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UNIT 5 The Empirical Formula of an Oxide

 

 

4.0
Introduction

5.0
Objective

6.0
Main Content

  1. Theory  
  2. Concept of the experiment
  3. Procedure

 

4.0  Conclusion

5.0  Summary

6.0  Tutor Marked Assignments

7.0  References/Further Reading

 

1.0
Introduction

Chemical formulas indicate the composition of compounds. A formula that gives only the simplest ratio of the relative number of atoms in a compound is the empirical formula or simplest formula. The ratio usually consists of small whole numbers. A formula that gives the actual numbers of each type of atom in a compound is called the molecular formula. The numbers in a molecular formula will be whole number multiples of the numbers in an empirical formula. To determine the molecular formula of a compound, we need to know both the empirical formula and the molar mass of the compound.

 

2.0 Objectives

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At the end of this unit, you will be able to:

  • Describe the concept of empirical formula.
  • Determine the empirical formula of a metal oxide.
  • Determine the mass and mole ratios of elements.
  • Perform basic a laboratory measurements of mass and volume using appropriate glassware and equipment.

 

  1. Main Content
  2. Theory

A chemical formula is the simplest way to express information about the atoms that constitute any given chemical compound. A chemical compound is composed of atoms of two or more elements chemically combined in definite proportions. For example, H2O and CO are compounds whereas H2
is a diatomic element.

The atoms in a compound are held together by chemical bonds. The chemical formula helps identify each constituent element by its chemical symbol and indicates the number of atoms of each element in the compound. If a molecule contains more than one atom of a particular element, this number is indicated using a subscript after the chemical symbol. For example Copper (I) sulfide has a chemical formula of Cu2S. In this compound there are two atoms of copper and one atom of sulfur. Copper and sulfur are combined in a ratio of 2:1. This formula also means that there are 2 moles of copper for every mole of sulfur.

The total mass of each element in a compound depends on the number of its atoms. For example, in compounds, CO and CO2, the number of oxygen atoms are respectively, one and two. In other words, the mass of oxygen in CO2
is twice as much as in CO. Yet another way to describe this is that one mole of carbon combines with one mole of oxygen in CO (mole ratio: 1:1) and one mole of carbon combines with two moles of oxygen in CO2
(mole ratio: 1:2). This suggests that elements always combine in definite proportions by mass or moles. Accordingly, ratio of moles of the constituent elements in a compound is nearly always a ratio of small, whole numbers. The formula containing the lowest possible whole number ratio is known as the empirical formula. The empirical formula does not necessarily indicate the exact number of atoms in a single molecule. This information is given by the molecular formula, which is always a simple multiple (n) of the empirical formula, i.e., Molecular formula (MF) = Empirical formula (EF) x n. The value of “n” can be derived by dividing the actual molar mass of the compound and empirical formula mass. Molar masses of the compounds are obtained from various experimental procedures including mass spectrometry technique.

To understand the behavior of a chemical compound one must first know its chemical formula. The process of the discovery of chemical formula of a compound begins with the determination of the ratio of the individual elements in a known mass of the compound (determination of empirical formula). For most ionic compounds, empirical formula is the same as the compound’s molecular formula. However, for covalent compounds the empirical formula is not necessarily the same as the molecular formula of the compound. For example, the empirical formula of water, H2O, is the same as its molecular formula. The empirical formula of hydrogen peroxide is HO, but its molecular formula is H2O2. To determine whether or not a covalent compound’s empirical and molecular formulas are the same, we need to know the molar mass of the compound.

To find the empirical formula we must combine the elements to form the compound under conditions that allow us to determine the mass of each element. From these data, the moles of each element may be determined. By dividing the moles obtained for each element by the smallest number of moles, we obtain quotients that are in a simple ratio of integers or they are easily changed to a simple ratio by multiplication process.

Sample calculations: Determination of an empirical formula of a compound based on experimental data Image From EcoleBooks.comis shown below.

Example: A strip of aluminum weighing 0.690 g is ignited yielding an oxide that weighs 1.300 g.

Calculate the empirical formula of the compound formed.

We will follow the following strategy:

  1. Determine the mass of each element.
  2. Convert masses of elements to their respective moles.
  3. Divide each of the moles by the smallest mole number.
  4. Multiply the mole ratios with whole number integers until a whole number is obtained for each element.

 

  1. We must first determine the mass of aluminum and oxygen in the oxide formed. Given we started with

0.690 g of aluminum, it yielded 1.300 g of an oxide.

Image From EcoleBooks.com

i.e. Image From EcoleBooks.com

Image From EcoleBooks.com

  1. Now that we know the mass of aluminum and oxygen, we can calculate moles of each element.

 

 

Image From EcoleBooks.com

Image From EcoleBooks.com

  1. Since we now know the moles of each element, ratio of atoms may be determined by dividing the moles of each element by the smallest number of moles.

Image From EcoleBooks.com
Image From EcoleBooks.com

 

The ratio is 1.00 atom Al: 1.49 atoms O

This ratio can be changed to a whole number ratio by multiplying by a factor of 2. i.e. (1.00 atom : 1.49 atoms O) x 2

2 atom : 2.98 atoms O;

Simple rounding gives 2 atom Image From EcoleBooks.com: 3 atoms O

Therefore the empirical formula of Aluminum oxide formed Image From EcoleBooks.com.

 

3.2 Concept of the experiment

 

In this experiment, after a known mass of magnesium is burned, the product will consist of magnesium oxide together with a small amount of magnesium nitride (Mg3N2). Next, we add water to convert this small amount of nitride to magnesium hydroxide, Mg(OH)2
with the liberation of ammonia, NH3
(the water will not react with the magnesium oxide). Further heating will cause conversion of the hydroxide to the oxide with the loss of gaseous water. In summary, the following reactions take place:

Reaction of magnesium with molecular oxygen:

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Reaction of magnesium with molecular nitrogen

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Reaction of magnesium nitride with water

Image From EcoleBooks.com

Reaction upon heating magnesium hydroxide:

Image From EcoleBooks.com

In the end, all magnesium will have been converted to magnesium oxide. You can then calculate the mass of oxygen that is present in the oxide from its mass and the original mass of the magnesium. The laws of conservation of mass, as well as the concept of a mole, will lead you to the method by which you can determine the empirical formula of this oxide.

 

Materials and Equipment

Balance, furnace, crucible, crucible tongs, wire gauze, magnesium ribbon, sandpaper, medicine dropper, tweezers

 

 

 

 

 

3.3 Procedure

  1. Obtain a crucible, wash, rinse, and dry it.
  2. Heat the crucible for about 5 min in a furnace at 200°C. Remove the crucible from the furnace and allow the crucible to cool (5-10 min).

 

CAUTION: Avoid burning your fingers. Do not touch the crucible at any time during this experiment.

  1. While waiting, weigh approximately 0.2 g of magnesium ribbon (if it is not bright, clean the surface with sandpaper). Record the mass.
  2. When the crucible is cool, transfer it to the pan of a balance using crucible tongs while holding a wire gauze under the crucible, but do not put the wire gauze on the pan. If you must wait to use the balance, do not place the crucible directly on the bench.
  3. Measure and record the mass of the crucible.
  4. Repeat steps 2, 4 and 5 until two consecutive masses differ by no more than +0.01 g or any other precision that is stipulated by your laboratory instructor. Record the mean or average value of these two masses. You will use this result in subsequent calculations.
  5. Fold the magnesium ribbon into a loose ball that will fit completely inside the crucible.
  6. Weigh and record the mass of the crucible with the magnesium.
  7. Return the crucible to the furnace, using crucible tong.
  8. Heat the crucible with its contents in the furnace at 800 °C for 45 min.

 

During this waiting period, you will write the first written exam!

  1. Take out crucible and put it on a wire gauze and let it cool to room temperature. The contents should be white or slightly gray.
  2. Add a few drops of distilled water from a medicine dropper directly on the contents. The smell of ammonia may be evident at this point.
  3. Heat the crucible until the contents are dry. Next heat the crucible strongly for 8-10 min in a furnace to convert the hydroxide to the oxide.
  4. Allow the covered crucible and its contents to cool.
  5. Weigh and record the new mass of the crucible withImage From EcoleBooks.com.
  6. Calculate the empirical formula from your recorded masses.
  7. Clean the crucible and all other items used.

 

 

 

 

 

 

Laboratory Data Sheet
Results

Make sure you report your answer using the correct number of significant figures for each mass! a. Image From EcoleBooks.com

Image From EcoleBooks.com

  1. Average mass of empty crucible (g) ___________
  2. Mass of crucible with Mg before heating (g) ___________
  3. Mass of Mg (g) (c – b) ___________
  4. Mass of crucible and oxide after heating(g) ___________
  5. Mass of magnesium oxide (g) (e – b) ___________
  6. Mass of oxygen (O) = mass magnesium oxide – mass Mg (f – d) ___________

 

 

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Write balanced chemical equations along with their physical states for the following reactions: a. Magnesium with molecular oxygen: _______________________________________

 

  1. Magnesium with molecular nitrogen: _______________________________________

 

  1. Magnesium nitride with water: _____________________________________

 

  1. Heating magnesium hydroxide: _____________________________________

 

4.0  Conclusion

The empirical formula is the simplest formula for a compound. A molecular formula is the same as or a multiple of the empirical formula, and is based on the actual number of atoms of each type in the compound.

5.0
Summary
In this you have been able to Describe the concept of empirical formula, determine the empirical formula of a metal oxide, determine the mass and mole ratios of elements and finally perform a basic laboratory measurements of mass and volume using appropriate glassware and equipment. 6.0
Tutor Marked Assignment (TMA)

 

  1. Image From EcoleBooks.comThe empirical formula of a compound was determined to be “CH” and its molar mass was found to be 78.12 amu, what is the value of the simplest multiplier (n) of empirical formula to be used to determine its molecular formula? HintImage From EcoleBooks.com

 

Image From EcoleBooks.com

 

 

  1. What is the empirical formula of a compound containing 47.37% carbon, 10.59% hydrogen and 42.04% oxygen?

     

  2. What is the empirical formula for a compound containing 26.57% potassium, 35.36% chromium, and 38.07% oxygen?
  3. A sample of an oxide of nitrogen is found to contain 30.4% nitrogen. What is its empirical formula?

     




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EcoleBooks | UNIT 5  The Empirical Formula of an Oxide - INORGANIC CHEMISTRY PRACTICALS

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