Mass Relationships of Atoms - Dr. Marten’s AP...
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Mass Relationships of Atoms
Transcript of Mass Relationships of Atoms - Dr. Marten’s AP...
Mass Relationships of Atoms
Atomic number and mass number
Atomic number (Z) = the number of protons in the nucleus.
Mass number (A) = the sum of the number of protons + neutrons in the nucleus.
ZA X
Some isotopes
11
H 126
C 168
O 23592
U
21H 13
6C 17
8O 238
92U
31H 14
6C
188O
Atomic masses- synonymous with atomic weight
- is a relative scale- the isotope of carbon with 6 p+ and 6
no(carbon-12) is the reference atom and assigned an atomic mass of exactly 12
- one atomic mass unit (amu) is defined as a mass exactly equal to 1/12th the mass of one
carbon-12 atom- so 1 amu is about the mass of 1 ___ or 1 ___p+ no
Average Atomic Mass and the Periodic Table
Average Atomic Mass: The Story of Boron
• Any chunk of Boron you dig up is a mixture of 2 isotopes: on average ~20% will be 10B, ~80% will be 11B
• Say we get a small chunk of B.
• 100 atoms.
• 20 are 10B with a mass of 10 amu each
• 80 are 11B with a mass of 11 amu each
• What is the average mass of Boron?
The Story of Boron (cont.)• Average mass of a B atom
• = (total mass)/(total # of B atoms)
• Total mass for these 100 B atoms
• = 20 x 10 amu + 80 x 11 amu
• Average mass of a B atom
• = (20 x 10 amu + 80 x 11 amu)/100
• = 20 x 10 amu + 80 x 11 amu
• = 0.20 x 10 amu + 0.80 x 11 amu = 10.8 amu100 100
So to get the average atomic mass, multiply % abundance times each isotope mass and add up!
20% 80%
The Mole- the fundamental SI measure of “amount of
substance”- the amount of substance that contains as
many elementary entities as there are atoms in exactly 12 g of carbon-12
- this number of atoms is 6.022045 x 1023
Avogadro’s number
The Mole vs. The Dozen
The Dozen - the amount of substance that contains 12 entities.
The Mole - the amount of substance that contains Avogadro’s number (6.022x1023) of
entities.
Dozen Apples = 10 Lbs.
Mole of Helium atoms = 4.0026g
Dozen Apples = 12 Apples
Mole of Helium atoms = 6.022x1023 atoms
Example
1.3 Lbs
How many dozens of apples are represented by 1.3 Lbs. of apples.
Converting to Dozens
10 Lbsx
1 dozen.13 dozen=
Example
How many moles of He are in 6.46 g of He?
Converting to Moles
1.61 mol=6.46 gHe4.003 g1 molx
Mass Relationships of Atoms - Extended
ExampleCalculate the number of grams of lead (Pb)
In 12.4 moles of lead.
12.4 mol 2569 g=207.2 g
1molx
ExampleWhat is the mass in grams of one silver atom?
107.9 g
1 mol
= 17.91 x 10–23 g 1 atom
6.022 x 1023 atoms x
1 molSolution: Convert grams per mole to grams per atom.
Experimental DeterminationOf Atomic & Molecular Masses
Atomic mass is measured by mass spectrometry
An atom is ionized in the instrument.
+
e–
Ion is deflected by magnetic field
+
amount of deflection depends on mass to charge
ratio
highest m/z ratiodeflected ________
lowest m/z ratiodeflected ________
N
Sleast
most
Ions are detected after passagethrough magnetic field
+ ++ ++
+
N
S
target
Ions are detected after passagethrough magnetic field
++++ ++mixture of ions of
different mass givesseparate peak for each m/zintensity of peak proportionalto percentage of each atom of
different mass in mixtureseparation of peaks
depends on relative mass
N
S
target
19 20 21 22 23
90.92%
0.26%8.82%
The mass spectrum ofthe three isotopes of
neon.
Percent Composition of Compounds
Percent composition is the percent by mass of each element the compound contains.
Obtained by dividing the mass of each element in one mole of the compound by the
molar mass of the compound and multiplying by 100%
Example
Calculate the percent composition by mass of H,P and O for phosphoric acid (H3PO4)
Molar mass = 3(1.008g) + 30.97g + 4(16.00) = 97.99
Solution: Assume you have 1 mole of the compound.
3.086%
Molar mass = 3(1.008g) + 30.97g + 4(16.00) = 97.99
31.61%
65.31%
x 100% =%H = 97.99g
3(1.008g)
100% =%P = x30.97g
97.99g
100% =%O = x4(16.00)
97.99g
Determining Formula
Determining the Formula with a CHN Analyzer
Determining the Formula with a CHN Analyzer
0.1156 g compound containing only
C,H,&N H2O absorber CO2 absorber
O2
O2compound + H2O + CO2 + other gasses
Burning the sample completely
CO2 , H2O, O2 , and N2 O2, and other gases
Reacting the C,H,N sample with O2 , we assume all of the hydrogen and carbon present in the 0.1156g
sample is converted to H2O and CO2 respectively
Determining the Formula
0.1638g CO2
0.1676g H2O
grams collected
x12.01g C
44.009 g CO2
x2.016g H
18.015 g H2O
= 0.04470 g C
0.01876 g H=
(27.3% C in CO2)
(11.1% H in H2O)
Remembering the original mass of the sample the percentages of the components can be determined
Determining the Formula
38.67% C + 16.23% H + % N = 100%
x 100%0.1156 g sample
0.04470 g C
0.01876 g H
0.1156 g samplex 100%
=
= 16.23% H
38.67% C
% N = 45.10%
Elemental Composition
Levels of Structure
Empirical Formula
Molecular Formula
Constitution
Configuration
Conformation
Examples:
Elemental Composition
Formaldehyde Glucose
C: 40.00% C: 40.00%
H: 6.73% H: 6.73%O: 53.27% O: 53.27%
Elemental Composition
Empirical Formula
Molecular Formula
Constitution
Configuration
Conformation
Levels of Structure
√
The empirical formula tells us which elements are present and the simplest
whole-number ratio of their atoms.
Empirical Formula
Examples: Formaldehyde and Glucose
Empirical Formula
Elemental Composition
C: 40.00%H: 6.73%O: 53.27%
assume a 100g samplecalculate atom ratios by dividing by atomic
weight
40.00 g6.73 g
53.27 g
40.00 g x 1 mol
12.01 g= 3.33 mol
6.73 g x 1 mol
1.00 g= 6.73 mol
53.27 g x 1 mol
16.0 g= 3.33 mol
100 g
Calculating Empirical Formula
C:
H:
O:
Examples: Formaldehyde and Glucose
Empirical Formula
Elemental CompositionC: 40.00%H: 6.73%O: 53.27%
assume a 100g sample
calculate atom ratios by dividing by atomic weight
40.00 g6.73 g
53.27 g
3.33 mol6.73 mol3.33 mol
determine the smallest whole number ratio by dividing by the smallest molar value
40.00 g x 1 mol
12.01 g= 3.33 mol
6.73 g x 1 mol
1.00 g= 6.73 mol
53.27 g x 1 mol
16.0 g= 3.33 mol
100 g
Calculating Empirical Formula
3.33 mol
3.33 mol
3.33 mol
= 1.00
= 2.02
= 1.00
C:
H:
O:
Examples: Formaldehyde and Glucose
Empirical Formula
Elemental CompositionC: 40.00%H: 6.73%O: 53.27%
40.00 g6.73 g
53.27 g
3.33 mol6.73 mol3.33 mol
12
1
Empirical Formula: CH2O
A 1.723 g sample of aluminum oxide (which consists of aluminum and oxygen only) contains 0.912g of Al. Determine the empirical formula of the compound.
Example
0.912 g Al = 0.811 g O 1.723 g sample -
0.912 g Al x 1 mol Al26.98 g Al
0.811 g O x 1 mol O16.0 g O
= 0.0338 mol
= 0.0507 mol0.0338 mol
0.0338 mol
= 1.0
1.5=
x 2 =
x 2 =
Al2O3
2
3
Elemental Composition
Empirical Formula
Molecular Formula
Constitution
Configuration
Conformation
Levels of Structure
√√
determined from empirical formula and experimentally determined molecular mass
Molecular Formula (the REAL formula)
Compound EmpiricalFormula
True molar mass
formaldehyde CH2Oglucose CH2O
30180
Calculation of empirical mass1 mol C = 12.01 g
2 mol H x 2 = 2.016 g1 mol O = 16.00g
30.026g
30g 30g
= 1formaldehyde
180g 30g
= 6glucoseEmpirical mass
Molecular mass
determined from empirical formula and experimentally determined molecular mass
Molecular Formula
Compound EmpiricalFormula
Molar mass
Molecularformula
formaldehyde CH2Oglucose CH2O
30180
CH2OC6H12O6
Mass Relationships
Stoichiometry
mass relationships
how much reactant is needed to yield a certain amount of product
Amounts of Reactantsand Products
The mole method
1. Write and balance the equation.2. Convert to moles.
3. Use the coefficients in the balanced equation to relate the number of moles of known
substances to the unknown one.
4. Convert to desired units (g, L, etc.).5. Check your answer.
Stoichiometry
X YMass of X Mass of Y
Moles of X
(using molar
ratio of Y to X)
nynx
Moles of Y
Example
How many grams of nitrogen dioxide can be formed by reaction of 1.44 g of nitrogen
monoxide with oxygen?
NO + O2 NO22 2
Stoichiometry
1.44 g of NO Mass of NO2
Moles of X
Molar ratio of Y to X nynx
Moles of Y
NO + O2 NO22 2
Stoichiometry
1.44 g of NO Mass of NO2
0.048 Moles NO
Molar ratio of Y to X nynx
Moles of Y
NO + O2 NO22 2
Stoichiometry
1.44 g of NO Mass of NO2
0.048 Moles NO
Molar ratio NO2 to NO 22
Moles of Y
NO + O2 NO22 2
Stoichiometry
1.44 g of NO Mass of NO2
0.048 Moles NO
Molar ratio NO2 to NO 22
0.048 Moles NO2
NO + O2 NO22 2
Stoichiometry
1.44 g of NO 2.21 g of NO2
0.048 Moles NO
Molar ratio NO2 to NO 22
0.048 Moles NO2
NO + O2 NO22 2
Example
1.44g NO
NO + O2 NO22 2
= 2.21g NO2
x 2mol NO2 2mol NO
x46g NO2
1mol NO2
1mol NO
30g NO x
Limiting Reagents
Limiting Reagent
Reactants are not always present (or available) in “stoichiometric” quantities.
One reactant may be present in quantities such that it is completely consumed while excess
amounts of other reactants remain.
- called “limiting reactant” or “limiting reagent”
The limiting reagent will limit the amount of product produced.
Example
How many moles of MgCl2 will be produced?
Mg + Cl2 MgCl2
Start 1 mol 1 mol 0
Finish 0 0 1 mol
Example
How many moles of MgCl2 will be produced?
Mg + Cl2 MgCl2
Start 1 mol 2 mol 0
Finish 0 1 mol 1 mol
magnesium is the limiting reagent1 mol of chlorine will be left unchanged
W + X YMass of X Mass of Y
Moles of X
Molar ratio Y to X nynx
Moles of Y
Mass of W
Moles of W
Limiting Reagent
W + X YMass of X
Moles of X
Mass of W
Moles of W
Compare molar ratio W to X to theircoefficients in balancd equation; identify LR
Molar ratio Y to LR
Mass of Y
Moles of Y
ExampleDetermine the limiting reagent and the amount of PI3
produced when 6.00g P4 reacts with 25.0g of I2.
P4 + 6I2 4PI3
6.00g P4 = .0484mol P41mol P4
124g P4 x
25.0g I2 = .0984mol I21mol I2
254g I2 x
Example cont...Determine how much I2 would be needed to react
completely with the available amount of P4.
P4 + 6I2 4PI3
.0484mol P4 x 6mol I2
1mol P4 0.290mol I2 =
but only .0984mol I2 is availableI2 is the limiting reagent
amount of I2
needed
Example cont...…the amount of PI3 produced from the limiting
reagent...
.0984mol I2 x 4mol PI3
6mol I2 x 412g PI3
1mol PI3
=
27.0g PI3
another methodDetermine the limiting reagent and the amount of PI3
produced when 6.00g P4 reacts with 25.0g of I2.
P4 + 6I2 4PI3
Compute the moles of product formed 2 ways (assuming P4 is completely consumed and then assuming I2 is
completely consumed. See which reactant, if used up completely, gives the least amount of product. It’s limiting.
Reaction Yield
Theoretical yield the amount of product that would result if
all the limiting reagent reacted
Actual yield the amount of product actually obtained
from the reaction
Almost always less than the theoretical yield
Percent Yield
%yield = x 100%Theoretical yield
Actual yield
Determines how efficient a reaction is
Example
In a certain industrial operation 3.54 x 107g of TiCl4 is reacted with 1.13 x 107g of Mg. (a) Calculate the
theoretical yield of Ti in grams. (b) Calculate the percent yield if 7.91 x 106g are actually obtained.
TiCl4 + 2Mg Ti + 2MgCl2
= 1.87 x 105mol TiCl4
= 4.65 x 105 mol Mg
1.13 x 107g Mg1mol Mg
24.31g Mg x
1mol TiCl4
187.7g TiCl4
x3.54 x 107g TiCl4
1.87 x 105mol TiCl4
2mol Mg
1mol TiCl4
x = 3.74 x 105 mol Mg
TiCl4 is limiting
Calculate theoretical yield
there is more than enough Mg
1mol TiCl4
187.7g TiCl4
x3.54 x 107g TiCl4
8.93 x 106g Ti=
x1mol Ti
1mol TiCl4
x47.88g Ti
1mol Ti
%yield = x 100%
Theoretical yield
Actual yield
8.93 x 106g Ti
7.91 x 106g Ti100% x =
= 88.6%
