Chapter 3 Formulas, Equations, and Moles. Balancing Chemical Equations Alphabet – elemental...

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Chapter 3 Formulas, Equations, and Moles

Transcript of Chapter 3 Formulas, Equations, and Moles. Balancing Chemical Equations Alphabet – elemental...

Page 1: Chapter 3 Formulas, Equations, and Moles. Balancing Chemical Equations Alphabet – elemental symbols Words – chemical formulas Sentences – chemical equations.

Chapter 3

Formulas, Equations, and Moles

Page 2: Chapter 3 Formulas, Equations, and Moles. Balancing Chemical Equations Alphabet – elemental symbols Words – chemical formulas Sentences – chemical equations.

Balancing Chemical Equations• Alphabet – elemental symbols• Words – chemical formulas• Sentences – chemical equations (chemical reactions)

reactants products

limestone quicklime + gas

Calcium carbonate calcium oxide + carbon dioxide

CaCO3(s) CaO(s) + CO2(g)

Page 3: Chapter 3 Formulas, Equations, and Moles. Balancing Chemical Equations Alphabet – elemental symbols Words – chemical formulas Sentences – chemical equations.

Balancing Chemical Equations

• Chemical reactions include– Reactants– Products– Balanced – Law of Conservation of Mass

• # of atoms of an element on the reactant side must equal the # of atoms of that element on the product side.

– Indicate the state of matter of each chemical in the reaction (Chapter 4)

Page 4: Chapter 3 Formulas, Equations, and Moles. Balancing Chemical Equations Alphabet – elemental symbols Words – chemical formulas Sentences – chemical equations.

Balancing Chemical Equations

• Write the equation without coefficients• List the elements in each equation

– Secret: if the same polyatomic ion exists on both sides, keep it together

• Determine the # of each kind of atom on both sides• Balance atoms one element at a time by adjusting

coefficients– DO NOT ALTER THE FORMULA OF THE COMPOUND!!!!!

• Only coefficients can be altered– Secret:

• Balance atoms appearing only once on each side first.• Save compounds comprised of only one type of element till last.

• Reduce to lowest terms if necessary

Page 5: Chapter 3 Formulas, Equations, and Moles. Balancing Chemical Equations Alphabet – elemental symbols Words – chemical formulas Sentences – chemical equations.

Examples

• Balance the following equations:– Al(s) + Fe2O3(s) → Al2O3 (s) + Fe (l)

– Solid copper reacts with aqueous silver nitrate to form aqueous copper (II) nitrate and silver solid

– H3PO4 (l) → H2O (l) + P4O10 (s)

– C4H10(g) + O2 (g) → CO2(g) + H2O (g)

Page 6: Chapter 3 Formulas, Equations, and Moles. Balancing Chemical Equations Alphabet – elemental symbols Words – chemical formulas Sentences – chemical equations.

Avogadro’s Number and the Mole

• Meaning of a chemical reaction

2 C4H10(g) + 13 O2(g) → 8 CO2(g) + 10 H2O (g)

– 2 molecules of C4H10(g) reacts with 13 molecules of O2(g)

to form

8 molecule of CO2(g) and 10 molecules of H2O(g)

Page 7: Chapter 3 Formulas, Equations, and Moles. Balancing Chemical Equations Alphabet – elemental symbols Words – chemical formulas Sentences – chemical equations.

Avogadro’s Number and the Mole

• Molecule’s mass = the sum of the atomic

masses of the atoms making up the

molecule.

• m(C2H4O2) = 2·mC + 4·mH + 2·mO

» = 2·(12.01) + 4·(1.01) + 2·(16.00)

» = 60.06 amu

Page 8: Chapter 3 Formulas, Equations, and Moles. Balancing Chemical Equations Alphabet – elemental symbols Words – chemical formulas Sentences – chemical equations.

Avogadro’s Number of the Mole

• One mole (mol) of any substance contains 6.02 x 1023 (Avogadro’s Number) units of that substance.

• One mole (mol) of a substance is the gram mass value equal to the amu mass of the substance.– Calculated the same as amu’s for a molecule

Page 9: Chapter 3 Formulas, Equations, and Moles. Balancing Chemical Equations Alphabet – elemental symbols Words – chemical formulas Sentences – chemical equations.

Avogadro’s Number and the Mole

• Calculate the molar mass of the following:

– Fe2O3 (Rust)

– C6H8O7 (Citric acid)

– C16H18N2O4 (Penicillin G)

Page 10: Chapter 3 Formulas, Equations, and Moles. Balancing Chemical Equations Alphabet – elemental symbols Words – chemical formulas Sentences – chemical equations.

Avogadro’s Number and the Mole

• Methionine, an amino acid used by organisms to make proteins, is represented below. Write the formula for methionine and calculate its molar mass. (red = O; gray = C; blue = N; yellow = S; ivory = H)

Page 11: Chapter 3 Formulas, Equations, and Moles. Balancing Chemical Equations Alphabet – elemental symbols Words – chemical formulas Sentences – chemical equations.

Stoichiometry

• 4 Conversion units– Chemical formula– Balanced chemical equation

• Coefficients can read as;– # of molecules

– # of moles of that molecules

• Allows conversion between compounds in an equation

– Avogadro’s # - 6.02 x 1023 of X = 1 mole of X– Molar mass – how many grams of a substance = 1

mole of that substance

Page 12: Chapter 3 Formulas, Equations, and Moles. Balancing Chemical Equations Alphabet – elemental symbols Words – chemical formulas Sentences – chemical equations.

Stoichiometric Calculations

Page 13: Chapter 3 Formulas, Equations, and Moles. Balancing Chemical Equations Alphabet – elemental symbols Words – chemical formulas Sentences – chemical equations.

Avogadro’s Number and the Mole

• How many grams of oxygen are present in 5.961 x 1020 molecules of KClO3? How many atoms of oxygen are present?

Page 14: Chapter 3 Formulas, Equations, and Moles. Balancing Chemical Equations Alphabet – elemental symbols Words – chemical formulas Sentences – chemical equations.

Avogadro’s Number and the Mole

• Calculate the number of oxygen atoms in 29.34 g of sodium sulfate, Na2SO4. 

– A.  1.244 × 1023 O atoms– B.  4.976 × 1023 O atoms– C.  2.409 × 1024 O atoms– D.  2.915 × 1024 O atoms– E.  1.166 × 1025 O atoms

Page 15: Chapter 3 Formulas, Equations, and Moles. Balancing Chemical Equations Alphabet – elemental symbols Words – chemical formulas Sentences – chemical equations.

Problem

• Potassium dichromate, K2Cr2O7, is used in tanning leather, decorating porcelain and water proofing fabrics. Calculate the number of chromium atoms in 78.82 g of K2Cr2O7. 

– A.  9.490 × 1025 Cr atoms– B.  2.248 × 1024 Cr atoms– C.  1.124 × 1024 Cr atoms– D.  3.227 × 1023 Cr atoms– E.  1.613 × 1023 Cr atoms

Page 16: Chapter 3 Formulas, Equations, and Moles. Balancing Chemical Equations Alphabet – elemental symbols Words – chemical formulas Sentences – chemical equations.

Stoichiometry: Chemical Arithmetic

Page 17: Chapter 3 Formulas, Equations, and Moles. Balancing Chemical Equations Alphabet – elemental symbols Words – chemical formulas Sentences – chemical equations.

Stoichiometry: Equation Arithmetic

• Balance the following, and determine how

many moles of CO will react with 0.500

moles of Fe2O3.

Fe2O3(s) + CO(g) → Fe(s) + CO2(g)

Page 18: Chapter 3 Formulas, Equations, and Moles. Balancing Chemical Equations Alphabet – elemental symbols Words – chemical formulas Sentences – chemical equations.

Stoichiometry: Chemical Arithmetic

• Aqueous sodium hydroxide and chlorine

gas are combined to form aqueous sodium

hypochlorite (household bleach), aqueous

sodium chloride and liquid water.

– How many grams of NaOH are needed to react

with 25.0 g of Cl2?

Page 19: Chapter 3 Formulas, Equations, and Moles. Balancing Chemical Equations Alphabet – elemental symbols Words – chemical formulas Sentences – chemical equations.

Problem• Sulfur dioxide reacts with chlorine to produce thionyl chloride (used

as a drying agent for inorganic halides) and dichlorine monoxide (used as a bleach for wood, pulp and textiles).

SO2(g) + 2Cl2(g) → SOCl2(g) + Cl2O(g)

If 0.400 mol of Cl2 reacts with excess SO2, how many moles of Cl2O are formed? 

– A.  0.800 mol– B.  0.400 mol– C.  0.200 mol– D.  0.100 mol– E.  0.0500 mol

Page 20: Chapter 3 Formulas, Equations, and Moles. Balancing Chemical Equations Alphabet – elemental symbols Words – chemical formulas Sentences – chemical equations.

Problem

• Nitrogen gas and hydrogen gas are combined to form ammonia (NH3), an important source of fixed nitrogen that can be metabolized by plants, using the Haber process.

How many grams of nitrogen are needed to produce 325 grams of ammonia? 

– A.  1070 g– B.  535 g– C.  267 g– D.  178 g– E.  108 g

Page 21: Chapter 3 Formulas, Equations, and Moles. Balancing Chemical Equations Alphabet – elemental symbols Words – chemical formulas Sentences – chemical equations.

Lab Homework

• MISC 486 Problem Set 2 – Due

Page 22: Chapter 3 Formulas, Equations, and Moles. Balancing Chemical Equations Alphabet – elemental symbols Words – chemical formulas Sentences – chemical equations.

Yields of Chemical Reactions

• Yields of Chemical Reactions: If the actual

amount of product formed in a reaction is less

than the theoretical amount, we can calculate

a percentage yield.

100% yieldproduct lTheoretica

yieldproduct Actual yield%

Page 23: Chapter 3 Formulas, Equations, and Moles. Balancing Chemical Equations Alphabet – elemental symbols Words – chemical formulas Sentences – chemical equations.

Yield of Chemical Reactions

• Dichloromethane (CH2Cl2) is prepared by reaction of

methane (CH4) with chlorine (Cl2) giving hydrogen

chloride as a by-product. How many grams of

dichloromethane result from the reaction of 1.85 kg of

methane if the yield is 43.1%?

Page 24: Chapter 3 Formulas, Equations, and Moles. Balancing Chemical Equations Alphabet – elemental symbols Words – chemical formulas Sentences – chemical equations.

Problem

• What is the percent yield for the reaction

PCl3(g) + Cl2(g) → PCl5(g)

if 119.3 g of PCl5 ( MM = 208.2 g/mol) are formed when 61.3 g of Cl2 (  MM = 70.91 g/mol) react with excess PCl3? 

– A.  195%– B.  85.0%– C.  66.3%– D.  51.4%– E.  43.7%

Page 25: Chapter 3 Formulas, Equations, and Moles. Balancing Chemical Equations Alphabet – elemental symbols Words – chemical formulas Sentences – chemical equations.

Reactions with Limiting Amounts of Reactants

• Limiting Reagents: The extent to which a

reaction takes place depends on the reactant

that is present in limiting amounts—the limiting

reagent.

• Process– Convert each reactant into a single product– The one that forms the least is the limiting reactant– Complete all other calculations using the limiting reactant

Page 26: Chapter 3 Formulas, Equations, and Moles. Balancing Chemical Equations Alphabet – elemental symbols Words – chemical formulas Sentences – chemical equations.

Reactions with Limiting Amounts of Reactants

• Limiting Reagent Calculation: Lithium oxide is a drying

agent used on the space shuttle. If 80.0 kg of water is to

be removed and 65 kg of lithium oxide is available,

which reactant is limiting?

Li2O(s) + H2O(l) 2 LiOH(s)

• MM(Li2O) = 29.88 g/mol

• MM(H2O) = 18.02 g/mol

Page 27: Chapter 3 Formulas, Equations, and Moles. Balancing Chemical Equations Alphabet – elemental symbols Words – chemical formulas Sentences – chemical equations.

Reactions with Limiting Amounts of Reactants

• Limiting Reagent Calculation: Cisplatin is an anti-cancer agent prepared as follows:

K2PtCl4 + 2 NH3 Pt(NH3)2Cl2 + 2 KCl

If 10.0 g of K2PtCl4 and 10.0 g of NH3 are allowed to react: (a)

which is the limiting reagent? (b) How many grams of the excess reagent are consumed? (c) How many grams of cisplatin are formed?

MM(K2PtCl4) = 415.08 g/mol MM(NH3) = 18.04 g/mol

MM [Pt(NH3)2Cl2] = 299.9 g/mol

Page 28: Chapter 3 Formulas, Equations, and Moles. Balancing Chemical Equations Alphabet – elemental symbols Words – chemical formulas Sentences – chemical equations.

Problem

• What is the percent yield of H2O if 97.2 g CH4S reacts with 183 g of O2 to produce 58.5 g H2O according to the following chemical reaction? (Reaction may or may not be balanced.)

CH4S + O2 → CO2 + H2O + SO3

– Write the balanced chemical reaction?– Calculate the molar mass of CH4S? – Calculate the molar mass of H2O? – Calculate the molar mass of O2?– Calculate the theoretical yield of H2O?– What is the limiting reagent?– Calculate the percent yield of H2O?

Page 29: Chapter 3 Formulas, Equations, and Moles. Balancing Chemical Equations Alphabet – elemental symbols Words – chemical formulas Sentences – chemical equations.

Lab Homework

• MISC 486 Problem Set 4 - Due

Page 30: Chapter 3 Formulas, Equations, and Moles. Balancing Chemical Equations Alphabet – elemental symbols Words – chemical formulas Sentences – chemical equations.

Concentrations of Reactants in Solution: Molarity

• Molarity: The most common way of expressing the amount of a substance dissolved in a solution– Conversion factor between moles and volume

• It is important to note that the final volume of solution must be used, not volume of solvent.

solution of Liters

solute of Moles(M)Molarity

Page 31: Chapter 3 Formulas, Equations, and Moles. Balancing Chemical Equations Alphabet – elemental symbols Words – chemical formulas Sentences – chemical equations.

Solution Stoichiometry

• Known: molarity of solution, volume of solution, and balanced chemical equation

Page 32: Chapter 3 Formulas, Equations, and Moles. Balancing Chemical Equations Alphabet – elemental symbols Words – chemical formulas Sentences – chemical equations.

Concentrations of Reactants in Solution: Molarity

• How many moles of solute are present in

125 mL of 0.20 M NaHCO3?

• How many grams of solute would you use

to prepare 500.0 mL of 1.25 M NaOH?

Page 33: Chapter 3 Formulas, Equations, and Moles. Balancing Chemical Equations Alphabet – elemental symbols Words – chemical formulas Sentences – chemical equations.

Problem

• A 0.150 M sodium chloride solution is referred to as a physiological saline solution because it has the same concentration of salts as normal human blood. Calculate the mass of solute needed to prepare 275.0 mL of a physiological saline solution. 

– A.  41.3 g– B.  31.9 g– C.  16.1 g– D.  8.77 g– E.  2.41 g

Page 34: Chapter 3 Formulas, Equations, and Moles. Balancing Chemical Equations Alphabet – elemental symbols Words – chemical formulas Sentences – chemical equations.

Concentrations of Reactants in Solution: Molarity

• Solution Preparation– Determine the mass of solid needed to obtain the

desired # of moles– Mass the solid– Add it to a volumetric flask– Add water to the volumetric flask until it is about half-

full– Cap and shake to dissolve the solid– Add water to the line– Cap and shake

Page 35: Chapter 3 Formulas, Equations, and Moles. Balancing Chemical Equations Alphabet – elemental symbols Words – chemical formulas Sentences – chemical equations.

Diluting Concentrated Solutions

• Dilution: process of reducing a solution’s concentration by adding more solvent.– Key concept: # of moles remains constant

Page 36: Chapter 3 Formulas, Equations, and Moles. Balancing Chemical Equations Alphabet – elemental symbols Words – chemical formulas Sentences – chemical equations.

Diluting Concentrated Solutions

Concentrated solution + Solvent Dilute solution

Moles of solute (mol) = molarity (M) x volume (V)

Mconcentrated x Vconcentrated = Mdilute x Vdilute

only use if the initial solution and the final solution are the

same substance

Page 37: Chapter 3 Formulas, Equations, and Moles. Balancing Chemical Equations Alphabet – elemental symbols Words – chemical formulas Sentences – chemical equations.

Diluting Concentrated Solutions

• What volume of 18.0 M H2SO4 is required to

prepare 250.0 mL of 0.500 M aqueous H2SO4?

• What is the final concentration if 750 mL of 3.50

M glucose is diluted to a volume of 400.0 mL?

Page 38: Chapter 3 Formulas, Equations, and Moles. Balancing Chemical Equations Alphabet – elemental symbols Words – chemical formulas Sentences – chemical equations.

Problem

• Calcium chloride is used to melt ice and snow on roads and sidewalks and to remove water from organic liquids. Calculate the molarity of a solution prepared by diluting 165 mL of 0.688 M calcium chloride to 925.0 mL. 

– A.  3.86 M– B.  0.743 M– C.  0.222 M– D.  0.123 M– E.  0.114 M

Page 39: Chapter 3 Formulas, Equations, and Moles. Balancing Chemical Equations Alphabet – elemental symbols Words – chemical formulas Sentences – chemical equations.

Titration

• Titration: A technique for determining the concentration of a solution.– Process –

• a carefully measured volume of an unknown solution is allowed to react with of a standard solution (concentration is known).

• The volume of the known is measured.

• Stoichiometry calculations are performed

Page 40: Chapter 3 Formulas, Equations, and Moles. Balancing Chemical Equations Alphabet – elemental symbols Words – chemical formulas Sentences – chemical equations.

Titration

• What is the molarity of a sulfuric acid solution if a

25.0 mL sample is titrated to equivalence with

50.0 mL of 0.150 M potassium hydroxide solution?

H2SO4(aq) + KOH(aq) K2SO4(aq) + H2O(l)

Page 41: Chapter 3 Formulas, Equations, and Moles. Balancing Chemical Equations Alphabet – elemental symbols Words – chemical formulas Sentences – chemical equations.

Problem

• How many milliliters of 1.58 M HCl are needed to react completely with 23.2 g of NaHCO3 ( MM  = 84.02 g/mol)?

HCl(aq) + NaHCO3(s) → NaCl(s) + H2O(l) + CO2(g) 

– A.  638 mL– B.  572 mL– C.  536 mL– D.  276 mL– E.  175 mL

Page 42: Chapter 3 Formulas, Equations, and Moles. Balancing Chemical Equations Alphabet – elemental symbols Words – chemical formulas Sentences – chemical equations.

Lab Homework

• MISC 486 Problem Set 5 - Due

Page 43: Chapter 3 Formulas, Equations, and Moles. Balancing Chemical Equations Alphabet – elemental symbols Words – chemical formulas Sentences – chemical equations.

Percent Composition and Empirical Formulas

• Percent Composition: Identifies the elements

present in a compound as a mass percent of the

total compound mass.

• The mass percent is obtained by dividing the

mass of each element by the total mass of a

compound and converting to percentage.

Page 44: Chapter 3 Formulas, Equations, and Moles. Balancing Chemical Equations Alphabet – elemental symbols Words – chemical formulas Sentences – chemical equations.

Percent Composition and Empirical Formulas

• Sugar is 42.1% C, 6.4% H, and 51.5% O

• Meaning – out of 100 g of sugar 42.1 g of it is due to C

Page 45: Chapter 3 Formulas, Equations, and Moles. Balancing Chemical Equations Alphabet – elemental symbols Words – chemical formulas Sentences – chemical equations.

Percent Composition and Empirical Formulas

• From the percent composition empirical formulas are developed

– empirical formula gives the smallest whole

number ratio of the atoms of each element in

a compound.

• Same as ionic formula for ionic compounds

Page 46: Chapter 3 Formulas, Equations, and Moles. Balancing Chemical Equations Alphabet – elemental symbols Words – chemical formulas Sentences – chemical equations.

Percent Composition and Empirical Formulas

• Compound Formula Empirical Formula

• Hydrogen H2O2 OH

peroxide

• Benzene C6H6 CH

• Ethylene C2H4 CH2

• Propane C3H8 C3H8

Page 47: Chapter 3 Formulas, Equations, and Moles. Balancing Chemical Equations Alphabet – elemental symbols Words – chemical formulas Sentences – chemical equations.

Percent Composition and Empirical Formulas

• A compound’s empirical

formula can be determined

from its percent composition.

• A compound’s molecular

formula is determined from the

molar mass and empirical

formula.

Page 48: Chapter 3 Formulas, Equations, and Moles. Balancing Chemical Equations Alphabet – elemental symbols Words – chemical formulas Sentences – chemical equations.

Percent Composition and Empirical Formulas

• A compound was analyzed to be 82.67% carbon

and 17.33% hydrogen by mass. An osmotic

pressure experiment determined that its molar

mass is 58.11 g/mol.

What is the empirical formula and molecular

formula for the compound?

Page 49: Chapter 3 Formulas, Equations, and Moles. Balancing Chemical Equations Alphabet – elemental symbols Words – chemical formulas Sentences – chemical equations.

Problem

• Gadolinium oxide, a colorless powder which absorbs carbon dioxide from the air, contains 86.76 mass % Gd. Determine its empirical formula. 

– A.  Gd2O3

– B.  Gd3O2

– C.  Gd3O4

– D.  Gd4O3

– E.  GdO

Page 50: Chapter 3 Formulas, Equations, and Moles. Balancing Chemical Equations Alphabet – elemental symbols Words – chemical formulas Sentences – chemical equations.

Problem

• Hydroxylamine nitrate contains 29.17 mass % N, 4.20 mass % H, and 66.63 mass O. If its molar mass is between 94 and 98 g/mol, what is its molecular formula? 

– A.  NH2O5

– B.  N2H4O4

– C.  N3H3O3

– D.  N4H8O2

– E.  N2H2O4

Page 51: Chapter 3 Formulas, Equations, and Moles. Balancing Chemical Equations Alphabet – elemental symbols Words – chemical formulas Sentences – chemical equations.

Determining Empirical Formulas: Elemental Analysis

• Combustion analysis is one of the most common methods for determining empirical formulas.

• A weighed compound is burned in oxygen and its products analyzed by a gas chromatogram.

• It is particularly useful for analysis of hydrocarbons.

Page 52: Chapter 3 Formulas, Equations, and Moles. Balancing Chemical Equations Alphabet – elemental symbols Words – chemical formulas Sentences – chemical equations.

Determining Empirical Formulas: Elemental Analysis

• Terephthalic acid, used in the production of polyester fibers and films, is composed of carbon, hydrogen, and oxygen. When 0.6943 g of terephthalic acid was subjected to combustion analysis it produced 1.471 g CO2 and 0.226 g H2O. What is its empirical formula? 

– A.  C2H3O4

– B.  C3H4O2

– C.  C4H3O2

– D.  C5H12O4

– E.  C2H2O

Page 53: Chapter 3 Formulas, Equations, and Moles. Balancing Chemical Equations Alphabet – elemental symbols Words – chemical formulas Sentences – chemical equations.

Optional Homework

• Text –3.30, 3.32, 3.34, 3.36, 3.40, 3.42, 3.54, 3.58, 3.62, 3.64, 3.70, 3.72, 3.74, 3.76, 3.80, 3.82, 3.86, 3.88, 3.90, 3.92, 3.94, 3.96, 3.100, 3.106, 3.108, 3.112, 3.116

• Chapter 3 Homework – from website

Page 54: Chapter 3 Formulas, Equations, and Moles. Balancing Chemical Equations Alphabet – elemental symbols Words – chemical formulas Sentences – chemical equations.

Required Homework

• MISC 486 Problem Set 3 – Due

• Assignment #3