Determination of chemical formulas

INTRODUCTION

This presentation is to show the student how to calculate the mass composition of a compound and the mole composition of a compound.

The student show understand that mass measurements give the gram amount of a substance and that moles give the number amount of a substance.

DETERMINATION OF CHEMICAL FORMULAS
The student at this point has the tools to understand how chemical formulas are experimentally determined.

A chemical formula tells the number of atoms of each element in the compound.

A chemical formula could be interpreted to be the number of moles of each element in the compound.

The composition of a compound can be given as a fraction or mass percent of the elements or as a mole ratio of elements.

An example will show how to calculate both the ratio and percent mass composition of the elements in a compound and the mole composition (ratio) of the elements in a compound.  View the video at Magnesium reacting with oxygen.

In this video, the element Mg reacted with the element O2 from the air.   Mg is a LIMITING REAGENT; O2 is an EXCESS REAGENT. This means the amount of Mg determined the amount of white product. There is O2 remaining in the air when the reaction is completed. Hence O2 is the excess agent.

Information for the reaction in the video. The  mass of the Mg oxidized  was 2.75 g.   Mass of white compound formed was 4.56 g.

What accounted for the increase in mass? Oxygen

How much O2 reacted?  To determine the grams of oxygen in the white compound, one subtracts the mass of Mg from the mass of the white compound formed?

             4.56 g compound – 2.75 g Mg = 1.81 g O

The next step is to calculate the mass ratio and % composition of Mg and O in the compound.

Mass ratio of Mg:

Mass ratio Mg

To express this as a %, multiply by 100 to get 60.3 %

Mass ratio of O

Mass ratio O

To express this as a %, multiply by 100 to get 39.3 %
Moles of Mg in the white compound?

mol Mg

How many moles of O are in the white compound?

mol O

The ratio of moles of Mg to moles of O is 1 to 1.  Therefore the formula for the white compound is MgO.

Additional examples. In the following examples, data for two compounds is given. Compound #1 is a gas and compound #2 is a liquid. Use the steps presented in the example just completed, to determine the mole ratio (chemical formula) for both compounds.

Data for compound #1.   1.500 g of compound #1 had 1.384 g of C and 0.115 g of H.

Calculate the mole composition of Compound #1:
Moles of C:

mol C#1

Moles of H:

 

mol H#1

The mole ratio between C and H is 1 to 1.  The formula for this compound is CH

Data for compound #2.     0.750 g of compound #2 contained 0.692 g of C and 0.058 g of H.

Calculate the mole composition of Compound #2 is:
Moles of C:

mol C #2

Moles of H:

mol H#2

The mole ratio between C and H is 1 to 1 and the formula for this compound is CH

How can these two compounds have the same formula? They look different (one is a gas and the other is liquid) and if we were to compare their chemical properties, they would react different.

When chemical formulas are calculated from experimental data (they always are calculated from experimental data), the mole ratio calculated is always the smallest mole ratio of the elements in the compound.

Chemists refer to this type calculation as the EMPIRICAL FORMULA calculation. The empirical formula for a compound has an empirical molecular mass.

The actual (real) chemical formula cannot be determined until the gram molecular mass of the compound is known. This value must also be determined by an experiment.

In the example calculation for the two compounds, Compound #1 has a gram molecular mass of 26 g/mole. Compound #2 has a gram molecular mass of 78 g/mole.

The empirical formula mass for Compound #1 is 13 g/mole. It’s molar mass is 26 g/mole. Therefore, the chemical formula mass is 2 times the empirical formula mass This means the formula for this compound is C2H2.

The empirical molecular mass for Compound #2 is 13 g/mole. It’s molar mass is 78 g/mole. Therefore, the molecular formula mass is 6 times the empirical formula mass. This means the formula for this compound is C6H6.

ADDITIONAL EXAMPLES

Try to work these. If you cannot, the answers are given at the bottom of this page

Example 1
A blue solid was analyzed and determined to have O, S, and Cu. A 3.47 g sample contained 1.39 g of O, 0.70 g of S, and 1.39 g of Cu. What is the mass % composition? What is the empirical formula? If the molar mass of the blue solid if 159.60, what is it’s real formula?

Example 2
A clear liquid with an odor resembling vinegar was analyzed and found to contain 40.0 % C, 6.7 % H, and 53.3 % O. What is the empirical formula of this compound? The molar mass of the compound is 60 g/mole. What is the real formula of the compound.

 

Answer to A: 39.81 % Cu, 20.09 % S, 40.10 % O; CuSO4
Answer to B: CH2O; C2H4O2