Chapter 3 Atoms: the Building Blocks of Matter. The parts that make up an atom are called subatomic...
-
Upload
ilene-blake -
Category
Documents
-
view
225 -
download
0
Transcript of Chapter 3 Atoms: the Building Blocks of Matter. The parts that make up an atom are called subatomic...
Chapter 3Chapter 3
Atoms: the Building Blocks Atoms: the Building Blocks of Matterof Matter
The parts that make up an atom are called The parts that make up an atom are called subatomic particlessubatomic particles..
1) Protons (p1) Protons (p++) positively charged particle) positively charged particle
2) Neutron (n2) Neutron (noo) neutral particle (uncharged)) neutral particle (uncharged)
3) Electrons (e3) Electrons (e--) negatively charged particle) negatively charged particle Neutrons and Protons Neutrons and Protons
are located in the are located in the nucleusnucleus of an atom. of an atom.
Electrons orbit around Electrons orbit around the nucleus.the nucleus.
Q- How are atoms of different elements distinguished from Q- How are atoms of different elements distinguished from
one another? In other words, how do we distinguish a one another? In other words, how do we distinguish a
helium atom from a carbon atom?helium atom from a carbon atom?
A- Their number of A- Their number of pprotonsrotons, indicated by the , indicated by the atomic numberatomic number
Let’s look at helium, He.It has an atomic number of 2,
which means that is has 2protons in it’s nucleus.
Atomic StructureAtomic StructureHere are the basics; you need to know these.Here are the basics; you need to know these.
11
HH1.00761.0076
HydrogenHydrogen
Atomic Number (Atomic Number (ZZ)): the number of : the number of protonsprotons (p (p++))
Atomic MassAtomic Mass: the number of : the number of protonsprotons (p (p++) + the number of ) + the number of neutronsneutrons (n (n00))
▪ ▪ measured in measured in atomic mass unitsatomic mass units (amu) which is one twelfth (amu) which is one twelfth
the mass of a carbon-12 atom.the mass of a carbon-12 atom.
▪ ▪ the mass of electrons (1/1860 the mass of electrons (1/1860 pp++) is negligible.) is negligible.
Number of NeutronsNumber of Neutrons: the : the atomic massatomic mass - the - the atomic numberatomic number
Atomic Number
Atomic Symbol
Atomic Mass
Lets practice! Find the missing Lets practice! Find the missing information?information?
ElementElement Atomic Atomic ##
AtomicAtomic
MassMassProtonsProtons ElectronElectron
ssNeutronNeutronss
ArAr 1818 39.948 39.948 amuamu
HeHe 22 22
OO 15.999 15.999 amuamu
88
The Famous Gold Foil The Famous Gold Foil ExperimentExperiment
This showed us that the atom is made of mostlyThis showed us that the atom is made of mostly
empty spaceempty space..
IsotopesIsotopesAtoms of the same element with different Atoms of the same element with different
number of number of neutronsneutrons
Because they have the same number of protons, all Because they have the same number of protons, all isotopes of an element have the same chemical isotopes of an element have the same chemical
properties.properties.
Mass Numbers of Hydrogen IsotopesMass Numbers of Hydrogen Isotopes
What would the masses be?What would the masses be?
The Mole: A The Mole: A Measurement of Measurement of
MatterMatterAt the end of this section, you shouldAt the end of this section, you should
be able to:be able to:Describe how Avogadro’s Describe how Avogadro’s
number is related to a mole of number is related to a mole of any substanceany substance
Calculate the mass of a mole Calculate the mass of a mole of any substanceof any substance
The Mole The Mole (aka (aka Avagadro’s Number)Avagadro’s Number)::
6.02 x 1023
The Mole and Avogadro’s The Mole and Avogadro’s NumberNumber
SI unit that measures the amount of SI unit that measures the amount of substancesubstance
1 mole = 6.02 x 101 mole = 6.02 x 102323 representative representative particlesparticles
Representative particles are usually Representative particles are usually atoms, molecules, or formula units (ions) atoms, molecules, or formula units (ions)
But Why the Mole?But Why the Mole?Just as 12 = 1 dozen, or 63,360 inches = 1 mile,Just as 12 = 1 dozen, or 63,360 inches = 1 mile,
the mole allows us to count microscopic itemsthe mole allows us to count microscopic items
(atoms, ion, molecules) on a macroscopic scale.(atoms, ion, molecules) on a macroscopic scale.
So, So, 1 mole1 mole of any substance is a set number of of any substance is a set number of
Items, namely: Items, namely: 6.02 x 106.02 x 102323. .
Chemistry = awesomeChemistry = awesome
Examples:Examples:SubstanSubstancece
RepresentativRepresentative Particlee Particle
Chemical Chemical FormulaFormula
RepresentatiRepresentative Particles ve Particles in 1.00 molin 1.00 mol
Atomic Atomic nitrogennitrogen
AtomAtom NN 6.02 x 106.02 x 102323
WaterWater MoleculeMolecule HH22OO 6.02 x 106.02 x 102323
Calcium Calcium ionion
IonIon CaCa2+2+ 6.02 x 106.02 x 102323
SolveSolveSubstancSubstancee
Representative Representative
ParticleParticleFormulaFormula
UnitUnitRepresentatiRepresentative Particles ve Particles in 1.00 molin 1.00 mol
Nitrogen Nitrogen gasgas
NN22 MoleculeMolecule
Calcium Calcium FluorideFluoride
CaFCaF22 MoleculeMolecule
SucroseSucrose CC1212HH2222OO1111 MoleculeMolecule
CarbonCarbon CC MoleculeMolecule
AnswersAnswers
Nitrogen gas-molecule-NNitrogen gas-molecule-N22
Calcium fluoride-formula unit-CaFCalcium fluoride-formula unit-CaF22
Sucrose-molecule-CSucrose-molecule-C1212HH2222OO1111
Carbon-atom-C Carbon-atom-C
All have 6.02 x 10All have 6.02 x 102323 representative representative particles in 1.00 molparticles in 1.00 mol
How many atoms are in a How many atoms are in a mole?mole?
Determined from the chemical formulaDetermined from the chemical formula List the elements and count the atomsList the elements and count the atoms Solve for COSolve for CO2 2
C - 1 carbon atomC - 1 carbon atom
O - 2 oxygen atoms O - 2 oxygen atoms
Add: 1 + 2 = 3 Add: 1 + 2 = 3 Answer: 3 times Avogadro’s number of Answer: 3 times Avogadro’s number of
atomsatoms
Solve: How many atoms are in Solve: How many atoms are in a mole ofa mole of
1. Carbon monoxide – CO1. Carbon monoxide – CO 2. Glucose – C2. Glucose – C66HH1212OO66
3. Propane – C3. Propane – C33HH88
4. Water – H4. Water – H22OO
How many moles of How many moles of magnesium is 1.25 x 10magnesium is 1.25 x 102323
atoms of magnesium?atoms of magnesium? Refer to page 174 in textRefer to page 174 in text Divide the number of atoms or Divide the number of atoms or
molecules given in the example by molecules given in the example by 6.02 x 106.02 x 102323
Divide (1.25 x 10Divide (1.25 x 1023)23) by (6.02 x 10 by (6.02 x 1023)23)
Express in scientific notationExpress in scientific notation Answer = 2.08 x 10Answer = 2.08 x 10-1-1 mol Mg mol Mg
ObjectivesObjectives
Use the molar mass to convert Use the molar mass to convert between mass and moles of a between mass and moles of a substancesubstance
Use the mole to convert among Use the mole to convert among measurements of mass, volume, and measurements of mass, volume, and number of particlesnumber of particles
Molar massMolar mass
Mass (in grams) of one mole of a Mass (in grams) of one mole of a substancesubstance
Broad term (can be substituted) for Broad term (can be substituted) for gram atomic mass, gram formula gram atomic mass, gram formula mass, and gram molecular massmass, and gram molecular mass
Can be unclear: What is the molar Can be unclear: What is the molar mass of oxygen?mass of oxygen?
O or OO or O22 ? - element O or molecular ? - element O or molecular compound Ocompound O22 ? ?
Calculating the Molar Mass of Calculating the Molar Mass of Compounds (Molecular and Compounds (Molecular and
Ionic)Ionic) 1. List the elements1. List the elements 2. Count the atoms2. Count the atoms 3. Multiply the number of atoms of 3. Multiply the number of atoms of
the element by the atomic mass of the element by the atomic mass of the element (atomic mass is on the the element (atomic mass is on the periodic table)periodic table)
4. Add the masses of each element4. Add the masses of each element 5. Express to tenths place5. Express to tenths place
What is the molar mass (gfm) What is the molar mass (gfm) of ammonium carbonate of ammonium carbonate
(NH(NH44))22COCO33?? N 2 x 14.0 g = 28.0 gN 2 x 14.0 g = 28.0 g H 8 x 1.0 g = 8.0 gH 8 x 1.0 g = 8.0 g C 1 x 12.0 g = 12.0 gC 1 x 12.0 g = 12.0 g O 3 x 16.0 g = 48.0 gO 3 x 16.0 g = 48.0 g Add ________Add ________ Answer 96.0 g Answer 96.0 g
Practice ProblemsPractice Problems 1. How many grams are in 9.45 mol 1. How many grams are in 9.45 mol
of dinitrogen trioxide (Nof dinitrogen trioxide (N22OO33) ?) ? a. Calculate the grams in one molea. Calculate the grams in one mole b. Multiply the grams by the number b. Multiply the grams by the number of molesof moles 2. Find the number of moles in 92.2 2. Find the number of moles in 92.2
g g
of iron(III) oxide (Feof iron(III) oxide (Fe22OO33).). a. Calculate the grams in one mole a. Calculate the grams in one mole b. Divide the given grams by the b. Divide the given grams by the
grams in one mole grams in one mole
AnswersAnswers
1. 718 g N1. 718 g N22OO3 3 (one mole is 76.0g) (one mole is 76.0g)
2. 0.578 mol Fe2. 0.578 mol Fe22OO33 (one mole is (one mole is 159.6 g)159.6 g)
Volume of a Mole of GasVolume of a Mole of Gas
Varies with a change in temperature Varies with a change in temperature or a change in pressureor a change in pressure
At STP, 1 mole of any gas occupies a At STP, 1 mole of any gas occupies a volume of 22.4 Lvolume of 22.4 L
Standard temperature is 0Standard temperature is 0°°CC Standard pressure is 101.3 kPa Standard pressure is 101.3 kPa
(kilopascals), or 1 atmosphere (atm)(kilopascals), or 1 atmosphere (atm) 22.4 L is known as the molar volume22.4 L is known as the molar volume
22.4 L of any gas at STP contains 22.4 L of any gas at STP contains 6.02 x 106.02 x 102323 representative particles representative particles of that gasof that gas
One mole of a gaseous element and One mole of a gaseous element and one mole of a gaseous compound one mole of a gaseous compound both occupy a volume of 22.4 L at both occupy a volume of 22.4 L at STP (Masses may differ)STP (Masses may differ)
Molar mass (g/mol) = Density (g/L) x Molar mass (g/mol) = Density (g/L) x Molar Volume (L/mol) Molar Volume (L/mol)
ObjectivesObjectives
Define the termsDefine the terms Calculate the percent composition of Calculate the percent composition of
a substance from its chemical a substance from its chemical formula or experimental dataformula or experimental data
Derive the empirical formula and the Derive the empirical formula and the molecular formula of a compound molecular formula of a compound from experimental data from experimental data
Terms to KnowTerms to Know
Percent composition – relative Percent composition – relative amounts of each element in a amounts of each element in a compoundcompound
Empirical formula – lowest whole- Empirical formula – lowest whole- number ratio of the atoms of an number ratio of the atoms of an element in a compoundelement in a compound
An 8.20 g piece of magnesium An 8.20 g piece of magnesium combines completely with combines completely with 5.40 g of oxygen to form a 5.40 g of oxygen to form a
compound. What is the compound. What is the percent composition of this percent composition of this
compound?compound?1. Calculate the total mass1. Calculate the total mass
2.2. Divide each given by the total Divide each given by the total mass mass
and then multiply by 100%and then multiply by 100%
3.3. Check your answer: The Check your answer: The
percentages should total 100%percentages should total 100%
AnswerAnswer
The total mass is 8.20 g + 5.40 g = The total mass is 8.20 g + 5.40 g = 13.60 g13.60 g
Divide 8.2 g by 13.6 g and then multiply Divide 8.2 g by 13.6 g and then multiply by 100% = 60.29412 = 60.3%by 100% = 60.29412 = 60.3%
Divide 5.4 g by 13.6 g and then multiply Divide 5.4 g by 13.6 g and then multiply by 100% = 39.70588 = 39.7%by 100% = 39.70588 = 39.7%
Check your answer: 60.3% + 39.7% = Check your answer: 60.3% + 39.7% = 100%100%
Calculate the percent Calculate the percent composition of propane (Ccomposition of propane (C33HH88))
1. List the elements1. List the elements 2. Count the atoms2. Count the atoms 3. Multiply the number of atoms 3. Multiply the number of atoms
of the element by the atomic of the element by the atomic mass of the element (atomic mass of the element (atomic mass is on the periodic table)mass is on the periodic table)
4. Express each element as a 4. Express each element as a percentage of the total molar percentage of the total molar massmass
5. Check your answer5. Check your answer
AnswerAnswer
Total molar mass = 44.0 g/molTotal molar mass = 44.0 g/mol 36.0 g C = 81.8%36.0 g C = 81.8% 8.0 g H = 18.2%8.0 g H = 18.2%
Calculate the mass of carbon Calculate the mass of carbon in 52.0 g of propane (Cin 52.0 g of propane (C33HH88))
1.1. Calculate the percent composition Calculate the percent composition using the formula (See previous using the formula (See previous problem)problem)
2. Determine 81.8% of 82.0 g2. Determine 81.8% of 82.0 g
Move decimal two places to the Move decimal two places to the
left (.818 x 82 g)left (.818 x 82 g)
3. Answer = 67.1 g3. Answer = 67.1 g
1) Find the percent composition of 1) Find the percent composition of
Aluminum Oxide (AlAluminum Oxide (Al33OO22))
2) How much of a 5-g piece of Iron2) How much of a 5-g piece of Iron
Bromide (FeBrBromide (FeBr33) is iron?) is iron?
Calculating Empirical Calculating Empirical FormulasFormulas
Microscopic – atomsMicroscopic – atoms Macroscopic – moles of atomsMacroscopic – moles of atoms Lowest whole-number ratio may not Lowest whole-number ratio may not
be the same as the compound be the same as the compound formula formula
Example: The empirical formula of Example: The empirical formula of hydrogen peroxide (Hhydrogen peroxide (H22OO22) is HO) is HO
Empirical FormulasEmpirical Formulas The first step is to find the mole-to-mole The first step is to find the mole-to-mole
ratio of the elements in the compoundratio of the elements in the compound If the numbers are both whole numbers, If the numbers are both whole numbers,
these will be the subscripts of the these will be the subscripts of the elements in the formulaelements in the formula
If the whole numbers are identical, If the whole numbers are identical, substitute the number 1substitute the number 1
Example: CExample: C22HH22 and C and C88HH88 have an have an empirical formula of CHempirical formula of CH
If either or both numbers are not whole If either or both numbers are not whole numbers, numbers in the ratio must be numbers, numbers in the ratio must be multiplied by the same number to yield multiplied by the same number to yield whole number subscripts whole number subscripts
What is the empirical What is the empirical formula of a compound formula of a compound that is 25.9% nitrogen that is 25.9% nitrogen
and 74.1% oxygen?and 74.1% oxygen? 1. Assume 100 g of the compound, so 1. Assume 100 g of the compound, so
thatthat
there are 25.9 g N and 74.1 g Othere are 25.9 g N and 74.1 g O 2. Convert to mole-to-mole ratio:2. Convert to mole-to-mole ratio:
Divide each by mass of one mole Divide each by mass of one mole
25.9 g divided by 14.0 g = 1.85 mol N25.9 g divided by 14.0 g = 1.85 mol N
74.1 g divided by 16.0 g = 4.63 mol O74.1 g divided by 16.0 g = 4.63 mol O 3. Divide both molar quantities by the 3. Divide both molar quantities by the
smaller number of molessmaller number of moles
4. 1.85/1.85 = 1 mol N4. 1.85/1.85 = 1 mol N 4.63/1.85 = 2.5 mol O4.63/1.85 = 2.5 mol O 5. Multiply by a number that 5. Multiply by a number that
converts each to a whole number converts each to a whole number (In this case, the number is 2 (In this case, the number is 2 because 2 x 2.5 = 5, which is the because 2 x 2.5 = 5, which is the smallest whole number )smallest whole number )
2 x 1 mol N = 22 x 1 mol N = 2 2 x 2.5 mol O = 52 x 2.5 mol O = 5 Answer: The empirical formula is Answer: The empirical formula is
NN22OO55
Determine the Empirical Determine the Empirical FormulasFormulas
1. H1. H22OO22
2. CO2. CO2 2
3. N3. N22HH44
4. C4. C66HH1212OO66
5. What is the empirical formula of a 5. What is the empirical formula of a compound that is 3.7% H, 44.4% C, compound that is 3.7% H, 44.4% C, and 51.9% N?and 51.9% N?
AnswersAnswers
Compound Empirical FormulaCompound Empirical Formula 1. H1. H22OO22 HO HO
2. CO2. CO2 2 COCO2 2
3. N3. N22HH4 4 NHNH22
4. C4. C66HH1212OO66 CH CH22OO
5. HCN5. HCN
Calculating Molecular Calculating Molecular FormulasFormulas The molar mass of a compound The molar mass of a compound
is a simple whole-number is a simple whole-number multiple of the molar mass of the multiple of the molar mass of the empirical formulaempirical formula
The molecular formula may or The molecular formula may or may not be the same as the may not be the same as the empirical formula empirical formula
Calculate the molecular Calculate the molecular formula of the compound formula of the compound
whose molar mass is 60.0 g whose molar mass is 60.0 g and empirical formula is and empirical formula is
CHCH44N.N. 1. Using the empirical formula, calculate 1. Using the empirical formula, calculate
the empirical formula mass (efm)the empirical formula mass (efm) (Use the same procedure used to calculate (Use the same procedure used to calculate
molar mass.) molar mass.) 2. Divide the known molar mass by the 2. Divide the known molar mass by the
efm efm 3. Multiply the formula subscripts by this 3. Multiply the formula subscripts by this
value to get the molecular formulavalue to get the molecular formula
AnswerAnswer
Molar mass (efm) is 30.0 gMolar mass (efm) is 30.0 g 60.0 g divided by 30.0 g = 260.0 g divided by 30.0 g = 2 Answer: CAnswer: C22HH88NN22
Practice ProblemsPractice Problems 1) What is the empirical formula of a 1) What is the empirical formula of a
compounds that is 25.9% nitrogen and 74.1% compounds that is 25.9% nitrogen and 74.1% oxygen?oxygen?
2) Calculate the empirical formula of a 2) Calculate the empirical formula of a compound that is 32.00% C, 42.66% O, compound that is 32.00% C, 42.66% O, 18.67% N, and 6.67% H.18.67% N, and 6.67% H.
3) Calculate the empirical formula of a 3) Calculate the empirical formula of a compound that is 42.9% C and 57.1% O.compound that is 42.9% C and 57.1% O.
Practice ProblemsPractice Problems 4) What is the molecular formula for each 4) What is the molecular formula for each
compound:compound:
a) CHa) CH22O, 90 g/molO, 90 g/mol
b) HgCl, 472.2 g/molb) HgCl, 472.2 g/mol
c) Cc) C33HH55OO22, 146 g/mol, 146 g/mol