Your Name:

Group Activity: More Moles
More Chemical Equations


In this activity you will be learning a few more skills that put the mole to good use. You will learn about how to find the percent composition of a compound, how to use percent composition to find an empirical formula and

how to use empirical formulas to find molecular formulas. You will also spend some time working with the chemical ratios found in chemical reaction equations. Finally you will practice balancing equations and molar mass convsersions.

Percent Composition

One of the most important issues of our time concerns the addition of large amount of carbon dioxide to our atmosphere through the use of fossil fuels. The amount of carbon per gram of compound is different for different fuels. Along with determinations of energy density (how much energy per unit volume of fuel) and burner effiency the amount of carbon released when a fuel is burned is a very important topic. One way to express this is as the percent by mass of carbon in a compound.

For example, take propane (C3H8). What is the percent by mass of carbon in propane fuel?
Step by step:

  1. Find the molar mass of the compound. For this calculation we will pretend we have a sample of exactly one mole.
  2. Find the number of grams that each element contributes to the total molar mass.
  3. Divide the number of grams contributed by an element by the molar mass of the compound and multiply by 100%.
1. (C: 3 × 12.011 g) + (H: 8 × 1.0079 g) = 44.096 g/mol
2. (C: 3 × 12.011 g) = 36.033 g C
   (H: 8 × 1.0079 g) = 8.063 g H
    36.033 g C                        8.063 g H
3. ————————————— x 100% = 81.715%   ————————————— x 100% = 18.285%
   44.096 g C3H8                     44.096 g C3H8

Use all of the significant figures given in your periodic table for these calculations or else your percent composition answers will be incorrect. It is OK to round your answers to 3 places after the decimal but do not round any number until the whole calculation is done. Notice that the percentages add up to 100%. We could have found the percentage of the compound that is hydrogen by subtraction: 100% - 81.715% = 18.285%

As an exercise, calculate the percentage of octane (ordinary gasoline) that is carbon and compare it to the percentage of propane that is carbon. Octane is C8H18.

Empirical Formulas

The empirical formula of a compound is an expression in lowest terms of the ratio of each of the elements in the compound to each other. For example, the empirical formula of benzene (C6H6) is just CH. The empirical formula of glucose (C6H12O6 a simple sugar molecule) is CH2O. Many molecules have the same empirical formula and molecular formula: H2O, CO2, C3H8, etc.

Empirical formulas are important because they are used as the first step in figuring out the chemical formula of a compound. As you are well aware, atoms and molecules are too small to be able to tell what something is made of just by looking at it. More complicated techniques are required. One way to do it is to burn a substance under controlled conditions with an excess of oxygen and to collect the products of combustion. This works especially well with compounds made of only carbon, hydrogen and oxygen. As you know, the products of the combustion of a hydrocarbon are CO2 and H2O. If the amount of CO2 and the amount of H2O are carefully measured it is possible to figure out the ratio of carbon to hydrogen in the compound you burn.

page break

To do this you might use an apparatus like the one at below. A good compound for the absorption of water is magnesium percholrate. Sodium hydroxide is effective at absorbing carbon dioxide.

combustion_analysis apparatus collects CO2 and H2O from burning a substance (4k)

Say you have a compound that you know is made only of carbon, hydrogen and oxygen. If you burn a sample in the apparatus shown above (after carefully measuring the mass of the two absorbing units) you can find the mass of CO2 and H2O that is produced.

For example, when you burn 0.452 g of the compound the increase in mass of the CO2 absorber is 0.663 g and the increase in mass of the H2O absorber is 0.271 g. To figure out the mass of carbon that came from the compound being analyzed you have to figure out the mass of carbon in the CO2 without the oxygen. Carbon dioxide has a molar mass of 44.01 g/mol. Atoms of carbon have a molar mass of 12.01 g/mol. Therefore, carbon is 12.01/44.01 = 0.2729 × 100% = 27.29% of the mass of CO2. That means that of the 0.663 g of CO2 collected only 0.2729 × 0.663 g = 0.181 g is carbon from the compound being analyzed. So now we know that the compound is 0.181 g/0.452 g × 100% = 40.0% carbon.

In the same experiment the mass of the water absorber increases by 0.271 g. Hydrogen only makes up 2.016 g/18.02 g × 100% = 11.19% of the mass of water. Therefore, 0.1119 × 0.271 g = 0.030 g of hydrogen came from the compound being analyzed. This mass is 0.030 g/0.452 g × 100% = 6.64% of the whole compound.

Now it should be easy to figure out what percentage of the compound is oxygen. Just add the two percentages found so far and subtract the total from 100%. Oxygen is therefore 100% - (40.0% + 6.64%) = 53.4% of the compound.

So how do you build the empirical formula of a compound from this? You find molar ratios. Say you have 100.0 g of the compound instead of 0.452 g. This is convenient because then the compound contains 40.0 g of carbon, 6.64 g of hydrogen and 53.4 g of oxygen. Find out how many moles of each of these this is:

             1 mol C
40.0 g C x ————————— = 3.33 mol C
            12.01 g
            1 mol H
6.64 g H x ————————— = 6.59 mol H
            1.008 g
            1 mol O
53.4 g O x ————————— = 3.34 mol O
            16.00 g

Remember that an empirical formula is the smallest whole-number ratio of the elements in a compound. To find the smallest ratio just divide the number of moles of each element found in 100.0 g of the compound by the number of moles of the element present in the smallest amount:

    3.33 mol C             6.59 mol H             3.34 mol O
   ——————————— ≈ 1        ——————————— ≈ 2        ——————————— ≈ 1
      3.33                    3.33                    3.33
So the empirical formula of the compound is CH2O!

page break

One more hint about finding the empirical formula. If you do not get numbers close to whole numbers by dividing by the smallest number of moles then try multiplying by integers until you do get whole numbers. For example, round numbers like 9.92 and 1.08 but do not round numbers like 2.25, 4.33, or 2.72.

Is CH2O the real molecular formula of the compound? Unless you know the molar mass there is no way to tell. An empirical formula stands for all the molecular formulas that have the same molar ratios of the component elements. In this case CH2O could be C2H4O2, C3H6O4, C6H12O6 (glucose), or even C20H40O20.

But there is an easy way to find out which formula is correct for the compound you analyzed. Just do another experiment to find the approximate molar mass of the compound. Even with a precision of ± 1 g you could decide between CH2O and C2H4O2. So if the experimental molar mass is approximately 60 g/mol, what is the correct molecular formula of the compound? (Hint: Find the molar mass of the empirical formula and compare it to the molar masses of the possible molecular formulas). Now that you know the correct formula, what is the actual molar mass (using four significant figures)?

Limiting Reagents

When you use the chemical ratios found in chemical equations to predict amounts of reactants needed and amounts of product expected you are doing an exercise in purely theoretical chemistry. To bring things a bit closer to practical reality imagine a case in which you have a limited amount of fuel to burn but an unlimited amount of oxygen. This is usually the case in the open air on the surface of the Earth. The atmosphere contains so much oxygen, and plants are producing more all the time, that running one engine is not enough to use up all the oxygen available. As you may know, the situation is different in enclosed spaces. Inside a closed-up room of a house it can be dangerous to burn fuel because the oxygen in the room will quickly be used up.

When there is more than enough oxygen to burn all of the available fuel then the fuel is what is called the limiting reagent by chemists. The limiting reagent can be thought of as the chemical in a reaction that gets to decide how much product forms. An example will help.

When ethanol burns the reaction looks like this:

C2H6O + 3O2 Arrowsngl 2CO2 + 3H2O

If you have 2.5 mol C2H6O and an unlimited supply of O2 then you will get

                  2 mol CO2                                  3 mol H2O
2.5 mol C2H6O x ————————————— = 5 mol CO2   2.5 mol C2H6O x ————————————— = 7.5 mol H2O
                  1 mol C2H6O                                1 mol C2H6O

But suppose you only have 1 mol of O2 available? What happens then? The amount of oxygen is not enough to burn all of the ethanol. To burn all of the 2.5 moles of ethanol you need at least 7.5 moles of oxygen (2.5 mol C2H6O × 3 mol O2/1 mol C2H6O = 7.5 mol O2). First, you have some ethanol left over which does not burn. Second, the amount of carbon dioxide and water produced depends only on the amount of oxygen present, not on the amount of ethanol.

             2 mol CO2                                3 mol H2O
1 mol O2 x ————————————— = 0.67 mol CO2   1 mol O2 x ————————————— = 1 mol H2O
             3 mol O2                                 3 mol O2

And how much ethanol will be left over? One mole of O2 will react with 1 mol O2 × (1 mol C2H6O/3 mol O2) = 0.33 mol C2H6O. Since we started with 2.5 mol C2H6O there are 2.5 - 0.33 = 2.17 mol ethanol leftover.

This type of problem is quite simple. A somewhat harder problem requires you to figure out which reactant is the limiting reagent. Say you have 1 mol ethanol and 45 g of oxygen. Which of these is the limiting reagent? To find out just convert the mass of oxygen to moles: 45 g O2 × (1 mol/32.0 g) = 1.4 mol. Since you know from the chemical equation that exactly 3 moles of oxygen are required for every 1 mole of ethanol this calculation shows you that oxygen is the limiting reagent. That is, since 1.4 mol is less than the 3 mol required, the oxygen will be used up before the ethanol is.

page break

Questions and Problems

Comprehension Questions

Answer the following questions using one or more complete sentences. Everyone in the group must write down complete answers. Discuss among your group members what the best way to answer the question is and then write it down. All members must write down the answer or risk losing participation points.

  1. Definitions Write definitions in your own words using what you have learned about the concepts listed. Do not copy a definition from somewhere else!
    1. percent composition:
    2. molar ratio: (with reference to chemical formulas)
    3. molar ratio: (with reference to chemical equations)
    4. empirical formula:
    5. molecular formula:
    6. combustion analysis:
    7. limiting reagent:
  2. Explain how combustion analysis leads to the determination of the chemical formula of a compound.
  3. Design an experiment to determine the empirical formula of iron oxide (common rust).
  4. You have a recipe that normally makes 30 cookies. The recipe calls for 2 cups of butter but you only have 1.33 cups. How many cookies can you make, assuming you have enough of all the other ingredients?

page break

Chemical Equations

Balancing Chemical Equations

Balance the following chemical equations. Please write neatly and clearly indicate the final answer for each problem. Show your work. One hint: balance elements in this order: MINOH. Metals, Ions, Non-metals, Oxygen, Hydrogen. For complicated equations this sequence can help.

  1.    As +    NaOH -->    Na3AsO3 +    H2
  2.    KClO3 -->    KCl +    O2
  3.    KClO3 -->    KClO4 +    KCl
  4.    Al +    Fe2O3 --->    Al2O3 +    Fe
  5.    Li2O +    H2O -->    LiOH
  6.    N2O5 +    H2O -->    HNO3
  7.    H2SO4 +    HI -->    H2S +    I2 +    H2O
  8.    Al +    O2 -->    Al2O3

  1.    H3PO4 -->    H4P2O7 +    H2O
  2.    C6H6 +    O2 -->    CO2 +    H2O
  3.    Ca3P2 +    H2O -->    Ca(OH)2 +    PH3
  4.    P4O10 +    H2O -->    H3PO4
  5.    SiC +    Cl2 -->    SiCl4 +    C
  6.    CO2 +    NH3 -->    OC(NH2)2 +    H2O
  7.    Ca(OH)2 +    H3PO4 -->    CaHPO4 +    H2O
  8.    Al2(SO4)3 +    Ca(OH)2 -->    Al(OH)3 +    CaSO4

page break

Molar Mass and Molecular Mass

For each of the following compounds compute the molar mass and the mass of one molecule.

For example, H2O has a molar mass of (2 × 1.008 g/mol) + (1 × 16.00 g/mol) = 18.01 g/mol.

       g          1 mol
18.01 ——— × ——————————————————————————  = 3.0 × 10-23 g/molecule
      mol     6.02 × 1023 H2O molecules

So one molecule of water has a mass of 3.0 × 10-23 g.

  1. magnesium hydroxide
  2. sodium sulfate
  3. lithium fluoride
  1. sodium chloride
  2. carbon dioxide
  3. ethanol

Percent Composition

For each of the following compounds calculate the percent composition for all elements in the compound.

For example:

The molar mass of NaCl is 58.44 g/mol.
The molar mass of Na is 22.99 g/mol.
The molar mass of Cl is 35.45 g/mol.
% of NaCl that is Na       % of NaCl that is Cl
 22.99                     35.45
——————— × 100% = 39.34%   ——————— × 100% = 60.66%
 58.44                     58.44
  1. H2O
  2. Cu2O
  3. AlCl3
  1. KH2PO4
  2. PbSO4
  3. Pb(ClO2)2

Empirical Formulas

Find the empirical formulas using the given information.

  1. Find the empirical formula of a compound that is 74.9% carbon, 25.1% hydrogen.
  2. A common compound used as a paint pigment is made up of titanium and oxygen. It is 59.9% Ti by mass. What is the empirical formula?
  1. A compound of only N and O consists of 30.4% N by mass. Its molar mass is 92 g/mol. What are the empirical and molecular formulas of this compound?
  2. The empirical formula of a compound is NPCl2. The molar mass of this compound is 347.64 g/mol. What is the molecular formula?

page break

  1. A compound contains only carbon, hydrogen, and oxygen. Combustion of 10.68 mg of the compound yields 16.01 mg CO2 and 4.37 mg H2O. The molar mass of the compound is 176.1 g/mol. What are the empirical and molecular formulas of the compound?

Chemical Ratios (Stoichiometry)

Chemical equations confirm the law of the conservation of mass because they show that the number and kinds of atoms on the reactant side are equal to the number and kinds of atoms on the product side. But they also do more than this. Each correctly balanced equation shows the molar ratios among all the reactants and products in that reaction.

The key to understanding how to use these chemical ratios is to remember that the chemical unit of the mole is central.

  1. Mercury and bromine will react with each other to produce mercury(II) bromide. Answer the questions below about the following balanced chemical equation.
    Hg + Br2Arrowsngl HgBr2

    1. How many moles of mercury(II) bromide are produced when 30 g of mercury is reacted with an excess of bromine?
    2. If you have 1 mole of mercury and 0.5 moles of bromine, which one is the limiting reagent? How many moles of mercury(II) bromide can be made from reacting these two amounts?
    3. What mass of HgBr2 can be produced from the reaction of 10.0 g of mercury and 9.00 g of bromine?
    4. If you have 500 g of mercury and 500 g of bromine, which one is the limiting reagent? How many moles of mercury(II) bromide can be made from reacting these two amounts? What mass of which reagent is left unreacted?

Many problems and some techniques in this activity are inspired by the Chem Team at
I also used the Chang and Zumdahl texts for examples and problem ideas. See the Bibliography.
Previous Group Activity: Moles, Balancing Chemical Equations, Stoichiometry
First Homework Assignment: Molar Mass, Balancing Chemical Equations
Second Homework Assignment: Stoichiometry (aka Chemical Ratios)
Last updated: May 24, 2007 Home