Unit 1 review
Chapter 1, chapter 2.1-2.2, 2.4
The Organization of Matter Matter
Mixtures:a) Homogeneous (Solutions)b) Heterogeneous
Pure Substances
Compounds Elements
Atoms
Nucleus Electrons
Protons Neutrons
Quarks Quarks
Pure Substances• Each have a fixed composition
• Each a unique set of properties– Elements –
• a type of matter that cannot be broken down into two or more pure substances
– Compounds –• A pure substance that contains more that one element.
Mercury, Hg
Chlorine gas, Cl2
Sodium metal, Na
Sodium Chloride, NaCl
Silver metal, Ag
Water, H2O
Phase Differences
Solid – definite volume and shape; particles packed in fixed positions.Liquid – definite volume but indefinite shape; particles close together but not in fixed positionsGas – neither definite volume nor definite shape; particles are at great distances from one anotherPlasma – high temperature, ionized phase of matter as found on the sun.
Properties of Substances• Chemical properties– Observed when substances take part in a chemical
reaction –• A change that converts it to a new substance.
• Physical properties– Observed without changing the chemical identity of a
substance.• Melting point• Boiling point• Color• Texture• Density• others
Properties of Gold are:
Physical: Melting point of 1063 oC (intensive)
Color gold (intensive)
Amount in weight (extensive)
Chemical: Gold can be stored in air without reacting chemically with oxygen
Separation of a CompoundThe Electrolysis of water
Water → Hydrogen + Oxygen2 H2O → 2 H2 + O2
Reactant → Products
Compounds must be separated by chemical means.
With the application of electricity, water can be separated into its elements
Mixtures• Contain two or more substances combined in such a
way that each substance retains its chemical identity.– Homogeneous
• Uniform mixture in which the composition is the same throughout.
– Heterogeneous• Nonuniform, different in composition throughout.
Granite
HeterogeneousBrass
Homogeneous
Copper Sulfate, heterogeneous
Separation of a Mixture
The constituents of the mixture retain their identity and may be separated by physical means.
Separation of a Mixture
The components of dyes such as ink may be separated by paper chromatography.
Separation of a Mixture by Distillation
Measurements• Scientific measurements are expressed in the
metric system.
• You will need to review this system if you are unfamiliar with it. Pages 14 in you text.
Fundamental SI units(systeme international-units agreed upon by the science community)
• Mass is measured in grams
• Length is measured in meters
• Time is measure in seconds
• Temperature is measured in Kelvin
• Electric currents is measure in amperes
• Amount of substance is measured in moles
All other units are derived from these basic units
TemperatureKelvin = celsius+ 273
Derived units
• Volume measured in cm3
– Derived from finding the volume of a cube
– Length (cm) x width (cm) x height (cm) = cm3
– Common unit (but not the SI) is liters
– The AP test will use liters or milliliters
Density
the ratio of mass to volume.
D = m/V
SI unit: g/cm3
common in g/mL
AP test uses: g mL-1
Significant Figures• Every measurement carries with it a degree
of uncertainty.
• This depends upon what instrument is being used to measure.
• We do not use the + in significant figures however it is understood that there is an uncertainty of at least one unit in the last digit place.
Rules of significant figures• If it is a number 1-9 it is significant.
• If it is a zero between two numbers, it is significant.
• If it is a zero that tells how well something was measured, it is significant. (zeros at the end with a decimal in the number)
• If it is a zero that just tells how big or how small a number is, it is NOT significant. (zeros at the beginning and at the end with no decimal in the number)
Examples
How many significant figures are in each ?•5.00•0.0090•2010•400.0•0.0609•1.50 x 103
•Answers•3•2•3•4•3•3
• When multiplying or dividing significant figures:– Count how many significant figures in each of the numbers
being used. Then use the smallest amount of significant figures of all the numbers when reporting the answer.
• When adding or subtracting significant figures:– Line up the decimals, the last place that is significant in
both numbers is where you draw a line. Add or subtract the numbers, then look at the number just past the line if it is bigger or equal to 5 round up if not just drop after the line.
Examples
Calculate and record the correct number of significant figures.1. 5.60 x 0.0232. 6.03 x 10 23/1.0 x1012
3. 5.1 + 54.674. 455 -22.0
Answers1. 0.132. 6.0 x 1011
3. 59.84. 433
Dalton’s Atomic Theory (1808)
❑ Atoms cannot be subdivided, created, or destroyed❑ Atoms of different elements combine in simple whole-number ratios to form chemical compounds❑ In chemical reactions, atoms are combined, separated, or rearranged
❑ All matter is composed of extremely small particles called atoms❑ Atoms of a given element are identical in size, mass, and other properties; atoms of different elements differ in size, mass, and other propertiesJohn
Dalton
Modern Atomic TheorySeveral changes have been made to Dalton’s theory.Dalton said:
Atoms of a given element are identical in size, mass, and other properties; atoms of different elements differ in size, mass, and other properties
Modern theory states:
Atoms of an element have a characteristic average mass which is unique to that element.
Modern Atomic Theory #2
Dalton said:
Modern theory states:
Atoms cannot be subdivided, created, or destroyed
Atoms cannot be subdivided, created, or destroyed in ordinary chemical reactions. However, these changes CAN occur in nuclear reactions
Discovery of the ElectronIn 1897, J.J. Thomson used a cathode ray tube to deduce the presence of a negatively charged particle.
Cathode ray tubes pass electricity through a gas that is contained at a very low pressure.
Thomson’s Atomic Model
Thomson believed that the electrons were like plums embedded in a positively charged “pudding,” thus it was called the “plum pudding” model.
J.J. Thomson
Mass of the Electron
1909 – Robert Millikan determines the mass of the electron.
The oil drop apparatus
Mass of the electron is 9.109 x 10-31 kg
Conclusions from the Study of the Electron
❑ Cathode rays have identical properties regardless of the element used to produce them. All elements must contain identically charged electrons.❑Atoms are neutral, so there must be positive particles in the atom to balance the negative charge of the electrons❑ Electrons have so little mass that atoms must contain other particles that account for most of the mass
Rutherford’s Gold Foil Experiment
❑ Alpha particles are helium nuclei❑ Particles were fired at a thin sheet of gold foil❑ Particle hits on the detecting screen (film) are recorded
Try it Yourself!In the following pictures, there is a target hidden by a cloud. To figure out the shape of the target, we shot some beams into the cloud and recorded where the beams came out. Can you figure out the shape of the target?
The Answers
Target #1 Target #2
Rutherford’s Findings
❑ The nucleus is small❑ The nucleus is dense❑ The nucleus is positively charged
❑ Most of the particles passed right through❑ A few particles were deflected❑ VERY FEW were greatly deflected
“Like howitzer shells bouncing off of tissue paper!”
Conclusions:
Atomic Particles
The Atomic Scale
▪ Most of the mass of the atom is in the nucleus (protons and neutrons)▪ Electrons are found outside of the nucleus (the electron cloud)▪ Most of the volume of the atom is empty space
Helium-4
Image: User Yzmo Wikimedia Commons.
Atomic NumberAtomic number (Z) of an element is the number of protons in the nucleus of each atom of that element. This identifies the atom.
Mass NumberMass number is the number of protons and neutrons in the nucleus of an isotope.
Mass # = p+ + n0
8 8 18
18Arseni
c75
33
75Phosphor
us15
31
16
Electrons mass are so much smaller than a proton and neutron that they don’tContribute much to the overall mass of the atom ,therefore they are notCounted in the mass number of the atom.
Periodic TableMetals – good conductors of heat and electricityNonmetals- nonconductorsMetalloids -semiconductorsPeriods-horizontal rowsFamilies/groups-vertical columns
SilverSulfur
Periodic tableGroup names- elements in the same group will react
similarly
1- alkali metals (except Hydrogen) -most reactive metals
2- alkaline earth metals
3-12- transitions metals
17- halogens- most reactive nonmetals
18- noble gases (inert or unreactive)
Main group elements- groups 1,2,13-18
Percent error
• Percentage error = accepted value – experimental value
accepted value
Unit 1.1
Chapters 2.3, 3.3
Crash Course: chapter 2
IsotopesIsotopes are atoms of the same element having different masses due to varying numbers of neutrons.
Atomic MassesAtomic mass is the average of all the naturally isotopes of that element.
Carbon = 12.011
Purpose of Mass Spectrometry▪ Produces spectra of masses from the molecules in a
sample of material, and fragments of the molecules.▪ Used to determine
▪ the elemental composition of a sample▪ the masses of particles and of molecules▪ potential chemical structures of molecules by
analyzing the fragments▪ the identity of unknown compounds by determining
mass and matching to known spectra▪ the isotopic composition of elements in a molecule
StagesThe ionizer converts some of the sample into ions.
Mass analyzers separate the ions according to their mass-to-charge ratio.
The detector records either the charge induced or the current produced when an ion passes by or hits a surface
Interpreting Mass Spectra
The height of each peak is proportional to the amount of each isotope present (i.e. it’s relative abundance). The m/z ratio for each peak is found from the accelerating voltage for each peak. Many ions have a +1 charge so that the m/z ratio is numerically equal to mass of the ion.
Calculating the relative atomic mass from mass spectrometry
1. Measure the height of each peak.
2. Calculate the percentage relative abundance
% abundance = amount of isotope x 100
total amount of all isotopes
3. Calculate the average mass mass of isotope A x abundance + mass of isotope B x
abundance
100
Atomic and molecular weights
Average atomic mass is the weighted average of all isotopes of an atom
H = 1.0078 amu amu=atomic mass unit
O = 15.9949 amu
Average atomic mass is on the periodic table and takes into account all isotopes of an atom and their abundance
Unit 1.2
Chapter 3.4
Crash course: chapter 2
The Mole (mol)1 dozen =
1 gross =
1 ream =
1 mole =
12
144
5006.02 x 1023
There are exactly 12 grams of carbon-12 in one mole of carbon-12.
Avogadro’s Number6.02 x 1023 is called “Avogadro’s Number” in honor of the Italian chemist Amadeo Avogadro (1776-1855).
Amadeo Avogadro
I didn’t discover it. Its just named after me!
Molar mass1 mol of book and 1 mol of feather
– Same number of items, NOT the same mass
Molar mass: mass in 1 mole of a substance (g/mol or g mol-1)– Molar mass = formula mass or average atomic
mass
1 mol a carbon = 6.02 x 10 23 atoms = 12.01 g
1 mol NaOH = 6.02 x 10 23 formulas = 40 g
1 mol CO = 6.02 x 10 23 molecules = 28 g
Calculations with Moles:Converting grams to moles
How many moles of lithium are in 18.2 grams of lithium?
18.2 g Li= mol Li
6.94 g Li
1 mol Li 2.62
Calculations with Moles:Converting moles to grams
How many grams of lithium are in 3.50 moles of lithium?
3.50 mol Li= g Li
1 mol Li6.94 g Li 24.3
Calculations with Moles:Using Avogadro’s Number
How many atoms of lithium are in 3.50 moles of lithium?
3.50 mol atoms1 mol
6.02 x 1023 atoms= 2.11 x 1024
Calculations with Moles:Using Avogadro’s Number
How many atoms of lithium are in 18.2 g of lithium?
18.2 g Li
= atoms Li
1 mol Li 6.022 x 1023 atoms Li
1.58 x 1024
6.94 g Li 1 mol Li
(18.2)(6.022 x 1023)/6.94
Moles
• Moles will convert to grams using molar mass
• Moles will convert to atoms/molecules/compounds using Avogadro's number
Calculating Formula MassCalculate the formula mass of carbon dioxide, CO2.
12.01 g + 2(16.00 g) = 44.01 g
Formula mass = mass in one chemical formula
Add the mass of all atoms in the formula
Calculating Formula MassCalculate the formula mass of carbon dioxide, CO2.
12.01 g + 2(16.00 g) = 44.01 g
Formula mass = mass in one chemical formula
Add the mass of all atoms in the formula
Conversions
• How many grams are in 2.50 mol of oxygen gas?
• (2.50 mol) (32 g O2/ 1mole)
• 80.0 g
• How many molecules are in 25.0 g of sulfuric acid?
• (25.0 g) (1 mol/98.1 g H2SO
4) (6.02x1023 molecules/1
mol)
• 1.53 x 10 23 molecules
• How many hydrogen atoms are in 25.0 g of sulfuric acid
• 2(1.53 x 10 23 molecules)
Ibuprofen, C13
H18
O2 has a molar mass of 206.29 g/mol.
If a bottle of ibuprofen contains 33 g of it, how many moles of ibuprofen are in the bottle and how many molecules are there?
0.16 moles, 9.6 x 1022 molecules
Unit 1.3
Chapter 3.5
Crash course: chapter 2
Percent composition
• We can find the mass percent of each element in a compound.
mass of element in compound x 100
total mass of compound
Calculating Percentage Composition
Calculate the percentage composition of magnesium carbonate, MgCO3.
Formula mass of magnesium carbonate:24.31 g + 12.01 g + 3(16.00 g) = 84.32 g
100.00
Formulas
❑ molecular formula = (empirical formula)n [n =
integer]
❑ molecular formula = C6H
6 = (CH)
6
❑ empirical formula = CH
Empirical formula: the lowest whole number ratio of atoms in a compound.
Molecular formula: the true number of atoms of each element in the formula of a compound.
Formulas (continued)
Formulas for ionic compounds (metals bonded to nonmetals) are ALWAYS empirical (lowest whole number ratio).Examples:
NaCl
MgCl2 Al2(SO4)3 K2CO3
Ionic compounds to not form molecules (they form crystals) so the formula doesn’t show the exact number of atoms in the compound but instead a ratio of how they bond.
Formulas (continued)
Formulas for molecular compounds (nonmetals bonded to nonmetals to form molecules) MIGHT be empirical (lowest whole number ratio).
Molecular:
H2O
C6H12O6 C12H22O11
Empirical:
H2O
CH2O C12H22O11
Empirical Formula Determination
1. Base calculation on 100 grams of compound.
2. Determine moles of each element in 100 grams of compound.
3. Divide each value of moles by the smallest of the values.
4. Multiply each number by an integer to obtain all whole numbers.
73.9% Hg and 26.1% Cl
73.9 g Hg and 26.1 g Cl
73.9/200.6 = 0.368 mol Hg
26.1/35.5 = 0.735 mol Cl
0.735/0.368 = 1.99
0.368/0.368 = 1
HgCl2
Make sure to go at least two places past decimal when finding molar mass
If you get a 1.5:1 ratio, double both 3:2
If you get a 1.33 : 1 ratio, triple both 4:3
What about 1.25: 1 ?
Empirical Formula Determination
Adipic acid contains 49.32% C, 43.84% O, and 6.85% H by mass. What is the empirical formula of adipic acid?
Empirical Formula Determination(part 2)
Divide each value of moles by the smallest of the values.
Carbon:
Hydrogen:
Oxygen:
Empirical Formula Determination(part 3)
Multiply each number by an integer to obtain all whole numbers.
Carbon: 1.50 Hydrogen: 2.50 Oxygen: 1.00x 2 x 2 x 2
3 5 2
Empirical formula: C3H5O2
Finding the Molecular FormulaThe empirical formula for adipic acid is C3H5O2. The molecular mass of adipic acid is 146 g/mol. What is the molecular formula of adipic acid?
1. Find the formula mass of C3H5O2
3(12.01 g) + 5(1.01) + 2(16.00) = 73.08 g
Finding the Molecular FormulaThe empirical formula for adipic acid is C3H5O2. The molecular mass of adipic acid is 146 g/mol. What is the molecular formula of adipic acid?
3(12.01 g) + 5(1.01) + 2(16.00) = 73.08 g
2. Divide the molecular mass by the mass given by the emipirical formula.
Finding the Molecular FormulaThe empirical formula for adipic acid is C3H5O2. The molecular mass of adipic acid is 146 g/mol. What is the molecular formula of adipic acid?
3(12.01 g) + 5(1.01) + 2(16.00) = 73.08 g
3. Multiply the empirical formula by this number to get the molecular formula.
(C3H5O2) x 2 = C6H10O4
example
• Empirical formula C3H
4
Molecular mass = 121 amu
What’s the molecular formula?
Mass of empirical formula = 40 amu
121/40 = 3.02
Molecular formula: C9H
12
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