# Ch. 10 Chemical Quantities - PowerPoint PPT Presentation

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Ch. 10 Chemical Quantities. 3 Methods of Measuring. Counting Mass Volume. Example 1. If 0.20 bushel is 1 dozen apples, and a dozen apples has a mass of 2.0 kg, what is the mass of .050 bushel of apples?. Example 1. Count: 1 dozen apples = 12 apples Mass: 1 dozen apples = 2.0 kg apples

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Ch. 10 Chemical Quantities

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## Ch. 10 Chemical Quantities

• Counting

• Mass

• Volume

### Example 1

• If 0.20 bushel is 1 dozen apples, and a dozen apples has a mass of 2.0 kg, what is the mass of .050 bushel of apples?

### Example 1

• Count: 1 dozen apples = 12 apples

• Mass: 1 dozen apples = 2.0 kg apples

• Volume: 1 dozen apples = 0.20 bushels

apples

Conversion Factors:

• 1 dozen2.0 k.g1 dozen

12 apples 1 dozen 0.20 bushels

### Example 1

• 0.50 bushel x 1 dozen x 2.0 kg =

0.20 bushel 1 dozen

= 5.0 kg

• Named after the Italian scientist Amedo Avogadro di Quaregna

• 6.02 x 10 23

### Mole (mol)

• 1 mol = 6.02 x 10 23 representative particles

• Representative particles: atoms, molecules ions, or formula units (ionic compound)

### Mole (mol)

• Moles= representative x 1 mol

particles 6.02 x 10 23

### Example 2 (atoms  mol)

• How many moles is 2.80 x 10 24 atoms of silicon?

### Example 2

• 2.80 x 10 24 atoms Si x 1 mol Si

6.02 x 10 23 atoms Si

= 4.65 mol Si

### Example 3 (mol  molecule)

• How many molecules of water is 0.360 moles?

### Example 3

• 0.360 mol H2O x 6.02 x 10 23 molecules H2O1 mol H2O

=2.17 molecules H2O

### The Mass of a Mole of an Element

• The atomic mass of an element expressed in grams = 1 mol of that element = molar mass

Molar mass S

Molar mass Hg

Molar mass C

Molar mass Fe

6.02 x 10 23 atoms S

6.02 x 10 23 atoms Hg

6.02 x 10 23 atoms C

6.02 x 10 23 atoms Fe

### Example 4 (mol  gram)

• If you have 4.5 mols of sodium, how much does it weigh?

### Example 4

• .45 mol Na x 23 g Na = 10.35 g Na = 1.0 x 10 2 g Na

1 mol Na

### Example 5 (grams  atoms)

• If you have 34.3 g of Iron, how many atoms are present?

### Example 5

• 34.3 g Fe x 1 mol Fe x 6.02 x 10 23 atoms

55.8 g Fe 1 mol Fe

=3.70 x 10 23 atoms Fe

### The Mass of a Mole of a Compound

• To find the mass of a mole of a compound you must know the formula of the compound

• H2O  H= 1 g x 2

O= 16 g

18 g = 1 mole = 6.02 x 10 23

molecules

### Example 6 (gram  mol)

• What is the mass of 1 mole of sodium hydrogen carbonate?

### Example 6

• Sodium Hydrogen Carbonate = NaHCO3

• Na=23 g

• H=1 g

• C=12 g

• O=16 g x3

• 84 g NaHCO3 = 1 mol NaHCO3

### Mole-Volume Relationship

• Unlike liquids and solids the volumes of moles of gases at the same temperature and pressure will be identical

• States that equal volumes of gases at the same temperature and pressure contain the same number of particles

• Even though the particles of different gases are not the same size, since the gas particles are spread out so far the size difference is negligible

### Standard Temperature and Pressure (STP)

• Volume of a gas changes depending on temperature and pressure

• STP= 0oC (273 K)

101.3 kPa (1 atm)

### Standard Temperature and Pressure (STP)

• At STP, 1 mol = 6.02 X 1023 particles = 22.4 L of ANY gas= molar volume

### Conversion Factors

• AT STP

• 1 mol gas22.4 L gas

22.4 L gas 1 mol gas

### Example 7

• At STP, what volume does 1.25 mol He occupy?

### Example 7

• 1.25 mol He x 22.4 L He = 28.0 L He

1 mol He

### Example 8

• If a tank contains 100. L of O2 gas, how many moles are present?

### Example 8

• 100. L O2 X 1 mol O2 = 4.46 mol O2

22.4 L O2

### Calculating Molar Mass from Density

• The density of a gas at STP is measured in g/L

• This value can be sued to determine the molar mass of gas present

### Example 9

• A gaseous compound of sulfur and oxygen has a density of 3.58 g/L at STP. Calculate the molar mass.

### Example 9

• 1 mol gas x 22.4 L gas X 3.58 g gas =

1 mol gas 1 L gas

Molar Mass= 80.2 g

### Percent Composition

• The relative amounts of the elements in a compound

• These percentages must equal 100

### Percent Composition

• %element = mass of element x 100

mass of compound

### Example 10

• Find the percentage of each element present in Al2 (CO3)3

### Example 10

• Al2(CO3)3

• Al= 27 g x 2 = 54 g / 234 g x 100=23%

• C= 12 g x 3 = 36 g/ 234 g x 100= 15%

• O = 16 g x 9 = 144 g / 234 g x 100=62%

234 gAl2(CO3)3