Chapter 6 – Chemical Composition Counting Atoms

1. Chapter 6 – Chemical Composition Counting Atoms • Atoms are too small and too numerous to count individually. • A unit of measure called the mole has been established for use in counting atoms. • One mole of something consists of 6.022 X 1023 units of that substance: 1 mol = 6.022 X 1023

2. Moles • A mole is often referred to as “the chemist’s dozen” because it gives the number of “things” in one mole of anything • 1 mole eggs = 6.022 x 1023 eggs • 1 mole pencils = 6.022 x 1023 pencils • 1 mole of an element = 6.022 x 1023 atoms • 1 mole carbon = 6.022 x 1023 carbon atoms • 1 mole of a compound = 6.022 x 1023 molecules • 1 mole water = 6.022 x 1023 water molecules

3. Example #1: How many sodium atoms are in 4.5 mol of sodium?

4. Example #2: How many moles of copper are in 1.38 X 1025 copper atoms?

5. Moles and Molar Mass • The SI definition of the mole is the amount of a substance that contains as many entities as there are in exactly 12 grams of carbon-12. 1 mol carbon = 6.022 x 1023 C atoms = 12.01 g C • The definition of a mole provides a relationship between mass (grams of a substance) and the number of atoms (Avogadro’s number) • The molar mass of any substance is the mass (in grams) of 1 mol of the substance. • The atomic mass gives the number of grams in one mole of an element. • The formula mass gives the number of grams in one mole of a compound.

6. Example #3: Calculate the number of moles of gold in 26.2 g of gold.

7. Example #4: Calculate the number of atoms present in 2.34 g of calcium.

8. Example #5: Calculate the mass (in grams) of 7.9 X 1021 uranium atoms.

9. Example #6: Calculate the number of moles in 8.91 g of potassium permanganate.

10. Example #7: Calculate the mass (in grams) of 1.27 mmol of carbon dioxide.

11. Example #8: Calculate the mass (in grams) of 4.03 X 1022 molecules of benzene, C6H6.

12. Counting Moles Since we count atoms and molecules in mole units, we can find the number of moles of a constituent element if we know the number of moles of the compound. For example, 1 mol of CH4 molecules consists of 1 mol of carbon atoms and 4 mol of hydrogen atoms. Moles of Compound Moles of Constituents 1 mol NaCl 1 mole Na, 1 mole Cl 1 mol H2O 2 mol H, 1 mole O 1 mol CaCO3 1 mol Ca, 1 mol C, 3 mol O 1 mol C6H12O6 6 mol C, 12 mol H, 6 mol O

13. Example #9: Determine the number of moles of oxygen in 3.5 mol of Ca(OH)2. Given: 3.5 mol of Ca(OH)2 Find: mol O Conversion Factor: 1 mol of Ca(OH)2 molecules contains 2 mol of O atoms Solution Map: mol Ca(OH) 2 -> mol of O Solution: 3.5 mol Ca(OH)2 X 2 mol of O 1 mol of Ca(OH)2

14. Example #10: A 15.5 gram sample of diphosphorous pentoxide contains how many grams of phosphorous? Given: 15.5 g of P2O5 Find: g of P Conversion Factors: formula mass of P2O5 = 2(30.97 g/mol) + 5(16.00 g/mol) = 141.94 g/mol 1 mol of P2O5 molecules contains 2 mol of P atoms 1 mol of P = 30.97 g of P Solution Map: g P2O5 -> mol P2O5 -> mol P -> g P Solution: 15.5 g P2O5 X 1 mol P2O5 X 2 mol P X 30.97 g P 141.94 g P2O5 1 mol P2O5 1 mol of P = 6.76 g P

15. Example #11: Find the grams of iron in 79.2 g of iron (III) oxide. Given: 79.2 g of Fe2O3 Find: g of Fe Conversion Factors: formula mass of Fe2O3 = 2(55.85 g/mol) + 3(16.00 g/mol) = 159.70 g/mol 1 mol of Fe2O3 molecules contains 2 mol of Fe atoms 1 mol of Fe = 55.85 g of Fe Solution Map: g Fe2O3 -> mol Fe2O3 -> mol Fe -> g Fe Solution: 79.2 g Fe2O3 X 1 mol Fe2O3 X 2 mol Fe X 55.85 g Fe 159.70 g Fe2O3 1 mol Fe2O3 1 mol of Fe

16. Percent Composition • Percentage of an element in a compound. • Can be determined from either the formula of the compound or the experimental mass analysis of the compound. The percent composition or mass percent of an element can be determined from experimental data using the following formula: Mass percent of element X = Mass of X in a sample of compound X 100% Mass of the sample of compound X • The mass percent tells you the mass of a constituent element in 100 g of the compound. For example, the fact that Al2O3 is 52.9% aluminum by mass means that 100 g of Al2O3 contains 52.9 g of Al. • The mass percent can be used as a conversion factor. For example, g Al2O3 X 52.9 g Al = g Al 100 g Al2O3

17. Example #12: A 4.67 g sample of iron completely reacts with oxygen to form 6.67 g of iron (III) oxide. Calculate the mass percent composition of iron in iron (III) oxide. Given: sample of 6.67 g Fe2O3 4.67 g of iron in sample Find: mass percent composition of Fe in Fe2O3 Equation: Mass percent of element X = Mass of Fe in the Fe2O3 sample X 100% Mass of the Fe2O3 sample Solution: Mass % Fe = 4.67 g Fe X 100% 6.67 g Fe2O3

18. Example #13: Copper (I) bromide contains 44.30% copper by mass. Calculate the mass of copper in grams contained in 32.5 g of copper (I) bromide. Given: 32.5g CuBr and mass % Cu in CuBr is 44.30% Find: g Cu Conversion Factor: 44.30 g Cu in every 100 g CuBr Solution Map: g CuBr -> g Cu Solution: 32.5 g CuBr X 44.30 g Cu 100 g CuBr

19. Example #14: Calculate the mass percent composition of phosphorous in potassium phosphate. Find: mass percent composition of phosphorous in K3PO4 Equation: Mass percent of element X = Atomic Mass of X X 100% Formula Mass of Compound Solution: Atomic Mass of P = 30.97 g/mol Formula Mass of K3PO4 = 3(39.10 g/mol) + 1(30.97 g/mol) + 4(16.00 g/mol) = 212.27 g/mol Mass % P = 30.97 g/mol P 212.27 g/mol K3PO4

20. Example #15: Calculate the mass percent of each element in carvone, C10H4O.

21. Example #16 - Problem 108 in the text: A leak in the air conditioning of an older car releases 55 g CF2CL2 per month. How much Cl is emitted into the atmosphere each year by this car?

22. Chapter 6 Homework Due March 14th 3, 4, 5, 6, 8, 18, 19, 20, 24, 26 30, 34, 40, 44, 45 (d) only, 51 (d) only, 52 (b) and (d), 56, 58, 64, 72, 74, 78, 80, 98, 100, 110