### What is a Mole

 Chemists use the mole to count microscopic particles.

 How many socks come in a pair?

 2  How many eggs are in a dozen?

 12  How many eggs come in a gross?

 144  How many pencils come in a ream?

 500

### So how many atoms come in a mole?

 602,213,670,000,000,000,000,000.

 Seriously  This number was created by an Italian physicist and lawyer named Amedeo Avogadro.

 Is there an easier way to write it?

 6.02 x 10 23 items = 1 mole

### Converting between moles and number of particles

 Think about eggs  1 dozen eggs = 12 eggs  Conversion factor:  12 eggs/1 dozen eggs  OR  1 dozen eggs/12 eggs  So if you have 3.5 dozen eggs how many eggs do you have?

 3.5 dozen x (12 eggs)/1 dozen = 42 eggs

 If you have 3.5 moles of sugar how many particles of sugar do you have?

 1 mole of sugar = 6.02 x 10 23 particles of sugar.

 Set up the problem:  3.5 moles x (6.02 x 10 23 particles/1 mole)  Solve:  2.107 x 10 24

        Zinc (Zn) is used to form a corrosion-inhibiting surface on galvanized steel. Determine the number of Zn atoms in 2.50 moles of Zn.

1.51 x 10 24 atoms Calculate the number of molecules in 11.5 mol of water.

6.92 x 10 24 molecules Silver nitrate Ag(NO Ag(NO 3 ) 2 ?

3 ) 2 is used to make several different compounds used in photographic films. How many molecules of silver nitrate are there in 3.25 moles of 1.96 x 10 24 How many atoms of oxygen are there in 5.0 mol of oxygen gas?

6.02 x 10 24 atoms

### Converting from number of particles to moles

 Calculate the number of moles of Zinc that contains 4.50 x 10 24 atoms.

 0.914 moles  How many moles can be made up from 5.57 x 10 24 atoms of Al?

 9.55 moles  How many moles can be made up from 2.5 x 10 25 of Fe?

atoms  41.5 moles

 What would have more mass a dozen eggs or a dozen elephants?

 Why would a dozen elephants have more mass?

 Just like elephants and eggs certain atoms are bigger than others.  For example Neon is much bigger than Helium.

 Which would have more mass 1 mole of Neon or 1 mole of Helium?

### Molar Mass

 The molar mass of an element is the mass in grams of one mole of that element.

 Why is this important?

 Molar mass can be used to calculate the number of atoms with out using a microscope.

### Using molar mass

 If one dozen jelly beans has a mass of 35 g how much mass does 5 dozen jelly beans have?

 175 g  To convert from Moles to mass you multiply by the molar mass.

 What is the mass of 3.oo moles of copper?

 191 g cu.

 To convert from mass to moles you divide by the molar mass.

### Example Problems

 Chromium (Cr), a transition element, is a component of chrome plating. Chrome plating is used on metals and steel alloys to control corrosion. Calculate the mass in grams of 0.045 moles of Cr.

 2.34 g Cr

 Calcium (Ca), the fifth most abundant element on earth, is always found combined with other atoms because of its high reactivity. How many moles of calcium are in 525 g Ca?

 13.1 moles Ca

### Converting between mass and atoms

 The next step is converting a given mass of an element into a number of atoms.

 If we have 550 g of jelly beans and there are 35 g of jelly beans in a dozen how many dozen jelly beans do we have?

 16 dozen  How many jelly beans are there in 16 dozen?

 192

### Mass-to-Atom Conversion

 Gold (Au) is one of a group of metals called the coinage metals. How many atoms of gold are there in a U.S. gold eagle coin with a mass of 31.1 g?

 9.51 x 10 22

### Atom-to-Mass Conversion

 Helium (He) is an unreactive noble gas found underground. A party balloon contains 5.5 x 10 22 of helium gas. What is the mass, in grams, of the atoms helium?

 0.366 g

### Homework

 Textbook (sorry)  P. 328 # 15(a&b), 16(a&b)  P. 329 # 17(a&b), 18(a&b)  P. 331 # 19(a&b), 20(a&b)

### Chemical Formulas and the Mole

 In the compound CCl 2 F 2 how many atoms do we have?

 C: 1  Cl: 2  F: 2  So if we have one mole of CCl 2 F 2 we have one mole of carbon, two moles of Chlorine and one mole of Fluorine.

### Molar Mass of a Compound

 Determine the molar mass of each of the following compounds:  NaOH  CaCl 2  KC 2 H 3 O 2  C 2 H 5 OH  HCN   CCl 4 (NH 4 ) 3 PO 4

### Converting moles to mass

 The characteristic odor of garlic is due to allyl sulfide (C 3 H 5 ) 2 S. What is the mass of 2.5 moles of allyl sulfide?

 What is the mass of 3.25 mol of H 2 SO 4 ?

### Converting Mass to Moles

 Calculate the number of moles in 325 g of Calcium hydroxide Ca(OH) 2 ?

 Calculate the number of moles in 22.6 g of AgNO 3 .

### Mass to Number of Particles

 Aluminum chloride (AlCl 3 ) is used in refining petroleum and manufacturing rubber. How many Aluminum Chloride molecules are present in 35.6 g of AlCl 3  How many Al atoms are there?

 How many Cl atoms are there?

### Example

 Aluminum oxide (Al 2 O 3 ) when dissolved in water breaks apart into ions (Al 3+ and O 2 ) how many moles of Al 3+ will be produced when 1.5 moles of Al 2 O 3 dissolved in water?

are

### Examples

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2.

 What is the molar mass of ethanol C 2 H 5 OH?

How many ethanol molecules are present in 45.6 g?

 How many carbon atoms are there in 45.6 g of ethanol?

 How many hydrogen atoms are there in 45.6 g of ethanol?

 How many oxygen atoms are there in 45.6 g of ethanol?

 A sample of sodium sulfite Na 2 SO 3 has a mass of 2.25 g How many molecules of sodium sulfite are there?

   How many atoms of sodium are there?

How many atoms of Sulfur are there?

How many atoms of oxygen are there?

### Percent Composition

 A compound is made up of one or more atoms bonded together.

 Each of the atoms contributes mass to the compound.  Example:  The molar mass of NaOH is  Na – 22.98 g/mol  O – 16 g/mol  H – 1 g/mol  Total: 39.98 g/mol

### Calculating Percent Composition

 Percent composition is calculated by dividing the mass of an individual element by the mass of the whole compound and then multiplying by 100.

 NaOH:  Percent Composition of Na:  (22.98/39.98) x 100 = 57.5%  Percent Composition of O:  (16/39.98) x 100 = 40%  Percent Composition of H:  (1/39.98) x 100 = 2.5%

### Compounds with multiple atoms of the same element

  NaHCO 3 Molar Mass: 84.01 g/mol  Percent Composition of Na:  (22.98/84.01) x 100 = 27.37%  Percent Composition of H:  (1/84.o1) x 100 = 1.2%  Percent Composition of C:  (12.01/84.01) x 100 = 14.3%  Percent Composition of O:  (48/84.01) x 100 = 57.14%

      

### Empirical Formula

When we know a compounds percent composition we can determine it’s formula. The empirical formula of a compound is the formula with the smallest whole-number ratio of elements. Find the empirical formula for a compound that is 40.05% S and 59.95% O.

If we assume that we have 100 g of this compound: How many grams of Sulfur do we have?

 40.05 g How many grams of Oxygen do we have?

 59.95 g Convert from grams to moles:   40.05 g S x 1 mol S/32.07 g S = 1.249 mol 59.95 g O x 1 mol O/16.00 g O = 3.747 mol

 S 1.249

O 3.747

 SO 3  Sometimes the empirical formula is not the actual formula for the compound.

 Find the empirical formula for hydrogen peroxide which has a molar mass of 34 g/mol and is 5.9% Hydrogen and 94.1% Oxygen

### Molecular Formula

 A compounds molecular formula is its actual formula (not always the same as empirical formula)  To determine the true molecular formula for a compound its molar mass must be known (either looked up or found experimentally).

 Once we know the molar mass and the empirical formula we can find the true formula of the compound.

 Example:          Acetylene is a highly flammable compound used in blow torches. Acetylene has a molar mass of 26.04 g/mol. It is 92.2% Carbon and 7.8% Hydrogen. What is the chemical formula of Acetylene?

Step 1: Find empirical formula: CH Step 2: Calculate molar mass of empirical formula. (If the molar mass is the same as the actual molar mass then the empirical formula is the true formula.) 13.02 g/mol Step 3: Divide real molar mass by the molar mass of the empirical formula.

26.04/13.02 = 2.000

Step 4: Multiply all atoms in the empirical formula by the answer to the true formula.

C 2 H 2