Question

A compound with molar mass \(180.1 \mathrm{g} / \mathrm{mol}\) has the following composition by mass: $$\begin{array}{|ll|}\hline C & 40.0 \% \\\H & 6.70 \% \\\O & 53.3 \% \\\\\hline\end{array}$$ Determine the empirical and molecular formulas of the compound.

Short answer

The empirical formula of the compound is CH2O and the molecular formula is C6H12O6.

Step-by-step solution

Convert percentages to grams

Assume that we have 100 grams of the compound. In this case, the mass percentages will be equivalent to the masses in grams for each element. So we have: - C: 40.0 g - H: 6.70 g - O: 53.3 g

Find moles for each element

Next, we will convert the masses of each element to moles by dividing their mass by their respective molar mass: - Moles of C: \(\frac{40.0\ \mathrm{g}}{12.01\ \mathrm{g/mol}} = 3.33\ \mathrm{mol}\) of C - Moles of H: \(\frac{6.70\ \mathrm{g}}{1.01\ \mathrm{g/mol}} = 6.63\ \mathrm{mol}\) of H - Moles of O: \(\frac{53.3\ \mathrm{g}}{16.00\ \mathrm{g/mol}} = 3.33\ \mathrm{mol}\) of O

Calculate the ratio of moles

To find the empirical formula, we need to determine the ratio of moles of each element. Divide the number of moles of each element by the smallest value: - For C: \(\frac{3.33}{3.33} = 1\) - For H: \(\frac{6.63}{3.33} \approx 2\) - For O: \(\frac{3.33}{3.33} = 1\) The empirical formula is CH2O.

Calculate the empirical formula molar mass

Now we need to find the molar mass of the empirical formula: - 1 mol of C: 1 × 12.01 g/mol = 12.01 g/mol - 2 mol of H: 2 × 1.01 g/mol = 2.02 g/mol - 1 mol of O: 1 × 16.00 g/mol = 16.00 g/mol Empirical formula molar mass: 12.01 g/mol + 2.02 g/mol + 16.00 g/mol = 30.03 g/mol

Determine the molecular formula

To find the molecular formula, we need to determine the ratio of the molecular molar mass to the empirical molar mass. Divide the molecular molar mass (180.1 g/mol) by the empirical molar mass (30.03 g/mol): \(n = \frac{180.1\ \mathrm{g/mol}}{30.03\ \mathrm{g/mol}}\approx 6\) This ratio means that the molecular formula is 6 times the empirical formula. So, the molecular formula is: C6H12O6.

Key concepts

Molar Mass

Molar mass is an essential concept in chemistry. It's the mass of one mole of a given substance and is usually expressed in grams per mole (g/mol). To calculate the molar mass, you need to sum up the atomic masses of all the atoms in a molecule. These atomic masses are found on the periodic table, specific to each element.
For example, when we have a compound like carbon dioxide (CO₂), we need to find the molar mass of both carbon (C) and oxygen (O). Carbon weighs 12.01 g/mol, and oxygen weighs 16.00 g/mol. Since CO₂ has one atom of carbon and two atoms of oxygen, you calculate its molar mass as follows:

• Carbon: 12.01 g/mol
• Oxygen: 2 × 16.00 g/mol = 32.00 g/mol
• Total: 12.01 g/mol + 32.00 g/mol = 44.01 g/mol

In the original exercise, knowing the molar mass of a compound (180.1 g/mol) helps us deduce its molecular formula, as we'll explore further.

Composition by Mass

Composition by mass of a compound indicates what percentage of each element makes up the compound. It can be a crucial aspect when determining the empirical formula. To determine the composition by mass, assume a certain mass of the compound, often 100 grams, makes conversion easier. Thus, the percentage values translate directly to grams.
For instance, in the exercise provided, the compound consists of:

• 40.0% carbon, therefore 40.0 g when assuming a 100 g sample
• 6.70% hydrogen, resulting in 6.70 g
• 53.3% oxygen, resulting in 53.3 g

Understanding these percentages enables the calculation of the amount of moles for each element, which then guides us to the empirical formula.

Empirical Formula

The empirical formula is the simplest form of a chemical formula. It shows the smallest whole number ratio of elements in a compound. This formula may not represent the actual number of atoms within a molecule, but it provides insight into the basic composition.
To find the empirical formula, convert mass percentages to grams, then to moles, and subsequently find simple ratios as outlined in the exercise:

• Calculate moles by dividing each element's mass by its molar mass.
• Find the smallest mole value among the elements.
• Divide each element's mole count by the smallest value to get the ratio.

For example, if a compound has the empirical formula CH₂O, it means that for every carbon atom, there are two hydrogen atoms and one oxygen atom. This step is important in chemistry as it lays the foundation for further molecular formula determination.

Molecular Formula

The molecular formula gives the actual number of atoms of each element in a molecule and is a multiple of the empirical formula. To derive the molecular formula, you need both the empirical formula and the compound's molar mass.
Follow these steps:

• First, calculate the molar mass of the empirical formula.
• Then, divide the given molar mass of the compound by the empirical formula's molar mass to find a scaling factor or multiplier.
• Finally, multiply each subscript in the empirical formula by this multiplier to obtain the molecular formula.

In the exercise, the compound had an empirical formula of CH₂O and a molar mass of 180.1 g/mol. The scaling factor was found to be 6, leading to a molecular formula of C₆H₁₂O₆. This step crucially transforming the simplest formula to a comprehensive one that reflects the compound's actual structure.