Question

The molar masses and empirical formulas of several compounds containing carbon and nitrogen are listed here. Find the molecular formula of each compound. (a) \(163.26 \mathrm{~g} / \mathrm{mol}, \mathrm{C}_{11} \mathrm{H}_{12} \mathrm{~N}\) (b) \(186.24 \mathrm{~g} / \mathrm{mol}, \mathrm{C}_{6} \mathrm{H}_{2} \mathrm{~N}\) (c) \(312.29 \mathrm{~g} / \mathrm{mol}, \mathrm{C}_{3} \mathrm{H}_{2} \mathrm{~N}\)

Short answer

The molecular formulas are (a) \(C_{11}H_{12}N\), (b) \(C_{12}H_{4}N_{2}\), (c) \(C_{24}H_{16}N_{8}\).

Step-by-step solution

Calculate the empirical formula mass of compound (a)

To find the empirical formula mass of compound (a), which is \(C_{11}H_{12}N\), add the atomic masses of all the atoms in the empirical formula: \(11 \times \text{Atomic mass of C}) + (12 \times \text{Atomic mass of H}) + (1 \times \text{Atomic mass of N})\). Using the atomic masses (C = 12 g/mol, H = 1 g/mol, N = 14 g/mol), the calculation is \(11 \times 12) + (12 \times 1) + (14) = 146 \mathrm{g/mol}\).

Determine the molecular formula of compound (a)

Divide the molar mass of compound (a) by the empirical formula mass to find the multiple: \(163.26 \mathrm{g/mol} \div 146 \mathrm{g/mol}\). The result is roughly 1.12. Since the molecular formula is a whole number multiple of the empirical formula, and 1.12 is close to 1, the molecular formula is the same as the empirical formula \(C_{11}H_{12}N\).

Calculate the empirical formula mass of compound (b)

The empirical formula mass of compound (b) is calculated in a similar way: \(6 \times 12) + (2 \times 1) = 74 \mathrm{g/mol}\).

Determine the molecular formula of compound (b)

Divide the molar mass of compound (b) by its empirical formula mass: \(186.24 \mathrm{g/mol} \div 74 \mathrm{g/mol}\). This yields approximately 2.51, which suggests that the molecular formula is two times the empirical formula due to the value being close to 2. The molecular formula of compound (b) is thus \(C_{12}H_{4}N_{2}\).

Calculate the empirical formula mass of compound (c)

Again, calculate the empirical formula mass: \(3 \times 12) + (2 \times 1) = 38 \mathrm{g/mol}\).

Determine the molecular formula of compound (c)

The molar mass of compound (c) divided by its empirical formula mass is: \(312.29 \mathrm{g/mol} \div 38 \mathrm{g/mol}\) yielding approximately 8.22. The multiple should be a whole number, so we round to 8. The molecular formula of compound (c) is \(C_{24}H_{16}N_{8}\).

Key concepts

Empirical Formula Mass

Understanding the empirical formula mass is essential in the study of chemistry as it represents the simplest, integer ratio of atoms in a compound. It's a stepping stone in finding the molecular formula of a substance.

In our textbook exercise, the empirical formula mass is calculated by multiplying the number of atoms of each element by their respective atomic masses and then adding these up. For example, with compound (a) which has an empirical formula of \(C_{11}H_{12}N\), we calculate the empirical formula mass by adding together the atomic masses for carbon (C, 12 g/mol), hydrogen (H, 1 g/mol), and nitrogen (N, 14 g/mol) according to how many of each there are in the formula.

This step is foundational because the molecular formula is often a multiple of the empirical formula, and determining this multiple relies on knowing the correct empirical formula mass.

Molar Mass Determination

The molar mass of a compound is the weight of one mole of that substance, typically expressed in grams per mole (g/mol). It is essentially the mass of a given substance divided by the amount of that substance in moles. Determining the molar mass is vital for converting between the mass of a substance and the amount in moles, a fundamental concept in stoichiometry.

In the provided examples, such as compound (a) with a molar mass of 163.26 g/mol, comparing this value to the previously calculated empirical formula mass allows us to understand the relationship between the empirical and molecular formula. If the molar mass is higher than the empirical formula mass, as in compound (b), we know that the molecular formula contains a whole number multiple of the empirical units.

Atomic Mass

Atomic mass is the mass of a single atom, usually expressed in atomic mass units (amu). On a larger scale, when dealing with moles, atomic mass is utilized in grams per mole (g/mol), which is convenient for chemical calculations.

Each element's atomic mass can be found on the periodic table and is pivotal in calculating both the empirical formula mass and molar mass. The atomic mass is a weighted average that takes into account the masses of an element's various isotopes and their abundance. When we solved the textbook problem, for instance, we used the atomic masses of carbon, hydrogen, and nitrogen to calculate the empirical formula mass of each compound.

Stoichiometry

Stoichiometry is the section of chemistry that pertains to the calculation of the quantities of reactants and products involved in a chemical reaction. It is based on the conservation of mass and the concept of moles. Stoichiometry is not only crucial for reactions but also for understanding the composition of compounds and their formulas.

In the textbook solutions, stoichiometry principles are applied to deduce the molecular formula from the empirical formula using molar mass. When dealing with compound (c), the calculated multiple from dividing the molar mass by the empirical formula mass gives the stoichiometric number needed to find the exact molecular formula. It shows us how many times the empirical formula must be multiplied to reach the actual molecular composition of the substance.