Problem 61 - Chemistry: An Atoms-Focused Approach

Question

What is the empirical formula of the compound that is \(24.2 \% \mathrm{Cu}, 27.0 \% \mathrm{Cl},\) and \(48.8 \%\) O by mass?

Step-by-Step Solution

Answer

Answer: The empirical formula of the compound is CuCl₂O₈.

Step 1: Convert percentages into grams

Since we are given the percentage composition of the compound, we can assume we have 100 grams of the compound to work with. This will make it easier to convert the percentages into grams: Cu: \(24.2 \% \rightarrow 24.2 \, g\) Cl: \(27.0 \% \rightarrow 27.0 \, g\) O: \(48.8 \% \rightarrow 48.8 \, g\)

Step 2: Convert grams into moles

Now, we can use the molar masses of the elements to convert grams into moles: Cu: \(\frac{24.2 \, g}{63.55 \, g/mol} = 0.381 \, mol\) Cl: \(\frac{27.0 \, g}{35.45 \, g/mol} = 0.762 \, mol\) O: \(\frac{48.8 \, g}{16.00 \, g/mol} = 3.05 \, mol\)

Step 3: Determine the mole ratio

To find the empirical formula, we will calculate the mole ratios by dividing all the moles by the smallest number of moles amongst them: Cu: \(\frac{0.381 \, mol}{0.381 \, mol} = 1\) Cl: \(\frac{0.762 \, mol}{0.381 \, mol} = 2\) O: \(\frac{3.05 \, mol}{0.381 \, mol} ≈ 8\)

Step 4: Write the empirical formula

We can now write the empirical formula using the mole ratios we've determined: \(\mathrm{CuCl_2O_8}\) So, the empirical formula of the compound is \(\mathrm{CuCl_2O_8}\).

Key Concepts

Mole Ratio

The mole ratio is a key concept in chemistry that helps us understand the relationship between the amounts of different elements in a compound. It is derived by comparing the number of moles of each element present. By calculating the mole ratio, you can simplify the formula down to the simplest whole numbers, which is crucial in determining the empirical formula.

To find the mole ratio, first convert your element quantities from grams to moles using the molar mass of each element. Then, divide each element’s molar quantity by the smallest number of moles present among the elements.

This ratio provides the simplest way to describe how many atoms of each element combine to form the compound. For instance, if a calculation results in 1 mol of Cu, 2 mol of Cl, and 8 mol of O, the mole ratio 1:2:8 indicates the empirical formula \(\mathrm{CuCl_2O_8}\).

• This step helps reveal the simplest integer ratio of atoms.
• It’s important for identifying the empirical formula, which represents the composition of the compound on an atomic scale.

Percentage Composition

Percentage composition provides a detailed breakdown of a compound's mass into percentages that signify how much of each element is present. This is incredibly useful when trying to determine the empirical formula, as it gives a tangible sense of each element’s presence within the compound.

In the earlier exercise, we were provided with the mass percentages for copper (Cu), chlorine (Cl), and oxygen (O). By assuming 100 grams of total compound, these percentages directly convert to grams.

• Examples: \(24.2\%\ \mathrm{Cu}\) becomes \(24.2 \, g\) of Cu.
• This conversion is straightforward and useful for calculations, especially when using concepts like the mole ratio and molar mass in subsequent steps.

This technique simplifies complex chemical compositions into forms that can be easily calculated and converted into more insightful values like moles.

Molar Mass

Molar mass is the mass of one mole of a given substance, typically expressed in grams per mole (g/mol). This measure is crucial when converting between grams and moles, as it acts as the bridge between mass and amount of substance.

For the conversion, every element has its own unique molar mass, which is based on its atomic weight as listed in the periodic table. For example, copper (Cu) has a molar mass of approximately \(63.55 \, g/mol\), chlorine (Cl) is about \(35.45 \, g/mol\), and oxygen (O) is \(16.00 \, g/mol\).

• The calculation involves dividing the mass (in grams) of the element by its molar mass to find the number of moles.
• For example: \( \frac{24.2 \, g}{63.55 \, g/mol} = 0.381 \, mol \) for copper.

This conversion is vital for determining the mole ratio, which in turn leads to the correct empirical formula. Molar mass helps us connect the macroscopic measurements from a lab scale to a molecular scale.