the mole & chemical composition
DESCRIPTION
The Mole & Chemical Composition. Section 1 Avogadro’s Number & Molar Conversions. For Review. Mole SI unit for amount # of atoms in 12g of carbon-12 Avogadro’s Number # of particles in a mole 6.022 x 10 23 - # of (?) in 1.000 mole Used to count any kind of particle. - PowerPoint PPT PresentationTRANSCRIPT
The Mole &Chemical Composition
Section 1Avogadro’s Number &
Molar Conversions
For Review
MoleSI unit for amount# of atoms in 12g of carbon-12
Avogadro’s Number# of particles in a mole6.022 x 1023 - # of (?) in 1.000 moleUsed to count any kind of particle
The Mole is a Counting Unit
Ex. – 1 dozen = 12
The mole is used to count out a given number of particles, whether they are atoms, molecules, formula units, ions, or electrons.
Amount in Moles converted to # of Particles
6.022 x 1023 particles = 1 mol
1mol 1
10022.6 23
1
10022.6
mol 123
Choose a conversion factor that cancels the given units.
Example #1
Find the number of molecules in 2.5 mol of sulfur dioxide.
224
23
2 SO molecules 105.1mol 1
10022.6SO mol 5.2
Homework
Practice A p. 228#’s 2 - 4
Example #2
A sample contains 3.01 x 1023 molecules of sulfur dioxide, SO2. Determine the amount in moles.
2
223
22
23
SO mol 500.0
SO molecules 106.022
SO mol 1SO molecules 1001.3
Homework
Practice Bp. 229#’s 2 – 4#5 c – g
Molar Mass Relates Moles to Grams
The mass in grams of 1 mole of substance
= atomic mass of monatomic elements & formula mass of compounds & diatomic elements
ExamplesCarbon = 12g/molO2 = 16+16 = 32g/mol
CH4 = 12+1+1+1+1 = 16g/mol
Example #3
Find the mass in grams of 2.44 x 1024 atoms of carbon, whose molar mass is 12.01 g/mol.
C g7.48mol 1
C g01.12
106.022
mol 1atoms 1044.2
2324
Homework
Practice Cp. 231 #’s 2-4
Example #4
Find the number of molecules present in 47.5 g of glycerol, C3H8O3. The molar mass of glycerol is 92.11g/mol.
molecules 1011.3
mol 1
molecules 10022.6
OHC g 11.92
mol 1OHC g 5.47
23
23
383383
Homework
Practice Dp. 232#’s 2, 3
Quiz.7.1 Answer List
57.41 g695 g1.4 x 1024 atm1.195 mol0.0206 mol
The Mole &Chemical Composition
Section 2Relative Atomic Mass &
Chemical Formulas
For Review
Isotope – atoms of the same element with different #’s of neutrons (different mass #’s)
Average Atomic Mass – weighted average of atomic masses of elements isotopes
Calculating Average Atomic Mass
Need to know % abundance to calculate avg. atomic mass
Native copper is a mixture of two isotopes. Copper-63 contributes 69.17% of the atoms, and copper-65 the remaining 30.83%.
Example #1
The mass of Cu-63 atom is 62.94 amu, and that of a Cu-65 atom is 64.93 amu. Using the data for the previous figure, find the average atomic mass of Cu.
Isotope % Decimal
Contribution
Copper-63
69.17%
0.6917 62.94 x 0.6917
Copper-65
30.83%
0.3083 64.93 x 0.3083 amu 55.63)3083.0amu 93.64()6917.0amu 94.62(
Practice #1
1. Calculate the average atomic mass for gallium if 60.00% of its atoms have a mass of 68.926 amu and 40.00 % have a mass of 70.925 amu.
2. Calculate the average atomic mass of oxygen. Its composition is 99.76% of atoms with a mass of 15.99 amu, 0.038% with a mass of 17.00 amu, and 0.20% with a mass 18.00 amu.
To Understand
Chemical formula1. Which elements2. How much of each
Ionic compounds Show simplest ratio of cations & anionsKBr – 1:1 – 1 K+ & 1 Br-
Molar Mass of Compound
= sum of masses of all atoms in g/mol
Ex. H2O - H → 2 x 1.00 = 2.00
- O → 1 x 16.00 = 16.00 18.00
g/mol
Examples
ZnCl2
(NH4)2SO4
g/mol 29.136
90.7045.352Cl
39.6539.651Zn
g/mol 0.132
64164O
32321S
818H
28142N
Practice – Homework
Practice F p. 239 - 240
#’s 1 - 4
The Mole &Chemical Composition
Section 3Formulas & Percent
Composition
Definitions
Percent composition – percentage by mass of each element in a compound
Empirical formula – shows simplest ratio
a chemical formula that shows the composition of a compound in terms of relative #’s & kinds of atoms
Example – Find the empirical formula.C – 60.0%H – 13.4% O – 26.6%
1. Convert mass to moles.
Assume you have a 100g sample.
O mol 66.1O g 16
O mol 1 26.6g
H mol 4.13H g 1
H mol 113.4g
C mol 00.5C 12g
C mol 10.60
g
Example – Find the empirical formula.2. formulas are written using whole
#’s(to convert divide by smallest)
3. Write formula
C3H8O
166.1
66.107.8
66.1
4.1301.3
66.1
00.5
Practice – In Class
1. A compound is 63.52% Fe and 36.48% S.
2. 26.58% K, 35.35% Cr, 38.07% O
3. 32.37% Na, 22.58% S, 45.05% O
Practice – Homework
Practice G p. 243
#’s 1 - 4
Definitions
Molecular formulas – whole # multiple of empirical formula
Ex. Empirical Molecular
– CH2O x 1 = CH2O formaldehyde
x 2 = C2H4O2 acetic acid
x 6 = C6H12O6
glucose
Example- Find the molecular formula.
The empirical formula for a compound is P2O5. Its experimental molar mass is 284 g/mol. Determine the molecular formula of the compound.
g/mol 94.141
80165O
94.6197.302 P
00.294.141
284
10452 OP)OP(2
Practice – In ClassWhat is the molecular formula?1. Experimental MM = 232.41
g/molEmpirical formula = OCNCl
2. Experimental MM = 32.06 g/molEmpirical formula = NH2
Practice - Homework
Practice Hp. 245
#’s 1 - 3
Percent Composition
Calculate the percent composition of copper (I) sulfide, a copper ore called chalcocite.
g/mol 2.159SCu
32.132.071 S
127.163.552Cu
2
Cu %8.791002.159
1.127 S %2.20100
2.159
1.32
Check that everything adds up to 100%
Practice – In ClassDetermine the %
Composition1. NaClO
2. H2SO3
3. C2H5COOH
Answers to Practice
Na 1 x 23 = 23Cl 1 x 35.5 = 35.5
O 1 x 16 = 16
NaClO 74.5 g/mol
Na Cl O(23/74.5) x 100
= 30.9%(35.5/74.5) x 100 = 48.0%
(16/74.5) x 100 = 21.5%
Answers to Practice
H 2 x 1 = 2S 1 x 32.1 = 32.1
O 3 x 16 = 48
H2SO3 82.1 g/mol
H S O(2/82.1) x 100
= 2.5%(32.1/82.1) x 100 = 39.1%
(48/82.1) x 100 = 58.5%
Answers to Practice
C 3 x 12 = 36H 6 x 1 = 6
O 2 x 16 = 32
C2H5COOH 74.0 g/mol
C H O(36/74.0) x 100
= 48.6%(6/74.0) x 100
= 8.1%(32/74.0) x
100 = 43.2%
Qz.7.3Calculate the percent composition1. SrBr2
2. CaSO4
3. Mg(CN)2
4. Pb(CH3COO)2
Practice - Homework
Practice Ip. 248 #’s 1- 4