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Mass, the Mole, and Avogadro’s Number (Molar Mass)

Mass, the Mole, and Avogadro’s Number

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Mass, the Mole, and Avogadro’s Number. (Molar Mass). The Mass of a Mole C-12 Basis. Atomic masses are based on the mass of a C-12 atom. Molar masses are based on the mass of a mole of C-12 atoms. Molar Mass (MM). Definition The mass in grams of 1 mole of any substance. - PowerPoint PPT Presentation

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Page 1: Mass, the Mole, and Avogadro’s Number

Mass, the Mole, and Avogadro’s Number

(Molar Mass)

Page 2: Mass, the Mole, and Avogadro’s Number

The Mass of a MoleC-12 Basis

• Atomic masses are based on the mass of a C-12 atom.

• Molar masses are based on the mass of a mole of C-12 atoms.

Page 3: Mass, the Mole, and Avogadro’s Number

Molar Mass(MM)

• Definition– The mass in grams of 1 mole of any

substance.

– Has the units g / mol• Units

Page 4: Mass, the Mole, and Avogadro’s Number

• The molar mass (MM) of any element is numerically equal to its atomic mass.

Molar Mass of Elements

20

Ca  40.08

Ex: Molar mass of Ca

Ex: Molar mass of Cl 17Cl

35.45

Page 5: Mass, the Mole, and Avogadro’s Number

Ex: Molar mass of CaCl2

Avg. Atomic mass of Calcium = 40.08gAvg. Atomic mass of Chlorine = 35.45g

Molar Mass of Compounds

• The molar mass (MM) of a compound is determined the same way, except now you add up all the atomic masses for the molecule (or compound)

20

Ca  40.08

17Cl

35.45

Molar Mass of calcium chloride = 40.08 g/mol Ca + (2 X 35.45) g/mol Cl 110.98 g/mol CaCl2

Page 6: Mass, the Mole, and Avogadro’s Number

Molar MassPractice Problem

• Calculate the Molar Mass of calcium phosphate

Ca3(PO4)2Formula =

Masses of elements:

Molar Mass = =

Ca = 40.078 g

P = 30.974 g

O = 15.999 g

3 (40.078 g) + 2 (30.974 g) + 8 (15.999 g)

310.174 g Ca3(PO4)2 / mole

Page 7: Mass, the Mole, and Avogadro’s Number

• Find the molar mass of:– Copper– Sulfuric acid– Xenon tetrafluoride– Hypochlorous acid– Oxygen

Molar MassPractice

Problems

63.55 g / mol

98.06 g / mol

206.9 g / mol

52.45 g / mol

31.99 g / mol

Page 8: Mass, the Mole, and Avogadro’s Number

FlowchartRepresentative Particles

Moles

Mass (grams)

1 mol______6.02 X 1023 particles

6.02 X 1023 particles 1 mol

Page 9: Mass, the Mole, and Avogadro’s Number

FlowchartRepresentative Particles

Moles

Mass (grams)

1 mol______6.02 X 1023 particles

6.02 X 1023 particles 1 mol

molar mass 1 mol_____1 mol_____

molar mass

Page 10: Mass, the Mole, and Avogadro’s Number

Mole Mass

# of moles x # of grams = mass

1 mol

0.0450 mol Cr x 52.00 g Cr = 2.34 g Cr

1 mol Cr

Example:

Calculate the mass in grams of 0.0450 moles of chromium.

Page 11: Mass, the Mole, and Avogadro’s Number

• Calculate the mass of 0.625 moles of calcium.

Mole MassPractice Problems

• Determine the mass of 0.187 moles of tin (II) sulfate.

Page 12: Mass, the Mole, and Avogadro’s Number

• Calculate the mass of 0.625 moles of calcium.

Mass of Ca = mol Ca x grams Ca

1 mol Ca

Mass of Ca = 0.625 mol Ca x 40.078 grams Ca

1 mol Ca

= 25.0 g Ca

Mole MassPractice Problems

Page 13: Mass, the Mole, and Avogadro’s Number

• Determine the mass of 0.187 moles of tin (II) sulfate.

Mass of SnSO4 = mol SnSO4 x grams SnSO4

1 mol SnSO4

Mass of SnSO4 = 0.187 mol SnSO4 x 214.8 grams SnSO4

1 mol SnSO4

= 40.2 g SnSO4

Mole MassPractice Problems

Page 14: Mass, the Mole, and Avogadro’s Number

Mass Mole

Mass x ____1 mole____ = # of moles

# of grams

Example:

How many moles of calcium are in 525 grams of calcium?

525 g Ca x 1 mol Ca = 13.1 mol Ca

40.08 g Ca

Page 15: Mass, the Mole, and Avogadro’s Number

• A roll of copper wire has a mass of 848 g. How many moles of copper are in the roll?

Mass MolePractice Problems

• At 4.0°C, water has a density of 1.000 g/ml. How many moles of water are in 1.000 kg of water?

Page 16: Mass, the Mole, and Avogadro’s Number

• A roll of copper wire has a mass of 848 g. How many moles of copper are in the roll?

Mass MolePractice Problems

Mass Cu x ____1 mole_Cu___ = # of moles Cu

# of grams Cu

848 g Cu x ____1 mole_Cu___ = 13.3 moles Cu

63.546 g Cu

Page 17: Mass, the Mole, and Avogadro’s Number

Mass MolePractice Problems

• At 4.0°C, water has a density of 1.000 g/ml. How many moles of water are in 1.000 kg of water?

Mass H2O x ____1 mole_ H2O ___ = # of moles H2O

# of grams H2O

1.000 x 103 g H2O x ____1 mole_ H2O __ = 55.51 moles H2O 18.015 g H2O

Page 18: Mass, the Mole, and Avogadro’s Number

Mass Atoms

Mass x ____1 mole____ x 6.02 x 1023 atoms = # of atoms

# of grams 1 mole

Example:

How many atoms of gold are in a pure gold nugget having a mass of 25.0 grams?

25.0 g Au x 1 mol Au x 6.02 x 1023 atoms Au = 7.65 x 1022 atoms Au

196.69 g Au 1 mol Au

Page 19: Mass, the Mole, and Avogadro’s Number

Mass AtomsPractice Problems

Mass x ____1 mole____ x 6.02 x 1023 atoms = # of atoms

# of grams 1 mole

Practice:

How many atoms of lead are in 4.77 g of lead ?

How many molecules of benzene are in 17.2 g of benzene (C6H6) ?

Page 20: Mass, the Mole, and Avogadro’s Number

Mass AtomsPractice Problems

Mass x ____1 mole____ x 6.02 x 1023 atoms = # of atoms

# of grams 1 mole

How many molecules of benzene are in 17.2 g of benzene (C6H6) ?

17.2 g C6H6 x 1 mol C6H6 x 6.02 x 1023 molec C6H6 = 1.33 x 1023 molec

78.11 g C6H6 1 mol C6H6 C6H6

How many atoms of lead are in 4.77 g of lead ?4.77 g Pb x 1 mol Pb x 6.02 x 1023 atoms Pb = 1.38 x 1022 atoms Pb

207.2 g Pb 1 mol Pb

Page 21: Mass, the Mole, and Avogadro’s Number

Atoms Mass

# of atoms x ____1 mole____ x # of grams = Mass

6.02 x 1023 atoms 1 mole

Example:

A party balloon contains 5.50 x 1022 atoms of helium gas. What is the mass in grams of the helium?

5.50 x 1022 atoms He x 1 mol He x 4.00 g He = 0.366 g He

6.02 x 1023 atoms He 1 mol He

Page 22: Mass, the Mole, and Avogadro’s Number

Atoms MassPractice Problem

# of atoms x ____1 mole____ x # of grams = Mass

6.02 x 1023 atoms 1 mole

Practice:

How many grams of mercury are in 1.19 x 1023 atoms of mercury gas?

1.19 x 1023 atoms Hg x 1 mol Hg x 200.59 g Hg = 39.7 g

6.02 x 1023 atoms Hg 1 mol Hg Hg