1. Up until this point we have been talking about atoms and molecules. The problem with this approach is that atoms and molecules are very small things. In a single drop of water for example, there are trillions and trillions of water molecules. A reaction between a single molecule of hydrogen and a single molecule of oxygen, as we discussed above, would be undetectable. Instead of talking about single molecules in science, we talk about groups of molecules. You can think of it like buying eggs. You don't go to the store and buy an egg - you buy a dozen. Contained within that dozen are the individual eggs. Its the same thing when we talk about molecules. We don't talk about single units, we talk about groups. But even a dozen molecules is a tiny amount. What we need is a big number - a huge number! That number is the mole.

2. One mole equals 6.02 x 10 23 (also known as Avogadro's number). A 6 followed by 23 zeros. Now that's a pretty big number. But that's all it is, a number. You can't just have a mole, you have to have a mole of something. A mole of atoms. A mole of water molecules. A mole of pennies (which would make you richer than you can imagine). 600,000,000,000,000,000,000,000! A penny has approximately 6 x 10 23 molecules. Amadeo Avagadro

3. An Avogadro's number of standard soft drink cans would cover the surface of the earth to a depth of over 200 miles. If you had Avogadro's number of unpopped popcorn kernels, and spread them across the United States of America, the country would be covered in popcorn to a depth of over 9 miles. If we were able to count atoms at the rate of 10 million per second, it would take about 2 billion years to count the atoms in one mole. If marbles that have a diameter of one centimeter were to be lined up end-to-end in a straight line, the distance covered by this string of marbles can hold in about 500,000,000 of our Solar Systems placed end-to-end.

4. Why the mole? As it turns out, the mole has some interesting properties. One mole of hydrogen atoms (6.02 x 10 23 H atoms) weighs 1 g. In fact, one mole of any element is equal to the atomic mass of that element (in grams). Let's think about that for a second. If we know the molar mass of an element, and we know how many elements make up a specific molecule, then you can calculate the molar mass of a compound by adding up the atomic weights. Huh? Take water for example. How much does a mole of water weigh? Well, one mole of water contains one mole of oxygen atoms and two moles of hydrogen atoms. A mole of hydrogen weighs 1 g and a mole of oxygen weighs 16 g (look at the atomic mass in the periodic table). So to calculate the weight of one mole of water: (2 moles H * 1 g per mole) + (1 mole O * 16 g per mole) = 18 g One mole of water weighs 18 grams!

5. We often need to figure out the molecular mass of a compound, particularly when making solutions for the lab. To find the molecular weight of a compound, simply add the atomic weights of each atom in the compound. Example: CaCl 2 has one Ca and 2 Cls. According to the Periodic Table: atomic weight of Ca = 40 g/mol atomic weight of Cl = 35 g/mol 35 x 2 = 70 Therefore 1 mole (or 6.02 x 1023 molecules) of CaCl 2 weighs 40 g/mol + 70 g/mol = 110 g/mol.

6. Another example: Find the molecular weight of C 3 H 8. C 3 H 8 has 3 Cs and 8 Hs. According to the Periodic Table: The atomic weight of C is 12 g/mol 12 x 3 = 36 g/mol The atomic weight of H is 1 g/mol 1 x 8 = 8 g/mol Therefore, 1 mole (or 6.02 x 10 23 molecules) of C 3 H 8 weighs 36 g/mol + 8 g/mol = 44 g/mol. Try some: Calculate the molecular weight for 1 mole of: 1.HCl2. NaH 2 PO 4 3. CaBr 2 4. RbS 2 5.KMnO 4 6. NH 4 Cl7. C 5 H 12 8. (NH 4 ) 2 SO 3 9. Al 2 SO 3 10. C 6 H 12 O 6

7. Moles (mol) Mass (g) Atoms or Molecules Volume (L) Mass  molecular weight # moles  6.022  10 23 atoms/mol

8. Challenge: What if you wanted to know the molecular weight of a compound with more or less than 1 mole? Think about it and try: 1. 10 moles of CH 4 2. 0.1 moles of NH 3 3. 0.002 moles of H 2 O4. 86 moles of AlF 3 5. 1.15 moles of NaCl6. 15 moles of KMnO 4 Challenge: What if you know the weight, but need to find the number of mole. Create a way to figure out how many moles are in 24 grams of NaCl.

9. Calculating The Number of Atoms in a Specific Mass You have a 1.00 g sample of lead. How many atoms of lead are present?  1.00 g Pb1 amu 1.66 x 10 24 g 1 atom Pb 207.2 amu 

10. Calculating Mass Example Calculate the mass (in amu) of 1.0 х 10 4 carbon atoms 1) Given : 2) Plan: Convert from atoms to amu 3) CF 4) Set Up Problem

11. Formula Mass The sum of atomic masses of all atoms in its formula Important role in nearly all chemical calculations Can be calculated for compounds and diatomic elements

12. Calculating Formula Mass Calculate the formula mass of calcium chloride Write the formula from the name given Ca 2+ (from group II) and Cl - (from group VII) Formula is CaCl 2 due to charge balance Formula mass: Sum of the atomic masses of atoms in the formula (1 Ca atom + 2 Cl atoms) = 40.08 amu = 70.90 amu Formula mass of CaCl 2

13. Calculating the Number of Molecules in a Mole How many molecules of bromine are present in 0.045 mole of bromine gas? Given: 0.045 mol Br 2 Need: molecules of Br 2 Avogadro’s number Conversion factors: Equality: Set Up Problem:

14. Calculating the Moles of an Element in a Compound How many moles of carbon atoms are present in 1.85 moles of glucose? Plan: moles of glucose moles of C atoms subscript (One) mol C 6 H 12 O 6 = 6 mols C atoms Equality: Conversion Factors: Set Up Problem: 11.1 mol C atoms

15. Molar Mass of a Compound Calculate the molar mass of iron (II) sulfate Formula is FeSO 4 1) Calculate the molar mass of each element 2) Each element is multiplied by its respective subscript: (number of moles of each element) 3) The molar mass is calculated by the sum of the molar masses of each element Moles of Element in Compound Moles of Compound Formula Subscript

16. Molar Mass of a Compound Formula is FeSO 4 : 1) Formula is FeSO 4 : The molar masses of iron, sulfur, and oxygen are 2) Multiply each molar mass by its subscript 3) Find the molar mass of the compound by adding the mass of each element

17. Calculations Using Molar Mass The three quantities most often calculated – Number of particles – Number of moles – Number of grams Using molar mass as a conversion factor is one of the most useful in chemistry – Can be used for g to mole and mole to g conversions

18. Converting Mass of a Compound to Moles International Foods Coffee contains 3 mg of sodium chloride per cup of coffee. How many moles of sodium chloride are in each cup of coffee? Equality: 1 mol NaCl = 58.44 g 3 mg NaCl moles of NaCl = 0.003 g NaCl

19. Converting Grams to Particles Ethylene glycol (antifreeze) has the formula C 2 H 6 O 2. How many molecules are present in a 3.86 × 10 -20 g sample? Plan: convert g moles molecules of ethylene glycol Equality 1: Conversion Factor 1 Equality 2: Conversion Factor 2 375 molecules Molar mass Avog Number

20. Percent Composition It is the percent by mass of each element in a compound Can be determined – By its chemical formula – Molar masses of the elements that compose the compound The percent of each element contributes to the mass of the compound

21. Calculating Percent Composition Example What is the percent composition of each element in NH 4 OH? Determine the contribution of each element Molar mass

22. Empirical Formulas The simplest ratio of elements in a compound It uses the smallest possible whole number ratio of atoms present in a formula unit of a compound If the percent composition is known, an empirical formula can be calculated

23. Empirical Formulas To Determine the empirical formula: 1) Calculate the moles of each element  Use molar mass (atomic mass) 2) Calculate the ratios of the elements to each other 3) Find the lowest whole number ratio  Divide each number of moles by the smallest number of moles present

24. Empirical Formula: Converting Decimal Numbers to Whole Numbers The subscripts in a formula are expressed as whole numbers, not as decimals The resulting numbers from a calculation represent each element’s subscript If the number(s) are NOT whole numbers, multiply each number by the same small integer (2, 3, 4, 5, or 6) until a whole number is obtained

25. Relating Empirical and Molecular Formulas n represents a whole number multiplier from 1 to as large as necessary Calculate the empirical formula and the mass of the empirical formula Divide the given molecular mass by the calculated empirical mass – Answer is a whole number multiplier

26. Relating Empirical and Molecular Formulas  Multiply each subscript in the empirical formula by the whole number multiplier to get the molecular formula

27. Calculate Empirical Formula from Percent Composition Lactic acid has a molar mass of 90.08 g and has this percent composition: 40.0% C, 6.71% H, 53.3% O What is the empirical and molecular formula of lactic acid? Assume a 100.0 g sample size – Convert percent numbers to grams

28. Calculate Empirical Formula from Percent Composition Convert mass of each element to moles Divide each mole quantity by the smallest number of moles The ratio of C to H to O is 1 to 2 to 1 Empirical formula is Empirical formula mass = 12.01 + 2 (1.008) + 16.00 = 30.03 g/mol CH 2 O

29. Determination of the Molecular Formula Obtain the value of n (whole number multiplier) Multiply the empirical formula by the multiplier Molecular formula = n х empirical formula Molecular formula = 3 (CH 2 O) C3H6O3C3H6O3

30. Formulas for Compounds Empirical Formula – Smallest possible set of subscript numbers – Smallest whole number ratio – All ionic compounds are given as empirical formulas Molecular Formulas The actual formulas of molecules It shows all of the atoms present in a molecule It may be the same as the EF or a whole- number multiple of its EF Molecular formula = n х Empirical formula

31. Parts of Chemical Equations Flour + Water+ Egg+ Yeast -> Bread    Reactants yield Product Reactants are the starting chemicals in a chemical reaction. Products are the new chemicals made in a chemical reaction. H 2 + Cl 2 -> 2HCl C 12 H 22 O 6 -> 12C + 11H 2 + 3O 2 The chemical properties of the reactants changed when they are combined.

32. What is an equation? Think about math…what does an equation in math look like? Something like this: a + b = c + d Chemistry is very similar. The equations look a little different, but amount to the same thing. For example: 2H 2 + O 2  2H 2 O + means to add or combine the chemicals  is basically the same as =, but means “becomes” so H 2 and O 2 combine to become H 2 O The 2 is the subset and is the number of atoms in the molecule, in this case, there are 2 Hs and 2 Os to start with, and 2 Hs in the end. Remember, no subset number means you have one of that atom. What about the 2 (big)?

33. 2H 2 + O 2  2H 2 O The big 2 is a coefficient. This is how many molecules you have. You will learn how to figure out how many of each molecule you need in a chemical reaction later this week. In the case of making water, you start with 2 H 2 s and one O 2, and make 2 waters. You will learn how to “balance” equations this week, as well.

34. Chemical bonds are VERY important in chemical reactions. The valence electrons dictate the kinds of reactions that can and will happen. You must be comfortable using them. If you aren’t, please come for extra tutoring. Reactions creating covalent double bonds. Reactions creating single bonds. Reactions forming ionic compounds.

35. Balancing Chemical Equations 2H 2 + O 2 -> 2H 2 O (Water) H 2 + O 2 -> H 2 O 2 (NO!) (Hydrogen Peroxide) The twos in front of the Hydrogen are called Coefficients. Coefficients tell us how much of an atom or molecule is present without changing the formula. Never add a subscript to any formula. You will change the formula.

36. How to balance equations H 2 + O 2 -> H 2 O Step 1: Write the equation out without coefficients. Make sure the formulas are written correctly. Step 2: Count the oxygen. There are two on the left side, and one on the right. Add a coefficient of two to the water to make the oxygens even, do not add any subscripts. H 2 + O 2 -> 2H 2 O

37. How to balance equations Step 3: Count the hydrogen. There are 4 on the right. (2H 2, 2*2=4) and 2 on the left. Write a 2 in front of the hydrogen on the right. H 2 + O 2 -> 2H 2 O 2H 2 + O 2 -> 2H 2 O Step 4: Check your work. All elements should have the same number of atoms on both sides. If not, check your work again. 2H 2 = 2H 2 ? Yes O 2 = 2O? Yes

39. TIPS for balancing Equations To make this an easier process try doing it in this order: 1. Balance the metals first. 2. Balance all non metals except O and H. 3. Balance the oxygen 4. Balance the hydrogen Also, make sure the formulas are written correctly

40. Let’s Practice! 1. Cl 2 + NaBr -> NaCl + Br 2 2. Fe + O 2 -> FeO 3. C 3 H 8 + O 2 -> CO 2 + H 2 O 4. Al + Cl 2 -> AlCl 3 5. C 12 H 22 O 6 -> C + H 2 + O 2

41. Write balanced chemical equations for the following reactions: (a) Solid mercury(II) sulfide decomposes into its component elements when heated. (b) The surface of aluminum metal undergoes a combination reaction with oxygen in the air. (c) The reaction that occurs when ethanol, C 2 H 5 OH(l), is burned in air.

42. Calculate the percentage of carbon, hydrogen, and oxygen (by mass) in C 12 H 22 O 11. Without using a calculator, arrange the following samples in order of increasing numbers of carbon atoms: 12 g 12 C, 1 mol C 2 H 2, 9  10 23 molecules of CO 2.

43. (a)How many nitric acid molecules are in 4.20 g of HNO 3 ? (b) How many O atoms are in this sample? Ascorbic acid (vitamin C) contains 40.92% C, 4.58% H, and 54.50% O by mass. What is the empirical formula of ascorbic acid?

44. A 5.325-g sample of methyl benzoate, a compound used in the manufacture of perfumes, is found to contain 3.758 g of carbon, 0.316 g of hydrogen, and 1.251 g of oxygen. What is the empirical formula of this substance? Mesitylene, a hydrocarbon that occurs in small amounts in crude oil, has an empirical formula of C 3 H 4. The experimentally determined molecular weight of this substance is 121 amu. What is the molecular formula of mesitylene?