Learning Goal 4: Use molar relationships in a balanced chemical reaction to predict the mass of...

Learning Goal 4: Use molar relationships in a balanced chemical reaction to predict the mass of product produced in a simple chemical reaction that goes to completion.

Substance Symbol Unit Mass

Graham Cracker S 7.0 g

Marshmallow Mm 7.1 g

Chocolate Pieces Or 3.3 g

S’more S2MmOr3 __________ g

2S + Mm + 3Or S2MmOr3The equation above gives the recipe

to make one S’more.

The equation for a chemical reaction gives the ingredients needed to make a specific amount of product.

Copyright © Houghton Mifflin Company

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Table 9.1

Calculating Masses of Reactants Calculating Masses of Reactants and Productsand Products

1.1. Balance the equationBalance the equation..

2.2. Convert mass to moles. Convert mass to moles.

3.3. Set up mole ratios. Set up mole ratios.

4.4. Use mole ratios to calculate Use mole ratios to calculate moles of desired molecule. moles of desired molecule.

5.5. Convert moles to grams or to Convert moles to grams or to particles, if necessary.particles, if necessary.

Working a Stoichiometry Working a Stoichiometry ProblemProblem

6.50 grams of aluminum reacts with an excess of oxygen. How many grams of aluminum oxide are formed.

Identify reactants and products and write the balanced equation.

4Al + 3O2 2Al2O3

What are the reactants?

Every reaction needs a yield sign!

What are the products?

What are the balanced coefficients?

Working a Stoichiometry ProblemWorking a Stoichiometry Problem

6.50 grams of aluminum reacts with an excessof oxygen. How many grams of aluminum oxide are formed?

4 Al + 3 O2 2Al2O3

=6.50 g Al

? g Al2O3

1 mol Al

26.98 g Al 4 mol Al

2 mol Al2O3

1 mol Al2O3

101.96 g Al2O3

6.50 x 2 x 101.96 ÷ 26.98 ÷ 4 =

12.3 g Al2O3

Like all percent problems:

part

whole

1) Find the mass of each of the components (the elements),

2) Next, divide by the total mass of the compound; then x 100

x 100 % = percent

ExampleCalculate the percent composition of a compound that is made of 29.0 grams of Ag with 4.30 grams of S.

29.0 g Ag33.3 g total

X 100 = 87.1 % Ag

4.30 g S33.3 g total

X 100 = 12.9 % S

Total = 100 %

Theoretical Yield The amount of product predicted from the

amount of reactant used.

The maximum amount of product that could form.

Actual Yield The amount of product that is actually produced.

Percent Yield

Actual Yield X 100% = Percent Yield

Theoretical Yield

First write the balanced equation.

2H2 + CO CH3OH

Next Calculate moles of reactants.

68,500 g CO 1 Mol CO = 2450 mol CO

28.01 g CO

8600 g H2 1 Mol H2 = 4270 mol H2

2.016 g H2

Next determine the limiting factor

2,450 mol CO 2 mol H2 = 4,900 mol H2

1 mol CO

4270 mol H2 < 4,900 mol H2

Hydrogen is the limiting factor.

Next determine the maximum amount of methanol that can be produced in this reaction.

4270 mol H2 1 mol CH3OH = 2140 mol CH3OH

2 mol H2

Convert moles to grams.

2140 mol CH3OH 32.04 g CH3OH= 68600 g CH3OH

1 mol CH3OH

Percent Yield

35700 g CH3OH x 100 % = 52.0%

68600 g CH3OH

FormulasFormulasmolecular formula = (empirical formula)molecular formula = (empirical formula)nn

[[nn = integer] = integer]

molecular formula = Cmolecular formula = C66HH66 = (CH) = (CH)66

empirical formula = CHempirical formula = CH

Empirical formula: the lowest whole number ratio of atoms in a compound.

Molecular formula: the true number of atoms of each element in the

Limiting ReactantLimiting Reactant The The limiting reactantlimiting reactant is the is the

reactant reactant

that is that is consumed firstconsumed first,, limiting limiting the amounts of products the amounts of products formed.formed.


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Write and balance the equation for the reaction.

Convert known masses of reactants to moles.

Determine which reactant is limiting.

Using the amount of the limiting reactant and the appropriate mole ratios, compute the number of moles of the desired product.

Convert moles to mass (if required).

Write the balanced equation.

N2 + 3H2 2NH3

Convert known masses to moles.

25000 g N2 1 mol N2 = 892 mol N2

28.02 g N2

5000 g H2 1 mol H2 = 2,480 mol H2

2.016 g H2

Determine the Limiting reactant.

892 mol N2 3 mol H2 = 2676 mol H2

1 mol N2

2480 mol H2 < 2676 mol H2

Hydrogen is the limiting reactant. To get all of the Nitrogen to react you would need 2676 mol of hydrogen gas.

Using the amount of the limiting reactant and the appropriate mole ratios, compute the number of moles of the desired product.

2480 mol H2 2 mol NH3 = 1653 mol NH3

3 mol H2

Convert moles to mass.

Calculate the molar mass of NH3

N 14.01 x 1 = 14.01

H 1.008 X 3 = 3.024 +

17.034

1653 mol NH3 17.03 g NH3 = 28,150 g NH3

1 mol NH3

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