1. Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.. Chapter 9 Chemical Composition

2.  Average mass/counting by weighing  Experimental determination of atomic mass  Moles and Avogadro's number  Molar Mass  Convert between moles and mass  Mass percent of elements in compounds  Empirical formulas  Calculating molecular formulas

3.  Nylon, aspartame, Kevlar (bulletproof vests), PVC, Teflon  All originated in chemist's laboratory  Once they make it – they must determine what it is ◦ What is it's composition? ◦ What is it's chemical formula?

4. The Mole Concept Avogadro’s Number = 6.022 x 10 23

5.  How can you quantify the amount of sand in a sand sculpture? Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.. CHEMISTRY & YOU You could measure the amount of sand in a sculpture by counting the grains of sand. Is there an easier way to measure the amount of sand?

6.  There are terms used to equate a certain quantity of matter o For example 1 dozen = 12, 1 Ton = 2,000 LBs and 1 gross = 144.  Before we get to the chemical terms lets convert between units we already know

7.  For example – let’s say you want to buy some Bazooka Gum o You could buy it by the piece from the deli o You could buy it by the box from ShopRite o You could buy it by mass from the candy store

8.  So, the different ways to measure Bazooka gum o 1 dozen = 12 pieces o 1 dozen = 1.75 boxes o 1 dozen = 168 grams

9.  Based on those conversions, I can determine how much Bazooka gum I have, regardless of how it is measured.  I have 95 pieces of Bazooka gum, how many grams do I have?

10.  I have 95 pieces of Bazooka gum, how many grams to I have?

11.  What Is a Mole? ◦ How do chemists count the number of atoms, molecules, or formula units in a substance? ◦ Mole - the number equal to the number of carbon atoms in 12.01 grams of carbon Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved..

12.  In chemistry, we do not use a dozen. o Since atoms are so small, we need a quantity with a larger amount  We use the mole (mol) o 1 mole = 6.022 X 10 23 representative particles (atoms, molecules, formula units) o It is known as Avogadro’s number

13.  1 mole Al = 27 grams = 6.02 x 10 23 atoms  1 mole Au = 197 grams = 6.02 x 10 23 atoms  1 mole Fe = 55.6 grams = 6.02 x 10 23 atoms  1 mole S = 32 grams = 6.02 x 10 23 atoms

14.  An Italian chemist whose important contributions was his resolution of the confusion surrounding atoms and molecules. Avogadro believed that particles could be composed of molecules and that molecules could be composed of still simpler units, atoms. The number of molecules in a mole (one gram molecular weight) was termed Avogadro's number

15.  1 mole of any representative particle is always 6.02 X 10 23  Using this value we can convert between moles and atoms/ molecules/ formula units

16. These particles are much, much smaller than grains of sand, and an extremely large number of them are in a small sample of a substance. Obviously, counting particles one by one is not practical. Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.. Recall that matter is composed of atoms, molecules, and ions.

17.  We usually are dealing with three types of representative particles o Molecules  H 2 O and H 2 are molecules, covalently bonded o Atoms  Al and Na are atoms, not bonded o Formula units  CaCl 2 and NaOH are formula units, ionicly bonded

18.  For example: If a sample contains 0.476 mol of O 2, how many representative particles are present?

19.  For example: If a sample contains 5.2 X 10 21 atoms of aluminum, how many moles are present?

20.  CO 2 is an air pollutant created from vehicle exhausts and factories. How many moles of CO 2 is 1.25 × 10 23 representative particles of CO 2 ? Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.. Sample Problem 10.2 Converting Number of Particles to Moles

21. 1. How many particles can be found in 3.91 moles of xenon? 2. If you had a bottle that contained 5.69x10 24 particles of water, how many moles of water does the bottle hold? 3. How many particles are in 2.85 moles of lithium chlorate?

23.  We can determine the molar mass of representative particles o Molar mass is the mass of 1 mole of that substance  The atomic mass is equal to the molar mass for an element o 1 mol Na = 22.99 g Na

24.  For molecules and formula units, we must calculate the molar mass  Let’s calculate the molar mass of SO 3 S = 1 X 32.07 = 32.07 O = 3 X 16.00 = 48.00 = 80.07 g/mol

25.  We can use the molar mass of a substance to convert from moles to mass o Remember the molar mass is the mass of 1 mole of that substance o SO 3 = 80.07 g/mol = 80.07 g SO 3 1 mol SO 3

26.  Using this relationship, we can easily convert a mole amount to a mass amount  How many grams are there in a 0.750 mole sample of SO 3 ?

27.  We can also convert a mass amount to a mole amount  How many moles are there in a 28.5 gram sample of SO 3 ?

28. ◦ The mole allows chemists to count the number of representative particles in a substance. ◦ The atomic mass of an element expressed in grams is the mass of a mole of the element. ◦ To calculate the molar mass of a compound, find the number of grams of each element in one mole of the compound. Then add the masses of the elements in the compound. Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved..

29. Do Now Calculate the number of moles and the number of representative particles in a 10.0 g sample of aluminum.

30.  Percent composition of a compound is the relative amount of each element in the compound.

31.  Example: Calculate the % composition of Sodium chlorate  1st: Determine the Formula: NaClO 3  2nd: Determine the Molar Mass of the compound: 23.0g + 35.5g + 3(16.0g) = 106.5g  3rd: To determine the % composition of each element that makes up sodium chlorate, you will use the following equation 3 times.

32.  In this next example: you are to determine the amount of sulfur in the compound Aluminum Sulfate.  1st Determine the formula:  2nd Determine the Molar Mass:  3rd Determine the % of Sulfur in this compound by using:

33. Which of the following compounds contains the highest percent of Iron? Iron III Acetate, Iron II Hydroxide or Iron II Oxide.

34.  Sometimes, percent composition can include compounds called hydrates  Hydrates are compounds that bind water molecules to their structure  BaCl 2 6H 2 O Name -> barium chloride hexahydrate  MgSO 4 3H 2 O Name ->

35. Find the percentage of water in barium chloride hexahydrate H2OH2O

37.  Find the percent composition of water in potassium sulfate hexahydrate.

38.  Empirical formula is the simplest whole number ratio between elements in a compound

39. You are given % of an element. Change the % to grams (the number does not change). Example 1: You have 26.56% K, 35.41% Cr, and 38.03% O. Therefore, you have: 26.56 g K, 35.41 g Cr, and Now change Grams to Moles

40.  Divide each molar quantity by the smaller number of moles to get 1 mol for the element with the smaller number of moles.  Multiply each part of the ratio by the smallest whole number that will convert both subscripts to whole numbers.

41. Example 2: 63.52% Fe and 36.48% S Find the Empirical Formula.

42. 1. Change % to grams 2. Change grams to moles 3. Divide all moles by the lowest number of moles 4. If all the numbers are not whole numbers, you must multiply the lowest number possible to get a whole number.

43. Calculate the empirical formula with the following: 56.4% potassium, 8.7% carbon, 34.9% oxygen Answer: K 2 CO 3

44.  Molecular Formula – A multiple of an empirical formula.  Can also be the same as empirical in some cases

45. Example: Empirical formula N 2 O – 2:1 ratio Molecular Formula of N 2 O could be: N 4 O 2 – A multiple of 2 2(N 2 O) = N 4 O 2 N 6 O 3 – A multiple of 3 3(N 2 O) = N 6 O 3 How to determine the multiple you need: Molecular Formula Mass = Multiple Empirical Formula Mass Then multiply the multiple to the Empirical Formula

46.  Example 1: Calculate the molecular formula of a compound whose molar mass is 60.0 g/mol and empirical formula is CH 4 N. First calculate the empirical formula mass.

47. Use the formula Molecular Formula Mass = Multiple Empirical Formula Mass Multiply the formula subscripts by this multiple to determine the molecular formula.

48. Example 2  What is the molecular formula of a compound with the empirical formula CClN and a molar mass of 184.5 g/mol.

49.  Find the molecular formula of ethylene glycol, which is used as antifreeze. The molar mass is 62.0 g/mole, and the empirical formula is CH 3 O