THE MOLEChapter 10

10.1 MEASURING MATTER

1. Chemists use the mole to count atoms, molecules, and ions.1. Atoms – individual atoms2. Molecules – two or more

atoms covalently bonded3. ions – positively or

negatively charged particles

4. Formula Units – Ionic compounds

• A mole is used to compare large amounts of atoms, particles, or # of molecules

• 1 mole = 6.02 x 1023 atoms/particles/molecules

AVOGADRO

• Avagadro’s Number = 6.02 x 1023

• Named in Honor of

In 1811 he determined the volume

of 1 mol of a gas

HOW WAS AVOGODRO’S NUMBER FIRST CALCULATED?

• The first person to have calculated the number of molecules in any mass of substance seems to have been Josef Loschmidt, (1821-1895), an Austrian high school teacher, who in 1865, using the new Kinetic Molecular Theory (KMT) calculated the number of molecules in one cubic centimeter of gaseous substance under oridnary conditions of temperature of pressure, to be somewhere around 2.6 x 1020 molecules. This has always been known as the"Loschmidt Number."•  

HOW WAS AVOGODRO’S NUMBER FIRST CALCULATED?

•Use the following conversions to calculate Avagodro’s number• 100 ml /1litre• 22.414 l/1mole

HOW WAS AVOGODRO’S NUMBER FIRST CALCULATED?

• (2.68 x 10 19 molecules) x (1000 ml) x (22.414 Liters ) =

• 1cm3 1 litre 1 mole

(2.68 x 10 19molecules) x (1000 ml) x (22.414 Liters ) = molecules

• 1cm3 1 litre 1 mole 1 mole

• The mole is independent of mass and size. It only measures number of particles

A MOLE COMPARISON

HOW MANY ROSES?

CONVERTING BETWEEN REAL WORLD UNITS

• To convert use a conversion factor (aka a ratio in math) to convert or switch from one unit to another unit.

• Ex. How many roses are in 3 dozen?• The conversion factor is 1 dozen / 12 roses or 12 roses / 1 dozen• The set up• What you are given is written first.• Then use the correct conversion factor• 3 dozen x 12 roses = 36 roses 1 dozenCross out = the units that cancel = one on top and one on bottom

CONVERTING BETWEEN MOLES AND PARTICLES

• Ex. How many molecules are in 3.50 moles of sucrose?3.50 mol sucrose x 6.02 x 1023 molecules sucrose =

1 mol sucrose• =2.11 x 1024 molecules sucrose

• Do questions 1& 2 pg 307, 4&5 pg 309

CONVERTING BETWEEN MOLES AND PARTICLES

3.50 mol sucrose x 6.02 x 1023 molecules sucrose = 1 mol sucrose

• =2.11 x 1024 molecules sucrose

• Do questions 1& 2 pg 307, 4&5 pg 309

CONVERTING BETWEEN MOLES AND PARTICLES

• Practice 1. How many particles are in 2.5 moles of Zn?

2.5 moles Zn x 6.02 x 1023 atoms Zn = 1.51 x 1024 atoms Zn 1 mole Zn

CONVERTING BETWEEN MOLES AND PARTICLES

2.5 moles Zn x 6.02 x 1023 atoms Zn = 1.51 x 1024 atoms Zn

1 mole Zn

CONVERTING BETWEEN MOLES AND PARTICLES

2.5 moles Zn x 6.02 x 1023 atoms Zn = 1.51 x 1024 atoms Zn

1 mole Zn

MORE PRACTICE

• Silver nitrate (AgNO3) is used to make several different silver halides used in photographic films. How many formula units of AgNO3 are there in 3.25 moles of AgNO3.

CONVERTING PARTICLES TO MOLES

• Start with the number of Representative Particles (aka atoms, molecules, ion, or formula units)• Use the correct conversion factor to cancel out unwanted

units.• How many moles in 2.11 x 1024 molecules of sucrose?• Ex. 2.11 x 1024 molecules of sucrose x 1 mole sucrose

= 6.02x1023 mol sucrose

= 3.50 mols sucrose

PRACTICE PARTICLES TO MOLES

1. How many moles are in 4.50 x 1024 atoms Zn?2. How many moles in 2.50 x 1020 atoms Fe?

10.2 MASS AND THE MOLE

1. A mole always contains the same number of particles; however, moles of different elements/compounds have different masses

2. Molar mass: the mass in grams of one mole of any pure substance: units=g/mol

3. Atomic Mass = g/mol

REAL WORLD EXAMPLE

• Was is the mass of 5 dozen jelly beans• 1 dozen jelly beans = 35 g jelly beans• 5 dozen jelly beans x 35 g jelly beans = 175 g jelly

beans 1 dozen jelly beans

EXAMPLE

• What is the mass of 3.00 moles of Cu?• Identify the Molar mass of Cu from the periodic

table(round to 4 sig figs)• Cu = 63.55 g/mol• Calculation

3.00 moles Cu x 63.55g Cu = 191 g Cu 1 mol Cu

PRACTICE

• Pg. 328 # 15b• Determine the mass in grams of 42.6 mol Si• # 16• 2.45 x 10-2 mol Zn (answer in scientific notation)

MASS TO MOLE CONVERSION

• Ex. Convert 525 g of Ca to mol• 525 g Ca x 1 mol Ca = 13.1 mol Ca

40.08 g Ca

Do questions 7 & 8 pg 315 9,12-15

PRACTICE

• Pg. 329 • # 18b. How many moles in 300.0 g of S• #19b. How many moles in 1.00 Kg Fe

10.3 MOLES OF COMPOUNDS

• Identify the molar mass of a compound.• Add up the molar mass of the atoms.• Ex. CCl4

• C = 12.01 g/mol• Cl = 35.45 g/mol x 4 = 141.8 g/mol

• 141.8 g/mol + 12.01 g/mol = 153.81 g/mol = 153.8 g/mol CCl4

• Ex. How many moles are in 500.0 g of CCl4

• 500.0 g CCl4 x 1 mol CCl4 = 3.251 mol CCl4

153.8 g CCl4

PRACTICE

1. What is the Molar Mass of H2SO4?

2. How many grams are in 3.25 moles of H2SO4?

MASS TO MOLE

• Ex. How many moles are in 35.0g of HCl?

• 35.0 g HCl x 1 mol HCl = 0.960 mol HCl 36.46 g HCl

PRACTICE

• How many moles are in 145 g of AgNO3?

MASS TO PARTICLES

• Mass in g moles particles

• Two conversions

• Ex. How many particles are in 35.0 g of AlCl3

• 1st Identify the molar mass of AlCl3.

Al = 26.98 g/mol

Cl = 35.45 g/mol x 3 = 106.35 g/mol

AlCl3 = 26.98 g/mol + 106.35 g/mol = 133.3 g/mol

TO CONVERT FROM MASS TO PARTICLES

• The given value goes in the upper left spot

35.6 g of AlCl3 x 1 mol AlCl3 = 0.267 mol AlCl3

133.3 g AlCl3

• 0.267 mol AlCl3 x 6.02 x 1023 formula units = 1.61 x 1023 formula units AlCl3

1 mol AlCl3

• OR 35.6 g of AlCl3 x 1 mol AlCl3 x 6.02 x 1023 formula units = 1.61 x 1023 133.3 g AlCl3 1 mol AlCl3 Formula Units

AlCl3

HOW MANY IONS?

• 1.61 x 1023 formula units AlCl3

• Identify the number of ions that make up the compound and multiply by that number.

• 1.61 x 1023 formula units AlCl3 x 3 Cl- Ions = 4.83 x1023 Cl- ions

1 AlCl3 formula unit

• 1.61 x 1023 formula units AlCl3 x 1 Al3+ Ions = 1.61 x1023 Al3+ ions

1 AlCl3 formula unit

MASS OF 1 FORMULA UNIT/MOLECULE

• 133.33 g/mol AlCl3 (molar mass) divide by Avogadro’s number to get the mass of 1 Formula unit

• 133.33 g/1 mol AlCl3 x 1 mol/6.02 x 1023 formula units

• = 2.21 x 10-22 g AlCl3/formula unit

10.4 EMPIRICAL AND MOLECULAR FORMULAS

• Percent composition1. Synthetic chemists makes the new compounds2. Analytical chemists identifies what its composition is

• Percent composition from experimental data1. Identify the percent by mass of a compound2. A 100 g sample of XY contains 55 g of X and 45 g of Y.

CALCULATING PERCENT BY MASS

• Percent by mass (element) = (mass of element/mass of compound) x 100

• Ex (55g X/ 100g XY) x 100 = 55% X

• Ex (45g Y/ 100g XY) x 100 = 45% Y

• Percent by mass from the chemical formula

• Percent by mass = mass of element in 1 mol of compound x 100

molar mass of compound

EXAMPLE

• Compound NaHCO3

• Na = 22.99 g/mol

• H = 1.008 g/mol

• C = 12.01 g/mol

• O = 16.00 g/mol x 3 = 48.00 g/mol

• NaHCO3 = 84.01 g/mol

• Find the percent of Na in NaHCO3.

• Percent Na = (22.99 g/mol /84.01 g/mol) x 100 = 27.37%

PRACTICE

• Find the % composition for H, C, and O in NaHCO3.

EMPIRICAL FORMULA

• The opposite of % composition.• When you know the % you can determine the empirical

formula • Empirical Formula: the formula with the smallest whole-

number mole ratio of the elements• May or may not be the same as the molecular formula

(aka the actual chemical formula)

SOLVING FOR EMPIRICAL FORMULA

• How to solve for Empirical formula

1. Turn % into mass (g) by assuming you have 100% and therefore 100.0g

2. Convert all masses in g to mole using the molar mass of the compound.

3. Divide each mole value by the smallest mole value (this gives the molar ratio)

4. If decimals are present multiply the all mole values by the smallest number to equal all whole numbers

5 Write the empirical formula with symbols and numbers of atoms present

EXAMPLE

• Pg. 345 Example Problem 10.11•Methyl acetate is a solvent commonly used in some paints, inks, and adhesives. Determine the empirical formula for methyl acetate, which has the following chemical analysis: 48.64% C, 8.16% H, and 43.20% Oxygen

PRACTICE EMPIRICAL FORMULA

• # 58 pg. 346• The circle graph at the right gives the percent

composition for a blue solid. What is the empirical formula for this solid?• 63.16% O• 36.84% N

WARM UP

• Identify the % composition of H and O is H2O.

MOLECULAR FORMULA

• Empirical Formula = smallest whole number ratio. (many compounds have the same empirical formula

• Molecular formula = actual formula

1. Begin with the steps to identify the empirical formula!2. Identify the molar mass of the empirical formula by adding up all the molar masses3. Compare that to the molar mass of the compound given. Divide the given molar mass/empirical formula molar mass4. Distribute the value through your empirical formula to create the molecular formula

EXAMPLE OF MOLECULAR FORMULA

• Pg. 348 10.12• Succinic acid is a substance produced by lichens. Chemical analysis indicates it is composed of 40.68% C, 5.08% H, and 54.24 % O and has a molar mass of 118.1g/mol. Determine the empirical then the molecular formula.

PRACTICE MOLECULAR FORMULA

• Pg. 350 # 62

• A compound was found to contain 49.98 g of C and 10.47 g of H. The molar mass is 58.12 g/mol. What is the empirical formula and the molecular formula?

WARM UP

• Pg. 361 #164

• Vitamin D3: Your body’s ability to absorb calcium is aided by vitamin D3. Chemical analysis yields the data that D3 is made of 84.31% C, 11.53% H, and 4.16% O. Find the Empirical and Molecular Formulas

• Molar mass of D3 = 384 g/mol

• Pg. 361 # 186

• A 1.628 g sample of a hydrate of magnesium iodide is heated until its mass is reduced to 1.072 g and all water has been removed. What is the formula of the hydrate?

CHAPTER 10.5 HYDRATES

What is a Hydrate?a compound that has a specific

number of water molecules bound to its atoms

NAMING HYDRATES

• The number of water molecules is written following a dot for example• Na2CO3 * 10H2O = sodium carbonate decahydrate

• Na = Sodium• CO3 = Carbonate• 10 = Deca• H2O = Hydrate

• The first compound is named based on Ionic or covalent naming.

ANALYZING A HYDRATE

• When a hydrate is heated it loses the water molecules and becomes the anhydrous form.• How do you determine the formula of a hydrate? (how

many waters it has?)1. Find the number of moles of water associated with 1 mole of

the hydrate.2. Dehydrate your sample3. Calculate the water lost and convert to the value to moles4. Calculate the mass of the anhydrate and convert to moles5. Divide the moles of water by the moles of anhydrate

EXAMPLE

• Pg. 352

• How many water molecules are in the hydrate BaCl2 * xH2O?• A 5.00g sample of BaCl2 * xH2O is dehydrated. The resulting mass is 4.26 g

of anhydrous BaCl2. • Subtract: 5.00g – 4.26g = 0.74 g H2O lost. • Convert to moles:

• 4.26g x 1mol BaCl2/208.2g BaCl2 = 0.0205 mol BaCl2 • 0.740g x 1 mol H2O/ 18.02 g H2O =

0.0411 mol H2ODivide the water by barium chloride

.0411mol H2O/0.0205 mol BaCl2 =2

So 2 moles of water to every 1 mole barium chloride = BaCl2 * 2H20

PRACTICE

•A mass of 2.50 g of blue, hydrated copper (II) sulfate is placed in a crucible and heated. After heating, 1.59g of white anhydrous copper sulfate remains. What is the formula for the hydrate? (How many waters?) Name the hydrate.