2. Chemical formulas are not simply abbreviations for words. They represent precise quantities. What 2 “bits” of information do formulas give? Water for example is H 2 O. It indicates that 1 molecule of water consists of 2 hydrogen atoms and 1 oxygen atom. How could you represent 2 molecules of water? Use a coefficient This formula shows 4 hydrogen atoms and 2 oxygen atoms. Elements Elements Number of atoms of each element Number of atoms of each element 2 H 2 O

3. In daily life, pieces of matter are often measured either by counting them or by massing them; the choice is determined by convenience. If you buy eggs, you buy them by the dozen —that is, by number. Eggs are easy to count out. So are oranges and lemons. Other items, though countable, are more conveniently sold by mass. A dozen peanuts is too small a number to buy, and several hundred are too difficult to count. You buy peanuts by the pound or kilogram—that is, by mass. Chemists, too, are interested in quantity—the quantity of an element or compound, which, like grocery items, can be measured by number or by mass. Although you can easily mass a sample of a substance, yet the number of atoms or molecules in it is much too large to count. Nevertheless, chemists are interested in knowing such numbers.

4. In the laboratory, when we “run” a reaction we want to know the number of atoms, molecules, or formula units in a substance because these are the entities that react with each other. However, these entities are much too small to count individually, so chemists use a unit called the mole to count by massing them. The mole was derived from the Latin word moles, meaning “ heap ” or “ pile.” The mole, whose symbol, mol, is the SI base unit for measuring the amount of substance. A mole is the amount of substance that is equal to the number of carbon atoms in exactly 12 g of the carbon-12 isotope. One mole always contains the same number of particles, no matter what the substance. 1 mole = 6.022 x 10 23 particles This value is referred to as Avogadro’s number (Amadeo Avogadro) in honor of the man who conceived the basic idea.

5. One mole each of various substances - Clockwise from top left: 1-octanol (C 8 H 17 OH); mercury(II) iodide (HgI 2 ); methanol (CH 3 OH); sulfur (S 8 ).

6. The central relationship between the mass of one atom and the mass of one mole of atoms is that the mass of an element is expressed in amu. The mass of one mole of atoms of an element is the same numeric value but measured in grams. 1 atom of sulfur has a mass of 32.07 amu. 1 mole of sulfur atoms has a mass of 32.07 g.

7. So, 1 mole S = 6.022 x 10 23 atoms S = 32.07 g S 1 mole of Fe 2+ = 6.022 x 10 23 ions= 55.85 g Fe 2+ 1 mole H 2 O = 6.022 x 10 23 molecules H 2 = 18.02 g H 2 O 1 mole NaCl = 6.022 x 10 23 formula units NaCl= 58.44 g NaCl Add up the individual masses of the elements. If we are talking about the mass of an atom we call it the atomic mass. If a compound is a molecule, we call this the molecular mass. If a compound is a formula unit, we call it the formula mass. How can I get the mass of one unit of a compound?

8. g-formula mass = 1 MOLE = 6.022 x 10 23 “particles” elementschargedcovalentionic elements charged covalentionic particlescompoundscompounds particles compounds compounds 1 mole substance of a substance = Avogadro’s constant = (6.022 x 10 23 ) mass of substance in grams

10. A 1-carat diamond has a mass of 0.200 g. moles How many moles of carbon? atoms How many atoms of carbon? USE FLM!!!!!!! Start with your given, identify units of answer, and write conversion factors. 0.200 g C1 mole C =moles C 12.01 g C 0.200 g C =moles C 0.200 g C = 1 mole C 0.0167 = 0.0167moles C 12.01 g C 0.200 g C1 mole C =atoms C 12.01 g C 0.200 g C =atoms C 0.200 g C1 mole C6.022 x 10 23 atoms C =atoms C 12.01 g C1 mole C 0.200 g C1 mole C6.022 x 10 23 atoms C 1.00 x 10 22 = 1.00 x 10 22 atoms C 12.01 g C1 mole C HINT: If moles of your substance does not appear in your given or in the desired units, then you must do at least 2 conversion factors.

11. A ring is constructed out of 3.06 x 10 22 atoms Au. How many grams? How many moles? 3.06 x 10 22 atoms Au1 mole Au =moles Au 6.022 x 10 23 atoms Au 3.06 x 10 22 atoms Au =moles Au 3.06 x 10 22 atoms Au1 mole Au196.97 grams Au =grams Au 6.022 x 10 23 atoms Au1 mole Au 3.06 x 10 22 atoms Au1 mole Au = 0.0508moles Au 6.022 x 10 23 atoms Au 3.06 x 10 22 atoms Au1 mole Au =grams Au 6.022 x 10 23 atoms Au 3.06 x 10 22 atoms Au1 mole Au196.97 grams Au = 10.0 grams Au 6.022 x 10 23 atoms Au1 mole Au

13. A raindrop contains about 0.050 g of water. How many molecules of water? How many moles of O? How many atoms of H?

14. = 1.7 x 10 21 molecules H 2 O How many molecules of water? How many moles of O? = 0.0028 moles O 0.050 grams H 2 O1 mole H 2 O 18.02 grams H 2 O 6.022 x 10 23 molecules H 2 O = 1 mole H 2 O 0.050 grams H 2 O = 1 mole H 2 O = 18.02 grams H 2 O 1 mole O = 1 mole H 2 O

15. How many atoms of H? 0.050 grams H 2 O1 mole H 2 O2 mole H6.022 x 10 23 atoms H = 18.02 grams H 2 O1 mole H 2 O1 mole H 0.050 grams H 2 O1 mole H 2 O 6.022 x 10 23 molecules H 2 O 2 atoms H = 18.02 grams H 2 O1 mole H 2 O1 molecule H 2 O = 3.3 x 10 21 atoms H

16. One teaspoon of table sugar (sucrose) is approximately 15 g of C 12 H 22 O 11. How many moles of sugar is this? 15 g C 12 H 22 O 11 1 mole C 12 H 22 O 11 = 342.34 g C 12 H 22 O 11 How many atoms of carbon are in this 15 g sample? 15 g C 12 H 22 O 11 1 mole C 12 H 22 O 11 6.022 x 10 23 molecules C 12 H 22 O 11 12 atoms C 342.34 g C 12 H 22 O 11 1 mole C 12 H 22 O 11 1 molecule C 12 H 22 O 11

17. What mass of CaCl 2 would be needed to furnish 1.00 mole of Cl 1- ions? 1.00 mole Cl 1- ions1 mole CaCl 2 110.98 g CaCl 2 = 2 moles Cl 1- ions1 mole CaCl 2

18. Compare the mass of each element present in one mole of a compound to the total mass of one mole of a compound. EXAMPLE: Find the % composition of Al 2 (SO 4 ) 3 Al=2x26.98= 53.96 S=3x32.07= 96.21 O=12x16.00=192.00 342.17 Al=x= S=x= O=x= =2x= S=3x= O=12x= Al=2x26.98= S=3x32.07= O=12x16.00= Al=2x26.98= 53.96 S=3x32.07= 96.21 O=12x16.00=192.00

19. Check the sum of the %’s. They should sum to very near EXAMPLE: Find the % composition of Al 2 (SO 4 ) 3 Al=2x26.98= 53.96 S=3x32.07= 96.21 O=12x16.00=192.00 342.17 % Al =x 100%= % S =x 100%= % O =x 100%= 53.96 342.17 15.77 % Al 96.21 28.12 % S 342.17 192.00 56.11 % O 342.17 100.00 %

20. EXAMPLE: Find the mass percent (% composition) of Ferrous nitrate Fe=x= N=x= O=x= % Fe =x 100%= % N =x 100%= % O =x 100%= 55.85 179.87 31.05 % Fe 28.02 15.58 % N 179.87 96.00 53.37% O 179.87 100.00 % Fe=1x= N=2x= O=6x= =x55.85= N=x14.01= O=x16.00= 55.85 28.02 96.00 179.87

21. If you had a 100.0 g sample of ferrous nitrate, how many grams of iron are there? 31.05 g Fe Nitrogen? 15.58 g N If you had a 540.0 g sample of ferrous nitrate, how many grams of iron are there? 540.0 x (.3105 % Fe) = 167.7 g Fe 540.0 g Fe(NO 3 ) 2 = 31.05 g Fe 100 g Fe(NO 3 ) 2 Nitrogen? 540.0 x (.1558 % N) = 540.0 g Fe(NO 3 ) 2 = 15.58 g N 100 g Fe(NO 3 ) 2 84.13 g N

22. Calcium fluoride 48.67% F, while sodium fluoride 45.25% F

23. We can also do the reverse and use percent composition data to determine empirical and molecular formulae. Empirical formulas simplest whole number ratio of atoms in a compound. Molecular formulas show the actual numbers of atoms of each element in a compound.

25. There are two ways to determine empirical formulas. Experimental Data % composition data Need to know mass of Assume the amount each element in the to be 100 g laboratory sample

26. EXAMPLE: 40.0% C 6.71% H 53.3% O 1. Assume 100 g sample 40.0 g C 6.71 g H 53.3 g O 2. Convert to moles 40.0 g C 1 mole C = 12.01 g C 6.71 g H1 mole H = 1.01 g H 53.3 g O1 mole O = 16.00 g O = 3.33 moles C= 6.64356 moles H= 3.33 moles O

27. = 3.33 moles C = 6.64356 moles H= 3.33 moles O 3.Find the mole ratio (divide by the smallest number of moles.) 3.33 moles C = 3.33 moles 6.64356 moles H = 3.33 moles 3.33 moles O = 3.33 moles = 1 C= 2 H= 1 O If the mole ratio ends in.20 (x5).25 (x4).33 or.66 (x3).50 (2) Otherwise round to one of the above decimals or to the nearest whole number. 4. Empirical Formula CH 2 O or CHOH

28. EXAMPLE: A binary compound consisting only of nitrogen and oxygen with a mass percent of 37.00% N 37.00 g N1 mole N = 14.01 g N 63.00g O1 mole O = 16.00 g O 2.6409707 mole N = 3.9375 mole O = 2.6409707 mole N = 2.6409707 mole 3.9375 mole O = 2.6409707 mole = 1 N = 1.490929 O

29. = 1 N = 1.490929 O multiply by 2 to eliminate and make whole number ratios. N2O3N2O3

30. EXAMPLE: 37.70% Na, 22.95% Si, 39.34% O 37.70 g Na 1 mole Na = 22.99 g Na 22.95 g Si1 mole Si = 28.09 g Si 39.34g O1 mole O = 16.00 g O 1.63984 mol Na = 0.8170167 mol Si = 2.45875 mol O = = 2.007 Na= 1 Si= 3.009 O 1.63984 mol Na = 0.8170167 mol 0.8170167 mol Si = 0.8170167 mol 2.45875 mol O = 0.8170167 mol Na 2 SiO 3

31. Molecular formulas show the actual number of atoms of each element in a molecule, as well as, the ratio of atoms. In order to solve these types of problems, 3 pieces of information are needed. molecular mass of the compound empirical mass of the compound number of empirical units present in the molecular formula

32. EXAMPLE: What is the formula of an unknown substance with a percent composition of 5.90% H and 94.10% O. It also has a molecular mass of 34.00 g/mol. 5.90% H and 94.10% Omeans5.90 g H and 94.10 g O 5.90 g H1 mole H = 1.01 g H 94.10 g O1 mole O = 16.00 g O 5.841584 mole H=5.88125 mole O= 5.841584 mole H = 5.841584 mole 5.88125 mole O = 5.841584 mole = 1.00 H = 1.006790 O

33. Empirical Formula is:HO The empirical mass is 17.01 g (1.01 + 16.00) We need to determine how much bigger the molecular formula is compared to the empirical. 34.00 = 17.01 34.00 = 2 17.01 Empirical Formula is:HO Molecular Formula is:H 2 O 2

34. Hydrates are crystals that contain water molecules in their crystal structure. The crystal has crystallized from a water solution with molecules of water adhering to the particles of the crystal. EXAMPLE: NiSO 3. 6 H 2 O 1 crystal of NiSO 3 with 6 molecules of H 2 O

35. To calculate the formula of a hydrate, we need: mass of the hydrated sample mass of dry sample (anhydrous, w/o water) after the H 2 O has been driven off. Mole ratio of the anhydrous compound to the H 2 O driven off. Finding the formula of a hydrate is very similar to finding empirical formulas.

36. EXAMPLE: We have a 10.407 g sample of hydrated barium iodide. The sample is heated to drive off the water. The dry sample has a mass of 9.520 g. What is the formula of the hydrate? The difference between the initial mass and that of the dry sample is the mass of water that was driven off. Mass of hydrate10.407 g Mass of dry sample- 9.520 g Mass of water 0.887 g

37. 9.520 g BaI 2 1 mole BaI 2 = 391.13 g BaI 2 0.887 g H 2 O1 mole H 2 O = 18.02 g H 2 O 0.02434 moles BaI 2 =0.0492 moles H 2 O= 0.02434 moles BaI 2 = 0.02434 moles 0.0492 moles H 2 O = 0.02434 moles = 1 BaI 2 = 2 H 2 O Formula: BaI 2. 2 H 2 O