Question

A compound was analyzed and was found to contain the following percentages of the elements by mass: lithium, \(46.46 \%\); oxygen, \(53.54 \%\). Determine the empirical formula of the compound.

Short answer

The empirical formula of the compound is Li₂O.

Step-by-step solution

Convert mass percentages to mass amounts

Considering 100 grams of the compound, the masses of lithium and oxygen can be directly taken as their respective mass percentages. - Mass of lithium, Li: \(46.46 g\) - Mass of oxygen, O: \(53.54 g\)

Convert mass amounts to moles

Divide the mass of each element by its respective molar mass to obtain the number of moles of each element. - Moles of lithium, Li: \(\frac{46.46 g}{6.94 g/mol} = 6.69 mol\) - Moles of oxygen, O: \(\frac{53.54 g}{16.00 g/mol} = 3.35 mol\)

Determine the mole ratio of the elements

Divide the moles of each element by the smallest number of moles among them to obtain their simplest whole number ratio. In this case, the smallest number of moles is 3.35 moles of oxygen. - Li: \(\frac{6.69}{3.35} \approx 2\) - O: \(\frac{3.35}{3.35} \approx 1\)

Write the empirical formula

Combine the elements and their respective simplest whole number ratios to obtain the empirical formula of the compound. The empirical formula of the compound is: Li₂O

Key concepts

Percent Composition

Understanding the percent composition of a compound involves determining the percentage by mass of each element within that compound. Imagine a new puzzle box showing a picture made up of different pieces; each piece represents an element, and its size denotes the percent composition. When dealing with chemical compounds, knowing this information is akin to knowing the contribution of each 'piece' to the overall 'picture' of the compound.

For instance, if we have a compound where lithium makes up 46.46% and oxygen makes up the remaining 53.54%, these percentages guide us like a recipe, showcasing how much of each ingredient or element is present in our compound 'dish'. It's the starting point for calculating the empirical formula, as it lets you start with a common base amount – often 100 grams – to easily convert these percentages into mass, which then sets the stage for further calculations needed to uncover the true formula of the compound.

Molar Mass

Molar mass is the weight of one mole of a substance, usually expressed in grams per mole (g/mol). It is the bridge between the microscopic world of atoms and molecules and the macroscopic world we can measure and feel. Think of it as a label on a bag of sugar that tells you how much one molecule 'weighs' when counted by the Avogadro number (approximately 6.022 x 10²³).

To proceed with our compound analysis, remember that each element in the periodic table has its own unique molar mass. Lithium has a molar mass of 6.94 g/mol, and oxygen's is 16.00 g/mol. Knowing these figures allows us to convert the mass of each element to an amount in moles. This conversion is key because chemical reactions occur on a mole-to-mole basis, not based on grams or kilograms.

Moles to Mass Conversion

Once we understand the concept of the molar mass, the next step is converting moles to mass and vice versa. The process is akin to translating different languages: moles speak the language of chemistry, while mass speaks the dialect of measurable quantities.

The calculation is straightforward: mass (in grams) divided by molar mass (in g/mol) gives us the number of moles. Conversely, multiplying the number of moles by the molar mass translates back to mass. In our example, dividing the mass of lithium by its molar mass allows us to see how many moles of lithium we're dealing with, and the same goes for oxygen. This conversion is crucial because it levels the playing field, allowing us to compare different elements fairly based on their quantities in moles, which leads to the correct stoichiometric ratios in the formula derivation.

Chemical Formula Derivation

The derivation of a chemical formula, particularly the empirical formula, is the grand reveal of our compound's identity, like uncovering the name of a mystery novel's protagonist. More specifically, it's the simplest whole number ratio of the elements in a compound. To find these ratios, we convert the mass of each element to moles and then compare these values to each other using the smallest number obtained as a reference.

Using the example from our exercise, the number of moles of each element is divided by the smallest number of moles found. Simplifying these ratios to the nearest whole numbers will give us the subscripts in the empirical formula. This step can involve a bit of mathematical finesse, especially if the ratios are not perfect wholes, but once we achieve it, we can present the compound in its simplest form. For lithium and oxygen, after normalization, we get a 2:1 ratio, leading us to the empirical formula Li₂O, which signifies for every two lithium atoms, there is one oxygen atom.