1. Counting Atoms

2. 1/25 Today you will need… A lab paper from side lab table, a calculator, a periodic table, and ONE partner. You can sit next to your partner even if its not your assigned seat. We will learn how to convert a measured mass to number of moles and amount of atoms in a sample.

3. On the back of the lab paper answer: how many grams and atoms are in the mole of silver (left pic) and how many grams and atoms are in the mole of carbon (right pic)?

4. During the lab…. You need to measure all ten elements and write down the exact number given to you on the scale. PLEASE DO NOT OPEN BAG. Make sure you take into consideration the mass of the baggie (1.0 g) when determining the mass of the element. When you and your partner have measured all ten elements, have a seat and start trying to convert the masses to number of moles. There is a sheet to help you determine how to calculate this conversion. Please read this sheet and try ON YOUR to calculate moles.

5. 1/27 HAPPY FRIDAY! Please pick up the paper from the side table. You will have ten minutes to complete the mole and atoms conversion lab from yesterday.

6. The pictures below show a mole of water and a mole of gold. Determine the mass of each. Also determine the number of atoms in the gold and the number of molecules in water.

7. Relating Mass to Number of Atoms How can the mass of an element from the periodic table be related to the actual number of atoms that are present in a sample? Use: The mole Avogadro’s number Molar mass

8. The Mole Definition-the amount of substance that contains as many particles as there are atoms in exactly 12 g of carbon-12. Huh? If you weighed exactly 12 g of carbon-12 and counted every single atom in your pile, the number of atoms present is defined as 1 mol. Think of it like exactly 12 eggs are in 1 dozen.

9. Avogadro’s Number So how many carbon-12 atoms would be in that pile? 6.0221367 X 10 23 Rounded to 6.022 X 10 23 This number is “nicknamed” Avogadro’s number after the Italian scientist Amedeo Avogadro. A mole of any substance contains 6.022 X10 23 particles, just like 1 dozen of any substance contains 12 particles.

10. If every person on earth (5 billion) counted continuously at 1 atom per second, it would take 4 million years for all atoms in one mole to be counted!

11.  The mass of 1 mole of a pure substance is called the molar mass  Equality that relates number of grams in a mole  Molar mass is equal to the atomic mass of an element  The unit for atomic mass is grams per mole (g/mol) Molar Mass

12. Atoms, Moles, & Molar Mass can be used as conversion factors! When performing conversion between atoms, moles, and grams (molar mass), remember: When performing conversion between atoms, moles, and grams (molar mass), remember: AGM

13. When using masses from the periodic table, round to two decimals for all calculations!

14. Conversion Example 1 What is the mass, in grams, of 2.67 mol of the element copper, Cu? What is the mass, in grams, of 2.67 mol of the element copper, Cu? Given: 2.67 mol Cu Given: 2.67 mol Cu Unknown: ? g Cu Unknown: ? g Cu

15. Conversion Example 2 How many moles of oxygen, O, are in 2,000,000 atoms of oxygen? How many moles of oxygen, O, are in 2,000,000 atoms of oxygen? Given: 2,000,000 atoms O Given: 2,000,000 atoms O Unknown: ? mol O Unknown: ? mol O

16. Molar Mass Used to Determine Percent Composition Molar mass-the mass of one mole of a compound

17. Why % composition? When a new compound is made or discovered, it is analyzed to determine its % composition. From this info, its empirical formula can be determined.

18. Percentage Composition Percentage by mass of each element in a compound Percentages should always total very close to 100! Always check!

19. Percentage Composition Example 1 Find the percentage composition of sodium nitrate. Formula of sodium nitrate: NaNO 3 Molar mass of NaNO 3 =85.00 g/mol Total100%

20. Percentage Composition Example 2 Find the percentage composition of silver sulfate. Formula of silver sulfate: Ag 2 SO 4 Molar mass of Ag 2 SO 4 =311.81 g/mol Total100.01%

21. Empirical Formula Simplified formula for a compound For an ionic compound, empirical formula is the compound’s formula unit because we always simplify the formula   Calcium sulfide: Ca 2 S 2  CaS For a molecular compound, may not represent actual number of atoms present in a molecule   Molecular formula: B 2 H 6   Empirical formula: BH 3

22. Empirical Formula C2H6C2H6 CH 3 reduce subscripts Smallest whole number ratio of atoms in a compound

23. How do you calculate empirical formula? 1. Find mass (or %) of each element. 2. Convert grams to moles of each element. 3. Divide moles by the smallest # to find subscripts. 4. When necessary, multiply subscripts by 2, 3, or 4 to get whole #’s.

24. Empirical Formula Example 1 A compound consist of 75% carbon and 25% hydrogen by mass. Determine the empirical formula for this compound 75 g 1 mol 12.01 g = 6.24 mol N 25 g 1 mol 1.01 g = 24.75 mol O 6.24 mol = 1 C = 4 H

25. Empirical Formula Example 2 Find the empirical formula for a sample of 25.9% N and 74.1% O. 25.9 g 1 mol 14.01 g = 1.85 mol N 74.1 g 1 mol 16.00 g = 4.63 mol O 1.85 mol = 1 N = 2.5 O

26. Empirical Formula Example N 1 O 2.5 Need to make the subscripts whole numbers  multiply by 2 N2O5N2O5

27. Molecular Formula “True Formula” - the actual number of atoms in a compound CH 3 C2H6C2H6 empirical formula molecular formula ?

28. How do you calculate molecular formula? 1. Find the empirical formula. 2. Find the empirical formula mass. 3. Divide the molecular mass by the empirical mass. 4. Multiply each subscript by the answer from step 3.

29. Molecular Formula Example The empirical formula for ethylene is CH 2. Find the molecular formula if the molecular mass is 28.1 g/mol. 28.1 g/mol 14.03 g/mol = 2.00 empirical mass (CH 2 ) = 14.03 g/mol (CH 2 ) 2  C 2 H 4

30. General pathway to solve problems Mass of each element Percent composition of compound Empirical formula for compound Molecular formula for compound

31. Molecular Formula Example A sample of a compound with a molar mass of 34.00 g/mol is found to consist of 0.44 g H and 6.92 g O. Find its molecular formula. 5.98 % H, 94.02 % O Empirical formula: HO Molecular formula: (HO) 2 =H 2 O 2