Chapter 11: Stoichiometry: Calculations Based on Chemical Equations

Stoichiometry

• Stoichiometry is the calculation of quantities of substances involved in chemical reactions.

• Balanced chemical equations, through their coefficients indicate the ratio of moles of reactants and products:

N2 + 3H2 2NH3

• Mole ratios:1 mol N2

3 mol H2

or 3 mol H2

1 mol N2

1 mol N2

2 mol NH3

or2 mol NH3

1 mol N2

3 mol H2

2 mol NH3

or 2 mol NH3

3 mol H2

Mole-Mole Calculations• Given the moles of one reactant or product,

calculate the moles required or produced in the other reactant(s) or product(s).

• Steps:– 1. Write the balanced chemical equation.– 2. Write the moles of the given reactant or product.– 3. Multiply by the mole ratio: – (moles desired)/(moles starting chemical)

What does this equation mean?

• N2 + 3H2 2NH3

1 molecule of nitrogen (each containing 2 atoms) reacts with

3 molecules of hydrogen (each containing 2 atoms) to form:

2 molecules of ammonia. ( Each molecule contains 1 atom of nitrogen and 3 atoms of hydrogen.)

1 mole of nitrogen (N2) reacts with

3 moles of hydrogen (H2) to form:

2 moles of ammonia (NH3)

Convert moles to moles

• moles wanted =

• moles known( coef wanted/coef known)

• or:

• Moles of desired chemical =

Given moles of starting chemical xMoles of desired chemical

Moles of starting chemical

For the reaction equation:aA + bB cC + dD

• Mole B = mol A x (b/a)

• Mole C = mol A x (c/a)

• Mole D = mol A x (d/a)

N2 + 3H2 2NH3

• Calculate the moles of NH3 that can be produced from 10.8 moles of H2.

• Mole C = mol A x (c/a)

• Mol NH3 = mol H2 x [(coef NH3)/(coef H2)]

• mol NH3 = 10.8 mol H2 [( 2 NH3)/(3 H2)]

Mol NH3 = 7.20 mol NH3

Steps in doing stoichiometry

• Converting Grams of A to Grams of B

• 1. Convert grams to moles

• 2. Convert moles to moles

• 3. Convert moles to grams

Convert grams known to moles

• Moles = g/MM

Convert moles to moles

• moles wanted =

• moles known( coef wanted/coef known)

Convert moles to grams

• g = mole x MM

Insert figure 11.1

Route to Calculating Mass Relationships

Stoichiometric Process in One Equation

• For the equation: aA bB

Grams B =Grams A

Molar Mass AX

b

aX (Molar Mass B)

Calculating Mass of Unknown Quantities from

Known Quantities

Example Problem• N2 + 3 H2 2NH3

• Start with 50 g N2. How many grams of NH3 will be produced?

• MM N2 = 2 x 14 = 28 g

• MM NH3 = 14 + 3x1 = 17 g

• mol N2 = g/MM = 50/28 = 1.78 mol

• mol NH3 = 1.78 mol N2 x (2 NH3/1 N2)= 3.56 mol

• g NH3 = mol x MM = 3.56x 17 = 60.53 g

Moles to Grams

• In doing a Stoichiometry calculation, you can start and end at any point in the process, depending upon what is given and what is desired:

• Grams A moles A moles B grams B• To Convert from moles A to grams B, follow

the sequence:• moles A moles B grams B

N2 + 3 H2 2NH3

• Calculate the grams of N2 that will react with 12.0 moles of H2.

• Moles N2 = (12.0 moles H2)x(1 N2/3 H2)

• = 4.00 moles N2

• Molar Mass N2 = 2(14.0) = 28.0 g/mol

• grams N2 = 4.00 moles x 28.0 g/mol

• = 112 g N2

Calculations Involving Molar Solutions

• Calculations of the volumes of solutions can be made by modifying the stoichiometry process slightly:

• 1. Convert grams or volume of A to moles A• 2. Convert moles A to moles B• 3. Convert moles B to volume B or grams B• Remember: Molarity = Moles/Liters

– Moles = Liters x Molarity

– Liters = Moles/Molarity

Insert figure 11.2

Stoichiometry Calculations Involving

Solutions

Stoichiometry with Volumes

• Calculate the volume of 0.987 M NaOH required to neutralized 50 mL of 0.670 M HCl:

• NaOH(aq) + HCl(aq) NaCl(aq) + H2O(l)

• mol HCl = MxL = 0.670(0.050) = 0.0335 mol• mol NaOH = mol HCl = 0.0335 mol (from the

equation)• L NaOH = mol/M = 0.0335/.987 = 0.0339 L =• 33.9 mL NaOH.

Limiting Reactants

• In a chemical reaction, there is an exact ratio of reactants, determined by the balanced chemical equation.

• If there is too much of one of the reactants, it is said to be in excess.

• The other reactant is said to be the limiting reactant (or reagent).

A Limiting Reagent Analogy

Molecular Model of Limiting Reagents

Steps in Identifying the Limiting Reagent

• 1. Determine the moles of each reactant (mol=g/MM)

• 2. Divide the moles of each reactant by its coefficient.

• 3. The lesser of the amounts in step 2 is the limiting reagent. The other reactants are said to be in excess.

Example LR problem• N2 + 3H2 2 NH3

• If you start with 50.0g of Nitrogen and 45.0g of Hydrogen, Determine which reactant is in Excess and which is Limiting.

• Moles N2 = 50/28 = 1.78

• Moles H2 = 45/2 = 22.5

• Divide by coeff. 1.78/1 = 1.78 ; 22.5/3 =7.5

• nitrogen is less,: LR, hydrogen is in excess.

• The theoretical yield would be the amount of product you would predict to be produced from a given quantity of reactant(s) from stoichiometric calculations.

• The actual yield is the amount of product actually obtained by weighing.

• %yield = (actual/theoretical) x 100%

Calculate theoretical, actual and percentage yield, given appropriate data.

Example Problem• When 5.00g of N2 are reacted with 7.00g of H2, 5.08g

of NH3 are produced. Calculate the theoretical and the percentage yield.

• The actual yield is 5.08g.

• The balanced equation is:

• N2(g) + 3H2(g) 2NH3(g)

• 1. Determine the Limiting Reagent:– a. Determine the moles of each reactant.

– Mol = g/M; M(N2) = 2x14 = 28

When 5.00g of N2 are reacted with 7.00g of H2, 5.08g of NH3 are produced. Calculate the

theoretical and the percentage yield.• N2(g) + 3H2(g) 2NH3(g)

– M(H2) = 2x1 = 2

– mol N2 = 5.00/28 = 0.179 mol

– mol H2 = 7.00/2 = 3.5 mol

– b. Divide each of the mol by its coefficient:• N2: 0.179/1 = 0.179

• H2: 3.5/3 = 1.17

– c. Choose the smaller of the answers in b. as the Limiting Reagent: N2 : Base further calculations on LR: N2

When 5.00g of N2 are reacted with 7.00g of H2, 5.08g of NH3 are produced. Calculate the

theoretical and the percentage yield.• N2(g) + 3H2(g) 2NH3(g)

• 2. Calculate theoretical yield based on the LR, N2:

– a. Mol N2 = 0.179 mol (see previous calc.)

– b. Mol NH3 = mol N2x (coef. NH3/coef. N2)

– = 0.179x(2/1) = 0.358 mol

– c. mass NH3 = molxM

– M = 14 + 3(1) = 17

– mass NH3 = 0.358 x 17 = 6.09 g = Th. Yield

Percentage Yield

• Percentage Yield = (Actual/Theoretical)x100

• = (5.08/6.09)x100

• = 83.4%

Energy Changes in Chemical Reactions

• Many reactions give off heat energy.• This energy is measured in Joules (J), kilojoules

(kJ), calories (cal), or kilocalories (kcal):– 1 cal = 4.184 J– 1 kcal = 4.184 kJ

• The heat energy given up may be treated as a product: In the combustion of methane:– CH4(g) + 2 O2(g) CO2(g) + H2O(g) + heat

Factors in Heat Energy

• Chemicals present• Amounts used• Conditions of reaction

– At 1 atm pressure, 25oC, for 1 mol CH4:

– CH4(g) + 2 O2(g) CO2(g) + H2O(g) + 802 kJ

– CH4(g) + 2 O2(g) CO2(g) + H2O(l) + 890 kJ

– More heat is given of if H2O is produced in the liquid state compared to the gas state.

Different Types of Energy May be Given Off or Absorbed

• If heat energy is given off, the reaction is exothermic.

• If other types of energy are given off, the reaction is exergonic.

• If heat energy is absorbed, the reaction is endothermic.

• If other types of energy is absorbed, the reaction is endergonic.

Enthalpy• The amount of energy given off in a chemical

reaction is equal to the difference in chemical energy between the reactants and products. When pressure is constant, this energy change is called the enthalpy of the reaction, H.

• The reaction of hydrogen with oxygen to form water is exothermic:– H2(g) + ½O2(g) H2O(l) + 283 kJ

• This can also be written:– H2(g) + ½O2(g) H2O(l) kJ

• Exothermic reactions have negative values.

Endothermic Reactions

• The reverse of this reaction corresponds to the electrolysis of water.– H2O(l) + 283 kJ H2(g) + ½O2(g)

– H2O(l) H2(g) + ½O2(g) = +283 kJ

• Endothermic Reactions have a positive enthalpy change.

• The reverse reaction has an enthalpy change of the same magnitude but opposite sign.

Quantitative Considerations

• H2(g) + ½O2(g) H2O(l) kJ

• Calculate the heat given off when 5.0 g of water is produced:

• The heat is proportional to the number of moles. The above reaction is for 1 mol.

• 5.0 g/ 18g/mol = 0.277 mol

• heat given off =

• (-283 kJ/mol) x 0.277mol= 78.6 kJ

Tendency for Reactions to Occur • First Law of Thermodynamics

– Energy is conserved. In giving off energy, compounds go to a lower energy state. Systems tend to move to lower energy states.

• Endothermic reactions can occur because systems also tend to become more random.– This randomness is called entropy. Change in entropy is

S. – According to the Second Law of Thermodynamics, the

entropy of the universe is always increasing. If in a chemical reaction the entropy of the universe increases, the reaction is spontaneous.