6: Determination of Empirical Formula
- Page ID
- 514168
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INTRODUCTION
Chemical formulas are a fundamental way to represent the composition of chemical compounds. The empirical formula is the simplest whole-number ratio of atoms in a compound. It is essential to distinguish this from the molecular formula, which specifies the number of atoms of each element in a molecule.
Many ionic compounds incorporate water molecules into their crystal structure, forming hydrates. Hydrates are crystalline compounds that incorporate water molecules into their structure in a specific ratio relative to the anhydrous salt. This water of hydration is an integral part of the crystal structure and can be removed by heating the hydrate. Without the water molecules, the resulting compound is known as an anhydrous compound.
In this experiment, you will investigate copper chloride hydrate, which is known for its green-blue color. The water molecules are removed by carefully heating the hydrate, which results in a color change to brown as the anhydrous copper chloride is formed. The mass difference between the original hydrate and the anhydrous compound allows us to determine the Mass of water lost.
We will use a single-displacement reaction with aluminum as the reducing agent to determine the ratio of copper to chloride in the anhydrous compound. The single displacement reaction involves displacing one element in a compound by another element. In our experiment, aluminum metal will reduce copper ions in solution, forming copper metal and aluminum ions. The copper metal is carefully separated, dried, and weighed. By determining the Mass of water lost and the Mass of copper and chloride in the original compound, we can calculate the moles of each component. By selecting the number of moles of water, copper, and chloride, we can establish this ratio and thus determine the empirical formula of the copper chloride hydrate. The empirical formula represents the simplest whole-number ratio of elements in a compound. From these values, the empirical formula of the copper chloride hydrate can be determined by finding the simplest whole-number ratio of the moles of each element and water, as shown in the example below.
A student heated a sample of a nickel(II) sulfate hydrate and collected the following data:
- Mass of empty crucible and lid: 25.668 g
- Mass of crucible, lid, and nickel sulfate hydrate: 28.225 g
- Mass of crucible, lid, and anhydrous nickel(II) sulfate: 27.389 g
Determine the empirical formula of the nickel(II) sulfate hydrate.
- Answer
-
Calculate the Mass of the hydrated salt:
o Mass of hydrated salt = Mass of crucible, lid, and hydrate - Mass of empty crucible and lid
o Mass of hydrated salt = 28.225 g - 25.668 g = 2.557 g
Calculate the Mass of the anhydrous salt:- Mass of anhydrous salt = Mass of crucible, lid, and anhydrous salt - Mass of empty crucible and lid
- Mass of anhydrous salt = 27.389 g - 25.668 g = 1.721 g
- Mass of water lost = Mass of hydrated salt - Mass of anhydrous salt
- Mass of water lost = 2.557 g - 1.721 g = 0.836 g
- Calculate the moles of anhydrous nickel(II) sulfate \(\ce{NiSO4}\):
- First, find the molar Mass of \(\ce{NiSO4}\):
- Ni: 58.69 g/mol
- S: 32.07 g/mol
- O: 16.00 g/mol x 4 = 64.00 g/mol
- Molar mass \(\ce{NiSO4}\) = 58.69 g/mol + 32.07 g/mol + 64.00 g/mol = 154.76 g/mol
- Moles of \(\ce{NiSO4}\) = Mass of anhydrous salt / Molar mass of \(\ce{NiSO4}\)
- Moles of \(\ce{NiSO4}\) = 1.721 g / 154.76 g/mol = 0.01112 mol
- [In your experiment, you will find the Mass and moles of copper and chlorine from experimental data obtained by a displacement reaction]
- First, find the molar Mass of \(\ce{NiSO4}\):
- Calculate the moles of water \(\ce{H2O}\):
- Molar mass of \(\ce{H2O}\) = (1.008 g/mol x 2) + 16.00 g/mol = 18.016 g/mol
- Moles of \(\ce{H2O}\) = Mass of water lost / Molar Mass of \(\ce{H2O}\)
- Moles of \(\ce{H2O}\) = 0.836 g / 18.016 g/mol = 0.04640 mol
- Determine the mole ratio of anhydrous salt to water:
- Divide both mole values by the smallest number of moles (0.01112 mol in this case):
- \(\ce{NiSO4}\): 0.01112 mol / 0.01112 mol = 1
- \(\ce{H2O}\): 0.04640 mol / 0.01112 mol = 4.17 ≈ 4
- Divide both mole values by the smallest number of moles (0.01112 mol in this case):
- Write the empirical formula of the hydrate:
- The ratio is approximately 1:4, so the empirical formula is \(\ce{NiSO4.4H2O}\)
- 6.1: Determination of Empirical Formula - Experiment
- This page covers vital safety precautions for experiments involving hydrochloric acid, highlighting the importance of wearing safety goggles and avoiding burns and open flames. It lists necessary equipment and chemicals and details a two-part experimental procedure focused on determining water mass through dehydration and copper mass via displacement. Additionally, it stresses proper chemical and waste disposal methods.
- 6.2: Determination of Empirical Formula - Pre-lab
- This page covers key terms in chemistry, including hydrates, anhydrous compounds, and empirical formulas. It provides examples of calculating empirical formulas using experimental data, specifically with tungsten oxides and compounds with given percentage compositions. The page also emphasizes the importance of proper heating techniques for copper chloride hydrates to prevent overheating, which can affect the accuracy of results.
- 6.3: Determination of Empirical Formula - Data and Report
- This page describes a laboratory exercise aimed at determining the mass of water in copper chloride hydrate through dehydration, calculating the mass of copper via a displacement reaction, and finding the mass of chlorine. It features structured data tables and calculations for moles of each component, ultimately leading to the determination of the hydrate's empirical formula.