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).

Example: For 100 g of unknown substance, there are 29. 3 g Na, 41. 1 g S, and 29. 6 g O.

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

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

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

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

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

1. 000 Na * 2 = 2. 000 Na 1. 006 S * 2 = 2. 012 S 1. 452 O * 2 = 2. 904 O

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.

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.

Example: For a compound that is 2 parts Na, 2 parts S, and 3 parts O, the empirical formula should be written as: Na2S2O3

On the other hand, if you are given the composition in percentages instead of grams, see the instructions on “Using Weight Percentages. " Example: Determine the empirical formula of an unknown substance made from 8. 5 g Fe (iron) and 3. 8 g O (oxygen. )

From a more technical perspective, you are actually multiplying the mass in grams by the mole ratio per atomic weight. Note that the atomic weight should be rounded to four significant places to maintain a certain degree of accuracy in your calculations. Example: When there are 8. 5 g Fe and 3. 8 g O: 8. 5 g Fe * (1 mol Fe / 55. 85 g Fe) = 0. 152 mol Fe 3. 8 g O * (1 mol O / 16. 00 g O) = 0. 238 mol O

8. 5 g Fe * (1 mol Fe / 55. 85 g Fe) = 0. 152 mol Fe 3. 8 g O * (1 mol O / 16. 00 g O) = 0. 238 mol O

Example: For this problem, the smallest amount of moles present is 0. 152 moles (the amount of Fe, iron, present). 0. 152 mol Fe / 0. 152 mol = 1. 000 Fe 0. 238 mol O / 0. 152 mol = 1. 566 O

0. 152 mol Fe / 0. 152 mol = 1. 000 Fe 0. 238 mol O / 0. 152 mol = 1. 566 O

For instance, if one element has an excess near 0. 25, multiply each element amount by 4. If an element has an excess near 0. 5, multiply each element amount by 2. Example: Since the ratio amount of oxygen equals 1. 566, you will need to multiply both ratio amounts by 2. 1. 000 Fe * 2 = 2. 000 Fe 1. 566 O * 2 = 3. 132 O

1. 000 Fe * 2 = 2. 000 Fe 1. 566 O * 2 = 3. 132 O

Example: The amount of Fe can be written as 2. The amount of O can be rounded down to 3.

Example: For a compound that is 2 parts Fe and 3 parts O, the empirical formula is: Fe2O3

Example: C8H16O8 On the other hand, if the subscripts do not all share a common factor, the molecular formula is also the empirical formula. Example: Fe3O2H7

Example: For C8H16O8, the subscripts are “16” and “8. " The factors of 8 are: 1, 2, 4, 8 The factors of 16 are: 1, 2, 4, 8, 16 The greatest common factor (GCF) between the two numbers is 8.

The factors of 8 are: 1, 2, 4, 8 The factors of 16 are: 1, 2, 4, 8, 16 The greatest common factor (GCF) between the two numbers is 8.

Example: For C8H16O8: Divide the subscript of 8 by the GCF of 8: 8 / 8 = 1 Divide the subscript of 16 by the GCF of 8: 16 / 8 = 2

Divide the subscript of 8 by the GCF of 8: 8 / 8 = 1 Divide the subscript of 16 by the GCF of 8: 16 / 8 = 2

Note that values of 1 are not usually indicated with subscripts. Example: C8H16O8 = CH2O