1. Topic 1: Stoichiometry 1.1 Nature of Substances 1.2 The Mole

2. Overview

3. Topic 1.1 Basic Concepts of Chemistry

4. We Are Here

5. Particle Theory Review SOLIDLIQUIDGAS Diagram Motion of particles Spacing of particles Shape Compressibility Sound conductivity Intermolecular forces

6. Word Equations Hydrogen + Oxygen  Water REACTANTS PRODUCTS The arrow means: ‘makes’ or ‘becomes’ not ‘equals’

7. Symbol Equations 2 H 2 + O 2  2 H 2 O vs H 2 + O 2  H 2 O 4 H 4 H 2 H 2 H 2 O 2 O 2 O 1 O The red numbers are called coefficients and tell you the number of each molecule involved in the reaction Required to balance the equation Without them the equation does not balance – each side of the reaction would have different numbers of each atom – which would break physics You can’t change the subscript numbers in the formulas as this changes the chemical If there is no coefficient, it is ‘1’

8. Construct equations and then balance them for each of the following: Magnesium (Mg) reacts with hydrochloric acid (HCl) to make magnesium chloride (MgCl 2 ) and hydrogen gas Ethane (C 2 H 6 ) reacts with oxygen gas to make carbon dioxide and water Lead nitrate (Pb(NO 3 ) 2 ) reacts with aluminium chloride (AlCl 3 ) to make aluminium nitrate (Al(NO 3 ) 3 ) and lead chloride (PbCl 2 ) Barium nitride (Ba 3 N 2 ) reacts with water to make barium hydroxide (Ba(OH) 2 ) and ammonia (NH 3 ) Ammonium perchlorate (NH 4 ClO 4 ) reacting with aluminium to make aluminium oxide (Al 2 O 3 ), aluminium chloride (AlCl 3 ), nitric oxide (NO) and water

9. Topic 1.1.2 Formulas and Composition by Mass

10. We Are Here

11. Formulas and Composition by Mass Objectives: Understand the difference between empirical and molecular formulas Use relative atomic mass to calculate relative formula mass Determine percentage composition by mass given an empirical or molecular formula Calculate empirical and molecular formulas from experimental data

12. Types of Formula Empirical: The ratio of the atoms of each element in a compound in its lowest terms Molecular: The number of atoms of each element in a molecule Note: because of their structure, ionic (and giant covalent) compounds do not form molecules so empirical formula is the only one relevant.…more on this in the bonding topic. You will meet other types in the organic chemistry unit including structural, displayed and skeletal.

13. Examples Oxygen (divide by 2) Molecular: O 2 Empirical: O Water (divide by 1) Molecular: H 2 O Empirical: H 2 O Sodium Chloride Molecular: n/a* Empirical: NaCl  Ethane (divide by 2)  Molecular: C 2 H 6  Empirical: ???  Glucose (???)  Molecular: C 6 H 12 O 6  Empirical: ???  Copper Sulphate (???)  Molecular: ???  Empirical: CuSO 4 *This does not have a molecular formula because it is ionic and ionic compounds don’t form molecules

14. Your turn… Chlorine gas Propane Ozone  Sulphuric acid  Ethanol  Sodium hydroxide Write empirical and molecular formulas for each of the following:

15. Relative Atomic Mass, A r The periodic table tells you the relative atomic mass of each element. This is the mass of an element relative to a 12 th of the mass of 12 C. It is a relative value, which means it has no units. Relative atomic mass has the symbol ‘A r ’ For example carbon: A r (C) = 12.01 The reason it isn’t a whole number is due to isotopes, which we will meet next topic

16. Relative Molecular or Formula Mass, M r Relative Molecular Mass is the mass of a molecule relative to a 12 th of the mass of 12 C. It is calculated by adding up the A r for each atom in a molecule (see next slide). The related term relative formula mass refers to the relative mass of one unit of a formula and is used to for empirical formulas

17. Calculating M r HCl A r (H) = 1.01 A r (Cl) = 35.45 M r = 1.01 + 35.45 = 36.46 C 2 H 4 A r (C) = 12.01 A r (H) = 1.01 M r = 2x12.01 + 4x1.01 = 16.06  H 2 SO 4  A r (H) = 1.01  A r (S) = 32.06  A r (O) = 16.00  M r = 2x1.01 + 32.06 + 4x16.00 = 98.08  Mg(OH) 2  Ar(Mg) = 24.31  Ar(O) = 16.00  Ar (H) = 1.01  M r = 24.31 + 2x16.00 + 2x1.01 = 58.33

18. Calculate M r for: Br 2 C 3 H 8  (NH 4 ) 2 SO 4  C 6 H 12 O 6

19. Percentage Composition by Mass If we divide the total mass of each element in a compound by the number of atoms For example ethanol, C 2 H 5 OH, M r = 46.08 CHO Number Present261 Multiply by A r 2 x 12.01 = 24.026 x 1.01 = 6.061 x 16.00 = 16.00 Divide by M r, convert to % 24.02/46.08 x 100 = 52.1% 6.06/46.08 x 100 = 13.1% 16.00/46.08 x 100 = 34.7%

20. Calculate % composition by mass for: 1. H 2 O 2. Mg(OH) 2 3. CuSO 4.5H 2 O 4. C 12 H 22 O 11 Answers: 1) H: 11.2%, O: 88.8%, 2) Mg: 41.7%, O: 54.7%, 3.46%, 3) Cu: 25.4%, S: 12.8%, O: 57.7%, H: 4.04%, 4) C: 42.1%, H: 6.49%, 51.4%

21. Determining Empirical and Molecular Formula from % Composition by Mass Write symbols as a ratio Divide each % by A r Write % composition below Multiply to remove fractions Divide M r by formula mass Multiply empirical formula by above Divide each by smallest Multiply everything to remove decimals that look like fractions ‘.5’ …..multiply by 2 ‘.33’ or ‘.67’….multiply by 3 ‘.25’ or ‘.75’….multiply by 4 With a mix of endings, you may need to do this a couple of times.

22. Example: a sample of a compound contains 20% hydrogen and 80% carbon by mass and M r = 30.08 C:H 80%:20% 80/12.01 = 6.67:20/1.01 = 20 6.67/6.67 = 1:20/6.67 = 3 n/a since no awkward decimals above Empirical formula = CH 3 : 30.08/(12.01 + 3 x 1.10) = 2 Molecular formula = CH 3 x 2 = C 2 H 6 Write symbols as a ratio Divide each % by A r Write % composition below Multiply to remove fractions Divide M r by formula mass Multiply empirical formula by above Divide each by smallest

23. Example: a sample of a compound contains 8.4% hydrogen, 65.2% carbon and 29.1% nitrogen by mass, and M r = 288.5 C:H:NC:H:N 62.5%:8.4%:29.1% 62.5/12.01= 5.20 8.4/1.01= 8.3129.1/14.01=2.08 5.20/2.08 = 2.5 8.31/2.08 = 4 2.08/2.08 = 1 2.5 x 2 = 5 4 x 2 = 8 1 x 2 = 2 Empirical formula = C 5 H 8 N 2 so 288.5/ 96.2 = 3 Molecular formula = C 5 H 8 N 2 x 3 = C 15 H 24 N 6 Write symbols as a ratio Divide each % by A r Write % composition below Multiply to remove fractions Divide M r by formula mass Multiply empirical formula by above Divide each by smallest

24. Lesson 3 Meet the Mole

25. We Are Here

26. Words and Numbers You are probably well used to the idea of words being used to represent numbers: A Pair = 2 A Trio = 3 A Dozen = 12 A Score = 20 A Gross = 144 A Grand = 1000 You are probably able to do some simple maths involving these numbers….

27. A Mole ‘Mole’ is another word used to describe a number The only difference is that it is a really large number A mole is: 6.02 x 10 23 602,000,000,000,000,000,000,000 Six hundred and two thousand quadrillion Given the symbol, L This number is called Avogadro’s number after the Italian scientist Amadeo Avogadro who first proposed it Picture – all the grains of sand, on all the beaches and in all the deserts in the world; that is about a tenth of a mole! We normally talk about moles of atoms or molecules, but could talk about moles of apples and bananas if we wanted

28. Moles and Quantity An amount of length is measure in metres An amount of time is measured in seconds An amount of force is measure in Newtons An amount of mass is measured in kilograms An amount of chemicals is measured in moles If we say we have the same amount of two chemicals, we mean we have the same number of moles.

29. Summary of all moles calculations: moles Number of particles Avogadro’s Constant = Mass of Substance Molar Mass = Pressure x Gas Volume Gas Constant x Temperature = Volume of Solution x Concentration =

30. Converting between numbers of particles and quantities in moles The left-hand equation lets us calculate a quantity in moles (n) from a number of particles (N) The right-hand equation lets us calculate a number of particles (N) from a quantity in moles (n). Where: n = quantity in moles N = number of particles L = Avogadro’s Constant (6.02x10 23 )

31. Example 1. You have 3.01x10 22 atoms of carbon. How many moles is this? n = N / L Since we are looking for a number of moles given a number of particles n(C) = N(C) / L No adjustment since no relevant sub-units n(C) = 3.01x10 22 / 6.02x10 23 n(C) = 0.0500 mol Choose your equation Sub in your numbers Adjust for the number of sub-units Evaluate the sum Note how brackets are used to indicate which substance we are referring to

32. Example 2. You have 6.02x10 24 molecules of water. How many moles of hydrogen atoms are present? n = N / L Since we are looking for a number of moles given a number of particles n(H) = (2 x N(H 2 O)) / L N multiplied by two, since each H 2 O molecule contains 2 H atoms n(H) = 2 x 6.02x10 24 / 6.02x10 23 n(H) = 20.0 mol Choose your equation Sub in your numbers Adjust for the number of sub-units Evaluate the sum