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The Mole

Once we know the mass of a sample we can use the mass of the atoms or molecules in the sample to determine how many atoms or molecules are in the sample.

How many 12C atoms are in a 1.00 kg block of pure 12C isotope?

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Obviously atoms and molecules are not a convenient unit of measure when we're working with macroscopic (i.e., human size) objects. For this reason chemist define a new unit of measure called the mole.

1 mole is defined as the number of carbon atoms in exactly 12.000000 grams of pure 12C.

From our definition of the atomic mass unit we get

1 mole = 6.0220943 x 1023 chemical units

The number 6.0220943 x 1023 is called Avagadro's number.

How many moles are in a 1.00 kg block of pure 12C isotope?

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Of course we know a 1.00 kg block of naturally occuring carbon will contain a mixture of 12C, 13C, and even some 14C isotopes. So the number of moles of carbon atoms in a 1.00 kg block of naturally occuring carbon is

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We can also calculate the mass of a mole of molecules.

What is the mass of a mole of methane (CH4) molecules?

mass of 1 mole of C is 12.011 g
mass of 4 moles of H is 4 X 1.00g
mass of 1 mole of CH4 is
16.043 g




In other words the molecular weight of CH4 is 16.043g.

Molecular weight = the mass in grams of 1 mole of a molecule.

For ionic compounds, which do not exist as individual molecules we use the term Formula weight

Formula weight = The mass in grams of 1 mole of the chemical formula.

What is the formula weight of CaCO3?

mass of 1 mole of Ca is 40.08 g
mass of 1 mole of C is 12.011 g
mass of 3 moles of O is 3 X 15.999 g
Formula weight of CaCO3 is
100.09 g

Let's look at another example.

How many hydrogen atoms are there in 2.50 g of NH3?

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Now, let's try some sample quiz questions on:

Avogadro's Number
  • Molecular Weight from Weight of Molecule:
  • Mass of Molecule:
  • Atoms Present in Mass of a Sample:
  • Atoms in a Shaped Material Given Density:
  • Molecules Present in Mass of a Sample:




The Mole Concept
  • Moles of One Type of Atom in a Mass of a Compound:
  • Total Number of Moles of Atoms in a Compound:
  • Comparison of Different Samples:
  • Moles of Water in Hydrates:
  • Grams of Sample from Number of Moles:
  • Moles from Grams of Sample:
  • Comparison of Moles of Atoms in Compounds:
  • Moles from Mass and Shape of An Object:
  • Molecular Weight Calculated from Molecular Formula:
  • Comparison of Molecular Weights of Several Compounds:




Mass Percent

The Mass Percent of a component is defined
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What is the mass percents of carbon, hydrogen, and oxygen in pure ethanol C2H6O?

First we calculate the mass of one mole of C2H6O...
mass of 2 moles of C is 2 X 12.011 g
mass of 6 moles of H is 6 X 1.008 g
mass of 1 mole of O is 15.999 g
Molecular weight of C2H6O is
46.069 g

Next we calculate the mass percents

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Note that the mass percentages should add up to 100%.





Combustion Analysis

When chemists make new compounds one of the first things they often do is determine the mass % for the different elements in the compound. To analyze the mass percent of carbon and hydrogen chemist use a combustion device.

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The sample is burned in the presence of excess oxygen which converts all the carbon to carbon dioxide and all the hydrogen to water. The CO2 and H2O produced are absorbed in two different stages and their masses determined by measuring the increase in weight of the absorbers.

Ascorbic acid (vitamin C) contains only C, H, and O. Combustion of 1.000 g of Ascorbic acid produced 40.9% C and 4.5% H. What is the empirical formula for Ascorbic Acid?

First we need to calculate the mass percent of Oxygen. Since the sample contains C, H, and O, then the remaining

100% - 40.9% - 4.5% = 54.6% is Oxygen

Now we need to express the composition in grams and determine the number of moles of each element:

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Next we divide by the smallest number of moles to obtain the mole ratio which is also the atom ratio. In this case carbon has the smallest number of moles, so...

C: 0.0340 moles/0.0340 moles = 1
H: 0.045 moles/0.0340 moles = 1.32 ~ 1 1/3
O: 0.0341 moles/0.0340 moles ~ 1

Finally we calculate the smallest whole integer ratios by multiplying each number above by 3 to get

C: 3 H:4 O:3

thus we obtain the empirical formula C3H4O3

Remember the empirical formula has the smallest whole integer ratios, the molecular formula can be different, e.g., C6H8O6, or C9H12O9, or C12H16O12, ... are all possible molecular formulas.

Now let's look at a related question.

What is the molecular formula if the molecular weight of Ascorbic Acid was formed to be 176 g/mole?

In this case we need to find the multiplication factor between the molecular formula and the empirical formula:

factor = (molecular weight)/(empirical formula weight)

The empirical formula weight of Ascorbic Acid is

mass of 3 moles of C is 3 X 12.011 g
mass of 4 moles of H is 4 X 1.008 g
mass of 3 moles of O is 3 X 15.999 g
mass of 1 mole of CH4 is
88.062g

Therefore the multiplicative factor is (176 g/mole)/(88.062 g/mole) ~ 2, and the molecular formula for Ascorbic Acid is C6H8O6





  • Empirical Formulas - Combustion Analysis of Compounds of C, H and O:
  • Empirical Formulas - Combustion Analysis of Compounds of C, H and N:
  • Empirical Formulas - Combustion Analysis of Three Element Compounds:
  • Empirical Formulas - Combustion Analysis of Complex Compounds:
  • Molecular Weight Determination:
  • Calculating the Percentage Composition from Formula:
  • Composition from Molecular Weight and Number of Atoms:




Homework from Chemisty, The Central Science, 10th Ed.

3.21, 3.23, 3.25, 3.29, 3.31, 3.33, 3.35, 3.37, 3.39, 3.41, 3.43, 3.45, 3.47, 3.49
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