1. How to count things that are way too small to see, like… Atoms Ions Molecules Chapter 8 The Mole Concept

2. Learning Objectives for Ch 8 1. Use the mole concept to convert between moles and particles. 2. Determine the molar mass for any element or compound. 3. Convert between moles, grams and particles. 4. Given a chemical formula calculate the percent composition for the compound. 5. Given composition data, calculate the empirical and molecular formulas for a compound.

3. In chemistry, we describe reactions as occurring between particles such as atoms or molecules. Ex: H 2 +O 2 → H 2 O However, under typical laboratory conditions we cannot detect or measure individual particles such as these. For example, the mass of a H atom is 1.7 x 10 -24 g And the size of the H atom is about 3 x 10 -11 m. To set this to scale, if a H atom was the size of a pea, Mr. J would be 3.7 million miles tall (that’s to the moon and back 8 times!) The mole concept provides a method to convert quantities of the sub-microscopic particles into measurable quantities.

4. A rather silly example to follow!

5. You receive a delivery of one ton (2000 lbs) of Gummy Bears. The Gummy Bears are in boxes that each contain 1000 Gummy Bears. Each box weighs 2.5 pounds. How would you calculate the total number of Gummy Bears that you have??? Remember, you are feeling lazy and don’t want to count all those boxes!!!

6. 2000 lbs1 box1000 GB 2.5 lbs1 box We know: Weight GummyBears = 2.5 lbs / 1 Box And : 1000 GummyBears / 1 Box So we can calculate the total number of GummyBears

7. But, What if different colors of GB’s had different weights?? One Box of GB ’ s contains 500 GB ’ s. Each color of GB ’ s comes in a different box but all contain 500 GB ’ s. 1 box of yellow GB ’ s = 2 lbs 1 box of blue GB ’ s = 3 lbs 1 box of red GB ’ s = 4 lbs 1 box of green GB ’ s = 5 lbs

8. So, how much would 1 GB of each color weigh? Yellow : 500 GB / 1 box 2 lb / 1 box Weight = For Red, weight = 1 GB1 box4 lb 500 GB1 box 1 GB1 box2 lb 500 GB1 box

9. A Brief History In 1811 Amadeo Avogadro proposed that the number of particles in a given volume of a gas (at a specific temperature and pressure) is always the same, regardless of the identity of the gas. In 1865, Loschmidt estimated a value for what Specific number this would be. Lorenzo Romano Amedeo Carlo Avogadro di Quaregna e di Cerreto

10. The mass of a H-1 atom is 1 amu. This is a relative unit that only tells us that the mass is 1/12 that of a C-12 atom. It cannot be “weighed” in the lab. However, the H-1 atom has a real mass in grams it is just much too small to measure. If we had some large number of H-1 atoms we could weigh them directly and get a mass in grams. If we had 100 atoms we would have some mass in grams 100(mass of H-1) If we had a million H-1 atoms, mass = 10 6 (mass of H-1) As it turns out, if we had 6.02 x 10 23 atoms of H-1, the mass we would have would be 1.00 grams!!!! This is the significance of the number 6.02 x 10 23. It converts the atomic mass of an element from amu to grams.

11. He-4 has a mass 4 times that of H-1. So if we had 6.02 x 10 23 atoms of He-4 they would have a mass 4 times that of the same number of H-1 atoms. This gives a mass of 4.00 g. Mole is defined as 6.022 x 10 23 particles. This can also be seen as an important conversion factor: 6.022 x 10 23 particles / 1 mole Just as:pair = 2 dozen = 12 gross = 144 The mass of 1 mole (6.022 x 10 23 particles) of any element will have a mass equal to the atomic mass of that element in GRAMS.

12. How to find the mass of 1 mole of an element 4.0026 g He 1 mole He 24.31g Mg 1 mole Mg 15.99g O 1 mole O 65.4g Zn 1 mole Zn

13. Essentially, the mole concept allows us to use the periodic table as a list of conversion factors to get the actual mass of atoms of any element in grams!!! A unit that we can actually measure in the lab!!

14. Practice: H = g / 1 mole N = g / 1 mole Cl = g / 1 mole W = g / 1 mole U = g / 1 mole

15. This means that the average atomic mass for an element of the periodic table also has a second meaning: the mass of 1 mole of that element in grams. This quantity is called the molar mass for an element and is always written as # g / 1 mole. The molar mass for Ge = _______________ The molar mass for Ti = _______________ With this concept, we have conversion factors between grams and moles for every element on the periodic table.

16. Calculations with the mole concept What is the mass of 2.0 moles of Ag? We want to convert from moles to grams. The conversion factor (molar mass) is Set up the conversion as usual: = 107.9 g Ag 1 mole Ag 2.0 mole Ag107.9 g Ag 1 mole Ag

17. 50.0g of F is how many moles? 3.5 moles of Ca is how many atoms? Of Au, H, Be? We want to convert from moles to atoms (particles). The conversion is: So: Atoms Au =Atoms H=Atoms Be= 6.02 x 10 23 atoms Ca 1 mole Ca 3.5 mole Ca6.02 x 10 23 atoms Ca 1 mole Ca

18. What is the mass of 1 atom of Pt? There is no direct conversion between atoms and grams!!! But we know And Therefore: 1 atom Pt1 mole Pt195.1 g Pt 6.02 x 10 23 atoms Pt1 mole Pt 6.02 x 10 23 atoms Pt 1 mole Pt 195.1 g Pt 1 mole Pt

19. Practice: Solve the following: How many moles are in 95.0 g of Bromine? 6.2 moles of Zn contains how many grams? What is the mass of 1000 atoms of tin? 0.005 g of Au is comprised of how many atoms?

20. The mole concept is not limited to elements only. A mole is a number that refer to anything, but in chemistry it will normally refer to atoms, molecules or ions. The formula for NaCl indicates that it contains 1 Na and 1 Cl for every NaCl molecule (not actually a molecule since it is an ionic compound). This means that 1 mole of NaCl will contain 1 mole of Na and 1 mole of Cl 1 mole of Na = 22.99 g1 mole of Cl = 35.45 g So, 1 mole of NaCl = 22.99 + 35.45 = 58.44 g This is the molar mass for a compound (often called formula weight, FW) Molar mass for NaCl = 58.44 g / 1 mole

21. Calculating Molar Mass of compounds CO 2 contains 1 C and 2 O C has a molar mass of 12.01 g/mol O has a molar mass of 15.99 g/mol CH 3 O - 1 C 3 H 1 O 1 C1 x 12.0112.01 2 O2 x 15.9935.98 Molar mass=43.99 g/mol 1 C1 x 12.0112.01 3 H3 x 1.0083.024 1 O1 x 15.9915.99 Molar mass=31.02 g/mol

22. H 2 SO 4 - 2 H 1 S 4 O 1 H2 x 1.008 2.016 1 S1 x 32.0732.07 4 O4 x 15.9963.96 Molar mass=98.05 g/mol

23. Practice: calculate the molar mass for the following C 3 H 8 – K 2 S – Al 2 O 3 – C 5 H 6 N 2 O -

24. 1 mole = 6.02 x 10 23 particles Atoms, molecules, ions Molar mass in grams for an element From periodic table Molar mass in grams for compounds Add up element atomic masses

25. Mixed Practice How many g in 2.5 moles of H 2 O? 47.0 g of KCl is how many moles?

26. Practice continued How many NO 2 molecules in 5 grams of NO 2 ? How many O atoms in 5 grams of NO 2 ?

27. Review: Remember, the masses given on the periodic table are average atomic masses that are weighted averages of the masses of each isotope and their natural abundances. Ex: Ag has 2 isotopes: Ag-107 = 106.905 abundance = 51.35% Ag-109 = 108.905 abundance = 48.65% To calculate Avg At Mass: A weighted average! 106.905 x 0.5135 = 54.90 Multiply the isotope mass 108.905 x 0.4865 = 52.98 by it’s natural abundance 107.88(move the decimal point 2 places). Add the results!

28. Percent Composition Percent composition of a chemical compound tells the percentage by mass of the components of a chemical compound. Imagine that you have a fruit basket that contains 5 apples and3 oranges Each apple weighs 3 pounds and each orange weighs 1 pound. How would you calculate the percent by weight of apples and oranges?

29. Remember that a percent is a part divided by the total times 100. The total weight of fruit is 5 apples x 3 pounds each=15 lbs 3 oranges x 1 pound each= 3 lbs Total weight =18 lbs The % by weight of apples is: x 100 = 83.3% The % by weight of oranges is : x 100 = 16.7% 15 lbs apples 18 lbs total 3 lbs oranges 18 lbs total

30. Calculating % composition for compounds works the same way. Find the molar mass of the compound. Divide the total mass of a given element by the total molar mass of the compound. Multiply by 100. Ex:H 2 O H: 2 x 1.008 = 2.016%H = x 100 = 11.19% O: 1 x 15.99 = 15.99 Molar mass = 18.01 % O = x 100 = 88.78% 2.016 18.01 15.99 18.01

31. Ex 2:C 2 H 6 O Molar mass = % C = %H = %O =

32. Practice: calculate the percent composition for all elements in the following Fe 2 O 3 Na 2 CO 3 C 12 H 22 O 11

33. Calculating Empirical Formulas How do we know that the formula for water is H 2 O? If we analyzed a sample of water in the lab, the analytical data we would get would be in weight percent. We could then calculate the formula from this data. The subscripts in a chemical formula are mole ratios. For water the formula indicates that there are 2 moles of H for every 1 mole of O. To calculate the formula we need to determine the mole ratios. Essentially, we are determining the subscripts. The empirical formula always gives the lowest ratio between the elements in a compound H 2 O not H 4 O 2 or H 6 O 3

34. In an empirical formula problem, you will be given % composition to calculate the formula. For example: A sample of a gas is found to contain 27.3% C and 72.7% O. Determine the empirical formula. Stepwise approach: 1. Convert the given percentages to grams. Simply change the “%” sign to g. 27.3% C = 27.3 g C 72.7% O = 72.7 g O 2. Convert grams of each element to moles using molar masses. = 2.273 mol C = 4.544 mol O 27.3 g C1 mol C 12.01 g C 72.7 g O1 mol O 16.00 g O

35. 3. To get a ratio, divide the number of moles for each element by the smallest number of moles. C: = 1.00 O: = 1.999 4. If the ratios are within 0.1 of a whole number then round to that whole number. C: 1.00 = 1O: 1.999 = 2 5. If a ratio is not near a whole number, multiply all ratios by an appropriate factor to get whole numbers. Ex: C: 1.00O: 1.49 = 1.50 x 2 x 2 2.00 = 2 3.00 = 3 2.273 mol 4.544 mol 2.273 mol

36. 6. Use the whole number ratios as the subscripts in the empirical formula. Since the example gave C:1, O:2, we write CO 2 Practice: Analysis of a gas sample gave the following % compositions15.93% B, 84.07% F Calculate the empirical formula.

37. Practice 2: A compound is 52.90% Al and 47.08% O. Calculate the empirical formula. Practice 3: A compound is 52.21% C, 13.04% H and 34.77% O. Calculate the empirical formula.

38. Deriving molecular formulas from empirical formulas. The molecular formula gives the exact number of moles of each element in a mole of compound while the empirical only gives the simplest ratio of moles. The molecular formula must be some multiple of the empirical formula. To find the molecular formula we must find that multiple. To find the molecular formula, you must be given the molar mass for the molecular formula. 1. Find the molar mass for the empirical formula. 2. Divide the molar mass for the molecular formula by the molar mass of the empirical formula. Round to the nearest whole number. This will give you the multiplying factor. 3. Multiply the empirical formula by the factor to get the molecular formula.

39. Ex: A compound has an empirical formula of C 2 H 3 and a molar mass of 81.13 g/mol. What is the molecular formula? Molar mass of C 2 H 3 = 27.05 factor = 81.13 / 27.05 = 2.996 = 3 molecular formula = 3 (C 2 H 3 ) = C 6 H 9 Practice: A compound is found to have a composition of 39.97% C, 6.67% H and 53.29% O and a molar mass of 60.05 g/mol. Calculate the empirical and molecular formulas for this compound.