Question

# An empirical formula is the lowest whole number ratio of the elements in a compound and...

An empirical formula is the lowest whole number ratio of the elements in a compound and provides information about the composition of a compound. For instance, the compound sodium sulfate decahydrate has the chemical formula Na2SO4•10H2O. This formula conveys the information that there are ten water molecules per two sodium ions per one sulfate ion. You might wonder: “How were these ratios determined?” To answer the question it is necessary to recognize that the ratio of atoms (or ions) is identical to the ratio of moles. That is, for sodium sulfate decahydrate, there are ten moles of water per two moles of sodium ions per one mole of sulfate. This conception of the molar ratio is essential since there is no way to count atoms in the lab but mass can be measured and mass can be converted to moles by the formula: Eq. 1 The goal of this lab is to determine x, y, and z in the chemical formula of the hydrate: CuxCly • zH2O Compounds with a “•zH2O” in the formula are called “hydrates” and have water molecules incorporated into the lattice structure of the ionic compound. When heated, hydrates decompose to produce water and an anhydrous salt. Anhydrous salt have no water molecules in their lattice structure. Removal of water molecules from the salt creates a change in the lattice structure, which is often accompanied by a change in color. For example, cobalt(II) chloride hexahydrate is pink. Heating the hydrate salt removes the six waters causing a color change from pink to a purple-blue. Figure 1: Hydrate to anhydrous color transition example. Chemistry: Structure and Properties by Tro, 1st ed, page 160. Under the conditions of the reaction, the evolved water will vaporize. CuxCly •zH2O → CuxCly + z H2O hydrate anhydrous salt evolved water 34 Thus, the difference in the mass before and after the reaction will be the mass of the water in the compound and the mass of water can be converted to moles of water by applying Eq. 1. The next step in the analysis is to determine the mass of copper in the sample. This is done by reducing the copper ion to elemental copper using aluminum as the reducing agent: 3 Cuy+ + y Al → y Al3+ + 3 Cu The mass of the copper metal produced is the same as the mass of the copper in the original sample. Since the sample contains only copper, water, and chloride, subtracting the mass of the water and copper from the mass of the sample or subtracting the mass of the copper from the mass of the anhydrous salt will yield the mass of the chloride.

Discussion Questions:

1. A student did not notice that after ten minutes some of the contents of the crucible remained blue. Would this cause your “z” value to be erroneously high, erroneously low, or would this error not affect “z”? Explain.

2. Would the procedural error described in Discussion Question #1 cause your “y” value to be erroneously high, erroneously low, or would this error not affect “y”? Explain.

3. A 5.123g sample of the hydrate MxCly•zH2O decomposed into 1.122g My+, 1.984g chloride, and 2.017g water. If the molar mass of “M” is 40.078 g/mol, what is the empirical formula of the compound? Show your calculations. What is the most likely identity of metal “M”? Explain your answer.

Discussion

1. If the student found some content still blue in the crucible, this error would be erronously low as the calculation is done on mole basis and the amount of salt evaporated to water would be higher in this case.

2. The value of "y" would be erroneously high, as the amount of mass lost also includes mass of leftover water in the crucible.

3.

M% = 1.122 x 100/5.123 = 21.90%

Cl% = 1.984 x 100/5.123 = 38.73%

H2O% = 2.017 x 100/5.123 = 39.37%

empirical formula becomes = M2Cl.H2O

M could be Li.