2.5A: Analytical Applications of Stoichiometry: Gravimetric Analysis

 

Figuring out what particular samples of matter are composed of is an age-old problem in chemistry. Questions like: How much cholesterol is in a sample of blood? How much vitamin C is in an orange? How much alcohol is in a sample of wine? The examples are endless. These types of questions are in the domain of analytical chemistry, the branch of chemistry that deals with quantifying the composition of complex mixtures.

We mentioned above that an important step in the investigation of materials is to simply measure their masses. Changes in mass upon various treatments can give important clues to its composition. To give a very simple example, suppose you want to know how much salt is in a sample of salt water. What do you do? The simplest approach would be to weigh the sample and then evaporate all of the water; the mass of the remaining salt can be compared to the initial mass and you’re done. This sort of analysis, called a gravimetric analysis, is based on mass measurements, usually several of them before and after particular chemical treatments of an original sample.

In the sections below, several examples of gravimetric analysis are described and what they all have in common is the simple idea that the mass changes undergone by a sample can be used to ascertain information about its composition, provided those mass changes are due to a chemical reaction with a known stoichiometry.

 

 

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Analysis of Hydrates

The salt-water example above involved a simple physical separation: there is no chemical reaction taking place. Most analyses involve chemical reactions of some type, in which the sample, or a component of the sample, undergoes a reaction that has a well-defined stoichiometry. By well-defined stoichiometry, we mean that we either know the balanced equation of the operative reaction, or the reaction generates a measurable amount of a pure material. Both instances allow us to use the concept of the mole to answer a practical question. 

One practical question concerns the composition of so-called hydrated compounds such a hydrous magnesium sulfate (MgSO₄•7H₂O). This white solid (which is sold in supermarkets and drugstores as Epsom salts) has water molecules incorporated into its crystalline structure. Despite the “H₂O” in the formula, it does not look or feel wet or moist because the water is, in effect, locked into the crystal lattice. The water molecules don’t flow so do not give the material the appearance or properties associated with liquid water. Another form of magnesium sulfate also exists which doesn’t have any water molecules in its structure. This is called anhydrous magnesium sulfate and it has the formula MgSO₄. 

There are many examples of hydrated compounds in nature. They usually form as crystals grown in aqueous environments, incorporating water molecules as they grow [1].  The amount of water is shown in the formula. In the case of MgSO₄•7H₂O, there are seven water molecules for each “unit” of magnesium sulfate [2]. 

 

 

Figure 2.9. Epsom salts have been used for a variety of medicinal applications for many generations. (Photo credit: "Pure Epsom Salts and Carters Senna Pods - The General Store - The Village Row - Gressenhall Farm & Workhouse, Dereham, Norfolk" by Glen Bowman is licensed under CC BY 2.0.)

 

 

 

How can we determine the number of water molecules in a given formula? The most direct way would heat a known mass of the hydrate with a Bunsen burner to drive off the water embedded in the crystal. It escapes as steam and, when the water is completely driven off, the mass of the remaining anhydrous compound is measured. We now have the two pieces of information we need:

• the number of moles of water in the original sample; this can be calculated from the decrease in mass decrease upon heating; and

• the number of moles of the anhydrous compound; this can be calculated from the mass of the sample after heating.

The above steps are illustrated in the box below using sample data. Note that in problems like this, you don’t want to prematurely round your answers; keep all of the digits your calculator gives you at each step and round only at the end.

 

Determining the Formula of a Hydrate: MgSO₄·x H₂O

 

We’ll approach this problem as if we didn’t know the composition of the hydrate MgSO₄•7H₂O. How do we experimentally determine that the formula should include 7 molecules of water per unit of MgSO₄?

The kernel of the problem is that we are really trying to find the ratio between the moles of two pure substances: the moles of H₂O and the moles of MgSO₄. The experimental steps (and some sample experimental data) are:

1. Measure the mass of a pure hydrate before heating: 5.325 g 

2. Heat the sample to completely drive off all the water molecules in the crystal lattice.

3. Measure the mass of sample after heating: 2.601 g

The second mass measurement corresponds to anhydrous MgSO₄. Assuming that the material is pure, we calculate the corresponding number of moles of MgSO₄ in the usual way:

\[ moles_{\ce{MgSO4}} = (2.601 \cancel{g\  \ce{MgSO4}}) (\dfrac{1\ mol\  \ce{MgSO4}}{120.361 \cancel{g\  \ce{MgSO4}}}) = 0.02161\ mol \  \ce{MgSO4} \nonumber \]

The difference in the two masses above is due to the loss of the water that was driven off. Water is also a pure material, so we can use the mass lost to find the moles of water.

\[ moles_{\ce{H2O}} = (5.325\ g - 2.601\ g) (\dfrac{1\  mol\ \ce{H2O}}{18.01\ g\ \ce{H2O}}) = 0.1512 mol\ \ce{H2O} \nonumber \]

Pulling it together, we simply need to find the ratio of moles of water to moles of MgSO₄:

\[ \dfrac{moles_{\ce{H2O}}}{moles_{\ce{MgSO4}}} = \dfrac{0.1512\ mol\  \ce{H2O}}{0.02161\  mol\   \ce{MgSO4}} = \dfrac{7.00\  mol\  \ce{H2O}}{1\  mol\   \ce{MgSO4}} \nonumber \]

This ratio is expressed in the formula as MgSO₄•7H₂O.

 

The general approach to finding the formula of hydrates is summarized in Equation 2.5.1.

 

 

 

 

Determining the Number of Water Molecules in the Formula of a Hydrate

The equation used to find the number of water molecules, x, in a hydrate with the general formula P_pQ_qR_r•xH₂O, is shown below.

\[ x=\dfrac{\text{moles of water}}{\text{moles of P}_{p}\text{Q}_{q}\text{R}_{r}} = \dfrac{[(\text{mass of }\text{P}_{p}\text{Q}_{q}\text{R}_{r}\cdot x \ce{H2O})-(\text{mass of }\text{P}_{p}\text{Q}_{q}\text{R}_{r})](\frac{1\ mol\ \ce{H2O}}{18.01 g\ \ce{H2O}})}{(\text{mass of }\text{P}_{p}\text{Q}_{q}\text{R}_{r})(\frac{1\ mol\ \text{P}_{p}\text{Q}_{q}\text{R}_{r}}{\text{molar mass}_{\text{P}_p\text{Q}_q\text{R}_r}})} \]

Here, P, Q, and R represent elements (or ions) in a molar ratio of p:q:r, respectively,

 

Exercise 

Problem 2.25) Cobalt (II) chloride exists in an anhydrous form and a hydrate. Interestingly, the hydrate, CoCl₂•xH₂O is a deep maroon color while the anhydrous form, CoCl₂, is a pale, powder blue, as shown in the following figure.

 

Figure 2.10: The hydrated (left) and anhydrous (right) forms of CoCl₂. The application of heat to the sample on the left causes the release of the water molecules of hydration that are part of the crystal lattice in the hydrate, resulting in the striking color change. (Image credits: both images by Wilco Oelen, licensed under a Creative Commons Attribution-ShareAlike 3.0 Unported License.)

 

 

 

To determine the formula for the hydrate, 6.932 g of CoCl₂•xH₂O is heated over a Bunsen burner until all the water is driven off as steam. The mass of the resulting CoCl₂ is 3.783 g. Calculate the value of x in the formula of the hydrate

Answer: 6

 

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Footnotes and References

[1]. Why is water sometimes incorporated into crystals? As we will see, water molecules have a high affinity for ions because of attractive electrostatic forces. A magnesium ion, which as a +2 charge, is usually surrounded by a "shell" of six water molecules which more or less act as a single unit as it condenses into a crystalline structure with the nagatively charged sulfate ion (SO₄^2-). 

[2] Why refer to a "unit" of MgSO₄ instead of a molecule. As we discuss in Chapter 4, many compounds between metallic and nonmetallic elements don’t exist as molecules, but pairs of oppositely charged ions. There are no “molecules” of magnesium sulfate in the compound but, rather, discrete ions, Mg²⁺ and SO₄^2-, in a 1:1 ratio.