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.

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 .

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.

Greenhouse effect

module

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

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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

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

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

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

Scherer.

Counting Objects of Fixed Relative Mass

12 red marbles @ 7g each = 84g12 yellow marbles @4e each=48g

55.85g Fe = 6.022 x 1023 atoms Fe32.07g S = 6.022 x 1023 atoms S

• 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 1023 particles

The Mole

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 1023 atoms

1 atom of Fe = 55.85 amu 1 mole of Fe = 55.85 g = 6.022 x 1023 atoms

1 atom of S = 32.07 amu 1 mole of S = 32.07 g = 6.022 x 1023 atoms

1 atom of O = 16.00 amu 1 mole of O = 16.00 g = 6.022 x 1023 atoms

1 molecule of O2 = 32.00 amu 1 mole of O2 = 32.00 g = 6.022 x 1023 molecule

1 molecule of S8 = 2059.52 amu 1 mole of S8 = 2059.52 g = 6.022 x 1023 molecules

CaCO3

100.09 gOxygen 32.00 gCopper 63.55 gWater 18.02 g

One Mole of Common Substances

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

•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

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: H2O 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 H2O = 6.022 x 1023 molecules of water = 1 mole H2O

• Calculate the formula weight of the following compounds from their formulas. Report your answers to three significant figures.

– calcium hydroxide, Ca(OH)2

– methylamine, CH3NH2

3 significant figures74.1 amu

3 significant figures31.1 amu

Total 74.095

2 O 2(16.00) = 32.00 amu

Ca(OH)2

1 Ca 1(40.08) = 40.08 amu

2 H 2(1.008) = 2.016 amu

CH3NH2

1 C 1(12.01) = 12.01 amu

5 H 5(1.008) = 5.040 amu1 N 1(14.01) = 14.01 amu

Total 31.060

Information Contained in the Chemical Formula of Glucose C6H12O6 ( M = 180.16 g/mol)

Carbon (C) Hydrogen (H) Oxygen (O)

Atoms/moleculeof compound

Moles of atoms/mole of compound

Atoms/mole ofcompound

Mass/moleculeof 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 1023) 12(6.022 x 1023) 6(6.022 x 1023) 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

• What is the mass in grams of the nitric acid molecule, HNO3?

First, find the molar mass of HNO3:

1 H 1(1.008) = 1.0081 N 1(14.01) = 14.013 O 3(16.00) = 48.00

63.018 (2 decimal places)63.02 g/mol

molecules10x6.02

mol1 x

mol

g 63.0223

g 10x41.04684385 22

figures)tsignifican(3

g 10x1.05 is molecule HNO one of mass The 223 .

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

• 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, NA

• The number of atoms in exactly 12 g of carbon-12

• NA = 6.02 × 1023 (to three significant figures)

• Molar Mass

• The mass of one mole of substance

• For example:

• Carbon-12 has a molar mass of 12 g or 12 g/mol

• A sample of nitric acid, HNO3, contains 0.253 mol HNO3. How many grams is this?

First, find the molar mass of HNO3:

1 H 1(1.008) = 1.0081 N 1(14.01) = 14.013 O 3(16.00) = 48.00

63.018 (2 decimal places)63.02 g/mol

mole1

g63.02xmole0.253

figures)tsignifican(3

g 15.9

Next, using the molar mass, find the mass of 0.253 mole:

= 15.94406 g

Mass - Mole Relationships of a Compound

Mass (g) of Element

Molesof Element

Atomsof Element

Mass (g)of compound

Amount (mol)of compound

Molecules(or formula units of compound)

Amount (mol)of compound

For an Element For a Compound

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 3680oC. 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 1020 atoms of Tungsten

1 mol W183.9 g W

6.022 x 1023 atoms 1 mole of W

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 Na3PO4. 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 Na3PO4 = 38.6 g Na3PO4 x (1 mol Na3PO4) 163.94 g Na3PO4

= 0.23545 mol Na3PO4 Formula units = 0.23545 mol Na3PO4 x 6.022 x 1023 formula units 1 mol Na3PO4= 1.46 x 1023 formula units

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 onemole of compound

M (g / mol) of X

Divide by mass (g) of one mole of compound

Multiply by 100

Calculating Mass Percentage and Masses of Elements in a Sample of a Compound - I

Problem: Sucrose (C12H22O11) 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%

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 Hmass of 1 mol sucrose 342.30 g

mol O x M of O 11 x 16.00 g Omass of 1 mol sucrose 342.30 g

0.421046 g C 1 g sucrose

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.

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.

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.

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 .

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.

Greenhouse effect

module