Question

A compound with empirical formula CH was found by experiment to have a molar mass of approximately 78 g. What is the molecular formula of the compound?

Short answer

The molecular formula of the compound with an empirical formula CH and a molar mass of 78 g/mol is C₆H₆.

Step-by-step solution

Determine the molar mass of the empirical formula

To find the molar mass of the empirical formula, CH, we need to add the molar masses of carbon (C) and hydrogen (H). The molar mass of carbon is 12.01 g/mol and the molar mass of hydrogen is 1.01 g/mol. Molar mass of CH = 12.01 g/mol + 1.01 g/mol = 13.02 g/mol

Calculate the ratio between the molecular formula and the empirical formula

To find the ratio, divide the given molar mass (78 g/mol) by the molar mass of the empirical formula (13.02 g/mol). Ratio = 78 g/mol / 13.02 g/mol ≈ 6 The ratio result should be a whole number as molecular formulas are whole-number multiples of the empirical formula.

Determine the molecular formula

Now that we have found the ratio between the molecular formula and the empirical formula, we can multiply the empirical formula by the ratio to find the molecular formula. Molecular formula = empirical formula × ratio Molecular formula = CH × 6 The molecular formula will then have 6 carbon atoms and 6 hydrogen atoms. Molecular formula = C₆H₆

Key concepts

Empirical Formula

The empirical formula is the simplest expression of the composition of a compound. It represents the simplest whole-number ratio of the different elements within a substance. For example, the empirical formula of a compound might be written as \( \text{CH} \), indicating that in the simplest form, the compound is composed of equal numbers of carbon (C) and hydrogen (H) atoms. It's important to note that the empirical formula does not provide the exact number of atoms in a molecule, but rather the simplest ratio. This simplicity can make it challenging to know the detailed makeup of the compound, especially when the compound may consist of multiple atoms in practice. To determine an empirical formula, you need to know the relative quantity (in moles) of each element in your compound. You can find this through chemical analysis or a combustion method. These methods allow chemists to create a straightforward yet essential representation of a compound's atomic makeup.

Molar Mass

Molar mass is a fundamental concept in chemistry that describes the mass of a given substance (molecule or compound) for one mole of that substance, expressed in grams per mole (g/mol). It takes into account the sum of the relative atomic masses of all atoms within a molecule. For instance, if the empirical formula of a compound is CH, you find the molar mass by adding the molar masses of one carbon atom, \( 12.01 \) g/mol, and one hydrogen atom, \( 1.01 \) g/mol, which equals \( 13.02 \) g/mol.Understanding molar mass is crucial because it relates the mass of a sample to the number of molecules or atoms it contains. This is particularly useful when converting between mass and moles in calculations. Additionally, when given a molar mass for the entire compound, such as \( 78 \) g/mol in this example, and knowing the molar mass of the empirical formula, it allows scientists to determine the molecular formula, which will be discussed further.

Ratio Calculation

Ratio calculation is a key process in identifying the molecular formula from the empirical formula. Given both the empirical formula mass and the compound's molar mass, we calculate the ratio by dividing the compound's molar mass by the empirical formula's molar mass.In our example, the compound's molar mass is \( 78 \) g/mol, and the molar mass of the empirical formula \( \text{CH} \) is \( 13.02 \) g/mol. By dividing these values, we obtain a ratio of approximately \( 6 \). This ratio is crucial as it denotes how many times the empirical formula is contained within the molecular formula.Thus, to derive the molecular formula, we multiply the number of each type of atom in the empirical formula by this ratio. Here, "CH" becomes "C₆H₆" through the multiplication. Ratios must be whole numbers because you can't have a fraction of an atom in a real-world context; hence the result \( 6 \) is perfect for our calculation. This step solidifies the bridge between basic empirical formula and the detailed molecular formula.