Download Chapter 3:Mass Relationships in Chemical Reactions

Chapter 3:Mass
Relationships in
Chemical Reactions
Chemistry 1411
Joanna Sabey
1
Atomic Mass
• Atomic Mass: The mass of the atom in atomic mass units(amu).
• Atomic Mass Unit: 1 atom 12C “weighs” 12 amu
• On This Scale:
• 1H = 1.008 amu
•
16O
= 16.00 amu
2
Average Atomic Mass
• Average Atomic Mass: the weighted average of all of the naturally
occurring isotopes of the element.
• Average atomic mass of natural carbon=
(0.9890)(12 amu) + (0.0110)(13.00335amu) =
12.01 amu
• The atomic mass on the periodic table is the
average atomic mass.
3
Average Atomic Mass
• Calculate the average atomic mass of copper given the following
information.
Isotopes
Mass
Percent
63Cu
62.93
69.09%
65Cu
64.9278
30.91%
• Convert the percent values to decimals.
• 0.6909
• 0.3091
• Add the contributions together to obtain the average atomic
mass.
• (0.6909)(62.93amu) + (0.3091) (64.9278 amu) =
• 43.48 + 20.07 = 63.55 amu
• Confirm that the average atomic mass matches that in the
periodic table.
4
Avogadro’s number and the mole
• Mole (mol): the amount of a substance that contains as many
elementary entities (atoms, molecules, or any other particles) as
there are atoms in exactly 12 g of the carbon-12 isotope.
• Avogadro’s number: the actual number of atoms in 12 g of
carbon-12. This number is 6.022 X 1023
• 1 mol of hydrogen atoms contains 6.022 X 1023 H atoms.
5
Molar Mass
• Molar Mass: the mass of 1 mol of a substance.
• The molar mass of an element is equivalent to its atomic mass
in amu. Ex: The atomic mass of Sodium, Na is 22.99 amu, and
its molar mass is 22.99 g/mol.
• Conversions from atoms to moles can occur with the
knowledge of Avogadro’s number and from mass to moles
using the atomic mass.
6
Molar Mass
• Helium(He) is a valuable gas used in industry. How many moles
of He atoms are in 6.46 g of He?
• Find the molar mass of Helium:
• He= 4.003 g/mol
• Set up conversion so that the values you are getting rid of cancel:
• 6.46 g He X
1 mol He
= 1.61 mol He
4.003 g He
7
Mole
• How many grams of Zn are in 0.356 mole of Zn?
• Find the molar mass of Zinc, Zn:
• 1 mol Zn = 65.39 g
• Set up the conversion so that the values you are getting rid of
cancel:
• 0.356 mol Zn X
65.39 g Zn
1 mol Zn
= 12.6 g Zn
8
Avogadro’s number
• How many atoms are in 16.3 g of sulfur, S?
• Find the molar mass of sulfur, S:
• 1 mol S = 32.07 g S
• Solve for the moles of sulfur:
• 16.3 g S X
1 mol S
= 0.508 mol S
32.07 g S
• Use Avogadro’s number to solve for the number of atoms:
• 1 mol S = 6.022 X 1023 atoms of S
•
6.022 X 1023 atoms S
0.508 mol S X
= 3.06 X 1023 atoms S
1 mol S
9
Molecular Mass
• Molecular mass: molecular weight, is the sum of atomic masses in
a molecule.
• H2O:
• H=1.008 X 2= 2.016 amu
• O=16.00 X 1 = 16.00 amu
• H2O = 18.016 amu
• C8H10N4O2:
•
•
•
•
•
C= 12.00 X 8 = 96.00 amu
H = 1.008 X 10 = 10.08 amu
N= 14.01 X 4 = 56.04 amu
O = 16.00 X 2 = 32.00 amu
C8H10N4O2 = 194.12 amu
10
Mole
• How many moles of CH4 are present in 6.07 g of CH4?
• Find the molecular mass of CH4:
•
•
•
•
C= 12.00 X 1 = 12.00 amu
H= 1.008 X 4 = 4.032 amu
CH4 = 16.032 amu
1 mol CH4 = 16.032 g CH4
• Set up the conversion so that the values you are getting rid of
cancel:
• 6.07 g CH4 X (1 mol CH4/ 16.032 g CH4) = 0.378 mol CH4
11
Avogadro’s number
• How many hydrogen atoms are present in 25.6 g of urea
[(NH2)2CO], MW=60.06g?
• Find molar mass of urea:
• Given: 60.06 g/mol
• Find how many moles of Hydrogen are in 1 mol of [(NH2)2CO]:
• 4 mole of Hydrogen in 1 mole of [(NH2)2CO]
• Set up the conversion so that the values you are getting rid of
cancel:
1.03 X 1024 atoms H
12
Percent Composition
• Percent composition by mass: Percent by mass of each element in a compound.
Percent composition of an element =
𝑛 X molar mass of element
molar mass of compound
X 100 %
• n is the number of moles of the element in 1 mole of the compound.
• What is the percent composition of Hydrogen and Oxygen in hydrogen peroxide
(H2O2)?
• Find the molar mass of H2O2:
• H2O2 = (1.008 X 2) + (16.00 X 2) = 34.016 amu
• Find the percent composition of H:
• 1 H has a mass of 1.008 and we have 2 H atoms
• %H=
2 X 1.008 amu
34.016 amu
𝑋 100 % = 5.926 %
• Find the percent composition of O:
• 1 O has a mass of 16.00 and we have 2 O atoms
• %O=
2 X 16.00 amu
34.016 amu
𝑋 100 % = 94.07 %
13
Percent Composition
• Chalcopyrite(CuFeS2) is a principal mineral of copper. Calculate
the number of kilograms of Cu in 3.71 X 103 kg of chalcopyrite.
• Find the percent composition of Cu in chalcopyrite:
• Mass of Cu= 63.55 amu
• Mass of chalcopyrite= (1 X 63.55) + (1 X 55.85)+ (2 X 32.07) = 183.5
amu
• % Cu =
63.55 amu
100 % = 34.63 %
183.5 amu
• Use the percent composition to solve for the mass in 3.71 X 103
kg of chalcopyrite:
• mass of Cu in CuFeS2 = 0.3463 × (3.71 × 103 kg)
= 1.28 × 103 kg
14
Determination of Empirical
Formula
• Ascorbic acid cures scurvy. It is composed of 40.92
% carbon, 4.58 % hydrogen and 54.50 % oxygen by
mass. Determine its empirical formula.
• Convert to grams and divide by molar mass:
•
40.92 g C
= 3.407 mol C
gC
12.01
mol
•
4.58 g H
=4.54 mol H
1.008 g H/mol
•
54.50 g O
=3.406 mol
16.00 g O /mol
O
• Divide each mole value by the smallest mole value:
• C = 3.407 mol / 3.406 mol = 1
• H= 4.54 mol / 3.406 mol = 1.33
• O= 3.406 mol / 3.406 mol = 1
• Change moles ratios to integer subscripts:
• For Hydrogen, find what value multiplied by 1.33 to
give a whole number
• 1.33 X 3 = 3.99 = 4
• Multiply all 3 values by 3
• C3H4O3
15
Empirical Formula
• A 1.640 g sample of Radium was heated with oxygen to form a
1.755 g sample of radium oxide. Find the empirical formula of
radium oxide.
• Step one: determine the mass of oxygen:
• Law of Conservation of mass
• 1.755 g- 1.640 g = 0.115 g O
• Step two: Find the moles of radium and oxygen:
• 1.640 g radium / (226 g/mol Ra) = 0.0073 mol Ra
• 0.115 g Oxygen/ (16.00 g/mol O) = 0.0072 mol O
• Step Three: Divide the mole number by the smallest mole number:
• Ra= 0.0073 mol/ 0.0072 mol = 1
• O=0.0072 mol /0.0072 mol = 1
• Determine the empirical formula by forming integers:
• RaO
16
Molecular Formula
• A sample contains 30.46 % nitrogen and 69.54 % oxygen by mass, the
molar mass of the compounds is found to be between 90 g and 95 g.
Determine the molecular formula and the accurate molar mass of the
compound.
• Find the empirical formula:
• Determine the moles:
• 30.46 g N/ (14.01 g / mol N) = 2.17 mol N
• 69.54 g O /(16.00 g/ mol O) = 4.246 mol O
• Determine the empirical formula:
• 2.17 mol N / 2.17 mol = 1
• 4.24 mol O / 2.17 mol = 2
• NO2
• Find the molecular formula:
• Determine the molar mass of the empirical formula:
• NO2 = (1 X 14.01) + (2 X 16.00) = 46.01 amu
• Determine the ratio between the molar mass and the empirical molar
mass:
• Determine the molecular formula
• (NO2)2
17
Chemical Equations
• Chemical Equation: The use of formulas and symbols to describe
a chemical reaction.
A + B

C + D
• A and B are the reactants, and C and D are the products.
• The physical state of the reactants and products are represented
by an abbreviation, (s) is solid, (l) is liquid, (g) is gas, and (aq) is
aqueous solution.
A (aq) + B (g)  C (s) + D(aq)
18
Chemical Equations
• Catalyst: A substance that speeds up a reaction without being
consumed or permanently altered. The presence is indicated by
placing its formula above the yield symbol (arrow).
S(s) + O2 (g)  SO2 (g)
Symbol
Interpretation of symbol

Produces, yield, gives
+
Reacts with, added to, plus
Heat is the catalyst for the reaction
Fe
Iron is the catalyst for the reaction
NR
No reaction
(s)
Solid substance or precipitate
(l)
Liquid substance
(g)
Gas substance
(aq)
Aqueous solution
19
Balancing Chemical Equations
• To balance an equation, a whole number coefficient is placed in
front of each substance.
H2 (g) + Cl2 (g)  HCl (g)
• Notice the subscript for H and Cl is 2, therefore we have 2 atoms of
each substance. In the products, we have HCl, 1 atom of each. We
can balance the equation by putting a 2 in front HCl.
H2 (g) + Cl2(g)  2 HCl (g)
20
Balancing Chemical Equations
• Always make sure that the formula is written correctly before
trying to balance an equation, this means subscripts.
• Start with the most complex formula first.
• Balance polyatomic ions as a single unit unless they breakdown.
• The coefficients must be whole numbers.
• After balancing an equation, check each symbol with its
corresponding number.
• Finally, Make sure the coefficients represent the smallest whole
number ratio.
21
Balancing Chemical Equations
• Al (s) + O2(g) Al2O3 (s)
• The most complicated would be O2 . On the left side, there are 2
oxygens, on the right, there are 3. The lowest common multiple is
6, therefore we need to make both sides 6.
• Al (s) + 3 O2 (g) 2 Al2O3 (s)
• Now, we look at Al, on the left side, we have 1 Al, on the right, we
have 4. in order to balance, we must have 4 atoms on both sides.
• 4 Al (s) + 3O2 (g) 2 Al2O3 (s)
22
Balancing Chemical Equations
• Al2(SO4)3(aq) + Ba(NO3)2(aq)  BaSO4(s) +Al(NO3)3(aq)
• First look at SO4 as a whole
• Al2(SO4)3(aq) + Ba(NO3)2(aq)  3 BaSO4(s) +Al(NO3)3(aq)
• Now, what element’s atom number changed?
• Al2(SO4)3(aq) + 3 Ba(NO3)2(aq)  3 BaSO4(s) +Al(NO3)3(aq)
• Now what element’s atom number changed?
• Al2(SO4)3(aq) + 3 Ba(NO3)2(aq)  3 BaSO4(s) +2 Al(NO3)3(aq)
• Is everything balanced?
23
Stoichiometry
• Stoichiometry: the quantitative study of reactants and products in
a chemical reaction.
• Mole Method: the stoichiometric coefficients in a chemical
equations can be interpreted as the number of moles of each
substance.
• N2(g) + 3 H2(g)  2 NH3(g)
1mol + 3 mol  2 mol
24
Stoichiometry
• Write the balanced equation.
• Convert the given amount of the reactant to number of moles.
• Use the mole ratio from the balance equation to calculate the
number of moles of product formed.
• Convert the moles of product to desired units of product.
25
Stoichiometry
• If 856 g of C6H12O6 is consumed by a person over a certain period, what is
the mass of CO2 produced?
C6H12O6(s) + 6 O2(g)  6 CO2(g) + 6 H2O(g)
• Convert grams of glucose to moles of glucose:
• MW of C6H12O6 = 180.2 g/mol
1 mol C H12 O6
=
6 H12 O6
• 856 g C6H12O6 X 180.2 g C6
4.750 mol C6H12O6
• Convert moles of glucose to moles of carbon dioxide:
• For every 1 mole of glucose, 6 moles of carbon dioxide are produced.
• 4.750 mol C6H12O6 X
6 mol CO2
=
1 mol C6 H12 O6
28.50 mol CO2
• Convert moles of carbon dioxide to grams of carbon dioxide:
• MW of CO2 = 44.01 g/mol
44.01 g CO2
• 28.50 mol CO2 X
1 mol CO2
=1.25 X 103 g CO2
26
Limiting Reagent
• Limiting Reagent: the reagent used up first in a reaction.
CO(g) + 2 H2(g)  CH3OH(g)
• Suppose this reaction takes place with 4 moles of CO and 6 moles
of H2 :
• Determine the moles of product formed from each reagent:
• 4 mol CO X (1 mol CH3OH / 1 mol CO) = 4 mol CH3OH
• 6 moles H2 X (1 mol CH3OH / 2 mol H2) = 3 mol CH3OH
• The smaller number of moles of product formed is the limiting
reagent:
• H2 is the limiting reagent for this reaction.
27
Limiting Reagent
• In a reaction 637.2 g of NH3 are treated with 1142 g of CO2. Which of the
two reactants is the limiting reagent and what is the mass of (NH2)2CO
formed in the reaction?
2 NH3(g) + CO2(g)  (NH2)2CO(aq) + H2O (l)
• Find the moles of (NH2)2CO produced from each reactant:
• MM NH3: 17.03 g/mol
• 637.3 g NH3 X
1 mol NH3
17.03 g NH3
X
• MM CO2: 44.01 g/mol
• 1142 g CO2 X
1 mol CO2
44.01 g CO2
X
1 mol (NH2 )2 CO
= 18.71 mol (NH2)2CO
2 mol NH3
1 mol (NH2 )2 CO
=25.95 mol (NH2)2CO
1 mol CO2
• Determine the limiting reactant:
• NH3 is the limiting reagent
• Using the limiting reactant, determine the mass of (NH2)2CO formed:
• MM (NH2)2CO =60.06 g/mol
66.06 g (NH2 )2 CO
• 18.71 mol (NH2)2CO X
= 1124 g (NH2)2CO
1 mol (NH2 )2 CO
28
Limiting Reagent
• How many grams of CH3Br and LiC4H9 will be needed to carry out
the preceding reaction with 10.0 g of CH3OH?
CH3OH +CH3Br + LiC4H9  CH3OCH3 + LiBr + C4H10
• Find the grams of CH3Br needed with CH3OH:
• MM CH3OH = 32.04 g/mol
• 10.0 g CH3OH X
1 mol CH3 OH
32.04 g CH3 OH
X
• MM CH3Br=94.93 g /mol
• 0.312 mol CH3Br X
1 mol CH3 Br
=
1 mol CH3 OH
94.93g CH3 Br
=
1 mol CH3 Br
0.312 mol CH3Br
29.6 g CH3Br
• Find the grams of LiC4H9 needed with CH3OH:
• MM CH3OH =32.04 g/mol
• 10.0 g CH3OH X
1 mol CH3 OH
32.04 gCH3 OH
• MM LiC4H9 = 64.05 g/mol
• 0.312 mol LiC4H9 X
X
1 mol LiC4 H9
=
1 mol CH3 OH
64.05 g LiC4 H9
=
1 mol LiC4 H9
0.312 mol LiC4H9
20.0 g LiC4H9
29
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 a
reaction.
% Yield =
Actual Yield
x 100%
Theoretical Yield
30
Reaction Yield
• In a certain industrial operation, 3.54 X 107 g of TiCl4 are reacted with
1.13 X 107 g of Mg. A) Calculate the theoretical yield of Ti in grams. B)
Calculate the percent yield if 7.91 X 106 g of Ti are actually obtained.
TiCl4(g) + 2 Mg(l)  Ti(s) + 2 MgCl2(l)
• Determine the moles of Ti formed from TiCl4:
• MM TiCl4= 189.7 g /mol
• 3.54 X 107 g TiCl4 X
1 mol TiCl4
189.7 g TiCl4
X
1 mol Ti
1 mol TiCl4
= 1.87 X 105 mol Ti
• Determine the moles of Ti formed from Mg:
• MM Mg=24.3 g /mol
• 1.13 X 107 g Mg X
1 mol Mg
24.3 g Mg
X
1 mol Ti
= 2.33 X 105 mol Ti
2 mol Mg
• Determine the limiting reagent:
• TiCl4 is the limiting reagent
• Determine the theoretical yield, mass, of Ti formed:
• MM= 47.9 g/mol
• 1.87 X 105 mol Ti X
4.79 g Ti
= 8.96 X 106 g Ti
1 mol Ti
31
Reaction Yield
• Use the formula to solve for percent yield:
• Theoretical yield = 8.96 X 106 g Ti
• Actual Yield = 7.91 X 106 g Ti
• % yield =
7.91 X 106 g Ti
8.96 X 106 g Ti
X 100 % =
88.3 % Ti
32