Chemistry 201
Lecture 2
Limiting Reagents
Solution Concentration Units
NC State University
Goals
• Define the common concentration units.
• Determine concentration given solution data.
• Determine how to make a solution of a given
concentration.
Definitions
Molality: moles of solute per kilogram of solvent.
Molarity: moles of solute per liter of solution.
Note the difference. We will use molality for
properties such a colligative properties (e.g.
freezing point depression) that depend on the
property of the solvent, but not the solute. For
most common laboratory solutions we use molarity
since the volume is easy to measure in all
situations.
Calculating molarity in ionic solutions
a) 0.1 M KCl
[K1+] =
[Cl1-] =
b) 0.1 M K3PO4
[K1+] =
[PO43-] =
c) 0.25 M Na2CO3
[Na1+] =
[CO32-] =
Calculating molarity in ionic solutions
a) 0.1 M KCl
[K1+] = 0.1
[Cl1-] = 0.1
b) 0.1 M K3PO4
[K1+] =
[PO43-] =
c) 0.25 M Na2CO3
[Na1+] =
[CO32-] =
Calculating normality in ionic solutions
When more than equivalent is present
We refer to normality rather than molarity
a) 0.1 M KCl
[K1+] = 0.1
[Cl1-] = 0.1
b) 0.1 M K3PO4
[K1+] = 0.3
[PO43-] = 0.1
c) 0.25 M Na2CO3
[Na1+] =
[CO32-] =
Calculating normality in ionic solutions
a) 0.1 M KCl
[K1+] = 0.1
[Cl1-] = 0.1
b) 0.1 M K3PO4
[K1+] = 0.3
[PO43-] = 0.1
c) 0.25 M Na2CO3
[Na1+] = 0.5
[CO32-] = 0.25
Calculating normality in ionic solutions
What is the molarity of Cl1- when 2.00 g of
CaCl2 are dissolved in enough water to
make 100.0 mL of solution?
Calculating normality in ionic solutions
What is the molarity of Cl1- when 2.00 g of
CaCl2 are dissolved in enough water to
make 100.0 mL of solution?
Solution: Step 1. Calculate the number of
moles of CaCl2.
Calculating normality in ionic solutions
What is the molarity of Cl1- when 2.00 g of
CaCl2 are dissolved in enough water to
make 100.0 mL of solution?
Solution: Step 1. Calculate the number of
moles of CaCl2.
Step 2. Determine the stoichiometry of Cl1- = 2
and the number of moles of Cl1- = 0.036 moles
Calculating normality in ionic solutions
What is the molarity of Cl1- when 2.00 g of
CaCl2 are dissolved in enough water to
make 100.0 mL of solution?
Solution: Step 1. Calculate the number of
moles of CaCl2.
Step 2. Determine the stoichiometry of Cl1- = 2
and the number of moles of Cl1- = 0.036 moles
Step 3. Scale the volume to 1 L to obtain
Molarity (normality) of [Cl1-] = 0.360 N
Making an ionic solution
How many grams of CaCl2 are needed to make
400.0 mL of 0.250 M solution?
Making an ionic solution
How many grams of CaCl2 are needed to make
400.0 mL of 0.250 M solution?
Solution: Step 1. Calculate the number of moles
of CaCl2 required.
Step 2. Use the molar mass of CaCl2 to obtain
the mass.
Making an ionic solution
How many grams of K3PO4 (Mm = 212.27 g/mol)
are needed to make 250.0 mL of solution that is
0.500 N in K1+?
Making an ionic solution
How many grams of K3PO4 (Mm = 212.27 g/mol)
are needed to make 250.0 mL of solution that is
0.500 N in K1+?
Solution: Step 1. Calculate the number of moles
of K3PO4 required.
Step 2. Use the molar mass of K3PO4 to obtain
the mass.
Calculating normality in ionic solutions
Normality is defined in terms of a particular ion.
We can say that a K3PO4 solution is 3 N in K+.
Since there are 3 equivalents of K+ ion in K3PO4
then the molarity is 1/3 of the normality in this
case. Hence the solution is 1 molar (1 M).
Goals
• Understand mass percentage and mole
fraction
• Convert ratios
• Learn the method for determining an
empirical formula and a molecular formula
Calculating percent mass in a solution
8.84 g of K3PO4 is added to 250.0 mL of H2O.
What is the mass % of K3PO4?
Calculating percent mass in a solution
8.84 g of K3PO4 is added to 250. mL of H2O.
What is the mass % of K3PO4?
Solution: Step 1. The mass of H2O is:
Calculating percent mass
8.84 g of K3PO4 is added to 250. mL of H2O.
What is the mass % of K3PO4?
Solution: Step 1. The mass of H2O is:
Step 2. Therefore
Calculating mole fraction
8.84 g of K3PO4 is added to 250. mL of H2O.
What is the mole fraction of K3PO4?
Solution: Step 1. Convert both masses into
moles.
Calculating mole fraction
8.84 g of K3PO4 is added to 250. mL of H2O.
What is the mole fraction of K3PO4?
Solution: Step 2. Calculate mole fraction:
From mass fraction to mole fraction
We need to recognize that the mass can be converted
Into mass fraction if we divide both top and bottom by
m1 + m2.
Since the mass fraction is a fraction it is true that
Therefore, we can always obtain on mass fraction from
the other (for a two-component mixture).
Empirical and molecular formulae
The molecular formula is given by the atom names
with subscripts the indicate how many such atoms are in the
molecule. For example, CO2 is the molecular formula for
carbon dioxide. Some molecular formulae have multiples of
atoms. For example, H2O2 consists of two unit that the formula
(HO). We call HO the empirical formula and H2O2 the
molecular formula.
Glucose (C6H12O6), ribose (C6H10O6), acetic acid (C2H4O2)
and formaldehyde (CH2O) have different molecular formulae,
But they all have the same empirical formula (CH2O).
If we know the empirical formula and the molecular mass
then we can readily determine the molecular formula.
Determining a molecular formula
A molecule with molecular weight of 180.18 g/mol is
analyzed and found to contain 40.00% carbon, 6.72%
hydrogen and 53.28% oxygen.
What are the empirical and molecular formulas of the
molecule?
Step 1: Find the number of moles of each element in a sample
of the molecule.
A 100 gram sample contains:
40.00 grams of carbon (40.00% of 100 grams)
6.72 grams of hydrogen (6.72% of 100 grams)
53.28 grams of oxygen (53.28% of 100 grams)
Find the number of moles
moles C = 40.00 g x 1 mol C/12.01 g/mol C = 3.33 moles C
moles H = 6.72 g x 1 mol H/1.01 g/mol H = 6.65 moles H
moles O = 53.28 g x 1 mol O/16.00 g/mol O = 3.33 moles O
Determining a molecular formula
A molecule with molecular weight of 180.18 g/mol is
analyzed and found to contain 40.00% carbon, 6.72%
hydrogen and 53.28% oxygen.
What are the empirical and molecular formulas of the
molecule?
Step 2: Find the ratios between the number of moles of each element.
Select the element with the largest number of moles in the sample. In this case,
the 6.65 moles of hydrogen is the largest. Divide the number of moles of each
element by the largest number.
3.33 mol C/6.65 mol H = 1 mol C/2 mol H
The ratio is 1 mole C for every 2 moles H
3.33 moles O/6.65 moles H = 1 mol O/2 mol H
The ratio between O and H is 1 mole O for every 2 moles of H
Determining a molecular formula
A molecule with molecular weight of 180.18 g/mol is
analyzed and found to contain 40.00% carbon, 6.72%
hydrogen and 53.28% oxygen.
What are the empirical and molecular formulas of the
molecule?
Step 3: Find the empirical formula.
For every 2 moles of hydrogen, there is one mole of carbon and one mole of
oxygen. The empirical formula is CH2O.
Step 4: Find the molecular weight of the empirical formula.
molecular weight of CH2O =
(1 x 12.01 g/mol) + (2 x 1.01 g/mol) + (1 x 16.00 g/mol)
molecular weight of CH2O = 30.03 g/mol
Determining a molecular formula
A molecule with molecular weight of 180.18 g/mol is
analyzed and found to contain 40.00% carbon, 6.72%
hydrogen and 53.28% oxygen.
What are the empirical and molecular formulas of the
molecule?
Step 5: Find the number of empirical formula units in the molecular
formula.
Number of empirical formula units in compound = 180.18 g/mol/30.03 g/mol
Number of empirical formula units in compound = 6
Step 6: Find the molecular formula.
molecular formula = 6 x CH2O
molecular formula = C6H12O6
Determining a molecular formula
A molecule with molecular weight of 180.18 g/mol is
analyzed and found to contain 40.00% carbon, 6.72%
hydrogen and 53.28% oxygen.
What are the empirical and molecular formulas of the
molecule?
Determining a molecular formula
A molecule with molecular weight of 180.18 g/mol is
analyzed and found to contain 40.00% carbon, 6.72%
hydrogen and 53.28% oxygen.
What are the empirical and molecular formulas of the
molecule?
Step 1: Find the number of moles of each element in a sample
of the molecule.
A 100 gram sample contains:
40.00 grams of carbon (40.00% of 100 grams)
6.72 grams of hydrogen (6.72% of 100 grams)
53.28 grams of oxygen (53.28% of 100 grams)
Find the number of moles
moles C = 40.00 g x 1 mol C/12.01 g/mol C = 3.33 moles C
moles H = 6.72 g x 1 mol H/1.01 g/mol H = 6.65 moles H
moles O = 53.28 g x 1 mol O/16.00 g/mol O = 3.33 moles O
Determining a molecular formula
A molecule with molecular weight of 180.18 g/mol is
analyzed and found to contain 40.00% carbon, 6.72%
hydrogen and 53.28% oxygen.
What are the empirical and molecular formulas of the
molecule?
Step 2: Find the ratios between the number of moles of each element.
Select the element with the largest number of moles in the sample. In this case,
the 6.65 moles of hydrogen is the largest. Divide the number of moles of each
element by the largest number.
3.33 mol C/6.65 mol H = 1 mol C/2 mol H
The ratio is 1 mole C for every 2 moles H
3.33 moles O/6.65 moles H = 1 mol O/2 mol H
The ratio between O and H is 1 mole O for every 2 moles of H
Determining a molecular formula
A molecule with molecular weight of 180.18 g/mol is
analyzed and found to contain 40.00% carbon, 6.72%
hydrogen and 53.28% oxygen.
What are the empirical and molecular formulas of the
molecule?
Step 3: Find the empirical formula.
For every 2 moles of hydrogen, there is one mole of carbon and one mole of
oxygen. The empirical formula is CH2O.
Step 4: Find the molecular weight of the empirical formula.
molecular weight of CH2O =
(1 x 12.01 g/mol) + (2 x 1.01 g/mol) + (1 x 16.00 g/mol)
molecular weight of CH2O = 30.03 g/mol
Determining a molecular formula
A molecule with molecular weight of 180.18 g/mol is
analyzed and found to contain 40.00% carbon, 6.72%
hydrogen and 53.28% oxygen.
What are the empirical and molecular formulas of the
molecule?
Step 5: Find the number of empirical formula units in the molecular
formula.
Number of empirical formula units in compound = 180.18 g/mol/30.03 g/mol
Number of empirical formula units in compound = 6
Step 6: Find the molecular formula.
molecular formula = 6 x CH2O
molecular formula = C6H12O6
Goals
• Understand units for concentration
- molality
- molarity
• Convert between concentration units
• Calculate concentration after dilution
• Learn how to prepare solutions by dilution
• Quantify reactions in solution
Symbolic representation of
molarity and molality
The molarity is defined as the number of moles of solute per
liter of solution.
The molality is defined as the number of moles of solute per
kilogram of solvent.
Concentration of the “solvent”
It is convenient to know the molarity of the solvent itself. We can
derive the molarity of the solvent starting with the molar volume. We
consider the most common solvent, H2O. We use the fact that density
is equal to the molar mass per unit molar volume. For H2O we have,
Therefore, the molarity of water is equal to
Since the density of H2O is 1 kg/L, the molality of pure H2O has
same numerical value.
Solutions: mass fraction to molality
A solution that is 20.0% by mass ethylene glycol
(C2H6O2) in water has a density of 1.026 g/mL.
What is the molality?
Solutions: mass fraction to molality
A solution that is 20.0% by mass ethylene glycol
(C2H6O2) in water has a density of 1.026 g/mL.
What is the molality?
Answer: if we imagine that there is 1 kg total mass, then
there is 200 grams of C2H6O2 and 800 grams of H2O.
Consequently, we can calculate the number of moles of
C2H6O2 (molar mass is 62).
Solutions: mass fraction to molality
A solution that is 20.0% by mass ethylene glycol
(C2H6O2) in water has a density of 1.026 g/mL.
What is the molality?
We scale the solvent so it has a mass of 1 kg since molality
is defined per kg of solvent.
Therefore, the number of moles of C2H6O2 is larger by this
factor and the molality is
Solutions: mass fraction to molarity
A solution that is 20.0% by mass ethylene glycol
(C2H6O2) in water has a density of 1.026 g/mL.
What is the molarity?
The molarity is defined per liter of solution. This means
that we can take the original calculation of 3.23 moles in 1
kg total mass and convert it to liters using the density.
Solutions: mass fraction to molarity
A solution that is 20.0% by mass ethylene glycol
(C2H6O2) in water has a density of 1.026 g/mL.
What is the molarity?
To obtain the concentration in 1 L of solution we divide the
number of moles of solute, 3.23 moles, by the total volume,
0.975 L.
Solutions: mass fraction to molarity
A solution that is 20.0% by mass ethylene glycol
(C2H6O2) in water has a density of 1.026 g/mL.
What is the molarity?
Solution: Use the combined formula (see text)
In this case,
Solutions: mole fraction to molarity
Harder problem
A solution that has a mole fraction of 0.2 of
ethylene glycol (C2H6O2) in water has a density of
1.026 g/mL. What is the molarity?
Solution: Step 1. The mole fraction is related to the mass of
the two components by the equation:
The total mass of one liter can be calculated from the
density to be 1026 grams. The total mass is given by:
Solutions: mole fraction to molarity
A solution that has a mole fraction of 0.2 of
ethylene glycol (C2H6O2) in water has a density of
1.026 g/mL. What is the molarity?
Solution: Step 2. Setup two equations and two unknowns
and solve.
Solutions: mole fraction to molarity
and solve for the mass of ethylene glycol in 1 liter (L).
Step 3. Convert to moles of ethylene glycol. This number is
equal to the molarity since the volume of solution is 1 L.
Solutions: mole fraction to molarity
Direct route
We can solve the problem for the general case and express
the molarities in terms of the mole fractions as follows:
One needs to express the density in units of grams/liter since
the molar masses are in units of grams/mole. Alternatively we
can express the density in standard units of grams/mL and
multiply by 1000.
Dilution
In this figure you see the optical effect of dilution.
We will show that the absorbance is proportional to
the concentration. This is convenient for
measuring the concentration (and dilution).
Dilution
The molarity is defined as,
or
so there is an inverse relationship leading to the
formula
Dilution
What is the concentration when 15 mL of 1.00 M
CuSO4 are diluted to 75 mL?
Dilution
What is the concentration when 15 mL of 1.00 M
CuSO4 are diluted to 75 mL?
Solution: The dilution factor is given by
We put in the values above
So that cfinal = 0.2 M.
Dilution factor
How many mL of 1.00 M CuSO4 are needed to
make 300.0 mL of 0.20 M CuSO4?
Dilution factor
How many mL of 1.00 M CuSO4 are needed to
make 300.0 mL of 0.20 M CuSO4?
Solution: We rearrange the formula for dilution
and plug in the numbers
which gives an initial volume of 60 mL.
Limiting reagent
We consider situations in which several species
are assumed to react completely. If their number
of moles is exactly equal to their stoichiometric
ratio then both compounds will react completely to
form products. However, if one starting reagent is
present in excess, then we say that the other
reagent is limiting. The limiting reagent determines
how much product can form.
Reactions in solution
What is the minimum volume of 0.150 M NaOH
needed to precipitate all the Cu2+ in 50.0 mL of
0.325 M CuSO4?
Reactions in solution
What is the minimum volume of 0.150 M NaOH
needed to precipitate all the Cu2+ in 50.0 mL of
0.325 M CuSO4?
Solution: We need to write the chemical reaction. Many
metal hydroxides are insoluble.
CuSO4 + 2 NaOH Na2SO4 + Cu(OH)2
We need to calculate the number of moles of CuSO4 in the
given solution
Reactions in solution
What is the minimum volume of 0.150 M NaOH
needed to precipitate all the Cu2+ in 50.0 mL of
0.325 M CuSO4?
The molar ratio is 2:1 so the number of moles NaOH is
twice the number of moles of CuSO4.
Now we calculate the volume
Reactions in solution
How many grams of Cu(OH)2 can be precipitated
when excess NaOH solution is added to 28.0 mL
of 0.325 M CuSO4?
Reactions in solution
How many grams of Cu(OH)2 can be precipitated
when excess NaOH solution is added to 28.0 mL
of 0.325 M CuSO4?
Solution: For this problem we can calculate the number of
Moles of CuSO4 and then obtain the mass of Cu(OH)2 from
the ratio of their molar masses (molecular weights).
Step 1. Calculate the mass of CuSO4.
Reactions in solution
How many grams of Cu(OH)2 can be precipitated
when excess NaOH solution is added to 28.0 mL
of 0.325 M CuSO4?
Step 2. Calculate the mass of Cu(OH)2 using molar mass
of 63.5 + 2(16) + 2 = 97.5 g/mol.
The mass of precipitated Cu(OH)2 is 0.887 grams.
Reactions in solution
Decomposition of 25.0 mL of H2O2 solution
produced 165 mL of O2 gas at 755 torr and 25 oC.
What was the molarity of H2O2?
Reactions in solution
Decomposition of 25.0 mL of H2O2 solution
produced 165 mL of O2 gas at 755 torr and 25 oC.
What was the molarity of H2O2?
Solution: Step 1. Calculate the number of moles of O2 gas.
n = 0.0067 moles of O2. Step 2. Given the volume of
0.025 L the molarity is 0.268 M, as calculated by