Download Mass and Moles of a Substance

Chapter 5
Chapter Three
Calculations with
Chemical Formulas
and Equations
Mass and Moles of a Substance
Chemistry requires a method for
determining the numbers of molecules
in a given mass of a substance.
This allows the chemist to carry out
recipes for compounds based on the
relative numbers of atoms involved.
The calculation involving the quantities of
reactants and products in a chemical
equation is called stoichiometry.
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1
Chapter 5
The molecular weight (mass) of a
substance is the sum of the atomic weights
of all the atoms in a formula of the
compound.
For, example, a molecule of H2O contains 2
hydrogen atoms (at 1.008 amu each) and 1
oxygen atom (16.000 amu), giving a molecular
weight of 18.016 amu.
For example, one formula unit of NaCl contains
1 sodium atom (22.999 amu) and one chlorine
atom (35.453 amu), giving a formula weight of
58.452 amu
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Mass and Moles of a Substance
The Mole Concept
A mole is defined as the quantity of a given
substance that contains as many molecules or
formula units as the number of atoms in exactly
12 grams of carbon-12.
The number of atoms in a 12-gram sample of
carbon-12 is called Avogadro s number (to
which we give the symbol Na). The value of
Avogadro s number is 6.02 x 1023.
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2
Chapter 5
Mass and Moles of a Substance
The molar mass (Mm) of a substance
is the mass of one mole of the
substance.
For all substances, molar mass, in grams
per mole, is numerically equal to the
formula weight in atomic mass units.
That is, one mole of any element weighs
its atomic mass in grams.
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Calculate the molar mass of sucrose,
C12H22O11
12 carbon atoms @ 12.01amu = 144.12
22 hydrogen atoms @ 1.01amu = 22.22
11 oxygen atoms @ 16.00amu = 176.00
Formula mass of sucrose
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= 342.34amu
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3
Chapter 5
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Mole calculations
Converting the number of moles of a
given substance into its mass, and vice
versa, is fundamental to understanding
the quantitative nature of chemical
equations.
mass of " A"
atomic (or molecular) mass of " A"
moles of " A"
Or
nA = mA/mmA
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Chapter 5
Mass and Moles of a Substance
Mole calculations
Suppose we have 100.0 grams of iron
(Fe). The atomic weight of iron is 55.8
g/mol. How many moles of iron does this
represent?
moles Fe
100.0 g Fe
55.8 g/mol
1.79 moles of Fe
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Mass and Moles of a Substance
Mole calculations
Conversely, suppose we have 5.75
moles of magnesium (atomic wt. = 24.3
g/mol). What is its mass?
mass Mg (5.75 moles) (24.3 g/mol)
140 grams of Mg
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5
Chapter 5
Mass and Moles of a Substance
Mole calculations
Conversely, suppose we have 3.25
moles of glucose, C6H12O6 (molecular wt.
= 180.0 g/mol). What is its mass?
mass C6 H12O 6
( 3.25 moles) (180.0 g/mol)
585 grams of C6 H12O 6
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1 mole of any substance
contains 6.022×1023
particles of that substance.
Avogadro s number
(NA) = 6.022×1023
The molar mass of a
compound is the formula
mass expressed in grams
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6
Chapter 5
Mass and Moles and Number of
Molecules or Atoms
The number of molecules or atoms in a sample
is related to the moles of the substance:
1 mole HCl 6.02 10 23 HCl molecules
1 mole Fe 6.02 10 23 Fe atoms
Suppose we have a 3.46-g sample of hydrogen
chloride, HCl. How many molecules of HCl does
this represent?
3.46g HCl
1 mole HCl
36.5g HCl
6.02 x 10 23 HCl molecules
1 mole HCl
= 5.71x1022 HCl molecules
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Which one of the following contains 1.20 1024
atoms?
1) 24.0 g O2
2) 4.00 g He
3) 42.0 g N2
4) 13.0 g C2H2
5) 8.0 g CH4
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7
Chapter 5
Determining Chemical Formulas
The percent composition of a
compound is the mass percentage of
each element in the compound.
We define the mass percentage of A
as the parts of A per hundred parts of
the total, by mass. That is,
mass % " A"
mass of " A" in whole
100%
mass of the whole
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Mass Percentages from
Formulas
Let s calculate the percent composition of
butane, C4H10.
First, we need the molecular mass of C4H10.
4 carbons @ 12.0 amu/atom 48.0 amu
10 hydrogens @ 1.00 amu/atom 10.0 amu
1 molecule of C4 H10 58.0 amu
Now, we can calculate the percents.
%C
%H
48.0 amu C
58.0 amu total
10.0 amu H
58.0 amu total
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100% 82.8%C
100% 17.2%H
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Chapter 5
Question
The mineral leadhillite, which is essentially
Pb4(SO4)(CO3)2(OH)2 (FW = 1079 amu), contains
____ % oxygen by weight.
1) 10.4
2) 11.9
3) 13.3
4) 14.8
5) 17.8
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Calculate the mass percent of nitrogen
in urea, CH4N2O.
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9
Chapter 5
Determining Chemical Formulas
1.
Determining the formula of a
compound from the percent
composition.
The percent composition of a compound
leads directly to its empirical formula.
An empirical formula (or simplest
formula) for a compound is the formula of
the substance written with the smallest
integer (whole number) subscripts.
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The empirical formula of a compound is the
simplest whole number ratio of the atoms.
The molecular formula is the actual number
of each type of element in the compound.
The molecular formula equals the empirical
formula or is a whole number multiple of the
empirical formula.
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Chapter 5
A compound contains 40.00% of sulfur and 60.00%
of oxygen. What is the empirical formula?
Step1
In 100 g of compound:
m S = 100 g x 0.4000 = 40.00 g S
m O = 100 g x 0.6000 = 60.00 g O
Step 2
ns = 40.00 g S x
1 mol S = 1.247
32.065 g S
nO = 60.00 g O x 1 mol O = 3.750
16.00 g O
Divide both mole values by smallest
1.247/1.247 = 1 S
3.750/1.247 = 3 O
Step 3
empirical formula is SO3
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The empirical formula of a carbohydrate is
CH2O. If the molar mass is 180 g/mol,
what is the molecular formula?
The empirical formula mass is:
12.0
2 1.0
16.0
30.0 g/mol
Dividing the empirical formula mass by
the molar mass:
180 g/mol
n=
6
30.0 g/mol
CH2 O
n
C 6H12O 6
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11
Chapter 5
Table 3.1
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2) Determining the formula of a compound
by a Combustion method. Measuring the
amount of hydrogen and carbon in a
compound
CxHy + excess O2
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y H2O + x CO2
2
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Chapter 5
Q? Ethylene glycol is used as an automobile antifreeze
and its formula contains only C,H and O. Combustion
of 6.38 mg of ethylene glycol gives 9.06 mg CO2 and
5.58 mg H2O.
a) Calculate the empirical formula of ethylene glycol.
b) If its molecular mass is 62.068 g/mole, what is its
molecular formula?
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Stoichiometry: Quantitative
Relations in Chemical
Reactions
Stoichiometry is the calculation of
the quantities of reactants and
products involved in a chemical
reaction.
Stoichiometry is based on the balanced
chemical equation and on the
relationship between mass and moles.
Such calculations are fundamental to
most quantitative work in chemistry.
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Chapter 5
Molar Interpretation of a
Chemical Equation
The balanced chemical equation can
be interpreted in numbers of
molecules, but generally chemists
interpret equations as mole-to-mole
relationships.
For example, the Haber process for
producing ammonia involves the reaction
of hydrogen and nitrogen.
N 2 (g ) 3 H 2 (g )
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2 NH 3 ( g )
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Haber Process
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Chapter 5
Molar Interpretation of a
Chemical Equation
This balanced chemical equation shows
that one mole of N2 reacts with 3 moles of
H2 to produce 2 moles of NH3.
N 2 (g)
3H 2 (g)
1 molecule N2
1 mol N2
+
+
3 molecules H2
3 mol H2
2 NH 3 ( g )
2 molecules NH3
2 mol NH3
Because moles can be converted to
mass, you can also give a mass
interpretation of a chemical equation.
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Mole calculations
Converting the number of moles of a
given substance into its mass, and vice
versa.
moles of " A"
mass of " A"
atomic (or molecular) mass of " A"
Using shorthand symbols:
nA = mA/MmA
or mA = nA x MmA
where n = # of moles
M= mass of compound A
Mm = molecular mass of A
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15
Chapter 5
Molar Interpretation of a
Chemical Equation
Suppose we wished to determine the
number of moles of NH3 we could
obtain from 4.8 mol H2.
N 2 (g) 3H 2 (g)
2 NH 3 ( g )
Because the coefficients in the balanced equation
represent mole-to-mole ratios, the calculation is
simple.
4.8 mol H 2
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2 mol NH 3
3 mol H 2
3.2 mol NH 3
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Ammonia, NH3, and oxygen can be reacted together
in the presence of a catalyst to form only nitrogen
Question
monoxide and water. The number
of moles of
oxygen consumed for each mole of NO that is
produced is
1) 0.625.
2) 1.25.
3) 2.50.
4) 3.75.
5) 5.00.
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Chapter 5
Mass Relationships in Chemical
Equations
Amounts of substances in a chemical
reaction by mass.
How many grams of HCl are required to
react with 5.00 grams manganese(IV)
oxide according to this equation?
4 HCl(aq)
MnO 2 (s )
2 H 2O(l) MnCl 2 (aq) Cl 2 (g )
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Steps in a Stoichiometric Calculation
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Chapter 5
Mass Relationships in Chemical
Equations
First, you write what is given (5.00 g MnO2)
and convert this to moles.
Then convert to moles of what is desired
(mol HCl).
Finally, you convert this to mass (g HCl).
5.00 g MnO 2
1 mol MnO 2
86.9g MnO 2
4 mol HCl
1 mol MnO 2
36.5 g HCl
1 mol HCl
8.40 g HCl
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Limiting Reactant
The limiting reactant (or limiting reagent)
is the reactant that is entirely consumed
when the reaction goes to completion.
The limiting reactant ultimately determines
how much product can be obtained.
For example, bicycles require one frame and
two wheels. If you have 20 wheels but only 5
frames, it is clear that the number of frames will
determine how many bicycles can be made.
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Chapter 5
Limiting Reactant
Zinc metal reacts with hydrochloric
acid by the following reaction.
Zn(s) 2 HCl(aq)
ZnCl 2 (aq) H 2 (g )
If 0.30 mol Zn is added to hydrochloric acid
containing 0.52 mol HCl, how many moles of
H2 are produced?
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Limiting Reactant
Take each reactant in turn and ask how
much product would be obtained if each
were totally consumed. The reactant that
gives the smaller amount is the limiting
reactant.
1 mol H 2
0.30 mol Zn
0.30 mol H 2
1 mol Zn
1 mol H 2
0.52 mol HCl
0.26 mol H 2
2 mol HCl
Because HCl is the limiting reactant, the
amount of H2 produced must be 0.26 mol.
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Chapter 5
If 50.0 g of O2 are mixed with 50.0 g of H2 and
the mixture is ignited, what mass of water is
produced?
1) 50.0 g
2) 56.3 g
3) 65.7 g
4) 71.4 g
5) 100.0 g
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Theoretical and Percent Yield
The theoretical yield of product is the
maximum amount of product that can be
obtained from given amounts of reactants.
The percentage yield is the actual yield
(experimentally determined) expressed as a
percentage of the theoretical yield
(calculated).
%Yield
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actual yield
100%
theoretical yield
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Chapter 5
Theoretical and Percent Yield
To illustrate the calculation of
percentage yield, recall that the
theoretical yield of H2 in the previous
example was 0.26 mol (or 0.52 g) H2.
If the actual yield of the reaction had
been 0.22 g H2, then
%Yield
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0.22 g H 2
100% 42%
0.52 g H 2
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Part B: If the reaction yielded 14,69 g of carbon disulfide
what is the % yield of the reaction?
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21
Chapter 5
Operational Skills
Calculating the formula
weight from a formula or
model.
Calculating the mass of an
atom or molecule.
Converting moles of
substance to grams and
vice versa.
Calculating the number of
molecules in a given mass.
Calculating the percentage
composition from the
formula.
Calculating the mass of an
element in a given mass of
compound.
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Calculating the percentages
C and H by combustion.
Determining the empirical
formula from percentage
composition.
Determining the molecular
formula from percentage
composition and molecular
weight.
Relating quantities in a
chemical equation.
Calculating with a limiting
reactant.
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Extra problems
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