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

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Presentation on theme: "Chapter 10 The MOLE. Describing Chemical Equations Balanced chemical equations give us a ratio of particles Balanced chemical equations give us a ratio."— Presentation transcript:

1 Chapter 10 The MOLE

2 Describing Chemical Equations Balanced chemical equations give us a ratio of particles Balanced chemical equations give us a ratio of particles 4 Fe + 3 O 2  2 Fe 2 O 3 4 Fe + 3 O 2  2 Fe 2 O 3 A ratio of particles is not practical from our perspective A ratio of particles is not practical from our perspective

3 Describing Chemical Equations  Atoms and molecules are very, very small, but we still need to “count” them.  Scientists instead weigh out an approximate number atoms

4 The Mole  The mole is a quantity; it is used to “count” atoms and molecules.  A mole contains 6.02 x 10 23 things  This is Avogadro’s number = 6.02214199 × 10 23

5 ê 1 mole of Al atoms = 6.02 x 10 23 Al atoms ê 1 mole H 2 O molecules = 6.02 x 10 23 H 2 O molecules ê 1 mole jelly donuts = 6.02 x 10 23 jelly donuts ê 1 mole of = 6.02 x 10 23

6 6.02 x 10 23 is a VERY large number

7 602,000,000,000,000,00 0,000,000

8 Avogadro’s Number facts Avogadro's number of inches is 1,616,434 light years, or across our galaxy and back 8 times. Avogadro's number of inches is 1,616,434 light years, or across our galaxy and back 8 times. Avogadro's number of seconds is about 19 quadrillion years, 4,240,666 times the age of the earth, or 954,150 times the age of the universe itself. Avogadro's number of seconds is about 19 quadrillion years, 4,240,666 times the age of the earth, or 954,150 times the age of the universe itself. Avogadro's number of cents could repay the United States National Debt 86 million times. Avogadro's number of cents could repay the United States National Debt 86 million times.United States National DebtUnited States National Debt Avogadro's number of kilograms is just over 20 times the mass of the earth Avogadro's number of kilograms is just over 20 times the mass of the earth

9 Measuring Numbers of Atoms We cannot count 6.02 x 10 23 Al atoms. We cannot count 6.02 x 10 23 Al atoms. We can, however, measure 26.98 grams of Al. We can, however, measure 26.98 grams of Al. 26.98 grams of Al contains 6.02 x 10 23 Al atoms 26.98 grams of Al contains 6.02 x 10 23 Al atoms 26.98 grams of Al contains 1 mole of Al atoms 26.98 grams of Al contains 1 mole of Al atoms

10 Molar Mass Molar mass of any element or compound = mass of 1 mole of that element or compound. Molar mass of any element or compound = mass of 1 mole of that element or compound. 26.98 grams is the molar mass of Al 26.98 grams is the molar mass of Al The molar mass of any element is its atomic mass expressed in grams The molar mass of any element is its 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 in amu. 26.98 is also the atomic mass of Al in amu.

11 What’s the molar mass of? Co Co U Pb Pb O 2 O 2

12 Molar Mass of a Compound The molar mass of any compound is the sum of the atomic masses of the all the atoms in that compound The molar mass of any compound is the sum of the atomic masses of the all the atoms in that compound Expressed in grams per mole (g/mol). Expressed in grams per mole (g/mol). The molar mass of H 2 O The molar mass of H 2 O 18.0 grams per mole. 18.0 grams per mole.

13 What’s the molar mass of? Sodium Chloride Sodium Chloride Magnesium Hydroxide Magnesium Hydroxide Iron (III) Oxide Iron (III) Oxide Sulfuric Acid Sulfuric Acid Glucose Glucose

14 Molar conversions: Use dimensional analysis to convert Use dimensional analysis to convert Moles to mass multiply by molar mass Moles to mass multiply by molar mass Mass to moles divide by molar mass Mass to moles divide by molar mass Check to make sure units are correct Check to make sure units are correct

15 Mass to moles Moles to mass If you are given 5 moles of carbon what is the mass? If you are given 5 moles of carbon what is the mass? If you are given 72g of Al how many moles is this? If you are given 72g of Al how many moles is this?

16 Particle and Mole Conversions: Moles to Particles multiply by Avogadro’s number. Moles to Particles multiply by Avogadro’s number. Particles to moles divide by Avogadro’s Number Particles to moles divide by Avogadro’s Number Check units and use significant digits Check units and use significant digits

17 Particles to moles Moles to particles If you are given 6.2 X 10 23 atoms of Al how many moles is this? If you are given 6.2 X 10 23 atoms of Al how many moles is this? If you are given 12 moles of C how many atoms is this? If you are given 12 moles of C how many atoms is this?

18 Multi-step Conversions Grams to Particles conversions are multi-step Grams to Particles conversions are multi-step Grams  Moles  Particles Grams  Moles  Particles How many particles are in an iron nail with a mass of How many particles are in an iron nail with a mass of Particles to grams Particles to grams Particles  Moles  Grams Particles  Moles  Grams

19 Mass to particles Particles to mass If you are given 52 grams of C, how many particles do you have? If you are given 52 grams of C, how many particles do you have? If you are given 6.02 X 10 23 atoms of Al, what is the mass? If you are given 6.02 X 10 23 atoms of Al, what is the mass?

20 Molar Conversions with Compounds When conversion are done with compound the same process is followed. When conversion are done with compound the same process is followed. However remember that you molar mass must be the sum of all atoms. However remember that you molar mass must be the sum of all atoms. The number of particles will be expressed in molecules NOT atoms! The number of particles will be expressed in molecules NOT atoms!

21 Practice Molar Conversions How many grams are in 15.5 moles of water? How many grams are in 15.5 moles of water? Find the mass of 0.650 moles of P 2 O 5 Find the mass of 0.650 moles of P 2 O 5 A large box of baking soda (sodium hydrogen carbonate) has a mass of 1.81 kg. How many moles are in the box? A large box of baking soda (sodium hydrogen carbonate) has a mass of 1.81 kg. How many moles are in the box?

22 Practice Particle and Mole Conversions: How many formula units are in 2.4 moles of methane? How many formula units are in 2.4 moles of methane? A piece of marble (calcium carbonate) contains 8.47 x 10 23 molecules. How many moles is this? A piece of marble (calcium carbonate) contains 8.47 x 10 23 molecules. How many moles is this?

23 Practice Multi-step Converstions There are approximately 8.27 x 10 23 formula units of glucose in the human body. How much mass does this have? There are approximately 8.27 x 10 23 formula units of glucose in the human body. How much mass does this have? How many particles are contained in in a 500.0 mL bottle of water? How many particles are contained in in a 500.0 mL bottle of water?

24 Molarity of Gases To determine moles of gases it is difficult measure mass. To determine moles of gases it is difficult measure mass. Instead chemists measure volume. Instead chemists measure volume. 1 Mole of gas has a volume of 22.4 Liters at standard temperature and pressure (STP). 1 Mole of gas has a volume of 22.4 Liters at standard temperature and pressure (STP).

25 Practice A 1.0 Liter bottle is filled with carbon dioxide at STP. How many moles is this? A 1.0 Liter bottle is filled with carbon dioxide at STP. How many moles is this? A chemical reaction requires 1.7 moles of nitrogen gas. How many liters should be used at STP? A chemical reaction requires 1.7 moles of nitrogen gas. How many liters should be used at STP?

26 Percent Composition Percent that each part of a substance is of the whole, by mass Percent that each part of a substance is of the whole, by mass %part = (mass of part/mass of whole)x100 %part = (mass of part/mass of whole)x100

27 Percent Composition Practice Hydrochloric acid Hydrochloric acid Glucose Glucose Decomposition results in 2.30 g Na, 1.60 g O, and 0.10 g H Decomposition results in 2.30 g Na, 1.60 g O, and 0.10 g H

28 Empirical Formula Empirical formula found by calculating the moles of each element in the compound. Empirical formula found by calculating the moles of each element in the compound. If given a percent convert the percent to grams. If given a percent convert the percent to grams. Divide each element in moles by the element with the least amount of moles Divide each element in moles by the element with the least amount of moles

29 Find the Empirical Formula A compound is found to be 20% hydrogen and 80% carbon. A compound is found to be 20% hydrogen and 80% carbon. 7.30 g Na, 5.08 g S, 7.62 g O 7.30 g Na, 5.08 g S, 7.62 g O 36.0 g C, 8.0 g O, 6.0 g H 36.0 g C, 8.0 g O, 6.0 g H

30 Molecular Formula Formula expressing the true (actual) number of atoms in the compound. Formula expressing the true (actual) number of atoms in the compound. True formula True formula A multiple of empirical formula A multiple of empirical formula C 6 H 6 = (CH) n where n = 6 C 6 H 6 = (CH) n where n = 6 n = molar mass of molecule molar mass of empirical formula n = molar mass of molecule molar mass of empirical formula

31 Find the Molecular Formula The empirical formula for  -carotene is C 5 H 7. The molar mass is 536 g/mol. Determine the molecular formula of  -carotene. The empirical formula for  -carotene is C 5 H 7. The molar mass is 536 g/mol. Determine the molecular formula of  -carotene. Ribose is 40.0% carbon, 6.67% hydrogen, and 53.3% oxygen. It has a molar mass of 150 g/mol. Determine the empirical and molecular formulas. Ribose is 40.0% carbon, 6.67% hydrogen, and 53.3% oxygen. It has a molar mass of 150 g/mol. Determine the empirical and molecular formulas.


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