1. Climate Change Climate change is a significant shift in temperature and weather patterns around the world. While some changes are normal, the vast majority of scientists agree that our activities are causing dramatic changes to the Earth’s climate.

2. A little bit of carbon dioxide is a good thing – it keeps the planet warm and habitable. Now, however, we are putting so much carbon dioxide into the atmosphere that the planet is getting too warm. This problem is also called the greenhouse effect.greenhouse effect Over the last three hundred years we have radically increased our use of energy sources like oil, coal and natural gas. Burning these fuels releases carbon dioxide into the atmosphere.

3. Greenhouse effect module

4. Effects The effects of climate change are already visible – melting glaciers, the migration of plant and animal species, and an increase in severe storms, to name a few. What will happen next? No one can say for sure, but here are some possibilities: Rising sea levels destroy coastal areas Frequent and intense heat waves More droughts and wildfires Extinction of millions of species Spreading of weather-sensitive disease

5. Carbon Calculator Estimate your personal carbon usage!! Here's a different sort of calculator Low Impact Living My footprint

6. Convert your average CO 2 consumption into grams There are 2000 lb/ton There are 454 g/lb

7. Contents and Concepts Mass and Moles of Substances Here we will establish a critical relationship between the mass of a chemical substance and the quantity of that substance (in moles). Molecular Mass and Formula Mass The Mole Concept

8. Determining Chemical Formulas Explore how the percentage composition and mass percentage of the elements in a chemical substance can be used to determine the chemical formula. 3.Mass Percentages from the Formula 4.Elemental Analysis: Percentages of C, H, and O 5.Determining Formulas

9. Stoichiometry: Quantitative Relations in Chemical Reactions Develop a molar interpretation of chemical equations, which then allows for calculation of the quantities of reactants and products. 6.Molar Interpretation of a Chemical Equation 7.Amounts of Substances in a Chemical Equation 8.Limiting Reactant: Theoretical and Percentage Yield

10. Figure 3.3: Reaction of zinc and iodine causing iodine to vaporize. Photo courtesy of James Scherer.

11. Counting Objects of Fixed Relative Mass 12 red marbles @ 7g each = 84g 12 yellow marbles @4e each=48g 55.85g Fe = 6.022 x 10 23 atoms Fe 32.07g S = 6.022 x 10 23 atoms S

12. The Mole is based upon the definition: The amount of substance that contains as many elementary parts (atoms, molecules, or other?) as there are atoms in exactly 12 grams of carbon -12. 1 Mole = 6.022045 x 10 23 particles The Mole

13. Mole - Mass Relationships of Elements Element Atom/Molecule Mass Mole Mass Number of Atoms 1 atom of H = 1.008 amu 1 mole of H = 1.008 g = 6.022 x 10 23 atoms 1 atom of Fe = 55.85 amu 1 mole of Fe = 55.85 g = 6.022 x 10 23 atoms 1 atom of S = 32.07 amu 1 mole of S = 32.07 g = 6.022 x 10 23 atoms 1 atom of O = 16.00 amu 1 mole of O = 16.00 g = 6.022 x 10 23 atoms 1 molecule of O 2 = 32.00 amu 1 mole of O 2 = 32.00 g = 6.022 x 10 23 molecule 1 molecule of S 8 = 2059.52 amu 1 mole of S 8 = 2059.52 g = 6.022 x 10 23 molecules

14. CaCO 3 100.09 g Oxygen 32.00 g Copper 63.55 g Water 18.02 g One Mole of Common Substances

15. Figure 3.2: One mole each of various substances. Photo courtesy of American Color.

16. Molecular Mass The sum of the atomic masses of all the atoms in a molecule of the substance Formula Mass The sum of the atomic masses of all atoms in a formula unit of the compound, whether molecular or not

17. Molecular Mass - Molar Mass ( M ) The Molecular mass of a compound expressed in amu is numerically the same as the mass of one mole of the compound expressed in grams. For water: H 2 O Molecular mass = (2 x atomic mass of H ) + atomic mass of O = 2 ( 1.008 amu) + 16.00 amu = 18.02 amu Mass of one molecules of water = 18.02 amu Molar mass = ( 2 x atomic mass of H ) + atomic mass of O = 2 ( 1.008 g ) + 16.00 g = 18.02 g 18.02 g H 2 O = 6.022 x 10 23 molecules of water = 1 mole H 2 O

18. Calculate the formula weight of the following compounds from their formulas. Report your answers to three significant figures. –c–calcium hydroxide, Ca(OH) 2 –m–methylamine, CH 3 NH 2

19. 3 significant figures 74.1 amu 3 significant figures 31.1 amu Total74.095 2 O2(16.00) =32.00 amu Ca(OH) 2 1 Ca1(40.08) =40.08 amu 2 H2(1.008) =2.016 amu CH 3 NH 2 1 C1(12.01) =12.01 amu 5 H5(1.008) =5.040 amu 1 N1(14.01) =14.01 amu Total31.060

20. Information Contained in the Chemical Formula of Glucose C 6 H 12 O 6 ( M = 180.16 g/mol) Carbon (C) Hydrogen (H) Oxygen (O) Atoms/molecule of compound Moles of atoms/ mole of compound Atoms/mole of compound Mass/molecule of compound Mass/mole of compound 6 atoms 12 atoms 6 atoms 6 moles of 12 moles of 6 moles of atoms atoms atoms 6(6.022 x 10 23 ) 12(6.022 x 10 23 ) 6(6.022 x 10 23 ) atoms atoms atoms 6(12.01 amu) 12(1.008 amu) 6(16.00 amu) =72.06 amu =12.10 amu =96.00 amu 72.06 g 12.10 g 96.00 g

22. What is the mass in grams of the nitric acid molecule, HNO 3 ? First, find the molar mass of HNO 3 : 1 H1(1.008) =1.008 1 N1(14.01) =14.01 3 O3(16.00) =48.00 63.018 (2 decimal places) 63.02 g/mol

23. Next, convert this mass of one mole to one molecule using Avogadro’s number:

24. Mole, mol The quantity of a given amount of substance that contains as many molecules or formula units as the number of atoms in exactly 12 g of carbon-12 Avogadro’s Number, N A The number of atoms in exactly 12 g of carbon-12 N A = 6.02 × 10 23 (to three significant figures)

25. Molar Mass The mass of one mole of substance For example: Carbon-12 has a molar mass of 12 g or 12 g/mol

26. A sample of nitric acid, HNO 3, contains 0.253 mol HNO 3. How many grams is this? First, find the molar mass of HNO 3 : 1 H1(1.008) =1.008 1 N1(14.01) =14.01 3 O3(16.00) =48.00 63.018 (2 decimal places) 63.02 g/mol

27. Next, using the molar mass, find the mass of 0.253 mole: = 15.94406 g

28. Mass - Mole Relationships of a Compound Mass (g) of Element Moles of Element Atoms of Element Mass (g) of compound Amount (mol) of compound Molecules (or formula units of compound) Amount (mol) of compound For an ElementFor a Compound

29. Calculating the Number of Moles and Atoms in a Given Mass of Element Problem: Tungsten (W) is the element used as the filament in light bulbs, and has the highest melting point of any element 3680 o C. How many moles of tungsten, and atoms of the element are contained in a 35.0 mg sample of the metal? Plan: Convert mass into moles by dividing the mass by the atomic weight of the metal, then calculate the number of atoms by multiplying by Avogadro’s number! Solution: Converting from mass of W to moles: Moles of W = 35.0 mg W x = 0.00019032 mol 1.90 x 10 - 4 mol NO. of W atoms = 1.90 x 10 - 4 mol W x = = 1.15 x 10 20 atoms of Tungsten 1 mol W 183.9 g W 6.022 x 10 23 atoms 1 mole of W

30. Calculating the Moles and Number of Formula Units in a Given Mass of Cpd. Problem: Sodium Phosphate is a component of some detergents. How many moles and formula units are in a 38.6 g sample? Plan: We need to determine the formula, and the molecular mass from the atomic masses of each element multiplied by the coefficients. Solution: The formula is Na 3 PO 4. Calculating the molar mass: M = 3x Sodium + 1 x Phosphorous = 4 x Oxygen = = 3 x 22.99 g/mol + 1 x 30.97 g/mol + 4 x 16.00 g/mol = 68.97 g/mol + 30.97 g/mol + 64.00 g/mol = 163.94 g/mol Converting mass to moles: Moles Na 3 PO 4 = 38.6 g Na 3 PO 4 x (1 mol Na 3 PO 4 ) 163.94 g Na 3 PO 4 = 0.23545 mol Na 3 PO 4 Formula units = 0.23545 mol Na 3 PO 4 x 6.022 x 10 23 formula units 1 mol Na 3 PO 4 = 1.46 x 10 23 formula units

31. Flow Chart of Mass Percentage Calculation Moles of X in one mole of Compound Mass % of X Mass fraction of X Mass (g) of X in one mole of compound M (g / mol) of X Divide by mass (g) of one mole of compound Multiply by 100

32. Calculating Mass Percentage and Masses of Elements in a Sample of a Compound - I Problem: Sucrose (C 12 H 22 O 11 ) is common table sugar. ( a) What is the mass percent of each element in sucrose? ( b) How many grams of carbon are in 24.35 g of sucrose? Mass Fraction of C = = (a) Determining the mass percent of each element: mass of C = 12 x 12.01 g C/mol = 144.12 g C/mol mass of H = 22 x 1.008 g H/mol = 22.176 g H/mol mass of O = 11 x 16.00 g O/mol = 176.00 g O/mol 342.296 g/mol Finding the mass fraction of C in Sucrose & % C : Total mass of C 144.12 g C mass of 1 mole of sucrose 342.30 g Cpd = 0.421046 To find mass % of C = 0.421046 x 100% = 42.105%

33. Calculating Mass Percents and Masses of Elements in a Sample of Compound - II (a) continued Mass % of H = x 100% = x 100% = 6.479% H Mass % of O = x 100% = x 100% = 51.417% O (b) Determining the mass of carbon: Mass (g) of C = mass of sucrose X( mass fraction of C in sucrose) Mass (g) of C = 24.35 g sucrose X = 10.25 g C mol H x M of H 22 x 1.008 g H mass of 1 mol sucrose 342.30 g mol O x M of O 11 x 16.00 g O mass of 1 mol sucrose 342.30 g 0.421046 g C 1 g sucrose

34. Chemical Formulas Empirical Formula - Shows the relative number of atoms of each element in the compound. It is the simplest formula, and is derived from masses of the elements. Molecular Formula - Shows the actual number of atoms of each element in the molecule of the compound. Structural Formula - Shows the actual number of atoms, and the bonds between them ; that is, the arrangement of atoms in the molecule.

35. Empirical and Molecular Formulas Empirical Formula - The simplest formula for a compound that agrees with the elemental analysis! The smallest set of whole numbers of atoms. Molecular Formula - The formula of the compound as it exists, it may be a multiple of the Empirical formula.

36. Climate Change Climate change is a significant shift in temperature and weather patterns around the world. While some changes are normal, the vast majority of scientists agree that our activities are causing dramatic changes to the Earth’s climate.

37. A little bit of carbon dioxide is a good thing – it keeps the planet warm and habitable. Now, however, we are putting so much carbon dioxide into the atmosphere that the planet is getting too warm. This problem is also called the greenhouse effect.greenhouse effect Over the last three hundred years we have radically increased our use of energy sources like oil, coal and natural gas. Burning these fuels releases carbon dioxide into the atmosphere.

38. Greenhouse effect module