Unit #1Unit #2Unit #3Unit #4Unit #5Unit #6Unit #7Unit #8Unit #9Unit #10
spacerUNIT # 8  
spacerspacerIntroduction
spacerspacerObjectives
spacerspacerReading


spacer
8.1 Conventions
spacerand Symbols
spacer8.2 Balancing
spacerChemical
spacerEquations

spacer8.3 Methods of
spacerBalancing
spacerEquations

spacerBalance by Inspection
spacerBalance algebraically
spacer8.4 Types of
spacerChemical
spacerReactions
spacer8.5 Calculations
spacerBased on
spacerChemical
spacerEquations
spacerMole-mole calculations
spacerMass-mass calculations
spacerMass-mole calculations
spacerLimiting Reagent calculations



spacerProblems
spacer1 | 2
| 3 | 4

Unit #8CHEMICAL REACTIONS

8.5 - Calculations based on Chemical Equations

8.5.2 - Mass-mass Calculations

In unit 7, we learned conversion between mass and moles (section 7.5.1) for a given compound. We will now combine this mass-to-mole conversion in stoichiometric calculations.

Using the same example from section 8.5.1, we now pose another question. This time you will be given the mass of one reactant (C2H5OH) and you will be asked to find the mass of the other reactant (oxygen gas).

How many grams of oxgen gas is required to burn 20.5 grams of ethyl alcohol, C2H5OH?

We use the basic steps for the example from section 8.5.1, and add extra steps for the conversions from moles to mass:
(Highlighted in red are the modifications needed in order to answer this question.)

Step 1: Identify the chemical equation involved: a combustion reaction.
Step 2: Write the balanced chemical equation (review balancing equations, section 8.3).
Step 2a: Convert the mass of C2H5OH to moles of C2H5OH.
Step 3: Determine the stoichiometric ratio of the substances that you are working with, specifically between ethyl alcohol and oxygen.
Step 4: Calculate the number of moles of oxygen by making the ratio specifically for the number of moles of C2H5OH from step 2a.
Step 5: Convert the number of moles of oxygen obtained in step 4 to mass of oxygen.

Let's analyze:

  1. Steps 1, and 2 are the same as the example in section 8.5.1.
  2. Step 2a is necessary because the mass of C2H5OH must be converted to moles since the coefficients in the balanced chemical reaction give the mole ratio of the reactants. To set up stoichiometric ratios using masses would be WRONG.
  3. In Step 4, we calculate the number of moles of oxygen that is required to burn the number of moles of C2H5OH. The number of moles of C2H5OH available is determined from the given mass of 20.5 grams C2H5OH in Step 2a.
  4. After we obtain the number of moles of oxygen from Step 4, we need to convert it to mass of oxygen in order to answer the question properly.

Summary:

Refer to section 8.5.1 for a detailed look at Step 1, 2, and 3. Click here for a detailed look at my work for Steps 2a, 3 and 5. If you need to see the specific steps, click on the links below.

mass of C2H5OH (20.5 grams)
moles of C2H5OH
moles of oxygen
mass of
oxygen


(Note: We use the calculator answer in our calculations up until the final answer. Then we follow significant figure rules to remove the excess digits.)

book iconSection 10.4
spacerThe Mass-Mass Stoichiometry Problems ..p251



back grey arrowBack



All contents copyrighted © 1996-2006
British Columbia Institute of Technology
Chemistry Department - 3700 Willingdon Avenue
Burnaby, B.C. Canada V5G 3H2