mass relationships in chemical reactions chapter 3 copyright © the mcgraw-hill companies, inc....
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Mass Relationships in Chemical Reactions
Chapter 3
Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
By definition: 1 atom 12C “weighs” 12 amu
On this scale
1H = 1.008 amu
16O = 16.00 amu
Atomic mass is the mass of an atom in atomic mass units (amu)
Micro Worldatoms & molecules
Macro Worldgrams
3.1
Natural lithium is:
7.42% 6Li (6.015 amu)
92.58% 7Li (7.016 amu)
7.42 x 6.015 + 92.58 x 7.016100
= 6.941 amu
3.1
Average atomic mass of lithium:
The mole (mol) is the amount of a substance that contains as many elementary entities as there
are atoms in exactly 12.00 grams of 12C
3.2
1 mol = NA = 6.0221367 x 1023
Avogadro’s number (NA)
1 O2 molecule1 mol O2 molecules
How many atoms of He are in 5.36 moles of He?
5.36 mol He x 6.022 x 1023 atoms He1 mol He
=
3.23 x 1024 atoms He
How many atoms of O are in 5.36 moles of O2?
5.36 mol O2
molecules x 6.022 x 1023 molecules O2x
6.46 x 1024 atoms O
2 atoms O =
Molar mass is the mass of 1 mole of in gramseggsshoes
marblesatoms
1 mole 12C atoms = 6.022 x 1023 atoms = 12.00 g
1 12C atom = 12.00 amu
1 mole 12C atoms = 12.00 g 12C
1 mole lithium atoms = 6.941 g of Li
For any element
atomic mass (amu) = molar mass (grams)
3.2
1 amu = 1.66 x 10-24 g or 1 g = 6.022 x 1023 amu
1 12C atom12.00 amu
x12.00 g
6.022 x 1023 12C atoms=
1.66 x 10-24 g1 amu
3.2
M = molar mass in g/mol
NA = Avogadro’s number
Do You Understand Molar Mass?
How many atoms are in 0.551 g of potassium (K) ?
1 mol K = 39.10 g K
1 mol K = 6.022 x 1023 atoms K
0.551 g K 1 mol K39.10 g K
x x 6.022 x 1023 atoms K1 mol K
=
8.49 x 1021 atoms K
3.2
Molecular mass (or molecular weight) is the sum ofthe atomic masses (in amu) in a molecule.
SO2
1S 32.07 amu
2O + 2 x 16.00 amu SO2 64.07 amu
For any molecule
molecular mass (amu) = molar mass (grams)
1 molecule SO2 = 64.07 amu
1 mole SO2 = 64.07 g SO2 3.3
Molecular mass (or molecular weight) is the sum ofthe atomic masses (in amu) in a molecule.
1N 14.0067 amu
3O + 3 x 16.00 amu HNO3 63.01 amu
1 molecule HNO3 = 63.01 amu
1 mole HNO3 = 63.01 g HNO3
3.3
HNO3
1H 1.008 amu
2 N 2 x 14.0067 amu
6 O + 6 x 16.00 amu Ba(NO3)2 261.33 amu
1 Ba(NO3)2 formula unit = 261.33 amu
1 mole Ba(NO3)2 = 261.33 g Ba(NO3)2
3.3
Ba(NO3)2
1Ba 137.327 amu
- OR -
Ba 137.327 amu
2 NO3 + 2 x 62.00 amu Ba(NO3)2 261.33 amu
How many atoms are in 1.00 mol of Ba(NO3)2 ?
1 Ba(NO3)21 mol Ba(NO3)2
x 6.022 x 1023 Ba(NO3)2x
5.42 x 1024 atoms
9 atoms =1.00 mol Ba(NO3)2
Do You Understand Molecular Mass?
How many H atoms are in 72.5 g of C3H8O ?
1 mol C3H8O = (3 x 12) + (8 x 1) + 16 = 60 g C3H8O
1 mol H = 6.022 x 1023 atoms H
5.82 x 1024 atoms H
3.3
1 mol C3H8O molecules = 8 mol H atoms
72.5 g C3H8O1 mol C3H8O
60 g C3H8Ox
8 mol H atoms
1 mol C3H8Ox
6.022 x 1023 H atoms
1 mol H atomsx =
KE = 1/2 x m x v2
v = (2 x KE/m)1/2
F = q x v x B3.4
Ligh
t
Ligh
t
Hea
vy
Hea
vy
Similar to JJ Thomson’s Experiment
+
Very small amount of
sample
Different masses deflect
differently
Can look at atoms, molecules or fragments
of molecules.
Can see isotopes of atoms.
Percent composition of an element in a compound =
n x molar mass of elementmolar mass of compound
x 100%
n is the number of moles of the element in 1 mole of the compound
C2H6O
%C =2 x (12.01 g)
46.07 gx 100% = 52.14%
%H =6 x (1.008 g)
46.07 gx 100% = 13.13%
%O =1 x (16.00 g)
46.07 gx 100% = 34.73%
52.14% + 13.13% + 34.73% = 100.0%
3.5
Empirical Formulas
Weight % of Elements Ratio of Elements
1 - Start with weight percent of elements (make sure they total to 100%).
2 - Assume 100 gram sample and convert weight percent of each element to grams of each element.
3 - Convert to moles of each element using atomic weights.
4 - Calculate relative ratio of elements by dividing by the smallest number of moles.
5 - Multiply to get rid of any fractions, if necessary.
When in doubt, convert to moles !!!
A compound is found to contain:20.2 wt. % Al and 79.8 wt. % Cl.
What is its empirical formula?1 - Make sure total is 100%. 20.2 % + 79.8 % = 100.0%
2 - Convert weight percent of each element to grams of each element. 20.2 g Al and 79.8 g Cl
3 - Convert to moles of each element using atomic weights.
20.2 g Al = 0.749 mol Al 79.8 g Cl = 2.25 mol Cl
26.98 g/mol Al 35.345 g/mol Cl
4 - Calculate relative ratio of elements by dividing by the smallest number of moles. => 0.749 mol Al is smaller
0.749 mol Al = 1.00 2.25 mol Cl = 3.01mol Cl / mol Al
0.749 mol Al 0.749 mol Al
=> AlCl3Round to 3
3.6
g CO2 mol CO2 mol C g C
g H2O mol H2O mol H g H
g of O = g of sample – (g of C + g of H)
Combust 11.5 g ethanol
Collect 22.0 g CO2 and 13.5 g H2O
6.0 g C = 0.5 mol C
1.5 g H = 1.5 mol H
4.0 g O = 0.25 mol O
Empirical formula C0.5H1.5O0.25
Divide by smallest subscript (0.25)
Empirical formula C2H6O
3.7
3 ways of representing the reaction of H2 with O2 to form H2O
A process in which one or more substances is changed into one or more new substances is a chemical reaction
A chemical equation uses chemical symbols to show what happens during a chemical reaction
reactants products
3.7
reactants productsStarting materials Substances formed
a A + b B c C + d D
plus plusto yield
Conservation of Mass
Must have the same number of atoms on each side of the reaction !!
How to “Read” Chemical Equations
2 Mg + O2 2 MgO
2 atoms Mg + 1 molecule O2 makes 2 formula units MgO
2 moles Mg + 1 mole O2 makes 2 moles MgO
48.6 grams Mg + 32.0 grams O2 makes 80.6 g MgO
IS NOT
2 grams Mg + 1 gram O2 makes 2 g MgO
3.7
Balancing Chemical Equations
1. Write the correct formula(s) for the reactants on the left side and the correct formula(s) for the product(s) on the right side of the equation.
Ethane reacts with oxygen to form carbon dioxide and water
C2H6 + O2 CO2 + H2O
2. Change the numbers in front of the formulas (coefficients) to make the number of atoms of each element the same on both sides of the equation. Do not change the subscripts.
3.7
2C2H6 NOT C4H12
Balancing Chemical Equations
3. Start by balancing those elements that appear in only one reactant and one product.
C2H6 + O2 CO2 + H2O
3.7
start with C or H but not O
2 carbonon left
1 carbonon right
multiply CO2 by 2
C2H6 + O2 2CO2 + H2O
6 hydrogenon left
2 hydrogenon right
multiply H2O by 3
C2H6 + O2 2CO2 + 3H2O
Balancing Chemical Equations4. Balance those elements that appear in two or
more reactants or products.
3.7
2 oxygenon left
4 oxygen(2x2)
C2H6 + O2 2CO2 + 3H2O
+ 3 oxygen(3x1)
multiply O2 by 72
= 7 oxygenon right
C2H6 + O2 2CO2 + 3H2O72
remove fractionmultiply both sides by 2
2C2H6 + 7O2 4CO2 + 6H2O
4. Remove all fractions (generally by multiplying everything by 2) and reduce all stoichiometric coefficients to try to get one equal to 1 (generally by dividing everything by 2).
Balancing Chemical Equations5. Check to make sure that you have the same
number of each type of atom on both sides of the equation.
3.7
2C2H6 + 7O2 4CO2 + 6H2O
Reactants Products
4 C12 H14 O
4 C12 H14 O
4 C (2 x 2) 4 C12 H (2 x 6) 12 H (6 x 2)14 O (7 x 2) 14 O (4 x 2 + 6)
Always show a check of your
solution on a quiz or exam !!
Balancing Chemical Equations1. Write the correct formula(s) for the reactants on the left
side and the correct formula(s) for the product(s) on the right side of the equation.
2. Change the numbers in front of the formulas (coefficients) to make the number of atoms of each element the same on both sides of the equation. Do not change the subscripts.
3. Start by balancing those elements that appear in only one reactant and one product.
4. Balance those elements that appear in two or more reactants or products.
4. Remove all fractions (generally by multiplying everything by 2) and reduce all stoichiometric coefficients to try to get one equal to 1 (generally by dividing everything by 2).
5. Check to make sure that you have the same number of each type of atom on both sides of the equation.
Stoichiometry
Quantitative study of chemical reactions.
How much in ? => How much out ?
You may have used these principles baking cookies, doing carpentry, ………..
Chemicals react in ratios of moles,
NOT in weight ratios !!
When in doubt, convert to moles !!!
CH4 + 2 O2 CO2 + 2 H2O
Get a set of relationships between all reactants and products:
1 mol CH4 = 2 mol O2 = 1 mol CO2 = 2 mol H2O
These are exact conversions !!!
1. Write balanced chemical equation
2. Convert quantities of known substances into moles
3. Use coefficients in balanced equation to calculate the number of moles of the sought quantity
4. Convert moles of sought quantity into desired units
Mass Changes in Chemical Reactions
3.8
When in doubt, convert to moles !!!
CH4 + 2 O2 CO2 + 2 H2O
1 mol CH4 = 2 mol O2 = 1 mol CO2 = 2 mol H2O
How many mole of CO2 can be generated from 10.31 moles of O2?
10.31 mol O2 x 1 mol CO2 = 5.155 mol CO2
2 mol O2
How many grams of CH4 are needed to produce 3.85 mol of CO2?
3.85 mol CO2 x 1 mol CH4 x 16.04 g CH4 =1 mol CO2
61.8 g CH4
1 mol CH4
MW CH4 = 12.01 + 4(1.008) = 16.04 g / mol CH4
Methanol burns in air according to the equation
2CH3OH + 3O2 2CO2 + 4H2O
If 209 g of methanol are used up in the combustion, what mass of water is produced?
grams CH3OH moles CH3OH moles H2O grams H2O
molar massCH3OH
coefficientschemical equation
molar massH2O
209 g CH3OH1 mol CH3OH
32.0 g CH3OHx
4 mol H2O
2 mol CH3OHx
18.0 g H2O
1 mol H2Ox =
235 g H2O
3.8
6 green used up6 red left over
Limiting Reagents
3.9
Who gets used up first?
When one reactant runs out the reaction stops.
Excess reagent is the reagent left over when the reaction is complete.
N2 + 3 H2 2 NH3
1 mol N2 = 3 mol H2 = 2 mol NH3
How many mole of NH3 can be generated by mixing 2.35 mol N2 and 6.50 mol of H2?
2.35 mol N2 x 2 mol NH3 = 4.70 mol NH3
1 mol N2
6.50 mol H2 x 2 mol NH3 = 4.33 mol NH3
3 mol H2
Lower number
H2 is limiting reagent and the amount of NH3 that can be generated is 4.33 mol
Do You Understand Limiting Reagents?
In one process, 124 g of Al are reacted with 601 g of Fe2O3
2Al + Fe2O3 Al2O3 + 2Fe
Calculate the mass of Al2O3 formed.
g Al mol Al mol Fe2O3 needed g Fe2O3 needed
OR
g Fe2O3 mol Fe2O3 mol Al needed g Al needed
124 g Al1 mol Al
27.0 g Alx
1 mol Fe2O3
2 mol Alx
160. g Fe2O3
1 mol Fe2O3
x = 367 g Fe2O3
Start with 124 g Al need 367 g Fe2O3
Have more Fe2O3 (601 g) so Al is limiting reagent3.9
Use limiting reagent (Al) to calculate amount of product thatcan be formed.
g Al mol Al mol Al2O3 g Al2O3
124 g Al1 mol Al
27.0 g Alx
1 mol Al2O3
2 mol Alx
102. g Al2O3
1 mol Al2O3
x = 234 g Al2O3
2Al + Fe2O3 Al2O3 + 2Fe
3.9
Theoretical Yield is the amount of product that wouldresult if all the limiting reagent reacted.
Actual Yield is the amount of product actually obtainedfrom a reaction.
% Yield = Actual Yield
Theoretical Yieldx 100 %
3.10
Did you get as much as you thought you were going to get ???
NOTE: Law of conservation of mass dictates that % Yield < 100 %
Can calculate using weight or moles?
N2 + 3 H2 2 NH3
If the theoreticall yield is 4.33 mol NH3 and 3.62 mol NH3 was recovered, what is the % yield ?
% Yield = Actual Yield
Theoretical Yieldx 100 %
3.62 mol NH3
4.33 mol NH3
X 100 % = 83.6 %
2 Mg + O2 2 MgO
If 26.5 g Mg was reacted with an excess of O2 and a 75.2 % yield was achieved, how much MgO was recovered ?
26.5 g Mg x 1 mol Mg = 1.09 mol Mg
24.30 g Mg
2 mol Mg = 2 mol MgO
1.09 mol Mg x 2 mol MgO = 1.09 mol MgO theoretical yield2 mol Mg
1.09 mol MgO x 0.752 = 0.820 mol MgO
MW MgO = (24.30 + 16.00) g/mol = 40.30 g/mol MgO
0.820 mol MgO x 40.30 g MgO = 33.0 g MgO1 mol MgO
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