Limiting reagents

Limiting Reagents Solution Concentration Units

Chemistry 201

NC State University

Lecture 2

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.

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 molarity 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+] = [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+] = 0.5 [CO32-] = 0.25


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.






moles of CaCl2.

Step 2. Determine the stoichiometry of Cl1- = 2

and the number of moles of Cl1- = 0.036 moles






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

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.

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+?


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.

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

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.


Solution: Step 1. The mass of H2O is:

Calculating percent mass in a solution



Solution: Step 1. The mass of H2O is:

Step 2. Therefore

Calculating percent mass


What is the mole fraction of K3PO4?

Solution: Step 1. Convert both masses into

moles.

Calculating mole fraction


What is the mole fraction of K3PO4?

Solution: Step 2. Calculate mole fraction:

Calculating 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).

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.

Empirical and molecular formulae

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

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



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



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



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





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



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



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



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

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.

Symbolic representation of molarity and molality

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.

Concentration of the “solvent”

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




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



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



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.




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.





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:

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

We can solve the problem for the general case and express

the molarities in terms of the mole fractions as follows:

Solutions: mole fraction to molarity Direct route

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?




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




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


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?


How many grams of Cu(OH)2 can be precipitated


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


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

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Limiting reagents