# Empirical Formula of Zinc Iodide

Topics: Oxygen, Oxide, Chemistry Pages: 8 (1849 words) Published: May 7, 2013
Experiment 9

Empirical Formula of Zinc Iodide

Objectives
Upon completion of this experiment, students should have learned:

1. The law of conservation of mass.
2. How to calculate an empirical formula.
3. The concept of limiting reagents.

Introduction
Synthesis and the determination of empirical formulas are two extremely important parts of chemistry. In this experiment, you will synthesize zinc iodide and determine its empirical formula. The molecular formula gives the actual number of atoms of each element in the molecule and the empirical formula gives the lowest whole number ratio of the atoms in the molecule. For instance, the molecular formula of glucose is C6H12O6 and its empirical formula is CH2O. For some molecules such as water, the empirical and molecular formulas are the same. One of the common techniques used to help determine the identity of a substance involves combustion of very small amounts of the substance. By measuring the amounts of water, carbon dioxide, and other gases produced, it is possible to determine the percentage by mass of each element in the compound. From the percentages, it is possible to calculate the empirical formula of the compound. Suppose, for example, that the combustion experiment for a compound that contains carbon, hydrogen, and oxygen gives the compositions as 40.00% carbon, 6.71% hydrogen, and 53.29% oxygen by mass. The goal is to convert mass percentages to mole or atom ratios. Whenever dealing with percentages in a calculation of this type, one of the easiest ways to proceed is to assume that you have 100 g of the substance. Thus 100 g of the substance contains 40.00 g of carbon, 6.71 g of hydrogen and 53.29 g of oxygen. As is common for many chemical calculations, the calculation of the number of moles is one of the first steps. Thinking ahead, once the number of moles is known, the ratio of moles is easily calculated.

40.00 g C x 1 mol C
------------- = 3.33 moles C 12.011 g C

6.71 g H x 1 mol H
------------- = 6.66 moles H 1.008 g H

53.29 g O x 1mol O
------------- = 3.33 moles O 15.999 g O

Dividing by the lowest number of moles gives a C:H:O ratio of 1:2:1 so the empirical formula is CH2O. This means the compound could be glucose but it could also be many other compounds including other sugars or formaldehyde. The determination of the molecular formula from the empirical formula requires an additional measurement of the molecular mass. If a molecular mass measurement for the above compound results in 150 ± 2 g/mole, the molecular formula would be calculated as C5H10O5. This means the compound could be a five carbon sugar such as ribose.

Molecular formula = molecular mass
-------------------------- x subscripts of empirical formula = empirical form. mass

150 x CH2O = C5H10O5. ---
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The empirical and molecular formulas can go a long way in helping to identify the compound.

Almost all ionic compounds are solids at room temperature. The crystal structure of the ionic solid consists of a repeating array, and it is not straightforward to define the smallest unit (a molecule) of the substance that still would have the properties of the substance. Thus it is better to refer to the formula mass of an ionic substance rather than molecular mass. For the same reasons, the formula calculations for ionic compounds should be reduced to the lowest whole number ratio or the empirical formula. Except for a few unusual cases, the term molecular formula is not applicable for ionic compounds.

For metals...