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THE MOLE Chapter 11

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THE MOLE. Chapter 11. Writing and balancing chemical equations. Chemical equation : identities and quantities of substances involved in chemical/physical change Balance using Law of Conservation of Mass and Law of Definite Composition same # of atoms on each side - PowerPoint PPT Presentation

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THE MOLEChapter 11

• Chemical equation: identities and quantities of substances involved in chemical/physical change– Balance using Law of Conservation of

Mass and Law of Definite Composition•same # of atoms on each side •fixed ratio of elements in

compound

Writing and balancing chemical equations

1. translate statement: reactants products

2. balance atoms using stoichiometric coefficients

3. Adjust stoichiometric coefficients (if necessary)– Smallest whole # preferred

4. Check5. Specify states of matter

Writing and balancing chemical equations

Samples

• Stoichiometry: study of quantitative aspects of chemical formulas/reactions

• Mole: unit chemists use to count chemical entities by weighing them

11.1 Measuring Matter

• The Mole– SI Unit for the amount of a substance– The number of particles equal to the

number of atoms in exactly 12.0 grams of carbon-12

– Also called Avogadro’s number• 1 mol = 6.02 x 1023 particles

mole - amount of substance that contains same # of entities as atoms in 12g of carbon-12.

1 mol contains 6.022x1023 entities

Avogadro’s number (N)

The mole

1 mole H2O contains 6.022 x 1023 H2O molecules

1 mole KNO3 contains 6.022x1023 KNO3 formula units

1 mole Hg contains 6.022x1023 Hg atoms

Mole represents large quantity of microscopic particles.

mass of 1 mass of 1Substance atom (molecule) mole of atoms (molecules)CaCO3 100.09 amu 100.09 gO2 32.00 amu 32.00 gH2O 18.02 amu 18.02 gCopper 63.55 amu 63.55 g

Can weigh out grams using scale

Use mass to ‘count’ entities.

6.022 x 1023 entities

•Molar mass (M) (gmw)- mass of 1 mole of entities–M (g/mol) numerically equal to formula weight (amu)

CH4 = 1(12.10 g/mol) + 4(1.008 g/mol) = 16.04 g/mol

The mole

Relating moles to chemical formulasGlucose C6H12O6 ( M = 180.16 g/mol)

Oxygen (O)

Mass/mole of compound

6 atoms

96.00 g

Table 3.2

Carbon (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

180.16 g/mole

Interconverting Moles, Mass, and # of Chemical Entities

Mass (g) = no. of moles x gmw

1 mol

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

1 mol

Mass % of element X =

Mass Percent from Chemical Formula

moles X in formula 1 mol compound

(molar mass of X)

molar mass of compoundx 100%

i.e. Mass % of H in H2O = 2 mol H 1.008 g H 1 mol H2O 1 mol H

18.02 g H2O 1 mol H2O

x 100%

= 11.19% H by mass

Particles

• Atoms – Single elements

• Formula units– Ionically bonded compounds

• Molecules– Covalent bonded compounds

Figure 3.6

macro

micro

Avogadro’s # takes us to/from macroscopic/microscopicLaw of Conservation of Mass

(6.02 x 1023

molecules)(6.02 x 1023

molecules)(1.20 x 1024

molecules)

Particle Mole Problems

• REMEMBER– The number of particles in 1 mole of ANY

substance is ALWAYS the same (6.02 x 1023)1 mol = 6.02 x 1023 particles

– How many molecules are in 2.2 moles of water?

1 mol H2O = 6.02 x 1023 particles

Use this as your conversion factor!!!Use this as your conversion factor!!!1 mol H2O

6.02 x 1023 molecules

6.02 x 1023 molecules 1 mol H2O

OROR

Calculating amounts of reactant and product

• Balanced equation needed for stoichiometric calculations– Ratios of reactants/products to calculate amounts of

reactants/products

Particle Mole Practice Continued

How many moles of sodium carbonate contain 7.9 x 1024 formula units?

2.2 mol H2O

1

Take your given value and put it

over ONE.

Multiply by the conversion factor that allows you to cancel out the

top unit. Leaving you with the unit you WANTED.6.02 x 1023 molecules 1 mol H2OX = 1.3244 x 1024

ANSWER: 1.3 x 1024 molecules of H2O

11.2 Mass and the Mole

• Atomic mass – The mass of an atom relative to the mass assigned

to carbon-12

• Molar mass– The mass in grams of one mole of any pure

substance– Use the average atomic mass off the periodic table

• Molar mass of an element– Atomic mass in grams per mole (g/mol)

• Molar mass of oxygen= 16.00 g/moL

• Molar mass of helium= 4.00 g/moL

• Calculate moles of O2 consumed when 10 mol of H2O are produced (using balanced equation from Table 3.5)?

– C3H8(g) + 5 O2(g) 3 CO2(g) + 4 H2O(g)

• Calculate mass of CO2 produced burning 1.00 g butane (C4H10).

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

• Practice, practice, practice!!!

Calculating amounts of reactant and product

Mass & Mole Problems

• How many moles are in 82.2g of aluminum?

1 mole of Al = 26.98g Al

Mass & Mole Problems

• How many grams are in 3.5 mol of neon?

1 mole of Ne = 20.18g Ne

11.3 Moles of Compounds

• Formula mass – the sum of the atomic masses of all the atoms in a compound

H2O

H: 2 x 1.01amu = 2.02 amuO: 1 x 16.00amu = 16.00 amu

18.02 amu

• Limiting reagent- reactant that forms fewer moles of product– Limits amount of product formed (chair

analogy)– To decide which reagent is limiting

reagent:1. If information given in g, convert to

moles2. Use ratio from balanced equation to

find moles of final product possibly produced3. Reagent that produces least possible

moles of product is limiting reagent

Chemical rxts. that involve limiting reagents

What mass of NH3 is produced from the rxt. of 1.00 g H2(g) w/ 1.00 g N2(g)?

(Info on both reactants given)

3H2(g) + N2(g) 2NH3(g)

• Is H2 or O2 the limiting reagent?– H2

Chemical rxts. that involve limiting reagents

Microscopic Picture

Moles of Compounds Continued

• Molar mass – The mass in grams of 1 mole of a

substance

– Formula mass of H2O = 18.02 amu

– Molar mass of H2O = 18.02g/mol

1 mol of H2O = 18.02 g = 6.02 x 1023 molecules

Mass Mole Problems

• Changing the mass to moles or vice versa using the molar mass

• How many moles are in 11.2g of NaCl?1. Determine the molar mass.

- Na: 1 x 23.00 = 23.00g- Cl: 1 x 35.45 = 35.45g

= 58.45g/mol

Therefore, 1 mol NaCl = 58.45g

Mass Mole Problems

2. Convert between the molar mass and the moles.

1 mol of NaCl = 58.45g1 mol of NaCl58.45g NaCl

58.45g NaCl 1 mol of NaCl

OROR

11.2g NaCl1

Take your given value and put it

over ONE.

Multiply by the conversion factor that allows you to cancel out the

top unit. Leaving you with the unit you WANTED.1 mol NaCl

58.45g NaClX =0.1916167665 mol NaCl

ANSWER: 0.192 mol NaCl

More Practice

• What is the mass of 2.50 mol of NaCl?– Find the molar mass…

1 mol of NaCl = 58.45g

1 mol of NaCl58.45g NaCl

58.45g NaCl 1 mol of NaCl

OROR

2.50 mol NaCl1

Take your given value and put it

over ONE.

Multiply by the conversion factor that allows you to cancel out the

top unit. Leaving you with the unit you WANTED.58.45g NaCl

1 mol NaClX = 146.125g NaCl

ANSWER: 146.13g NaCl

Multi-Step Conversions

• Mass-Particleg mol particles

• Particle-MassParticles mol g

• What is the mass of 8.2 x 1022 atoms of calcium?

1 mol Ca = 6.02 x 1023 atoms Ca1 mol Ca = 40.08g Ca

Mole Volume Problems

• Equal volumes of gases at the same temperature and pressure contain the same number of particles.

• Molar volume– The volume of 1 mol of gas at standard

conditions (STP)– STP

• standard temperature and pressure: 0oC and 1 atm

– 1 mol = 22.4 liters

Mole Volume Practice Problem

• What is the volume of 0.35 moles of helium gas at STP?

0.35 mol He1

Take your given value and put it

over ONE.

Multiply by the conversion factor that allows you to cancel out the

top unit. Leaving you with the unit you WANTED.22.4 L

1 mol HeX = 7.84L He

ANSWER: 7.84L He

Percent Composition

• Percent Composition– The percent by mass of each element

in a compound

mass of elementmass of compoundX 100= % composition

Percent Composition Problem

• Calculate the percent composition of hydrogen in water

H2O

H: 2 x 1.01amu = 2.02 amuO: 1 x 16.00amu = 16.00 amu

18.02 amu