Question

A compound with molar mass \(180.1 \mathrm{~g} / \mathrm{mol}\) has the following composition by mass: $$\begin{array}{ll}\mathrm{C} & 40.0 \% \\\\\mathrm{H} & 6.70 \% \\\\\mathrm{O} & 53.3 \%\end{array}$$ 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

Determine moles of each element based on mass percentages

Assume that we have 100 grams of the compound. In this case, the mass of carbon (C), hydrogen (H), and oxygen (O) are as follows: - Mass of C: \(40.0\%\) of 100g = 40.0g - Mass of H: \(6.70\%\) of 100g = 6.70g - Mass of O: \(53.3\%\) of 100g = 53.3g Now we will convert these masses to moles by dividing by molar mass of each element: - Moles of C: \(\frac{40.0 \text{g}}{12.01 \text{g/mol}}\) = 3.33 mol - Moles of H: \(\frac{6.70 \text{g}}{1.01 \text{g/mol}}\) = 6.63 mol - Moles of O: \(\frac{53.3 \text{g}}{16.00 \text{g/mol}}\) = 3.33 mol

Find the molar ratio and empirical formula

To find the molar ratio, divide the moles of each element by the moles of the lightest element. In this case, the lightest element is carbon, with 3.33 moles. So we have the following molar ratios: - Moles of C: \(\frac{3.33}{3.33}\) = 1 - Moles of H: \(\frac{6.63}{3.33}\) = 1.99 ≈ 2 - Moles of O: \(\frac{3.33}{3.33}\) = 1 From these molar ratios, we can determine the empirical formula: C₁H₂O₁ or simply CH₂O.

Determine the molecular formula

To find the molecular formula, we will first calculate the empirical formula mass: - Empirical formula mass: C (12.01 g/mol) + 2 × H (1.01 g/mol) + O (16.00 g/mol) = 30.03 g/mol Now, divide the compound's given molar mass (180.1 g/mol) by the empirical formula mass (30.03 g/mol): - Molecular formula multiple: \(\frac{180.1 \text{ g/mol}}{30.03 \text{ g/mol}}\) = 5.996 ≈ 6 Since the molecular formula multiple is 6, multiply the empirical formula by 6 to get the molecular formula: - Molecular formula: 6 × CH₂O = C₆H₁₂O₆

Key concepts

Molecular Composition

Understanding the molecular composition of a compound is fundamental in chemistry. It tells us which elements are present and in what ratio. For simpler substances, the molecular composition can be directly reflected in the empirical formula, which shows the simplest whole-number ratio of types of atoms in a compound. However, the true molecular formula may contain multiple units of the empirical formula.

In our example, the empirical formula, CH₂O, suggests a molecular composition where for every carbon atom, there are two hydrogen atoms and one oxygen atom. But this may not be the full picture of the actual substance. If the molar mass is significantly higher than the mass of the empirical formula, the molecular composition will be a multiple of this empirical formula. This is crucial because different molecular compositions can result in vastly different properties, even if the empirical formulas are the same.

Molar Mass

Molar mass is the weight of one mole (6.022 × 10²³ particles) of a given substance and it is expressed in grams per mole (g/mol). This physical property is a bridge between the micro world of atoms and molecules and the macro world we can measure in the laboratory. Each element's molar mass is found on the periodic table, while the molar mass of a compound is the sum of the molar masses of its constituent elements, each multiplied by their count in the molecular formula.

We've seen in the exercise that knowing the molar mass of the entire compound, alongside its empirical formula, allows us to deduce how many times the empirical formula must be multiplied to reach the actual molecular formula. This step is crucial because it provides an insight into the physical sample's quantities we might use in reactions or properties like boiling point, which depend on the size of the molecules.

Chemical Formula Determination

To determine the chemical formula of a compound, we combine the empirical formula with the molar mass. First, we deduce the empirical formula by translating mass percentages into moles, leveraging the molar masses of the individual elements. Then, we find the simplest whole-number ratio among these elements. However, that's halfway to the finish line.

The empirical formula may not reflect the true nature of the compound in question. By comparing the empirical formula mass to the given molar mass of the compound, we can identify how many times the empirical unit repeats within a molecule, giving us the molecular formula. In the example, dividing the compound's molar mass by the empirical formula's mass gave us approximately 6, providing a molecular formula of C₆H₁₂O₆, which presents the actual number of atoms of each element in a molecule of the compound.