1. Chemistry chapter 3 Sample Problems

2. Isotopes The number of neutrons is found by subtracting the atomic number from the mass number. Mass # (235)---atomic #(92) =# of nuetrons (143)

3. Sample Problem 3-1 How many protons, electrons, and neutrons are there in an atom of Chlorine-37? Given: name and mass # of chlorine-37 Unknown: #’s of protons, electron and neutrons (cont)

4. Sample Problem 3-1 Plan: Atomic # =# of protons = # of neutrons Mass # =# of neutrons + # of Protons Compute: Mass # of element - atomic number of element = # of neutrons in element (cont)

5. Sample Problem 3-1 Mass # -atomic # = 37-17 = 20 neutrons An atom of chlorine-37 contains 17 electons 17 protons and 20 neutrons

6. Calculating Average Atomic mass Background- the average atomic mass of an element depends on both the mass amd the relative abundance of each of the element’s isotopes. Copper.6917 (% natural abundance) X 62.929 599 amu + 0.3083 X 64.927 793 amu = 63.55 amu

7. The Mole A mole is the amount of substance that contains as many particles as there are atoms in exactly 12 g of Carbon 12

8. Avogadro’s # Is the number of particles in exactly one mole of a pure substance. = 6.022 X 10 23 / means divide

9. Relationships Amount of element in moles X 6.022 X 10 23 = Number of atoms of element Number of atoms of element X 1/ 6.022 X 10 23 X molar mass of element = Mass of element in grams Mass of element in grams X 1/molar mass of element = Amount 0f element in moles

10. Relationships Mass of element in grams X 1/ molar mass of element X 6.022 X 10 23 = Number of atoms of element. Amount of element in moles X molar mass of element = Mass of element in grams Number of atoms of element = 6.022 X 10 23 X Amount of element in moles

11. Sample Problem 3-2 What is the mass in grams of 3.50 mol of the element copper, Cu? Given: 3.50 mol Cu Unknown : mass of Cu in grams Amount of Cu in moles -------  mass of Cu in grams Moles of Cu X grams of Cu/moles of Cu = grams of Cu

12. Sample Problem 3-2 Compute: 3.50 moles Cu X 63.5 g Cu/mol Cu = 222g Cu

13. Sample Problem 3-3 A chemist produced 11.0 g of Al. How many moles of Al were produced? Given: 11.9 g Al Unknown: amount of Al in moles Grams Al X moles Al/grams of Al = moles of Al 11.9 grams Al X mol Al/26.98 g Al =0.441 mol Al

14. Sample Problem 3-4 How many moles of silver, Ag are in 3.01 X 10 23 atoms of Ag? Given 3.01 X 10 23 atoms of Ag Unknown: Amount of Ag in moles Ag atoms X moles Ag/Avogadro’s # of Ag atoms = moles of Ag 3.01 X 10 23 Ag atoms X mol Ag/6.022 X 10 23 Ag atoms = 0.50 mol Ag

15. Sample Problem 3-5 What is the mass in grams of 1.20 X 10 8 atoms of Cu? Given: 1.20 X 10 8 atoms of Cu Unknown: mass of Cu in grams Plan: number of atoms of Cu  amount of Cu in Moles  mass of Cu in Grams

16. Sample 3-5 Plan: Cu atoms X moles Cu/Avogadro’s # X grams Cu/moles Cu = grams Cu Compute: 1.20 X 10 8 Cu atoms X 1 mol Cu/ 6.022 X 10 23 Cu atoms X 63.55 g Cu/ mol Cu = 1.27 X 10 -14 g Cu