CHP 4 – Chemical

Quantities and

Aqueous Reactions

Chemical Reactions

Stoichiometry is the numerical relationship between chemical quantities

in a balanced chemical equation

Stoichiometry allows for predictions to be made on:

• The amount of products that form based upon the amount of reactants present

• How much of the reactants is needed to form a given amount of product

• How much of one reactant is needed to completely react with another reactant

These calculations are important in chemistry as it allows chemists to plan

and conduct chemical reactions to obtain products in the desired quantities

2

StoichiometryMoles of Elements from Compounds

Recall, how we used the chemical formula to determine the ratio of an element in a compound

Example: Calculate the mass, in grams, of oxygen in 126.5 g of sulfuric acid (H2SO4)?

Given: 1 mol H2SO4 = 98.078 g H2SO4

Molar Mass (M.M.) of O = 15.999 g/mol

Step 1: Determine the ratio of oxygen to sulfuric acid

• 4 mol O: 1 mol H2SO4

Step 2: Perform conversion

126.5 𝑔 𝐻2𝑆𝑂4 ×1𝑚𝑜𝑙 𝐻2𝑆𝑂4

98.078 𝑔 𝐻2𝑆𝑂4×

4 𝑚𝑜𝑙 𝑂

1 𝑚𝑜𝑙 𝐻2𝑆𝑂4×15.999 𝑔 𝑂

1 𝑚𝑜𝑙 𝑂= 82.54 𝑔 𝑂

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MOLES of

CompoundMOLES of

Element

MASS of

Compound

MASS of

Element

StoichiometryMoles of Reactants & Products

Now, we will use a balanced chemical equation to determine the amount of a particular reactant need or product that will be produced

A balanced chemical equation is like a “recipe” showing how much reactants are needed to form a desired product

Example: How many moles of NH3 is produced from the complete reaction of 2.4 mol of N2?

Write balanced chemical equation

3𝐻2 (𝑔) + 𝑁2(𝑔) → 2𝑁𝐻3(𝑔)

Determine molar ratio of N2 to NH3

• 1 mol N2 : 2 mol NH3

Perform calculation

2.4 𝑚𝑜𝑙 𝑁2 ×2 𝑚𝑜𝑙 𝑁𝐻3

1 𝑚𝑜𝑙 𝑁2= 4.8 𝑚𝑜𝑙 𝑁𝐻3

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MOLES of

Compound AMOLES of

Compound B

Stoichiometry:Mass of Reactants & Products

Since mole-to-mole conversions have been performed, now mass-to-mass conversions can also be performed

Example: How many grams of NH3 is produced from the complete reaction of 64.2 g of N2?

Write balanced chemical equation:

3𝐻2 (𝑔) + 𝑁2(𝑔) → 2𝑁𝐻3(𝑔)

Determine molar ratio of N2 to NH3

• 1 mol N2 : 2 mol NH3

Perform calculation

64.2 𝑔 𝑁2 ×1𝑚𝑜𝑙 𝑁2

28.014 𝑔 𝑁2×2 𝑚𝑜𝑙 𝑁𝐻3

1 𝑚𝑜𝑙 𝑁2×17.031 𝑔 𝑁𝐻3

1 𝑚𝑜𝑙 𝑁𝐻3= 78.1 𝑔 𝑁𝐻3

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MOLES of

Compound AMOLES of

Compound B

MASS of

Compound A

MASS of

Compound A

Limiting Reactant, Theoretical

Yield & Percent Yield

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𝑃𝑒𝑟𝑐𝑒𝑛𝑡 𝑌𝑖𝑒𝑙𝑑 =𝐴𝑐𝑡𝑢𝑎𝑙 𝑌𝑖𝑒𝑙𝑑

𝑇ℎ𝑒𝑜𝑟𝑒𝑡𝑖𝑐𝑎𝑙 𝑌𝑖𝑒𝑙𝑑× 100%

Limiting reactant

• The reactant

that makes the

least amount of

product in a

reaction

Theoretical yield

• The amount of

product that can

be made in a

chemical

reaction based

upon the

amount of the

limiting

reactant/reagent

Actual yield

• The amount of

product that is

actually

produced by the

chemical

reaction

Percent yield

• The percentage

of the

theoretical yield

that was actually

attained can be

calculated using

the formula

Limiting Reactant, Theoretical Yield &

Percent YieldExample

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Limiting reactant

Theoretical yield

Actual yield

Percent yield

• If I have 4 crusts and 10 c.

cheese, which is the

limiting reactant?

• How many pizzas can we

expect to make?

• We only make three

pizzas

•3 𝑝𝑖𝑧𝑧𝑎𝑠

4 𝑝𝑖𝑧𝑧𝑎𝑠× 100 = 75%

1 crust + 5 oz. tomato sauce + 2 c. cheese = 1 pizza

Limiting Reactant, Theoretical Yield &

Percent YieldWord Problem Starting with Moles

Example: What is the limiting reactant and theoretical yield of NH3 in moles if you begin with 2.6 mol of hydrogen and 1.4 mol of nitrogen?

1. Write balanced chemical equation:

3𝐻2 (𝑔) + 𝑁2(𝑔) → 2𝑁𝐻3(𝑔)

2. Determine molar ratio of N2 to NH3

• 3 mol H2 : 1 mol N2 : 2 mol NH3

3. Determine limiting reactant and theoretical yield

𝐻2: 2.6 𝑚𝑜𝑙 𝐻2 ×2 𝑚𝑜𝑙 𝑁𝐻3

3 𝑚𝑜𝑙 𝐻2= 1.7 𝑚𝑜𝑙 𝑁𝐻3

𝑁2: 1.4 𝑚𝑜𝑙 𝑁2 ×2 𝑚𝑜𝑙 𝑁𝐻3

1 𝑚𝑜𝑙 𝑁2= 2. 8 𝑚𝑜𝑙 𝑁𝐻3

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MOLES of

Compound AMOLES of

Compound B

Limiting Reactant, Theoretical

Yield & Percent YieldWord Problem Starting with Mass

Example: What is the limiting reactant and theoretical yield of NH3 in moles if you begin with 46.2 g of hydrogen and 76.9 g of nitrogen?

1. Write balanced chemical equation:

3𝐻2 (𝑔) + 𝑁2(𝑔) → 2𝑁𝐻3(𝑔)

2. Determine molar ratio of N2 to NH3

• 3 mol H2 : 1 mol N2 : 2 mol NH3

3. Determine limiting reactant and theoretical yield

𝐻2: 46.2 𝑔 𝐻2 ×1 𝑚𝑜𝑙 𝐻2

2.02 𝑔 𝐻2×2 𝑚𝑜𝑙 𝑁𝐻3

3 𝑚𝑜𝑙 𝐻2= 15.2 mol 𝑁𝐻3 ×

17.03 𝑔 𝑁𝐻3

1 𝑚𝑜𝑙 𝑁𝐻3= 260. 𝑔 𝑁𝐻3

𝑁2: 76.9 𝑔 𝑁2 ×1 𝑚𝑜𝑙 𝑁2

28.014 𝑔 𝑁2×2 𝑚𝑜𝑙 𝑁𝐻3

1 𝑚𝑜𝑙 𝑁2= 5.49 mol 𝑁𝐻3 ×

17.031 𝑔 𝑁𝐻3

1 𝑚𝑜𝑙 𝑁𝐻3= 93.5 𝑔 𝑁𝐻3

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Limiting Reactant

Theoretical Yield

MOLES of

Compound AMOLES of

Compound B

MASS of

Compound A

MASS of

Compound A

Solutions

Most chemical reactions are those

in which reactants are dissolved in

water

Terms:

Solution - a homogeneous mixture

of two or more substances

Solvent - the majority component

of a solution; amount varies

Solute - the minority component

of a solution; amount varies

Aqueous Solution - a solution

where water is the solvent

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Solution Concentration

Quantifying the amount of solute relative to the solvent is to determine

the concentration of the solution

Recall, that the amount of solute and solvent varies

Dilute Solution - contains a small amount of solute relative to the solvent

Concentrated Solution - contains a large amount of solute relative to the

solvent

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Solution ConcentrationMolarity

12

Quantify the amount of solute relative to the solvent is to determine

the concentration of the solution

One way to express concentration of a solution is Molarity (M or mol/L)

Molarity is moles of solute per liter of solution

𝑀𝑜𝑙𝑎𝑟𝑖𝑡𝑦 𝑀 =𝑎𝑚𝑜𝑢𝑛𝑡 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑒 (𝑚𝑜𝑙𝑒)

𝑣𝑜𝑙𝑢𝑚𝑒 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛 (𝐿)

Note: Molarity is a ratio of the amount of solute per liter of solution,

not per liter of solvent

Solution ConcentrationCalculating Molarity

Example: Determine the molarity of a solution that has 25.5 g KBr

dissolved to a volume of 1.75 L.

1. Determine formula mass of KBr and convert to moles

25.5 𝑔 𝐾𝐵𝑟 ×1 𝑚𝑜𝑙 𝐾𝐵𝑟

119.00 𝑔 𝐾𝐵𝑟= 0.214 𝑚𝑜𝑙 𝐾𝐵𝑟

2. Calculate the molarity of the solution

𝑀 =0.214 𝑚𝑜𝑙 𝐾𝐵𝑟

1.75 𝐿= 0.122 𝑀 𝐾𝐵𝑟 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛

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Preparing a Solution

Can be performed in two

ways:

1. Dissolve a solid solute

with a liquid solvent

2. Mixing a liquid solute

with a liquid solvent

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Preparing a SolutionSolid Solute and Liquid Solvent

Example: Prepare a 2.80 M CaCl2 solution in a 500.0 mL volumetric flask. How many grams of CaCl2 is needed?

1. Molarity is in mol/L so we need our volume in L

500.0 𝑚𝐿 ×1 𝐿

103 𝑚𝐿= 0.5000 𝐿

2. Determine moles of solute (CaCl2) to add to 0.5000 L of solution to make 2.80 M

2.80 𝑚𝑜𝑙 𝐶𝑎𝐶𝑙21 𝐿

× 0.5000 𝐿 = 1.40 𝑚𝑜𝑙 𝐶𝑎𝐶𝑙2

3. Determine amount of solute in grams

1.40 𝑚𝑜𝑙 𝐶𝑎𝐶𝑙2 ×110.984 𝑔

1 𝑚𝑜𝑙 𝐶𝑎𝐶𝑙2= 155 𝑔 𝐶𝑎𝐶𝑙2

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Preparing a Solution

Can be performed in two

ways:

1. Dissolve a solid solute

with a liquid solvent

2. Mixing a liquid solute

with a liquid solvent

16

Preparing a SolutionLiquid Solute and Liquid Solvent

Commonly, the solutions that are stored in a lab or purchased are highly concentrated

These solutions are referred to as stock solutions

Will need to make solutions of lower concentrations from stock solutions

To do this, you will need to add more solvent

Note: amount of solute, in moles, does not change, just the volume; thus, moles of solute in original solution is equal to the moles of solute in diluted (new) solution the only thing that changes is the concentration

Equation used to dilute a stock solution

𝑀1𝑉1 = 𝑀2𝑉2or

𝐶1𝑉1 = 𝐶2𝑉2

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M1 or C1 is the concentration of the original solution

V1 is the volume of M1 or C1 (solute) delivered

M2 or C2 is the concentration of the new solution you are trying to make

V2 is the volume of the new concentration (M2 or C2)

Preparing a solution

Example: A 0.50 M NaOH solution is prepared from 50.0 mL of 3.0 M

What is the volume (mL) of the diluted solution?

1. Input the given values into the dilution equation and solve for V2

𝑀1𝑉1 = 𝑀2𝑉2

3.0 𝑀 × 50.0 𝑚𝐿 = 0.50 𝑀 × 𝑉23.0 𝑀 × 50.0 𝑚𝐿

0.50 𝑀= 𝑉2 = 75 𝑚𝐿

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M1 or C1 is the concentration of the original solution

V1 is the volume of M1 or C1 (solute) delivered

M2 or C2 is the concentration of the new solution you are trying to make

V2 is the volume of the new concentration (M2 or C2)

Preparing a solution

Example: Ana prepared 250. mL of a diluted NaOH solution using 8 mL of

a 1.8 M NaOH solution. What was the concentration of her new solution?

1. Input the given values into the dilution equation and solve for V2

𝑀1𝑉1 = 𝑀2𝑉2

1.8 𝑀 × 8 𝑚𝐿 = 𝑀2 × 250 𝑚𝐿

1.8 𝑀 × 8 𝑚𝐿

250 𝑚𝐿= 𝑀2 = 0.06 𝑀

Note: The units used on one side of the dilution equation carry-over to

the other side of the equation

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Solution

stoichiometry

When doing calculations, molarity can be used to convert between the amount of reactants and/or products in a chemical reaction

Example: What volume (in L) of 0.150 M KCl solution is required to completely react with 0.150 L of a 0.175 M Pb(NO3)2 solution according to the following balanced equation?

2𝐾𝐶𝑙(𝑎𝑞) + 𝑃𝑏(𝑁𝑂3)2(𝑎𝑞)→ 𝑃𝑏𝐶𝑙2(𝑠) + 2𝐾𝑁𝑂3(𝑎𝑞)

1. Determine ration between of moles of Pb(NO3)2 and KCl

1 𝑚𝑜𝑙𝑒 𝑃𝑏(𝑁𝑂3)2∶ 2 𝑚𝑜𝑙𝑒 𝐾𝐶𝑙

2. Use the information given to determine the number of moles in Pb(NO3)2

𝑚𝑜𝑙𝑒𝑠 𝑜𝑓𝑃𝑏(𝑁𝑂3)2=0.175 𝑚𝑜𝑙𝑒

𝐿× 0.150 𝐿 = 0.02625 𝑚𝑜𝑙𝑒𝑠

3. Determine volume of KCl solution needed to get necessary moles

0.02625 𝑚𝑜𝑙𝑒𝑠 𝑃𝑏(𝑁𝑂3)2 ×2 𝑚𝑜𝑙𝑒 𝐾𝐶𝑙

1 𝑚𝑜𝑙𝑒 𝑃𝑏(𝑁𝑂3)2×

1 𝐿

0.150 𝑚𝑜𝑙 𝐾𝐶𝑙= 0.350 𝐿 𝐾𝐶𝑙

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MOLES of A MOLES of BVOLUME

of A

VOLUME

of B