chapter 12 stoichiometry. another analogy: (let’s get off the bike for a while and bake a cake!) ...

Chapter 12

Stoichiometry

Another Analogy: (let’s get off the bike for a while and bake a cake!)

Let’s say you want to bake a cake. Here’s a recipe: 2sticks of butter + 3 cups flour + 2 cups sugar +3 eggs 1 cake

Ask yourself these questions:

What are the ingredients? The goal?

How are the ingredients and cake measured (quantified)?

How much of each ingredient do you need to make 2 cakes?

What would happen if you have 2 eggs?

What would happen if you had 6 eggs?

12.1 The Arithmetic of Equations

Chemists use balanced chemical equations as a basis to calculate how much reactant is needed or product is formed in a reaction.

Stoichiometry – the calculation of quantities in chemical

reactions.

Interpreting Chemical Equations

Coefficients Represent

Coefficients Do Not Represent

moleculesmoles

volume

mass

•A balanced chemical equation can be interpreted in terms of different quantities, including atoms, molecules, moles, or volume.

* It is important to know what coefficients represent and what they do not represent in a chemical equation.

An example:

For the equation:

N2 (g) + 3H2 (g) 2NH3 (g)

(nitrogen) (hydrogen) (ammonia)

 Coefficients are: 1, 3, and 2

Meaning:1 mole of nitrogen gas reacts with 3 moles of hydrogen gas to produce 2 moles of ammonia.

12.2 Chemical Calculations

Remember conversion factors from chapter 10?

365 days/year

6.02 x 1023 molecules/mole

32.00 g O2/mole O2

Remember that algebra allows you to use these conversion factors to multiply an expression without changing the expression.

By carefully choosing conversion factors and canceling units, you can convert any measurement into one expressed in the desired units.

12.2 Chemical CalculationsFrom a balanced equation, mole ratios

can be determined that relate the moles of reactants to moles of product.

Mole ratios are conversion factors derived from the coefficients of a balanced chemical equation.

N2 (g) + 3H2 (g) 2NH3 (g)

From the equation above, 3 mole ratios can be derived (and their reciprocals):

1 mol N2 2 mol NH3 3 mol H2

3 mol H2 1 mol N2 2 mol NH3

Mole-to-mole calculations: If you know the number of moles of one substance,

the balanced chemical equation allows you to determine the number of moles of other substances in the reaction using those mole ratios (a type of conversion factor)

Example: using the previous equation, how many moles of ammonia (NH3) are produced when .60 mol of nitrogen reacts with hydrogen.

N2 + 3H2 2NH3

.60 mol N2 x 2mol NH3 = 1.2 mol NH3

1 mol N2

Mass-to-mass calculations

Step 1: Check that the equation is balanced.

Step 2: Start with the number given and multiply by a conversion factor with the molar mass to convert Mass to Moles.

 Step 3: Multiply by the next conversion factor using coefficient (mole) ratios to convert Moles to Moles

Step 4: Multiply by a third conversion factor using the molar mass of the new compound to convert Moles to Mass.

Mass to mass example:

1) 2AgNO3(aq) + Na2SO4(aq) Ag2SO4(s) + 2NaNO3(aq)

2) 17.0 g AgNO3 x 1 mol AgNO3 Molar mass

169.9 g AgNO3Ag = 107.9 x 1 = 107.9

N = 14.0 x 1 = 14.0

O = 16.0 x 3 = 48.0

TOTAL = 169.9

3) 17.0 g AgNO3 x 1 mol AgNO3 x 1 mol Ag2SO4

169.9 g AgNO3 2 mol AgNO3

4) 17.0 g AgNO3 x 1 mol AgNO3 x 1 mol Ag2SO4 x 311.8 g Ag2SO4 =15.6g 169.9 g AgNO3 2 mol AgNO3 1 mol Ag2SO4

Molar massAg = 107.9 x 2 = 215.8

S = 32.0 = 32.0

O = 16.0 x 4 = 64.0

TOTAL = 311.8

Other Stoichiometric Calculations Use the diagram on p797 to solve other

stoichiometric problems.

Summary: A balanced equation gives you mole ratios. With mole ratios you can calculate any measurement unit related to the mole, such as # of particles, mass, or volumes of gases (at STP)

12.3 Limiting Reactant

+ =

In any chemical reaction, it is possible to “run out” of one or another reactant. This has an impact on the amount of products that can result from a reaction

To solve these problems you must identify which of the reactants is going to run out first. This is the “limiting reactant”The other is the “excess reactant”

Example: – If you had three yellow blocks and four red blocks, you could only make three yellow/red combinations – because there is not enough yellow blocks to make four. The yellow block is the limiting reagent and the red block is the excess reagent

Container 1No Limiting Reactant

Zumdahl, Zumdahl, DeCoste, World of Chemistry 2002, page 269

Before and After Reaction 1

All the hydrogen and nitrogen atoms combine.

Before the reaction After the reaction

Zumdahl, Zumdahl, DeCoste, World of Chemistry 2002, page 269

In reaction 1 there is no limiting reactant

both N2 and H2 are completely used in the reaction.

In reaction 2 (next) H2 is runs out first and therefore limits the number of NH3 molecules produced.

Excess: Notice that two N2 molecules are left unreacted. This is called excess. If one reactant runs out the other reactant will be in excess.

Container 2 H2 as Limiting Reactant

Zumdahl, Zumdahl, DeCoste, World of Chemistry 2002, page 270

Before and After Reaction 2

Before the reaction After the reaction

Zumdahl, Zumdahl, DeCoste, World of Chemistry 2002, page 270

19

Determining Limiting Reactant

Ex: Synthesis of Ammonia (NH3)

N2 + 3H2 → 2NH3

If 5.0 g nitrogen gas and 5.0 g hydrogen gas react to form ammonia,

which is the limiting reactant?

Steps:

(1) Convert units of each reactant (i.e., grams) to moles

(2) Divide the moles of each reactant by the coefficient

for that reactant in the equation.

(3) The smaller reference number is the limiting reactant

N2 + 3H2 → 2NH3

The second step adjusts for the different moles

that react in the equation.20

Determining Limiting Reactant

21

Determining Limiting Reactant

Calculations:

83.03/48.202.2

10.5

18.01/18.002.28

10.5

22

22

22

22

HmolHg

HmolHg

NmolNg

NmolNg

Limiting Reactant = N2

33

3

2

3

2

22 08.6

1

04.17

1

2

02.28

10.5 NHg

NHmol

NHg

Nmol

NHmol

Ng

NmolNg

Stoichiometry

Use the limiting reactant to determine the mass of NH3 produced.

Determining the Mass of Excess Reactant

Use the limiting reactant (N2) to calculate the mass of excess reactant (H2) that was consumed in the reaction

Subtract this mass from the mass available before the reaction.

22

2

2

2

2

22 08.1

1

02.2

1

3

02.28

10.5 Hg

Hmol

Hg

Nmol

Hmol

Ng

NmolNg

222 92.308.10.5 HgHgHg

Another example:

Nitrogen gas can be prepared by passing gaseous ammonia over solid copper II oxide at high temperatures. The other products of the reaction are solid copper and water vapor. If a sample containing 18.1g of NH3 is reacted with 90.4g of CuO, which is the limiting reactant? How many grams of N2 will be formed?

Write the Balanced Equation:

Calculate Limiting Reactant:

Find the mass of N2 (g) formed:

)(3)(3)()(3)(2 223 gOHsCugNsCuOgNH

379.3/14.155.79

14.90

530.02/06.104.17

11.18 3

3

33

CuOmolCuOg

CuOmolCuOg

NHmolNHg

NHmolNHg

22

22 6.101

02.28

3

1

80

14.90 Ng

Nmol

Ng

CuOmol

Nmol

CuOg

CuOmolCuOg

LR = CuO

Percent Yield:

To calculate the percent yield, divide the actual yield by the theoretical (calculated Mass-Mass) and multiply by 100.

Example: Using the above problem and its theoretical yield, calculate the percent yield if only 10.0 g of N2 gas is formed.

10.0g/14.9g = .67 67%