1. The Mole 7.1 & 7.3

2. 7.1 Chemical Measurement: 2 1.Counting units: pair = 2 12 dozen = 12 20 score = 20 144 gross = 144 500 ream = 500 ? mole = ? 2. Counting atoms, molecules and grams are going to require much larger and much smaller numbers.

3. 3. Remember the mass of an atom is measured in amu ( a tomic m ass u nits) and the numbers on the Periodic Table are based on 1/12 of a carbon-12 atom. So from the table the mass of 1 carbon atom = 12.0 amu. This unit is very small, 1 atomic mass unit (amu) = 1.6605 x 10 -24 g. Determine how many carbon atoms equals 12.0 grams:

4. Determine how many phosphorus atoms equal 31.0 g:

5. Avogadro’s Number = 4. This value is called Avogadro’s Number = 6.02 x 10 23 6.02 x 10 23 = 1 mole(602,000,000,000,000,000,000,000!) amount carbon-12 5. A mole is the amount of a substance. It is based on the number of atoms of an element equal to the number of atoms in exactly 12.0g of carbon-12.

6. Mole Analogies An Avogadro's number of standard soft drink cans would cover the surface of the earth to a depth of over 200 miles. If you had Avogadro's number of unpopped popcorn kernels, and spread them across the United States of America, the country would be covered in popcorn to a depth of over 9 miles. If we were able to count atoms at the rate of 10 million per second, it would take about 2 billion years 6.02 X 10 23 Watermelon Seeds: Would be found inside a melon slightly larger than the moon.

7. Mole Analogies 6.02 X 10 23 Donut Holes: Would cover the earth and be 5 miles (8 km) deep. 6.02 X 10 23 Pennies: Would make at least 7 stacks that would reach the moon. 6.02 X 10 23 Grains of Sand: Would be more than all of the sand on Miami Beach. 6.02 X 10 23 Blood Cells: Would be more than the total number of blood cells found in every human on earth.

8. 6. Atomic mass : mass of one atom of an element measured in amu ( a tomic m ass u nits). 16.0 12.0 Ex #1) H = 1.0 amu O = 16.0 amu C = 12.0 amu 7. Formula mass : mass of all the atoms in a single molecule or formula unit of a compound. Ex #2) H 2 O = Ex #3) H 2 CO 3 = H: 1.0 x 2 = 2.0H: 1.0 x 2 = 2.0 O: 16.0 x 1 =16.0 + C: 12.0 x 1 = 12.0 18.0 amuO: 16.0 x 3 = 48.0 + 62.0 amu

9. g/mol tenths 8. Molar Mass : mass of one mole of an element The units used are g/mol. Round all elements’ masses to the tenths place. Ex #4) Elements: 63.5 Cu = 63.5 g/mol 6.02 x 10 23 1 mole Cu = 6.02 x 10 23 atoms 35.5 Cl = 35.5 g/mol 6.02 x 10 23 1 mole Cl = 6.02 x 10 23 atoms

10. Ex #5) Covalent Compounds: 18.0 H 2 O = 18.0 g/mol 6.02 x 10 23 1 mole H 2 O = 6.02 x 10 23 molecules 62.0 H 2 CO 3 = 62.0 g/mol 6.02 x 10 23 1 mole H 2 CO 3 = 6.02 x 10 23 molecules

11. Ex #6) Ionic Compounds: 58.5 NaCl = 58.5 g/mol 6.02 x 10 23 1 mole NaCl = 6.02 x 10 23 formula units 40.3 MgO = 40.3 g/mol 6.02 x 10 23 1 mole MgO = 6.02 x 10 23 formula units

12. 9. Examples: 28.0 6.02 x 10 23 N 2 = 28.0 g/mol = 6.02 x 10 23 molecules 56.0 12.04 x 10 23 2 N 2 = 56.0 g/mol = 12.04 x 10 23 molecules 10. Practice: Calculate the molar mass of sucrose, C 12 H 22 O 11. C: 12.0 x 12 = 144.0 H: 1.0 x 22 = 22.0 O: 16.0 x 11 = 176.0 + 342.0 g/mol

13. Avogadro’s Number 11. 6.02 x 10 23 1 mole = 6.02 x 10 23 particles = molar mass 6.02 x 10 23 atoms Ne20.2 1 mole Ne = 6.02 x 10 23 atoms Ne = 20.2 g 6.02 x 10 23 molecules CO 2 44.0 1 mole CO 2 = 6.02 x 10 23 molecules CO 2 = 44.0 g 6.02 x 10 23 formula units CaF 2 78.1 1 mole CaF 2 = 6.02 x 10 23 formula units CaF 2 = 78.1 g bridge 12. The mole is the bridge between calculations of # of particles, mass and volume of a gas.

14. 13. Conversion Factors:  1 mole = 6.02 x 10 23 particles ( atoms, molecules or formula units )  1 mole = molar mass ( grams ) from the periodic table May be used in a problem in one of two ways depending on the given units. Ex #1) Given 11.2 g of NaCl, how many moles does this represent?

15. Ex #2) Given 2.50 moles of NaCl, how many grams does this represent? Ex #3) How many particles are in 2.00 moles of H 2 O? How many atoms is this?

16. Ex #4) How many moles of CaCO 3 are 8.74 x 10 23 formula units? Multistep Problems: Ex #5) How many particles of copper are in 56.3 g of copper? We use atoms because copper is an element. There are 3 significant digits because 56.3 has 3.

17. Ex #7) How many sugar particles of sugar (C 12 H 22 O 11 ) are in 250 g of sugar? Moles & Gases: Molar Volume: 1 mole of any gas at S tandard T emperature and P ressure (STP = 0 o C and 1 atm) has a volume of 22.4 dm 3 or 22.4 L Ex #8) How many particles of CO 2 gas are in a 1.0L flask at STP?

18. Ex #9) How many atoms of radon gas are contained in a 6500. dm 3 basement room at STP? Ex #10) If a room has a volume of 4000. L, how many moles of air is this at STP?

19. 7.3 % Composition, Empirical & Molecular Formulas: % Composition: The mass of each element in a compound compared to the entire mass of the compound x by 100%. This can be calculated experimentally from the grams of each element in a compound, or the expected % can be calculated by using the molar masses of the elements in the compound compared to the molar mass of the compound. Ex #11) What is the % hydrogen in water? Use the molar masses when not provided with the lab data. The numbers should add up to 100% unless significant digits prevent it.

20. Ex #12) Find the % composition of a compound that contains 2.30 g of sodium, 1.60 g of oxygen, and 0.100 g of hydrogen. Ex #13) A sample of an unknown compound with a mass of 0.562 g has the following % composition: 13.0 % carbon, 2.20 % hydrogen, and 84.5 % fluorine. When this compound is decomposed into its elements, what mass of each element should be recovered?

21. whole Empirical Formula: A formula that gives the simplest whole-number ratio of the atoms of the elements in the compound. actual Molecular Formula: Gives the actual number of atoms of each element in a molecular compound. The molecular formula may be the same as the empirical formula. H 2 0 Ex #14) hydrogen peroxide = H 2 O 2 molecular formula 11 = HO empirical formula (a 1 : 1 ratio) Ex #15) glucose = C 6 H 12 O 6 molecular formula CH 2 O 111 = CH 2 O empirical formula (a 1 : 1 : 1 ratio)

22. Ex #16) Find the empirical formula of the compound if you have 80.g of carbon and 20.g of hydrogen.

23. Ex #17) An unknown compound is analyzed. It’s composition was determined to be 51.85 % carbon, 8.64 % hydrogen and 39.51 % oxygen. Find the empirical formula of the compound.

24. Ex #18) Find the molecular formula of ribose ( molar mass = 150.0 g/mol ). It has a chemical composition that is 40.0% carbon, 6.67% hydrogen and 53.3% oxygen. Assume a 100 g sample. Hint: First find the empirical formula!