How to Determine an Empirical Formula
Look at the data., Determine the number of grams for each element., Convert the mass of each element to moles., Divide each mole value by the smallest number of moles present., Multiply the ratio values to find near whole numbers., Round the values...
Step-by-Step Guide
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Step 1: Look at the data.
If you are given the elemental composition of an unknown compound in percentages rather than grams, you should assume that there are exactly
100.0 grams of the substance involved.
These are the instructions you should follow if the above is true.
If you are given the elemental composition of an unknown substance in grams, see the section on "Using Weight in Grams." Example:
Determine the empirical formula of a compound made from
29.3% Na (sodium),
41.1% S (sulfur), and
29.6% O (oxygen). -
Step 2: Determine the number of grams for each element.
Based on the assumption that there are 100 grams of the unknown substance, you can determine that the number of grams present for each element equals the percentage value of each element mentioned in the problem.
Example:
For 100 g of unknown substance, there are
29.3 g Na,
41.1 g S, and
29.6 g O. , The mass of each element in your composition, presently expressed in grams, will need to be converted into moles.
To do so, each mass must be multiplied by the mole ratio per their respective atomic weights.
In simpler terms, you will need to divide each mass by the atomic weight of that element.
Also note that the atomic weights used in this calculation should include at least four significant figures.
Example:
For a compound with
29.3 g Na,
41.1 g S, and
29.6 g O:
29.3 g Na * (1 mol S /
22.99 g Na) =
1.274 mol Na
41.1 g S * (1 mol S /
32.06 g S) =
1.282 mol S
29.6 g O * (1 mol O /
16.00 g O) =
1.850 mol O , You will need a stochiometric comparison between the elements in your compound, which essentially means that you need to calculate how much of an element you have in relation to the other elements present in your compound.
To do this, divide each number of moles by the smallest number of moles present.
Example:
The smallest number of moles present in the compound is
1.274 moles (the number of moles for Na, sodium).
1.274 mol Na /
1.274 mol =
1.000 Na
1.282 mol S /
1.274 mol =
1.006 S
1.850 mol O /
1.274 mol =
1.452 O , The amount of moles present for each element may not equal whole numbers.
For small values that are within a tenth value away from a whole number, this does not present a problem.
Once you have an excess value exceeding this amount, however, you should multiply the ratio values as needed to bring that value up to a whole number.
If one element has a value near
0.5, multiply each element by
2.
Similarly, if one element has a value near
0.25, multiply each element by
4.
Example:
Since the amount of oxygen (O) present is close to
1.5, you will need to multiply each value by “2” to bring the ratio of oxygen closer to a whole number.
1.000 Na * 2 =
2.000 Na
1.006 S * 2 =
2.012 S
1.452 O * 2 =
2.904 O , Even after the last step, the amount of moles present for each element may not be in exact whole numbers.
Since no decimals are used in empirical formulas, you will need to round each value to its nearest whole number.
Example:
For the ratio determined in the previous step:
2.000 Na can be written as 2 Na.
2.012 S can be rounded down to 2 S.
2.904 O can be rounded up to 3 O. , Translate the ratio of elements into the standard format used for empirical formula.
The molecular amount of each element should be indicated in subscript beside its respective element's symbol for all amounts greater than one.
Example:
For a compound that is 2 parts Na, 2 parts S, and 3 parts O, the empirical formula should be written as:
Na2S2O3 -
Step 3: Convert the mass of each element to moles.
-
Step 4: Divide each mole value by the smallest number of moles present.
-
Step 5: Multiply the ratio values to find near whole numbers.
-
Step 6: Round the values to nearest whole numbers.
-
Step 7: Write your final answer.
Detailed Guide
If you are given the elemental composition of an unknown compound in percentages rather than grams, you should assume that there are exactly
100.0 grams of the substance involved.
These are the instructions you should follow if the above is true.
If you are given the elemental composition of an unknown substance in grams, see the section on "Using Weight in Grams." Example:
Determine the empirical formula of a compound made from
29.3% Na (sodium),
41.1% S (sulfur), and
29.6% O (oxygen).
Based on the assumption that there are 100 grams of the unknown substance, you can determine that the number of grams present for each element equals the percentage value of each element mentioned in the problem.
Example:
For 100 g of unknown substance, there are
29.3 g Na,
41.1 g S, and
29.6 g O. , The mass of each element in your composition, presently expressed in grams, will need to be converted into moles.
To do so, each mass must be multiplied by the mole ratio per their respective atomic weights.
In simpler terms, you will need to divide each mass by the atomic weight of that element.
Also note that the atomic weights used in this calculation should include at least four significant figures.
Example:
For a compound with
29.3 g Na,
41.1 g S, and
29.6 g O:
29.3 g Na * (1 mol S /
22.99 g Na) =
1.274 mol Na
41.1 g S * (1 mol S /
32.06 g S) =
1.282 mol S
29.6 g O * (1 mol O /
16.00 g O) =
1.850 mol O , You will need a stochiometric comparison between the elements in your compound, which essentially means that you need to calculate how much of an element you have in relation to the other elements present in your compound.
To do this, divide each number of moles by the smallest number of moles present.
Example:
The smallest number of moles present in the compound is
1.274 moles (the number of moles for Na, sodium).
1.274 mol Na /
1.274 mol =
1.000 Na
1.282 mol S /
1.274 mol =
1.006 S
1.850 mol O /
1.274 mol =
1.452 O , The amount of moles present for each element may not equal whole numbers.
For small values that are within a tenth value away from a whole number, this does not present a problem.
Once you have an excess value exceeding this amount, however, you should multiply the ratio values as needed to bring that value up to a whole number.
If one element has a value near
0.5, multiply each element by
2.
Similarly, if one element has a value near
0.25, multiply each element by
4.
Example:
Since the amount of oxygen (O) present is close to
1.5, you will need to multiply each value by “2” to bring the ratio of oxygen closer to a whole number.
1.000 Na * 2 =
2.000 Na
1.006 S * 2 =
2.012 S
1.452 O * 2 =
2.904 O , Even after the last step, the amount of moles present for each element may not be in exact whole numbers.
Since no decimals are used in empirical formulas, you will need to round each value to its nearest whole number.
Example:
For the ratio determined in the previous step:
2.000 Na can be written as 2 Na.
2.012 S can be rounded down to 2 S.
2.904 O can be rounded up to 3 O. , Translate the ratio of elements into the standard format used for empirical formula.
The molecular amount of each element should be indicated in subscript beside its respective element's symbol for all amounts greater than one.
Example:
For a compound that is 2 parts Na, 2 parts S, and 3 parts O, the empirical formula should be written as:
Na2S2O3
About the Author
Richard Alvarez
Enthusiastic about teaching practical skills techniques through clear, step-by-step guides.
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