

MOLAR MASSES AND STOICHIOMETRIC CONVERSIONS 2 mol C (12.01 g/mol)= g 6 mol H (1.008 g/mol)= 6.048g 1 mol O (16.00 g/mol)=16.00 g g Calculate the molar mass of ethanol, C 2 H 5 OH 4I-1 (of 8) g C 2 H 5 OH = 1 mol C 2 H 5 OH

x 1 mol C 2 H 5 OH ______________________ g C 2 H 5 OH Calculate the number of ethanol molecules in 25.0 mL of pure ethanol. The density of the ethanol is g/mL mL C 2 H 5 OH x x molecules C 2 H 5 OH ________________________________________ 1 mol C 2 H 5 OH = 2.58 x molecules C 2 H 5 OH x g C 2 H 5 OH ____________________ 1 mL C 2 H 5 OH g C 2 H 5 OH = 1 mL C 2 H 5 OH 4I-2 (of 8)

x g C 2 H 5 OH ____________________ 1 mL C 2 H 5 OH x 1 mol C 2 H 5 OH ______________________ g C 2 H 5 OH Calculate the number of carbon atoms in a 10.0 mL sample of pure ethanol mL C 2 H 5 OH x 2 mol C __________________ 1 mol C 2 H 5 OH = 2.06 x atoms C x x atoms C ___________________________ 1 mol C 4I-3 (of 8)

x 40. mL C 2 H 5 OH ____________________ 100 mL vodka x 1 mol C 2 H 5 OH ______________________ g C 2 H 5 OH Calculate the number of ethanol molecules in 45.0 mL of 80. proof vodka. The density of the vodka is 0.92 g/mL mL vodka x x molecules C 2 H 5 OH ________________________________________ 1 mol C 2 H 5 OH = 1.9 x molecules C 2 H 5 OH x g C 2 H 5 OH ____________________ 1 mL C 2 H 5 OH 80. Proof vodka = 40.% C 2 H 5 OH by volume 100 mL vodka = 40. mL C 2 H 5 OH 4I-4 (of 8)

x g C 2 H 5 OH _______________________ mol C 2 H 5 OH Calculate the mass of one ethanol molecule, in grams. = x g x mol C 2 H 5 OH ________________________________________ x molecules C 2 H 5 OH 4I-5 (of 8) 1 molecule C 2 H 5 OH

x 1 mol O ____________ g O A metal oxide with the formula M 2 O 3 is 29.0% oxygen by mass. Calculate the molar mass of metal M g O x 2 mol M __________ 3 mol O = 58.8 g/mol 71.0 g M _________________ mol M 4I-6 (of 8) g M mol M Molar Mass of M = = 71.0 g M ? mol M = mol M

1 mol Na (22.99 g/mol)= g 1 mol N (14.01 g/mol)= g 3 mol O (16.00 g/mol)=48.00 g g THEORETICAL PERCENT COMPOSITION OF COMPOUNDS BY MASS NaNO 3 Calculate the percent composition by mass of sodium nitrate g Na  100 ___________________ g NaNO 3 % Na == % Na g O  100 ___________________ g NaNO 3 % O == % O g N  100 ___________________ g NaNO 3 % N == % N 4I-7 (of 8)

BaCl 2. 2H 2 O 1 mol Ba (137.3 g/mol)= g 2 mol Cl (35.45 g/mol)= g 2 mol H 2 O ( g/mol)=36.032g g Calculate the percentage by mass of water in barium chloride dihydrate g H 2 O  100 ____________________________ g BaCl 2. 2H 2 O % H 2 O == % H 2 O 4I-8 (of 8)

EMPIRICAL AND MOLECULAR FORMULA CALCULATIONS When a sample of a hydrocarbon is burned, 8.45 g CO 2 and 1.73 g H 2 O are produced. Calculate the empirical formula of the hydrocarbon. C x H y + O 2 →CO 2 + H 2 O 8.45 g CO 2 = g C x g C ________________ g CO g H 2 O= g Hx g H ________________ g H 2 O 4J-1 (of 9)

x 1 mol C _____________ g C = mol C g C x 1 mol H ____________ g H = mol H g H mol C _________________ mol H _________________ = 1.00 mol C= 1.00 mol H Empirical formula: CH EMPIRICAL AND MOLECULAR FORMULA CALCULATIONS When a sample of a hydrocarbon is burned, 8.45 g CO 2 and 1.73 g H 2 O are produced. Calculate the empirical formula of the hydrocarbon. 4J-2 (of 9)

A borane is any compound composed of boron and hydrogen. When a sample of a borane is burned, g B 2 O 3 and g H 2 O are produced. Calculate the empirical formula of the borane. B x H y + O 2 →B 2 O 3 + H 2 O g B 2 O 3 = g B x g B ________________ g B 2 O g H 2 O = g H x g H ________________ g H 2 O 4J-3 (of 9)

x 1 mol B _____________ g B = mol B g B x 1 mol H _____________ g H = mol H g H mol B ___________________ mol H _________________ = mol B= mol H Empirical formula: B 3 H 11 A borane is any compound composed of boron and hydrogen. When a sample of a borane is burned, g B 2 O 3 and g H 2 O are produced. Calculate the empirical formula of the borane. 4J-4 (of 9) x 3

When a 3.84 g sample of a compound containing C, H, and N is combusted, 7.34 g CO 2 and 2.51 g H 2 O are collected. Calculate the empirical formula of the compound. C x H y N z + O 2 →CO 2 + H 2 O + N a O b 7.34 g CO 2 = g C x g C ________________ g CO g H 2 O = g H x g H _____+__________ g H 2 O 3.84 g C x H y N z g C g H ________________________ g N 4J-5 (of 9)

When a 3.84 g sample of a compound containing C, H, and N is combusted, 7.34 g CO 2 and 2.51 g H 2 O are collected. Calculate the empirical formula of the compound. x 1 mol C _____________ g C = mol C g C x 1 mol H ____________ g H = mol H g H mol C _________________ mol H _________________ = 1.50 mol C= 2.51 mol H Empirical formula: C 3 H 5 N 2 x 1 mol N _____________ g N = mol N g N mol N ________________ = 1.00 mol N 4J-6 (of 9)

When a 2.75 g sample of a compound containing C, H, and O is combusted, 5.49 g CO 2 and 2.25 g H 2 O are collected. Calculate the empirical formula of the compound. C x H y O z + O 2 →CO 2 + H 2 O 5.49 g CO 2 = g C x g C ________________ g CO g H 2 O = g H x g H _________________ g H 2 O 2.75 g C x H y O z g C g H ________________________ g O 4J-7 (of 9)

When a 2.75 g sample of a compound containing C, H, and O is combusted, 5.49 g CO 2 and 2.25 g H 2 O are collected. Calculate the empirical formula of the compound. x 1 mol C ____________ g C = mol C g C x 1 mol H ____________ g H = mol H g H mol C _________________ mol H _________________ = 1.99 mol C= 4.00 mol H Empirical formula: C 2 H 4 O x 1 mol O _____________ g O = mol O g O mol O __________________ = 1.00 mol O 4J-8 (of 9)

C2H4OC2H4O 2 mol C (12.01 g/mol)= g 4 mol H (1.008 g/mol)=4.032g 1 mol O (16.00 g/mol)= g g 90 g/mol ________________ g/mol ≈ 2 Molecular formula: C 4 H 8 O 2 Calculate the molecular formula of the previous compound if it has a molar mass of about 90 g/mol. 4J-9 (of 9)