Question

An experimental catalyst used to make polymers has the following composition: Co, \(23.3 \% ; \mathrm{Mo}, 25.3 \%\); and \(\mathrm{Cl}, 51.4 \%\). What is the empirical formula for this compound?

Short answer

The empirical formula is \(\text{Co}_3 \text{Mo}_{2} \text{Cl}_{11} \).

Step-by-step solution

– Determine Moles of Each Element

Consider 100 g of the compound for simplicity. Therefore, the sample contains 23.3 g of Co, 25.3 g of Mo, and 51.4 g of Cl. Convert these masses to moles using the atomic masses: Co (59 g/mol), Mo (96 g/mol), and Cl (35.5 g/mol). \[ \text{Moles of Co} = \frac{23.3 \text{ g}}{59 \text{ g/mol}} = 0.395 \text{ mol} \] \[ \text{Moles of Mo} = \frac{25.3 \text{ g}}{96 \text{ g/mol}} = 0.2646 \text{ mol} \] \[ \text{Moles of Cl} = \frac{51.4 \text{ g}}{35.5 \text{ g/mol}} = 1.448 \text{ mol} \]

– Simplify the Mole Ratios

Find the mole ratio by dividing each number of moles by the smallest number of moles among the elements. Here, Mo has the smallest number of moles (0.2646).\[ \text{Ratio of Co} = \frac{0.395}{0.2646} \thickapprox 1.49 \] \[ \text{Ratio of Mo} = \frac{0.2646}{0.2646} = 1 \] \[ \text{Ratio of Cl} = \frac{1.448}{0.2646} \thickapprox 5.47 \]

– Adjust Ratios to the Nearest Whole Number

For the ratios determined, round each to the nearest whole number: Co rounds to 1.5, Mo is 1, and Cl rounds to 5.5. To get whole numbers, multiply all ratios by 2 (since 1.5 and 5.5 are rounded to halves).\[ \text{Ratios: Co} \times 2 = 3 \] \[ \text{Ratios: Mo} \times 2 = 2 \] \[ \text{Ratios: Cl} \times 2 = 11 \]

– Obtain the Empirical Formula

Using the simplified mole ratios, the empirical formula of the catalyst is determined by placing these values as subscripts to the respective elements: The empirical formula is \(\text{Co}_3 \text{Mo}_{2} \text{Cl}_{11} \)

Key concepts

Mole Ratio

Understanding the mole ratio is vital in chemistry, especially when calculating empirical formulas. The mole ratio represents the proportional relationship between the amounts of various elements in a compound.
To find the mole ratio, follow these simple steps:
1. Convert the given mass of each element to moles using their atomic masses.
2. Identify the smallest number of moles among the elements.
3. Divide each element's number of moles by this smallest number.
For instance, in our example, we determined the moles of Co, Mo, and Cl as 0.395, 0.2646, and 1.448, respectively. By dividing each by 0.2646 (the smallest value), we obtained the ratios
Co: 0.395 / 0.2646 ≈ 1.49
Mo: 0.2646 / 0.2646 = 1
Cl: 1.448 / 0.2646 ≈ 5.47.
These ratios guide us in identifying the composition of the compound in the simplest whole-number terms.

Atomic Mass

Atomic mass, also known as atomic weight, is the mass of an atom expressed in atomic mass units (amu). It is essential for converting between grams and moles of an element.
Each element has a specific atomic mass, which you can find on the periodic table:
1. Cobalt (Co) has an atomic mass of approximately 59 g/mol.
2. Molybdenum (Mo) has an atomic mass of about 96 g/mol.
3. Chlorine (Cl) has an atomic mass around 35.5 g/mol.
In practical applications, you use these values to convert the mass of each element to moles. For example, to convert 23.3 grams of Co to moles: moles of Co = 23.3 g / 59 g/mol ≈ 0.395 mol.
This conversion is crucial for determining the mole ratio and, ultimately, the empirical formula.

Stoichiometry

Stoichiometry is the branch of chemistry that deals with the quantitative relationships between the elements and compounds in chemical reactions. It's all about balancing equations and calculating the proportions of reactants and products.
To solve stoichiometric problems, you usually need to:
1. Write a balanced chemical equation.
2. Convert masses to moles using atomic or molecular masses.
3. Use the mole ratio (from the balanced equation) to find the amounts of other substances.
4. Convert the results back to masses, if required.
In our empirical formula calculation, stoichiometry helps us transition from the mass of each element to moles. Then, we use the mole ratio to find the simplest whole number ratio of atoms in the compound (the empirical formula). Understanding these conversions and relationships is essential for mastering stoichiometry.

Polymer Chemistry

Polymer chemistry involves the study of polymers, which are large molecules composed of repeating subunits called monomers. Polymers are essential in numerous applications, from plastics to DNA in living organisms.
When dealing with polymers, you might encounter empirical formula calculations to understand their composition. For instance, an experimental catalyst used in polymerization, as in our example, helps initiate or speed up the polymer formation.
Here's the process summarized:
1. Determine the percentage composition of the elements in the catalyst.
2. Convert these percentages to masses, assuming a 100 g sample.
3. Find the moles of each element.
4. Simplify the mole ratios to get the empirical formula.
The empirical formula gives insights into the structure and function of the polymer and its catalysts, aiding in the development of new materials and chemicals essential in various industries.