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Stoichiometry of Formulas and Equations

Chapter 3

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Mole - Mass Relationships in Chemical Systems

3.5 Fundamentals of Solution Stoichiometry

3.1 The Mole

3.2 Determining the Formula of an Unknown Compound

3.3 Writing and Balancing Chemical Equations

3.4 Calculating the Amounts of Reactant and Product

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mole(mol) - the amount of a substance that contains the same number of entities as there are atoms in exactly 12 g of carbon-12.

This amount is 6.022x1023. The number is called Avogadro’s number and is abbreviated as N.

One mole (1 mol) contains 6.022x1023 entities (to four significant figures)

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12 red marbles @ 7g each = 84g12 yellow marbles @4g each=48g 55.85g Fe = 6.022 x 1023 atoms Fe

32.07g S = 6.022 x 1023 atoms S

Figure 3.1 Counting objects of fixed relative mass.

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Water18.02 gCaCO3

100.09 g

Oxygen32.00 g

Copper63.55 g

One mole of common substances.

Figure 3.2

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Table 3.1 Summary of Mass Terminology

Term Definition Unit

Isotopic mass Mass of an isotope of an element amu

Atomic mass

Molecular (or formula) mass

(also called molecular weight)

Molar mass (M)

(also called atomic weight)

(also called gram-molecular weight)

amu

amu

g/mol

Average of the masses of the naturally occurring isotopes of an element weighted according to their abundance

Sum of the atomic masses of the atoms (or ions) in a molecule (or formula unit)

Mass of 1 mole of chemical entities (atoms, ions, molecules, formula units)

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Information Contained in the Chemical Formula of Glucose C6H12O6 ( M = 180.16 g/mol)

Oxygen (O)

Mass/mole of compound

6 atoms

96.00 g

Table 3.2Carbon (C) Hydrogen (H)

Atoms/moleculeof compound

Moles of atoms/mole of compound

Atoms/mole ofcompound

Mass/moleculeof compound

6 atoms 12 atoms

6 moles of atoms

12 moles of atoms

6 moles of atoms

6(6.022 x 1023) atoms

12(6.022 x 1023) atoms

6(6.022 x 1023) atoms

6(12.01 amu) =72.06 amu

12(1.008 amu) =12.10 amu

6(16.00 amu) =96.00 amu

72.06 g 12.10 g

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Interconverting Moles, Mass, and Number of Chemical Entities

Mass (g) = no. of moles x no. of grams

1 mol

No. of moles = mass (g) xno. of grams

1 mol

No. of entities = no. of moles x6.022x1023 entities

1 mol

No. of moles = no. of entities x 6.022x1023 entities

1 mol

g

M

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MASS(g)of elementMASS(g)

of element

AMOUNT(mol)of element

AMOUNT(mol)of element

ATOMSof elementATOMS

of element

Summary of the mass-mole-number relationships for

elements.

M (g/mol)

Avogadro’s number(atoms/mol)

Figure 3.3

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Sample Problem 3.1 Calculating the Mass and the Number of Atoms in a Given Number of Moles of an Element

PROBLEM: (a) A graduated cylinder contains 32.0 cm3 of mercury. If the density of mercury at 25 ºC is 13.534 g/cm3, how many moles of mercury are in the cylinder?(b) How many atoms of mercury are there?

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MASS(g)of compoundMASS(g)

of compound

AMOUNT(mol)of compound

AMOUNT(mol)of compound

MOLECULES(or formula units)

of compound

MOLECULES(or formula units)

of compound

AMOUNT(mol)of elements in

compound

AMOUNT(mol)of elements in

compound

Avogadro’s number(molecules/mol)

M (g/mol)

chemicalformula

Summary of the mass-mole-number relationships for

compounds.Figure 3.3

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Sample Problem 3.2 Calculating the Moles and Number of Formula Units in a Given Mass of a Compound

PROBLEM: You have 16.5 g of the common compound oxalic acid, C2H2O4.

(a) How many moles are represented by this mass?

(b) How many molecules of oxalic acid are in 16.5 g?

(c) How many atoms of carbon are in 16.5 g oxalic acid?

(d) Calculate the mass of one molecule of oxalic acid.

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Mass % of element X =

atoms of X in formula x atomic mass of X (amu)

molecular (or formula) mass of compound(amu)x 100

Mass % of element X =

moles of X in formula x molar mass of X (amu)

molecular (or formula) mass of compound (amu)x 100

Mass percent from the chemical formula

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Sample Problem 3.3 Calculating the Mass Percents of Elements in a Sample of Compound

PROBLEM: Calculate the percentage composition of C12H22O11

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Empirical and Molecular Formulas

Empirical Formula -

Molecular Formula -

The simplest formula for a compound that agrees with the elemental analysis and gives rise to the smallest set of whole numbers of atoms.

The formula of the compound as it exists, it may be a multiple of the empirical formula.

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Sample Problem 3.4 Determining the Empirical Formula from Masses of Elements

PROBLEM: Mercury forms a compound with chlorine that is 73.9 % mercury and 26.1 % chlorine by mass. What is the empirical formula of this chloride of mercury?

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Sample Problem 3.5 Determining a Molecular Formula from Elemental Analysis and Molar Mass

PROBLEM: Eugenol is the active compound of oil of cloves. It is composed of carbon, hydrogen and oxygen. Given the following information for eugenol:

C 73.14 %, H 7.37 %, molecular mass = 164.2 g/mol

(a) Determine the empirical formula of eugenol.

(b) Determine the molecular formula.

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Combustion train for the determination of the chemical composition of organic compounds.

Figure 3.4

CnHm + (n+ ) O2 = n CO(g) + H2O(g)m2

m2

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Sample Problem 3.6 Determining a Molecular Formula from Combustion Analysis

PROBLEM: Isopropyl alcohol (M = 60.09 g/mol), a substance sold as rubbing alcohol, is composed of C, H, and O. Combustion of 0.255 g of isopropyl alcohol produces 0.561 g CO2 and 0.306 g H2O. Determine the empirical and molecular formulas of isopropyl alcohol.

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Table 3.3 Some Compounds with Empirical Formula CH2O(Composition by Mass: 40.0% C, 6.71% H, 53.3% O)

NameMolecular Formula

Whole-Number Multiple

M(g/mol) Use or Function

formaldehyde

acetic acid

lactic acid

erythrose

ribose

glucose

CH2O

C2H4O2

C3H6O3

C4H8O4

C5H10O5

C6H12O6

12

3

4

5

6

30.03

60.05

90.09

120.10

150.13

180.16

disinfectant; biological preservative

acetate polymers; vinegar(5% soln)

part of sugar metabolism

sour milk; forms in exercising muscle

component of nucleic acids and B2

major energy source of the cell

CH2O C2H4O2 C3H6O3 C4H8O4 C5H10O5 C6H12O6

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Table 3.4 Two Pairs of Constitutional Isomers

Property Ethanol Dimethyl Ether

M(g/mol)

Boiling Point

Density at 200C

Structural formulas

46.07

78.50C

0.789 g/mL(liquid)

46.07

-250C

0.00195 g/mL(gas)

Butane 2-Methylpropane

C4H10 C2H6O

58.12 58.12

Space-filling models

-0.50C

0.579 g/mL(gas)

-11.060C

0.549 g/mL(gas)

C C C C

H H

H

H H

H

H

H

H

H

C C C

H H

H

HC

H

H

H

H H

H

C C

H H

H

H H

OH C O

H

H

H

C

H

H

H

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The formation of HF gas on the macroscopic and molecular levels.

Figure 3.6

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translate the statement

balance the atoms

specify states of matter

adjust the coefficients

check the atom balance

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Sample Problem 3.7 Balancing Chemical Equations

PROBLEM: Write balanced equations for the following reactions:

(a) P4(s) + Cl2(g) → PCl3(l)

(b) Combustion of propane, C3H8

(c) Combustion of butane, C4H10

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MASS(g)of compound A

MASS(g)of compound A

AMOUNT(mol)of compound A

AMOUNT(mol)of compound A

MOLECULES(or formula units)

of compound A

MOLECULES(or formula units)

of compound A

Avogadro’s number(molecules/mol)

M (g/mol) of compound A

molar ratio from

balanced equation

Summary of the mass-mole-number relationships in a chemical reaction.

Figure 3.8

MASS(g)of compound B

MASS(g)of compound B

AMOUNT(mol)of compound B

AMOUNT(mol)of compound B

MOLECULES(or formula units)

of compound B

MOLECULES(or formula units)

of compound B

Avogadro’s number(molecules/mol)

M (g/mol) of compound B

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Sample Problem 3.8 Calculating Amounts of Reactants and Products

PROBLEM: In the balanced reaction between P4(s) and Cl2(g) to give PCl3(l):

(a) calculate the mass of Cl2 required if all of 1.45 g of P4 is to react.

(b) What mass of PCl3(l) is formed?

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Sample Problem 3.9 Calculating Amounts of Reactants and Products

PROBLEM: An alloy has the composition: 93.7 % Al and 6.3 % Cu by mass and density = 2.85 g/cm3. A piece of alloy of volume 0.691 cm3 reacts with excess HCl(g) to afford aluminum chloride and hydrogen gas only. Assuming all of the Al, but no Cu, reacts, calculate the mass of H2 obtained.

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Sample Problem 3.10 Writing an Overall Equation for a Reaction Sequence

PROBLEM: Pure TiO2 is the pigment commonly used in white paint. It can be purified by reacting crude TiO2 with elemental carbon in the presence of chlorine. The resulting TiCl4 is a liquid, which can be purified by distillation and then reacted with oxygen to reproduce pure TiO2.

2TiO2(s) + 3C(s) + 4Cl2(g) → 2TiCl4(l) + CO2(g) + 2CO(g)

TiCl4(l) + O2(g) → TiO2(s) + 2Cl2(g)

(a) What is the overall balanced equation for this reaction?

(b) How much carbon is required to produce 1.00 kg of pure TiO2(s)?

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An ice cream sundae analogy for limiting reactions.Figure 3.10

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Table 3.5 Information Contained in a Balanced Equation

Viewed in Terms of

ReactantsC3H8(g) + 5O2(g)

Products3CO2(g) + 4H2O(g)

molecules 3 molecules CO2 + 4 molecules H2O 1 molecule C3H8 + 5 molecules O2

amount (mol) 1 mol C3H8 + 5 mol O2 3 mol CO2 + 4 mol H2O

44.09 amu C3H8 + 160.00 amu O2mass (amu) 132.03 amu CO2 + 72.06 amu H2O

mass (g) 44.09 g C3H8 + 160.00 g O2 132.03 g CO2 + 72.06 g H2O

total mass (g) 204.09 g 204.09 g

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Sample Problem 3.11 Calculating Amounts of Reactant and Product in Reactions Involving a Limiting Reactant

PROBLEM: In the balanced reaction between P4(s) (125 g) and Cl2(g) (323 g) to give PCl3(l), calculate the mass of PCl3(l) that can be formed.

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A + B(reactants)

C(main product)

D(side products)

The effect of side reactions on yield.Figure 3.11

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Sample Problem 3.12 Calculating Percent Yield

PROBLEM: In the balanced reaction between P4(s) (125 g) and Cl2(g) (323 g) to give PCl3(l), the mass of PCl3(l) that is actually isolated is 372 g. Calculate the percentage yield for the reaction.

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Sample Problem 3.13 Calculating the Molarity of a Solution

PROBLEM: Calculate the molarity of a solution made by dissolving 23.4 g of sodium sulfate (Na2SO4) in enough water to form 125 mL of solution.

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Summary of mass-mole-number-volume

relationships in solution.

Figure 3.12MASS (g)

of compoundin solution

AMOUNT (mol)of compound

in solution

MOLECULES(or formula units)

of compoundin solution

VOLUME (L)of solution

M (g/mol)

M (g/mol)Avogadro’s number(molecules/mol)

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Sample Problem 3.14 Calculating Mass of Solute in a Given Volume of Solution

PROBLEM: How many grams of Na2SO4 are required to make 0.350 L of 0.500 M Na2SO4?

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Laboratory preparation of molar solutions.Figure 3.12

A•Weigh the solid needed.•Transfer the solid to a volumetric flask that contains about half the final volume of solvent. B Dissolve the solid

thoroughly by swirling.

C Add solvent until the solution reaches its final volume.

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Converting a concentrated solution to a dilute solution.Figure 3.13

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Sample Problem 3.15 Preparing a Dilute Solution from a Concentrated Solution

PROBLEM: How many milliliters of 3.0 M H2SO4 are needed to make 450 mL of 0.1 M H2SO4?

MdilxVdil = #mol solute = MconcxVconc

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Sample Problem 3.16 Calculating Amounts of Reactants and Products for a Reaction in Solution

PROBLEM: The following is an example of an acid-base neutralization reaction that we will meet in the next chapter:

Ca(OH)2 (s) + 2HNO3 (aq) → Mg(NO3)2 (aq) + 2H2O (l)

How many grams of Ca(OH)2 are needed to neutralize (react with completely) 25.0 mL of 0.100 M HNO3?

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Sample Problem 3.17 Solving Limiting-Reactant Problems for Reactions in Solution

PROBLEM: A sample of 70.5 mg of potassium phosphate is added to 15.0 mL of 0.050 M silver nitrate, resulting in the formation of a precipitate of silver phosphate:

K3PO4 (aq) + 3AgNO3 (aq) → Ag3PO4 (s) + 2KNO3 (aq)

(a) Calculate the theoretical yield, in grams, of the precipitate that forms.

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Figure 3.15 An overview of the key mass-mole-number stoichiometry relationships.