Chapter 10 The MOLE. Describing Chemical Equations Balanced chemical equations give us a ratio of...

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Chapter 10 Chapter 10 The MOLE The MOLE

Transcript of Chapter 10 The MOLE. Describing Chemical Equations Balanced chemical equations give us a ratio of...

Page 1: Chapter 10 The MOLE. Describing Chemical Equations Balanced chemical equations give us a ratio of particles Balanced chemical equations give us a ratio.

Chapter 10Chapter 10The MOLEThe MOLE

Page 2: Chapter 10 The MOLE. Describing Chemical Equations Balanced chemical equations give us a ratio of particles Balanced chemical equations give us a ratio.

Describing Chemical EquationsDescribing Chemical Equations

Balanced chemical Balanced chemical equations give us a equations give us a ratio of particlesratio of particles

4 Fe + 3 O4 Fe + 3 O22 2 2 FeFe22OO3 3

A ratio of particles is A ratio of particles is not practical from not practical from our perspective our perspective

Page 3: Chapter 10 The MOLE. Describing Chemical Equations Balanced chemical equations give us a ratio of particles Balanced chemical equations give us a ratio.

Describing Chemical EquationsDescribing Chemical Equations

Atoms and molecules Atoms and molecules are very, very small, are very, very small, but we still need to but we still need to “count” them.“count” them.

Scientists instead Scientists instead weigh out an weigh out an approximate number approximate number atomsatoms

Page 4: Chapter 10 The MOLE. Describing Chemical Equations Balanced chemical equations give us a ratio of particles Balanced chemical equations give us a ratio.

The MoleThe Mole

The The molemole is a quantity; is a quantity; it is used to “count” it is used to “count” atoms and molecules.atoms and molecules.

A mole contains 6.02 x A mole contains 6.02 x 10102323 things things

This is Avogadro’s This is Avogadro’s number = 6.02214199 number = 6.02214199 × 10× 102323

Page 5: Chapter 10 The MOLE. Describing Chemical Equations Balanced chemical equations give us a ratio of particles Balanced chemical equations give us a ratio.

1 mole of Al atoms = 6.02 x 101 mole of Al atoms = 6.02 x 102323 Al atomsAl atoms

1 mole H1 mole H22O molecules = 6.02 x O molecules = 6.02 x 10102323 H H22O moleculesO molecules

1 mole jelly donuts = 6.02 x 101 mole jelly donuts = 6.02 x 102323 jelly donutsjelly donuts

1 mole of = 6.02 x 101 mole of = 6.02 x 102323

Page 6: Chapter 10 The MOLE. Describing Chemical Equations Balanced chemical equations give us a ratio of particles Balanced chemical equations give us a ratio.

6.02 x 106.02 x 102323 is a VERY is a VERY

large large numbernumber

Page 7: Chapter 10 The MOLE. Describing Chemical Equations Balanced chemical equations give us a ratio of particles Balanced chemical equations give us a ratio.

602,000,0602,000,000,000,0000,000,000,000,000,0,000,000,

000000

Page 8: Chapter 10 The MOLE. Describing Chemical Equations Balanced chemical equations give us a ratio of particles Balanced chemical equations give us a ratio.

Avogadro’s Number factsAvogadro’s Number facts

Avogadro's number of inches is Avogadro's number of inches is 1,616,434 light years, or across our 1,616,434 light years, or across our galaxy and back 8 times. galaxy and back 8 times.

Avogadro's number of seconds is about Avogadro's number of seconds is about 19 quadrillion years, 4,240,666 times 19 quadrillion years, 4,240,666 times the age of the earth, or 954,150 times the age of the earth, or 954,150 times the age of the universe itself. the age of the universe itself.

Avogadro's number of cents could Avogadro's number of cents could repay the repay the United States National DebtUnited States National Debt 86 million times. 86 million times.

Avogadro's number of kilograms is just Avogadro's number of kilograms is just over 20 times the mass of the earthover 20 times the mass of the earth

Page 9: Chapter 10 The MOLE. Describing Chemical Equations Balanced chemical equations give us a ratio of particles Balanced chemical equations give us a ratio.

Measuring Numbers of AtomsMeasuring Numbers of Atoms

We cannot count 6.02 x We cannot count 6.02 x 10102323 Al atoms. Al atoms.

We can, however, We can, however, measure 26.98 grams of measure 26.98 grams of Al.Al.

26.98 grams of Al contains 26.98 grams of Al contains 6.02 x 106.02 x 102323 Al atoms Al atoms

26.98 grams of Al contains 26.98 grams of Al contains 1 mole of Al atoms1 mole of Al atoms

Page 10: Chapter 10 The MOLE. Describing Chemical Equations Balanced chemical equations give us a ratio of particles Balanced chemical equations give us a ratio.

Molar MassMolar Mass

Molar mass of any element or Molar mass of any element or compound = mass of 1 mole of that compound = mass of 1 mole of that element or compound.element or compound.

26.98 grams is the molar mass of Al26.98 grams is the molar mass of Al The molar mass of any element is The molar mass of any element is

its atomic mass expressed in gramsits atomic mass expressed in grams Units are grams per mole (g/mol)Units are grams per mole (g/mol) 26.98 is also the atomic mass of Al 26.98 is also the atomic mass of Al

in amu.in amu.

Page 11: Chapter 10 The MOLE. Describing Chemical Equations Balanced chemical equations give us a ratio of particles Balanced chemical equations give us a ratio.

What’s the molar mass of?What’s the molar mass of?

CoCo UU PbPb OO22

Page 12: Chapter 10 The MOLE. Describing Chemical Equations Balanced chemical equations give us a ratio of particles Balanced chemical equations give us a ratio.

Molar Mass of a CompoundMolar Mass of a Compound

The molar mass of any The molar mass of any compound is the sum of the compound is the sum of the atomic masses of the all the atomic masses of the all the atoms in that compoundatoms in that compound

Expressed in grams per mole Expressed in grams per mole (g/mol).(g/mol).

The molar mass of HThe molar mass of H22OO 18.0 grams per mole.18.0 grams per mole.

Page 13: Chapter 10 The MOLE. Describing Chemical Equations Balanced chemical equations give us a ratio of particles Balanced chemical equations give us a ratio.

What’s the molar mass of?What’s the molar mass of?

Sodium Chloride Sodium Chloride Magnesium HydroxideMagnesium Hydroxide Iron (III) OxideIron (III) Oxide Sulfuric AcidSulfuric Acid GlucoseGlucose

Page 14: Chapter 10 The MOLE. Describing Chemical Equations Balanced chemical equations give us a ratio of particles Balanced chemical equations give us a ratio.

Molar conversions:Molar conversions:

Use dimensional analysis to Use dimensional analysis to convertconvert

Moles to mass multiply by molar Moles to mass multiply by molar massmass

Mass to moles divide by molar Mass to moles divide by molar massmass

Check to make sure units are Check to make sure units are correctcorrect

Page 15: Chapter 10 The MOLE. Describing Chemical Equations Balanced chemical equations give us a ratio of particles Balanced chemical equations give us a ratio.

Mass to molesMass to molesMoles to massMoles to mass

If you are given 5 moles of carbon If you are given 5 moles of carbon what is the mass?what is the mass?

If you are given 72g of Al how many If you are given 72g of Al how many moles is this?moles is this?

Page 16: Chapter 10 The MOLE. Describing Chemical Equations Balanced chemical equations give us a ratio of particles Balanced chemical equations give us a ratio.

Particle and Mole Particle and Mole Conversions:Conversions:

Moles to Particles multiply by Moles to Particles multiply by Avogadro’s number.Avogadro’s number.

Particles to moles divide by Particles to moles divide by Avogadro’s NumberAvogadro’s Number

Check units and use significant Check units and use significant digitsdigits

Page 17: Chapter 10 The MOLE. Describing Chemical Equations Balanced chemical equations give us a ratio of particles Balanced chemical equations give us a ratio.

Particles to molesParticles to molesMoles to particlesMoles to particles

If you are given 6.2 X 10If you are given 6.2 X 102323 atoms of atoms of Al how many moles is this?Al how many moles is this?

If you are given 12 moles of C how If you are given 12 moles of C how many atoms is this?many atoms is this?

Page 18: Chapter 10 The MOLE. Describing Chemical Equations Balanced chemical equations give us a ratio of particles Balanced chemical equations give us a ratio.

Multi-step ConversionsMulti-step Conversions

Grams to Particles conversions Grams to Particles conversions are multi-stepare multi-step

Grams Grams Moles Moles Particles Particles How many particles are in an How many particles are in an

iron nail with a mass ofiron nail with a mass of Particles to gramsParticles to grams Particles Particles Moles Moles Grams Grams

Page 19: Chapter 10 The MOLE. Describing Chemical Equations Balanced chemical equations give us a ratio of particles Balanced chemical equations give us a ratio.

Mass to particlesMass to particlesParticles to massParticles to mass

If you are given 52 grams of C, how If you are given 52 grams of C, how many particles do you have?many particles do you have?

If you are given 6.02 X 10If you are given 6.02 X 102323 atoms of atoms of Al, what is the mass?Al, what is the mass?

Page 20: Chapter 10 The MOLE. Describing Chemical Equations Balanced chemical equations give us a ratio of particles Balanced chemical equations give us a ratio.

Molar Conversions with Molar Conversions with CompoundsCompounds

When conversion are done with When conversion are done with compound the same process is compound the same process is followed.followed.

However remember that you molar However remember that you molar mass must be the sum of all atoms. mass must be the sum of all atoms.

The number of particles will be The number of particles will be expressed in molecules NOT atoms!expressed in molecules NOT atoms!

Page 21: Chapter 10 The MOLE. Describing Chemical Equations Balanced chemical equations give us a ratio of particles Balanced chemical equations give us a ratio.

Practice Molar ConversionsPractice Molar Conversions

How many grams are in 15.5 How many grams are in 15.5 moles of water?moles of water?

Find the mass of 0.650 moles of Find the mass of 0.650 moles of PP22OO55

A large box of baking soda A large box of baking soda (sodium hydrogen carbonate) (sodium hydrogen carbonate) has a mass of 1.81 kg. How has a mass of 1.81 kg. How many moles are in the box? many moles are in the box?

Page 22: Chapter 10 The MOLE. Describing Chemical Equations Balanced chemical equations give us a ratio of particles Balanced chemical equations give us a ratio.

Practice Particle and Mole Practice Particle and Mole Conversions:Conversions:

How many formula units are in How many formula units are in 2.4 moles of methane?2.4 moles of methane?

A piece of marble (calcium A piece of marble (calcium carbonate) contains 8.47 x 10carbonate) contains 8.47 x 102323 molecules. How many moles is molecules. How many moles is this?this?

Page 23: Chapter 10 The MOLE. Describing Chemical Equations Balanced chemical equations give us a ratio of particles Balanced chemical equations give us a ratio.

Practice Multi-step Practice Multi-step ConverstionsConverstions

There are approximately 8.27 x There are approximately 8.27 x 10102323 formula units of glucose in formula units of glucose in the human body. How much the human body. How much mass does this have?mass does this have?

How many particles are How many particles are contained in in a 500.0 mL bottle contained in in a 500.0 mL bottle of water?of water?

Page 24: Chapter 10 The MOLE. Describing Chemical Equations Balanced chemical equations give us a ratio of particles Balanced chemical equations give us a ratio.

Molarity of GasesMolarity of Gases

To determine moles of To determine moles of gases it is difficult gases it is difficult measure mass.measure mass.

Instead chemists Instead chemists measure volume.measure volume.

1 Mole of gas has a 1 Mole of gas has a volume of 22.4 Liters at volume of 22.4 Liters at standard temperature standard temperature and pressure (STP).and pressure (STP).

Page 25: Chapter 10 The MOLE. Describing Chemical Equations Balanced chemical equations give us a ratio of particles Balanced chemical equations give us a ratio.

PracticePractice

A 1.0 Liter bottle is filled with A 1.0 Liter bottle is filled with carbon dioxide at STP. How carbon dioxide at STP. How many moles is this?many moles is this?

A chemical reaction requires 1.7 A chemical reaction requires 1.7 moles of nitrogen gas. How moles of nitrogen gas. How many liters should be used at many liters should be used at STP?STP?

Page 26: Chapter 10 The MOLE. Describing Chemical Equations Balanced chemical equations give us a ratio of particles Balanced chemical equations give us a ratio.

Percent CompositionPercent Composition

Percent that each part of a Percent that each part of a substance is of the whole, by masssubstance is of the whole, by mass

%part = (mass of part/mass of %part = (mass of part/mass of whole)x100whole)x100

100% masswholemasspp

Page 27: Chapter 10 The MOLE. Describing Chemical Equations Balanced chemical equations give us a ratio of particles Balanced chemical equations give us a ratio.

Percent Composition Percent Composition PracticePractice

Hydrochloric acidHydrochloric acid GlucoseGlucose Decomposition results in 2.30 g Decomposition results in 2.30 g

Na, 1.60 g O, and 0.10 g HNa, 1.60 g O, and 0.10 g H

Page 28: Chapter 10 The MOLE. Describing Chemical Equations Balanced chemical equations give us a ratio of particles Balanced chemical equations give us a ratio.

Empirical FormulaEmpirical Formula

Empirical formula found by Empirical formula found by calculating the moles of each calculating the moles of each element in the compound.element in the compound.

If given a percent convert the If given a percent convert the percent to grams.percent to grams.

Divide each element in moles by Divide each element in moles by the element with the least the element with the least amount of moles amount of moles

Page 29: Chapter 10 The MOLE. Describing Chemical Equations Balanced chemical equations give us a ratio of particles Balanced chemical equations give us a ratio.

Find the Empirical FormulaFind the Empirical Formula

A compound is found to be 20% A compound is found to be 20% hydrogen and 80% carbon.hydrogen and 80% carbon.

7.30 g Na, 5.08 g S, 7.62 g O7.30 g Na, 5.08 g S, 7.62 g O 36.0 g C, 8.0 g O, 6.0 g H36.0 g C, 8.0 g O, 6.0 g H

Page 30: Chapter 10 The MOLE. Describing Chemical Equations Balanced chemical equations give us a ratio of particles Balanced chemical equations give us a ratio.

Molecular FormulaMolecular Formula

Formula expressing the true (actual) Formula expressing the true (actual) number of atoms in the compound.number of atoms in the compound.

True formulaTrue formula A multiple of empirical formulaA multiple of empirical formula CC66HH66 = (CH) = (CH)nn where n = 6where n = 6 n = molar mass of molecule n = molar mass of molecule

molar mass of empirical molar mass of empirical formulaformula

Page 31: Chapter 10 The MOLE. Describing Chemical Equations Balanced chemical equations give us a ratio of particles Balanced chemical equations give us a ratio.

Find the Molecular Formula Find the Molecular Formula The empirical formula for The empirical formula for -carotene -carotene

is Cis C55HH77. The molar mass is 536 . The molar mass is 536 g/mol. Determine the molecular g/mol. Determine the molecular formula of formula of -carotene.-carotene.

Ribose is 40.0% carbon, 6.67% Ribose is 40.0% carbon, 6.67% hydrogen, and 53.3% oxygen. It has hydrogen, and 53.3% oxygen. It has a molar mass of 150 g/mol. a molar mass of 150 g/mol. Determine the empirical and Determine the empirical and molecular formulas.molecular formulas.