7.5: Chemical Formulas as Conversion Factors
- Page ID
- 105309
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- Use chemical formulas as conversion factors.
Figure \(\PageIndex{1}\) shows that we need two hydrogen atoms and one oxygen atom to make 1 water molecule. If we want to make two water molecules, we will need four hydrogen atoms and two oxygen atoms. If we want to make five molecules of water, we need 10 hydrogen atoms and 5 oxygen atoms. The ratio of atoms we will need to make any number of water molecules is the same: 2 hydrogen atoms to 1 oxygen atom.
Using formulas to indicate how many atoms of each element we have in a substance, we can relate the number of moles of molecules to the number of moles of atoms. For example, in 1 mol of water (H2O) we can construct the relationships given in (Table \(\PageIndex{1}\)).
| 1 Molecule of \(H_2O\) Has | 1 Mol of \(H_2O\) Has | Molecular Relationships |
|---|---|---|
| 2 H atoms | 2 mol of H atoms | \(\mathrm{\dfrac{2\: mol\: H\: atoms}{1\: mol\: H_2O\: molecules}}\) or \(\mathrm{\dfrac{1\: mol\: H_2O\: molecules}{2\: mol\: H\: atoms}}\) |
| 1 O atom | 1 mol of O atoms | \(\mathrm{\dfrac{1\: mol\: O\: atoms}{1\: mol\: H_2O\: molecules}}\) or \(\mathrm{\dfrac{1\: mol\: H_2O\: molecules}{1\: mol\: O\: atoms}}\) |
The MOle is big
A mole represents a very large number! The number 602,214,129,000,000,000,000,000 looks about twice as long as a trillion, which means it’s about a trillion trillion.
(CC BY-SA NC; https://what-if.xkcd.com/4/).
A trillion trillion kilograms is how much a planet weighs. If 1 mol of quarters were stacked in a column, it could stretch back and forth between Earth and the sun 6.8 billion times.
| 1 Molecule of \(C_2H_6O\) Has | 1 Mol of \(C_2H_6O\) Has | Molecular and Mass Relationships |
|---|---|---|
| 2 C atoms | 2 mol of C atoms | \(\mathrm{\dfrac{2\: mol\: C\: atoms}{1\: mol\: C_2H_6O\: molecules}}\) or \(\mathrm{\dfrac{1\: mol\: C_2H_6O\: molecules}{2\: mol\: C\: atoms}}\) |
| 6 H atoms | 6 mol of H atoms | \(\mathrm{\dfrac{6\: mol\: H\: atoms}{1\: mol\: C_2H_6O\: molecules}}\) or \(\mathrm{\dfrac{1\: mol\: C_2H_6O\: molecules}{6\: mol\: H\: atoms}}\) |
| 1 O atom | 1 mol of O atoms | \(\mathrm{\dfrac{1\: mol\: O\: atoms}{1\: mol\: C_2H_6O\: molecules}}\) or \(\mathrm{\dfrac{1\: mol\: C_2H_6O\: molecules}{1\: mol\: O\: atoms}}\) |
|
2 (12.01 amu) C 24.02 amu C |
2 (12.01 g) C 24.02 g C |
\(\mathrm{\dfrac{24.02\: g\: C\: }{1\: mol\: C_2H_6O\: molecules}}\) or \(\mathrm{\dfrac{1\: mol\: C_2H_6O\: molecules}{24.02\: g\: C\: }}\) |
|
6 (1.008 amu) H 6.048 amu H |
6 (1.008 g) H 6.048 g H |
\(\mathrm{\dfrac{6.048\: g\: H\: }{1\: mol\: C_2H_6O\: molecules}}\) or \(\mathrm{\dfrac{1\: mol\: C_2H_6O\: molecules}{6.048\: g\: H\: }}\) |
|
1 (16.00 amu) O 16.00 amu O |
1 (16.00 g) O 16.00 g O |
\(\mathrm{\dfrac{16.00\: g\: O\: }{1\: mol\: C_2H_6O\: molecules}}\) or \(\mathrm{\dfrac{1\: mol\: C_2H_6O\: molecules}{16.00\: g\: O\: }}\) |
The following example illustrates how we can use the relationships in Table \(\PageIndex{2}\) as conversion factors.
Example \(\PageIndex{1}\): Ethanol
If a sample consists of 2.5 mol of ethanol (C2H6O), how many moles of carbon atoms does it have?
Solution
| Steps for Problem Solving |
If a sample consists of 2.5 mol of ethanol (C2H6O), how many moles of carbon atoms does it have? |
|---|---|
| Identify the "given"information and what the problem is asking you to "find." |
Given: 2.5 mol C2H6O |
| List other known quantities |
1 mol C2H6O = 2 mol C |
|
Prepare a concept map and use the proper conversion factor. |
 |
| Cancel units and calculate. |
Note how the unit mol C2H6O molecules cancels algebraically. \(\mathrm{2.5\: \cancel{mol\: C_2H_6O\: molecules}\times\dfrac{2\: mol\: C\: atoms}{1\: \cancel{mol\: C_2H_6O\: molecules}}=5.0\: mol\: C\: atoms}\) |
| Think about your result. | There are twice as many C atoms in one C2H6O molecule, so the final amount should be double. |
Exercise \(\PageIndex{1}\)
If a sample contains 6.75 mol of Na2SO4, how many moles of sodium atoms, sulfur atoms, and oxygen atoms does it have?
- Answer:
- 13.5 mol Na atoms, 6.75 mol S atoms, and 27.0 mol O atoms
The fact that 1 mol equals 6.022 × 1023 items can also be used as a conversion factor.
Example \(\PageIndex{2}\): Oxygen Mass
Determine the mass of Oxygen in 75.0g of C2H6O.
Solution
| Steps for Problem Solving |
Determine the mass of Oxygen in 75.0g of C2H6O |
|---|---|
| Identify the "given"information and what the problem is asking you to "find." |
Given: 75.0g C2H6O |
| List other known quantities |
1 mol O = 16.0g O 1 mol C2H6O = 1 mol O 1 mol C2H6O = 46.07g C2H6O |
|
Prepare a concept map and use the proper conversion factor. |
 |
| Cancel units and calculate. |
\(\require{cancel}\mathrm{75.0\: \cancel{g\: C_2H_6O}\times\dfrac{1\: \cancel{mol\: C_2H_6O}}{46.07\:\cancel{g\: C_2H_6O}}\times\dfrac{1\: \cancel{mol\:O}}{1\: \cancel{mol\:C_2H_6O}}\times\dfrac{16.00\: g\: O}{1\: \cancel{mol\:O}}=26.0\: g\: O}\) |
| Think about your result. |
Exercise \(\PageIndex{2}\)
- How many molecules are present in 16.02 mol of C4H10? How many C atoms are in 16.02 mol?
- How many moles of each type of atom are in 2.58 mol of Na2SO4?
- Answer a:
- 9.647 x 1024 C4H10 molecules and 3.859 x 1025 C atoms
- Answer b:
- 5.16 mol Na atoms, 2.58 mol S atoms, and 10.3 mol O atoms