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Chemistry-140 Lecture 6 . Chapter Highlights define molecules definitions of molecular formula & structural formula understand formation of cations & anions learn names & formula for polyatomic ions molar mass of compounds percent composition empirical formula. - PowerPoint PPT Presentation
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Chemistry-140 Lecture 6
Chapter 3: Molecules & Compounds
Chapter Highlights
define molecules
definitions of molecular formula & structural formula understand formation of cations & anions learn names & formula for polyatomic ions molar mass of compounds percent composition empirical formula
Molecules
Molecules are tightly bound assemblies of two or more atoms. This "package" behaves as a single unit.
Some elements exist as discrete molecules;
H2, C, O2 (O3), N2, F2, Cl2, Br2, I2, P4, S8
Remember: Although He, Ne, Ar, Kr, Xe, Rn are gases they are not really molecules since they are monatomic
Chemistry-140 Lecture 6
Formulas
Chemical formula: a collection of elemental symbols with subscripts that indicate the relative number of
atoms of each element in the substance
Molecular formula: the chemical formula of a molecular compound
Example:the molecular formula for sucrose is C12H22O11
Chemistry-140 Lecture 6
Molecular Compounds
Compounds are pure substances that can be decomposed into one or more different pure substances
Example:
1 molecule of sucrose
12 atoms of C + 11 molecules of water
side 4: frames 05413-05903
Chemistry-140 Lecture 6
Formulas Structural formula: emphasizes how atoms are connected and
shows any chemically active groups (functional group)
Example:the molecular formula for ethanol is C2H6O
the structural formula for ethanol is CH3CH2OH
OH (alcohol functional group) is an important chemically active group
C CH
H
H
O
H
H
H
Chemistry-140 Lecture 6
Molecular Models
Ball & stick model for ethanol:gray = carbon white = hydrogenred = oxygen
CH
H
H
HModels of methane CH4
ball &stick space filling perspective drawing
Chemistry-140 Lecture 6
Ions Atoms of almost all elements can gain or lose
electrons to form ions (charged species)
Compounds composed of ions are known as ionic compounds
Cations are positively charged ions Anions are negatively charged ions
Example:sodium chloride, NaCl is composed of sodium cations = Na+ and chloride anions = Cl-
Chemistry-140 Lecture 6
Chemistry-140 Lecture 6
Monatomic ions are comprised of single atoms, while polyatomic ions are comprised of several atoms
Examples:
monatomic: Na+ Ca2+ Fe3+ S2- Cl- N3-
polyatomic: NH4+ SO4
2- ClO3- PO4
3-
Monatomic and Polyatomic IonsChemistry-140 Lecture 6
The charge of an ion can be predicted from the elements position in the periodic table.
Monatomic ions: atoms either gain or lose electrons until they have the same number of electrons as the nearest noble gas.
metals lose electrons to form cationsnonmetals gain electrons to form anions
For example:Ar (18 electrons), K (19 electrons) and Cl (17
electrons).K K+
Cl Cl-
in order to have 18 electrons
Predicting the Charges on IonsChemistry-140 Lecture 6
The Monatomic Anionshydride
fluoride
chloride
bromide
iodide
carbide
nitride
phosphide
oxide
sulfideselenide
telluride
Chemistry-140 Lecture 6
Ionic compounds are those compounds formed from the combination of ions.
Ionic Compounds
Please Remember!!
total cationic charge = total anionic charge
overall the material is neutralthe TOTAL charge on the compound = ZERO
Chemistry-140 Lecture 6
Question:
What ionic compound would you expect from the combination of Mg and N?
Chemistry-140 Lecture 6
Answer:
Mg Mg2+(magnesium ion)
N N3- (nitride ion)
In order to obtain overall neutrality 3 Mg2+ combine
with 2 N3- to yield Mg3N2 (magnesium nitride)
Chemistry-140 Lecture 6
Positive ions.
Name plus ion, for example, aluminum ion.
Specify the charge on the ion, for example cobalt (II) and cobalt (III) ions
Ammonium, carbonium and oxonium ions all refer to different types of positive ions of ammonia, carbon and oxygen.
Naming CompoundsChemistry-140 Lecture 6
Negative ions
Simple anions end in –ide, for example chloride ion.
Polyatomic anions (often oxoanions) are not systematic and must be learned (see table 3.1):
perchlorate, chlorate, chlorite, hypochloritehydrogen phosphate, dihydrogen phosphate
carbonate, bicarbonate
ClO4- ClO3
- ClO2- ClO- HPO4
2- H2PO4-
Naming CompoundsChemistry-140 Lecture 6
The Mole
A convenient unit for matter that containsa known number of particles
Definition: the amount of substance that contains as many particles as their are atoms in exactly 12 g
of the carbon-12 isotope
1 mole = 6.022136736 x 1023
Avogadro’s Number (N)
Chemistry-140 Lecture 6
The Mole
How big is this number???
Popcorn kernels covering the continental US $$-Dollars-$$ a national debt ($ 3.6 trillion) computer counting at 10 million particles/second
1 mole = 6.022136736 x 1023
Avogadro’s Number (N)
Chemistry-140 Lecture 6
The Mole
What is the Chemical Significance?
A mole of any element (or compound) always contains the same number of atoms (or molecules)
Since each type of atom has a different atomic mass, a mole of atoms of one element has a different mass from the mass of a mole of a different element
Example: 1 mole of 16O has mass = 16.0 g while 1 mole of 19F has mass = 19.0 g
Chemistry-140 Lecture 6
Molar Mass The mass in grams of 1 mole of atoms of any element is the
molar mass of that element
Molar mass is conventionally shown as M and expressed in grams/mole (g/mol)
For any element, the molar mass in grams is equal to the atomic mass in atomic mass units (amu).
Example:Molar mass of Na = mass of 1 mol of Na atoms
= 22.98 g/mol = mass of 6.022 x 1023 Na atoms
Chemistry-140 Lecture 6
Mass Moles Conversion
The ability to convert from moles to mass and mass to moles is absolutely essential
MASS MOLES CONVERSION
Moles to Mass Mass to Moles
(Moles) = gramsgrams1 mole
(Grams) = moles
1 molegrams
molar mass 1/molar mass
Chemistry-140 Lecture 6
Question:
How many moles are in 454 g of silicon?
A Question of Conversion
Chemistry-140 Lecture 6
A Question of Conversion
Answer:
The molar mass of silicon is 28.09 g/mol (from the periodic table!).
Convert the mass of silicon to its equivalent in moles
(454 g Si) = 16.2 mol Si1 mole Si28.09 g Si
Chemistry-140 Lecture 6
Question:
What is the mass of 2.50 moles of lead (Pb)?
Another Question of Conversion
Chemistry-140 Lecture 6
Another Question of Conversion
Answer:
The molar mass of lead is 207.2 g/mol (where else but from the periodic table!).
Convert the moles of lead to its equivalent mass
(2.50 mol Pb) = 518 g Pb207.2 g Pb1 mol Pb
Chemistry-140 Lecture 6
Question:
A graduated cylinder contains 25.4 mL of mercury (Hg). If the density of mercury is 13.534 g/mL, how many moles of mercury are in the cylinder? How many atoms of Hg are in the cylinder?
A Real Calculation
Chemistry-140 Lecture 6
A Real Calculation
Method:
Requires certain conversion!
Volume, mL Mass, g Moles Atoms
density molar mass
Avogadro’s Number
x g/mL x atoms/molx mol/g
Chemistry-140 Lecture 6
A Real Calculation
Answer:
(25.4 mL Hg) = 344 g Hg
(344 g Hg) = 1.71 mol Hg
(1.71 mol Hg) = 1.03 x 1024 atoms Hg
13.534 g Hg1 mL Hg
1 mol Hg200.6 g Hg
6.022 x 10 atoms Hg1 mol Hg
23
Chemistry-140 Lecture 6
Chemistry-140 Lecture 7
Chapter 3: Molecules & Compounds
Chapter Highlights
define molecules
definitions of molecular formula & structural formula understand formation of cations & anions learn names & formula for polyatomic ions molar mass of compounds percent composition empirical formula
Molecules, Compounds & the Mole
Molar Mass M, is the mass of a mole of molecules of a particluar substance = Molecular Weight
Example:
The molecular weight of PBr3 = the atomic weight of P plus 3 x the atomic weight of Br
MW (PBr3) = AW (P) + 3[AW (Br)]= (30.97 amu) + 3(79.90 amu)= 270.7 amu
Thus: 1 mole of PBr3 has a mass of 270.7 g
Chemistry-140 Lecture 7
A Mole of ………….
Chemistry-140 Lecture 7
NaCl
CoCl2.6H2ONiCl2.6H2O K2Cr2O7
CuSO4.5H2O
Question (compare to example 3.9):
You have 23.2 g of ethanol, C2H6O.
How many moles are contained in this mass of ethanol?
How many molecules of ethanol are contained in 23.2 g?
How many atoms of carbon are contained in 23.2 g of ethanol?
What is the average mass of one molecule of ethanol?
Chemistry-140 Lecture 7
Method:
Conversions using molar mass & Avogadro’s number!
Mass, g Moles Molecules Number of C atomsmolar
massAvogadro’s
Number
x C atoms/moleculex molecules/molx mol/g
Chemistry-140 Lecture 7
Answer:Step 1: Calculate the molar mass of ethanol, C2H6O.
2 mol of C per mole of ethanol = (2 mol C)
= 24.02 g C
6 mol of H per mole of ethanol = (6 mol H)
= 6.048 g H
1 mol of O per mole of ethanol = (1 mol O)
= 16.00 g O
Molar mass of ethanol = (24.02 + 6.048 + 16.00) g =
12.01 g C1 mol C
1.008 g H1 mol H
16.00 g O1 mol O
46.07 g/mol
Chemistry-140 Lecture 7
Answer:
Step 2: Calculate the number of moles of C2H6O.
(23.2 g C2H6O) =
Number of moles of ethanol in 23.2 g is 0.504 mol.
1 mol C H O46.07 g C H O
2 6
2 6
0.504 mol C2H6O
Chemistry-140 Lecture 7
Answer:Step 3: Calculate the number of carbon atoms.
(0.504 mol C2H6O)
= 3.04 x 10 23 molecules of C2H6O
3.04 x 10 23 molecules of C2H6O
=
Number of carbon atoms in 23.2 g of ethanol is 6.07 x 10 23 .
6.022 x 10 molecules C H O1 mol C H O
232 6
2 6
6.07 x 10 23 atoms of C
2 C atoms1 molecule C H O2 6
Chemistry-140 Lecture 7
Answer:Step 4: Calculate the mass of 1 molecule of C2H6O.
=
The mass of 1 molecule of ethanol, C2H6O, is 7.650 x 10 -23 g.
46.07 g C H O1 mol C H O
2 6
2 6
7.650 x 10 -23 g/molecule of C2H6O
1 mol6.022 x 10 molecules23
Chemistry-140 Lecture 7
Percent Composition Composition can be given by the mass of each element
relative to the total mass of the compound = Mass Percentage
Mass percentage N in NH3 =
= x 100 % = Mass percentage H in NH3 =
= x 100 % =
mass of H in 1 mol of NHmass of 1 mole of NH
3
3
14.01 g N17.030 g NH3
mass of N in 1 mol of NHmass of 1 mole of NH
3
3
3.024 g H17.030 g NH3
82.27 %
17.76 %
Chemistry-140 Lecture 7
Empirical & Molecular Formulas
Percentage composition can be used to determine a simplest or empirical formula
Empirical or simplest formulas show the simplest ratio of the numbers of atoms of each element in a substance.
Example, C6H6 is the molecular formula showing the numbers of C and H atoms in the molecule benzene. CH is the empirical formula showing the simplest ratio of atoms.
Therefore, to convert an empirical formula to a molecular formula we need a molar mass!
Chemistry-140 Lecture 7
Example 3.10:
Eugenol has a molar mass of 164.2 g/mol and is73.14 % C and 7.37 % H with the remainder O. What are the molecular and empirical formulas for eugenol?
Chemistry-140 Lecture 7
findmole ratio
Method:
% B y mol B
AxBy
% A x mol Ax mol Ay mol B}
ratios givesformulaconvert weight
percentage to moles
Chemistry-140 Lecture 7
Answer:
Step 1: Find the number of moles of C and H in a 100 g sample of eugenol.
(73.14 g C) =
(7.37 g H) =
1 mol C12.011 g C
1 mol H1.008 g H
6.089 mol C
7.31 mol H
Chemistry-140 Lecture 7
Answer:
Step 2: Find the number of moles of O in a 100 g sample of vanillin by difference.
100.00 g = (73.14 g C + 7.37 g H) + mass of O
mass of O = 100.00 g - (73.14 g C + 7.37 g H)
= 19.49 g O
19.49 g O =
1 mol O15.999 g O
1.218 mol O
Chemistry-140 Lecture 7
Answer:Step 3: Calculate the ratio of moles = empirical formula.
= = 4.999
= = 6.00
= = 1.000
C4.999H6.00O
Mole CMole O
Mole HMole O
Mole OMole O
O mol 1.218C mol 6.089
O mol 1.218H mol 7.31
O mol 1.218O mol 1.218
C5H6Oempirical formula
Chemistry-140 Lecture 7
Answer:Step 4: Determine the molecular formula from the
empirical formula and the molar mass.
M(empirical formula) = [5(MC) + 6(MH) + (MO)]
= [5(12.011) + 6(1.008) + (15.999]
= [60 + 6 + 16]
= 82 g/mol
Determined molar mass of Eugenol is 164 g/mol.
C10H12O2
empirical formula
C5H6Omolecular formula
Chemistry-140 Lecture 7
X 2
Chemistry-140 Lecture 8
Chapter 3: Molecules & Compounds
Chapter Highlights
define molecules
definitions of molecular formula & structural formula definition of allotrope understand formation of cations & anions learn names & formula for polyatomic ions molar mass of compounds percent composition empirical formula
Determining & Using Formulas Empirical Formula can be determined by a
number of different experiments
Sn + I2 SnxIy
Chemistry-140 Lecture 8
Example 3.11:
1.056 g of tin metal and 1.947 g of solid iodine are allowed to react in 100 mL of ethylacetate. After the reaction is complete (all the iodine has reacted), 0.601 g of tin is recovered.
What is the empirical formula of the product formed from the reaction between Sn & I2 in this experiment ?
Determining & Using Formulas
Chemistry-140 Lecture 8
Answer:
Step 1: Calculate the mass of Sn that reacted with the I2.
(original mass of Sn) - (mass of Sn recovered after the reaction)
= (mass of Sn consumed in the reaction)
= (1.056 - 0.601) g = 0.455 g Sn
Chemistry-140 Lecture 8
Answer:
Step 1: Find the number of moles of Sn and I2 used to create the sample of SnxIy.
(0.455 g Sn) =
(1.947 g I2) =
BUT, remember there are 2 atoms of I in each molecule of I2,
therefore, moles of I = 2 x (7.671 x 10-3) =
1 mol Sn118.7 g Sn
1 mol I253.81 g I
2
2
3.83 x 10-3 mol Sn
7.671 x 10-3 mol I2
1.534 x 10-2 mol I
Chemistry-140 Lecture 8
Answer:
Step 3: Calculate the ratio of moles = empirical formula.
= = 4.01
= = 1.00
Mole IMole Sn
Mole SnMole Sn
1.534 x 10 mol I3.83 x 10 mol Sn
-2
-3
3.83 x 10 mol Sn3.83 x 10 mol Sn
-3
-3
SnI4The empirical formula is therefore
What is the molecular formula ?
Chemistry-140 Lecture 8
Example 3.12:
What mass of copper(I) sulfide, Cu2S, may be obtained
from 2.00 kg of copper ?
Using Chemical Formulas
Chemistry-140 Lecture 8
Answer:
Step 1: Find the number of moles of Cu in 2.00 kg of copper.
(2.00 kg Cu) =
1000 g Cu1 kg Cu
31.5 mol Cu
1 mol Cu63.55 g Cu
Chemistry-140 Lecture 8
Answer:
Step 2: Use the fact that there are 2 Cu atoms per molecule of Cu2S to determine the moles of Cu2S and then calculate the mass of Cu2S.
(31.5 mol Cu)
=
1 mol Cu S2 mol Cu
2
2510 g Cu2S
159.2 g Cu S1 mol Cu S
2
2
Chemistry-140 Lecture 8
Question:In the laboratory, you weigh out 1.023 g of hydrated copper (II) sulfate, CuSO4 . xH2O (blue). After heating in a porcelain crucible you are left with 0.603 g of anhydrous cobalt(II) sulfate, CuSO4 (white). What is the value of x in CuSO4 . xH2O ?
Determining the Formula of a Hydrated Compound
Chemistry-140 Lecture 8
Method:
Write out an equation to describe the reaction and assign values to known quantities and identify unknown quantities.
CuSO4 . xH2O + heat CuSO4 + xH2O
1.023 g 0.654 g + ? g
Chemistry-140 Lecture 8
Answer:
Step 1: Find out the mass of water removed by heating.
(mass of hydrated compound) - (mass of anhydrous compound)
= mass of water
(1.023 g - 0.654 g) =0.369 g H2O
Chemistry-140 Lecture 8
Answer:
Step 2: You want to know how many moles of H2O is associated with each mole of CuSO4; the
ratio! Simply convert the masses you have to moles!
Chemistry-140 Lecture 8
OH g 18.05
OH mol 1 OH g 936.02
22
4
42 CuSO g 159.6
CuSO mol 1 OH g 654.0
2.05 x 10-2 mol H2O
4.10 x 10-3 mol CuSO4
Answer:
Step 3: The ratio of moles of H2O to moles of CuSO4 is x.
=
=
The formula for hydrated copper (II) sulfate is
4
2
CuSO molesOH moles
43-
2-2
CuSO mol 10 x 4.10OH mol 10 x 2.05
CoSO4 . 5H2O
4
2
CuSO mol 1OH mol 5.00
Chemistry-140 Lecture 8
Question (Chapter 3, #105):
The weight percentage of oxygen in an oxide that has formula MO2 is 15.2 %. What is the element M ?
Some More Practice!!??!!
Chemistry-140 Lecture 8
Answer:
Step 1: Since the formula is MO2 we know that 1 mol of the compound contains 2 mol of O = (2 x 16.00 g) = 32.00 g of O.
Since this is 15.2 % of the total…...
=
(32 x 100) = (X x 15.2)
= X =210.5 g/mol
32.00 g / mol O g / mol MO2X
15.2 %100 0%.
320015 2.
Chemistry-140 Lecture 8
Answer:
Step 2: If the molar mass of the compound is 210.5 g/mol then …...
atomic mass of element M =
(210.5 g/mol - 32.00 g/mol) =
From the periodic table we identify the element M asHf (hafnium) with an atomic mass of 178.49 g/mol
178.5 g/mol
Chemistry-140 Lecture 8
Textbook Questions From Chapter #3Molecular Formulas: 14, 16 Ions & Ion Charges: 20, 23, 24Ionic Compounds: 28, 32Naming Compounds: 36, 42, Molar Mass & Moles: 46, 48, 50, 52, 62Percent Composition: 64, 66 Empirical & Molecular Formulas 68, 70, 74, 79, 81Extras 91, 105, 107, 111, 120
Chemistry-140 Lecture 8
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