Introduction
A look at the mass relationships in chemistry reveals little order or sense. The ratio of the masses of the elements in a compound, while constant, does not tell anything about the formula of a compound. For instance, while water always contains the same amount of hydrogen (11.11% by mass) and oxygen (88.89% by mass), these figures tell us nothing about how the formula H
2
O is obtained.
A chemical formula is a whole number ratio showing the relative numbers of the atoms present. It can be determined from the masses of each element present, if you take into account the differing masses of each kind of atom. The standard of chemical quantity is not mass but the number of moles (mol) of particles (ions, atoms, or molecules). The mass of one mole of any substance is numerically equal to its formula weight. Thus, one mole of carbon atoms has a mass equal to 12.011 grams. One mole of glucose (C
6
H
12
O
6
) has a mass equal to 180.156 grams. One mole of nitrate ions
(NO
3
-1 ) has a mass equal to 62.004 grams. You should verify these formula weights for yourself.
In many cases, it is a straightforward matter, experimentally, to find the mass of each element present in a sample of a compound. Dividing each mass by the mass of one mole (the atomic mass) of that element gives the number of moles of each element present. The resulting ratio, when simplified to small whole numbers, gives you the empirical formula of the compound, which is the smallest whole number atomic ratio.
The molecular formula may be a multiple of that and requires the molecular mass be determined. For instance, hydrogen peroxide has the empirical formula HO. Its molecular mass, however, is 34, which corresponds to a molecular formula of H
2
O
2
.
Example 1: A weighed sample of silver is converted to sliver chloride by a chemical reaction and the mass of silver chloride, Ag x
Cl y
, is determined. Typical results might be:
Mass Ag
Mass AgCl
By difference: Mass Cl:
1.007 g
1.338 g
0.331 g

The number of moles of each element is calculated by multiplying its mass by the conversion factor derived from the equality:
1 mol = atomic mass of the element expressed in grams
Ag: 1.007 𝑔 𝑥
1 𝑚𝑜𝑙 𝐴𝑔
107.870 𝑔 𝐴𝑔
= 0.009335 𝑚𝑜𝑙 𝐴𝑔
Cl: 0.331 𝑔 𝑥
1 𝑚𝑜𝑙 𝐶𝑙
35.453 𝑔 𝐶𝑙
= 0.00934 𝑚𝑜𝑙 𝐶𝑙
The numbers of moles of each element present in silver chloride are essentially the same, so the empirical formula must be AgCl (meaning Ag and Cl are present in a 1:1 ratio. Since this is an ionic compound, there are no molecules present and the question of molecular formula does not arise.
Since percent composition is the preferred method of reporting analytical results, you should know how to calculate it from analytical data. The fraction of each element present is the mass of that element in the sample, divided b yt he total mass of the sample.
% Ag = 1.007 𝑔 𝐴𝑔
1.338 𝑔 𝐴𝑔𝐶𝑙
𝑥 100 = 75.26%
% Cl = 0.331 𝑔 𝐶𝑙
1.338 𝑔 𝐴𝑔𝐶𝑙 𝑥 100 = 24.7%
Example 2: Some cases you might be supplied with a percent composition of a compound from an analytical laboratory. This requires an extra computational set to find the empirical formula. For instance, an organic compound is reported as consisting of 53.38% carbon, 11.18% hydrogen, and 35.53% oxygen.
Assume you have 100.00 grams of the compound. First find the moles of each element in the 100 grams.
Carbon: 53.38 𝑔 𝑥
1 𝑚𝑜𝑙 𝐶
12.011 𝑔 𝐶
= 4.444 𝑚𝑜𝑙 𝐶
Hydrogen: 11.18 𝑔 𝑥
1 𝑚𝑜𝑙 𝐻
1.008 𝑔 𝐻
= 11.09 𝑚𝑜𝑙 𝐻
Oxygen: 1 𝑚𝑜𝑙 𝑂
35.53 𝑔 𝑥
16.00 𝑔 𝑂
= 2.221 𝑚𝑜𝑙 𝑂

These can be converted to whole number ratios by dividing each by the smallest number present (2.221)
Carbon:
Hydrogen:
4.444
2.221
= 2.001 𝑚𝑜𝑙 𝐶 = 2 𝑚𝑜𝑙 𝐶
11.09
2.221
= 4.993 𝑚𝑜𝑙 𝐻 = 5 𝑚𝑜𝑙 𝐻
Oxygen: 2.221
2.221
= 1.000 𝑚𝑜𝑙 𝑂 = 1 𝑚𝑜𝑙 𝑂
Thus the empirical formula of the compound is C
2
H
5
O. This compound is found to have a molecular weight of 45.062 and its molecular formula is found to be the same as the empirical formula.
Pre-Lab
1.
A compound has a molecular formula Na
2
S
2
O
4
. What is its empirical formula?
2.
What is the mass of 0.986 moles of Na
2
S
2
O
4
?
3.
How many moles are there of 95.7 grams of Na
2
S
2
O
4
?
4.
What is the empirical formula of a compound that consists of 54.10% Ca, 2.70%
H and 43.20% O? Show your work.
5.
What is the percent composition of Na
2
S
2
O
4
? Show your work.
Lab
1.
Heat a clean, dry crucible with cover on a clay triangle, using a direct flame, for about 5 minutes. This is to make certain that the crucible is dry.
2.
Turn off the burner and let the crucible and cover cool. Leave the crucible and cover resting on the clay triangle.
3.
Weigh the empty, dry crucible and cover on the digital balance.
4.
Obtain a strip of magnesium ribbon that is about 30 to 40 cm long.
5.
Fold or coil the magnesium ribbon so that it will fit inside the crucible.
6.
Carefully weigh the crucible, cover, and magnesium ribbon together.
7.
Set up the crucible on the clay triangle with the cover very slightly ajar. With a very small flame, begin heating the crucible gently.
8.
If the crucible begins to smoke when heated, immediately cover the magnesium completely and remove the heat for 2-3 minutes. The smoke consists of the magnesium oxide product and must not be lost from the crucible.

9.
Heat the sample strongly for 5 minutes, with the cover on the crucible. Make sure the bottom of the crucible becomes red hot.
10.
Check to make sure all the shiny magnesium has been fully converted to the dull gray oxide. If not continue to heat for 5 minutes of full heat with the cover about half-open.
11.
Turn the burner off the let the crucible cool until it is only barely warm to the touch.
12.
To the cooled crucible, add about 10 drops of distilled water. Make sure to wet the entire surface of the sample, not just one spot.
13.
Warm the crucible with the damp sample using a gentle flame for a minute or so, then heat it moderately strongly for about 10 minutes. (The crucible need not become red hot for this phase of the experiment, and the cover is not needed for this heating unless it starts spattering.)
14.
Let the crucible and contents cool until the crucible is barely warm to the touch.
15.
Weigh the cooled crucible, cover, and contents of the crucible, recording this total mass.
Post-Lab
1.
In this experiment you determined the mass of oxygen (that combined with a weighted mass of magnesium) by subtraction: mass of product minus mass of original magnesium = mass of oxygen combined. As a result of this procedure, anything that was in the crucible at the end of the experiment, along with the magnesium oxide product, would cause an error in the figure that is recorded as
“mass of oxygen.” Would extra mass in the crucible cause the “mass of oxygen” to come out too high or low? Explain.
2.
The correct formula for magnesium oxide is MgO, a 1.0 to 1.0 ratio. But sometimes in this experiment the ratio of Mg to O comes out too low. (Example:
0.9 Mg to 1.0 O) In that case, it means that there was too much oxygen relative to the mass of magnesium. At other times it comes out that the ratio is too large.
An example would be: 1.2 to 1.0 (Mg to O). In such a case it must be that there has been too little oxygen (or too little weight at the end of the experiment, which registers as too little oxygen.) In each case below, decide whether the situation described would lead to a calculated ratio of too much oxygen, or too little oxygen, and explain your reasoning.
a.
Putting in more water than is needed and then not drying out this excess water.

b.
Having some Mg(OH)
2
left in the crucible instead of converting it to MgO.
c.
Having some Mg
3
N
2
in the crucible as product, instead of pure MgO.
d.
Forgetting to weight the cover along with the crucible and contents at the end.
e.
Letting a lot of the dense white smoke escape from the crucible during the burning.
3.
Here is some data from a similar experiment, to determine the empirical formula of an oxide of tin. Calculate the empirical formula according to this data.
a.
Mass of crucible, cover, and tin sample b.
Mass of empty crucible with cover c.
Mass of crucible, cover, and sample after heating
21.76 g
19.66 g
22.29 g