2. Dimensional Analysis Conversion factors Units Cancel to solve

3. Counting Atoms Mg burns in air (O 2 ) to produce white magnesium oxide, MgO. How can we figure out how much oxide is produced from a given mass of Mg?

4. Counting Atoms Chemistry is a quantitative science—we need a “counting unit.” 1 mole is the amount of substance that contains as many particles (atoms, molecules) as there are in 12.0 g of 12 C. MOLE

5. Particles in a Mole 6.02214199 x 10 23 Avogadro’s Number There is Avogadro’s number of particles in a mole of any substance. Amedeo Avogadro 1776-1856

6. Converting particles to moles moles = representative particles x 1 mole 6.02 x 10 23 representative particles (atoms/molecules/formula units) moles = 2.80 x 10 24 atoms Si x 1 mole Si 6.02 x 10 23 atoms Si (atoms/molecules/formula units) = 4.65 moles Si

7. Converting moles to particles Rep part = moles x 6.02 x 10 23 representative particles 1 mole Rep part = 1.14 mol SO 3 x 6.02 x 10 23 molecule 1 mole = 2.75 x 10 24 molecules SO 3

8. The mass of moles The atomic mass of an element expressed in grams is the mass of a mole of the element. Molar Mass - the mass of a mole of an element. Molar mass of a compound is the total mass of all the atoms making up the compound. Molar mass of a compound is calculated by adding the masses of each element in the compound. –Take the subscript and multiply it by the molar mass of that element then add all the elemental masses together.

9. Molar Mass 1 mol of 12 C = 12.00 g of C = 6.022 x 10 23 atoms of C 12.00 g of 12 C is its MOLAR MASS Taking into account all of the isotopes of C, the molar mass of C is 12.011 g/mol

10. One-mole Amounts

11. PROBLEM: What amount of Mg is represented by 0.200 g? How many atoms? Mg has a molar mass of 24.3050 g/mol. = 4.95 x 10 21 atoms Mg How many atoms in this piece of Mg?

12. Converting moles to mass mass = number of moles x mass (grams) 1 mole mass = 3.00 mol Na x 22.98977grams Na 1 mole Na = 69.0 grams Na

13. Converting mass to moles moles = mass in grams x 1 mole mass (grams) moles = 10.0 g Na 2 SO 4 x 1 mole Na 2 SO 4 142.1 grams Na 2 SO 4 = 0.070 g Na 2 SO 4

14. Mole to volume Avogadro’s hypothesis states equal volumes of gases at the same temperature and pressure contain equal numbers of particles. STP – 0 o C, 273 K / 1 atm, 101.3 kPa Molar volume at STP –22.4 L/mol

15. Converting mole to volume volume = moles of gas x 22.4 L 1 mol at STP = 8.40 L O 2 volume = 0.375moles of O 2 x 22.4 L 1 mol at STP

16. moles = 0.200 L H 2 x 1 mol 22.4 L at STP Converting volume to moles moles of gas = volume x 1 mol 22.4 L at STP = 8.93 x 10 -3 L H 2

17. Converting Molar Mass from Density Density is mass per unit volume (g/L) Molar mass grams per mol Need to cancel moles and insert liters Use molar volume – mol/liter Grams = grams 22.4 Lx Moles L 1 mol

18. Converting Molar Mass from Density Grams = grams 22.4 Lx Moles L 1 mol = 43.99 g/mol = 1.964 grams 22.4 Lx 1L 1 mol molar mass (g/mol)

19. Mol grams liters atoms    molar mass Avogadro’s number molar volume 22.4 L/mol 6.02 x 10 23 particles/mol Periodic table

20. Percent Composition Relative amounts (mass) of elements in a compound. Percent by mass. Mass of the element divided by the mass of the compound times 100. Use the subscripts to determine the mass of each element and the mass of the compound if the question calls for a percent composition and only the formula is given.

21. Percent Composition Consider NO 2, Molar mass = ? What is the weight percent of N and of O? What are the weight percentages of N and O in NO?

22. Determining Formulas In chemical analysis we determine the % by weight of each element in a given amount of pure compound and derive the EMPIRICAL or SIMPLEST formula. Remember formulas are mole to mole ratios. PROBLEM : A compound of B and H is 81.10% B. What is its empirical formula?

23. Because it contains only B and H, it must contain 18.90% H.Because it contains only B and H, it must contain 18.90% H. Assume a 100 gram sample.Assume a 100 gram sample. In 100.0 g of the compound there are 81.10 g of B and 18.90 g of H.In 100.0 g of the compound there are 81.10 g of B and 18.90 g of H. Calculate the number of moles of each constituents.Calculate the number of moles of each constituents. A compound of B and H is 81.10% B. What is its empirical formula?

24. Calculate the number of moles of each element in 100.0 g of sample. A compound of B and H is 81.10% B. What is its empirical formula?

25. Now, recognize that atoms combine in the ratio of small whole numbers. 1 atom B + 3 atoms H --> 1 molecule BH 3 or 1 mol B atoms + 3 mol H atoms ---> 1 mol BH 3 molecules Find the ratio of moles of elements in the compound. A compound of B and H is 81.10% B. What is its empirical formula?

26. But we need a whole number ratio. 2.5 mol H/1.0 mol B = 5 mol H to 2 mol B EMPIRICAL FORMULA = B 2 H 5 Take the ratio of moles of B and H. Always divide by the smaller number. A compound of B and H is 81.10% B. What is its empirical formula?

27. A compound of B and H is 81.10% B. Its empirical formula is B 2 H 5. What is its molecular formula ? Is the molecular formula B 2 H 5, B 4 H 10, B 6 H 15, B 8 H 20, etc.? B 2 H 6 is one example of this class of compounds. B2H6B2H6

28. Molecular formulas Molecular formulas are either the empirical formula or it is a whole number multiple of its empirical formula. Determine the empirical formula first, to determine the molar mass of the compound. Divide the experimental molar mass by the empirical molar mass. This give you the multiplier to convert the empirical formula to the molecular formula.

29. A compound of B and H is 81.10% B. Its empirical formula is B 2 H 5. What is its molecular formula ? We need to do an EXPERIMENT to find the MOLAR MASS. This experiment gives 53.3 g/mol Compare with the mass of B 2 H 5 = 26.66 g/unit = 26.66 g/unit Find the ratio of these masses. Molecular formula = B 4 H 10

30. Determine the formula of a compound of Sn and I using the following data. Reaction of Sn and I 2 is done using excess Sn.Reaction of Sn and I 2 is done using excess Sn. Mass of Sn in the beginning = 1.056 gMass of Sn in the beginning = 1.056 g Mass of iodine (I 2 ) used = 1.947 gMass of iodine (I 2 ) used = 1.947 g Mass of Sn remaining = 0.601 gMass of Sn remaining = 0.601 g See p. 104See p. 104

31. Find the mass of Sn that combined with 1.947 g I 2. Mass of Sn initially = 1.056 g Mass of Sn recovered = 0.601 g Mass of Sn used = 0.455 g Find moles of Sn used: Tin and Iodine Compound

32. Now find the number of moles of I 2 that combined with 3.83 x 10 -3 mol Sn. Mass of I 2 used was 1.947 g. How many mol of iodine atoms ? = 1.534 x 10 -2 mol I atoms

33. Tin and Iodine Compound Now find the ratio of number of moles of moles of I and Sn that combined. Empirical formula is SnI 4