Determine the molecular formulas to which the following empirical formulas and molar masses pertain. a. SNH (188.35 g/mol) c. \(\mathrm{CoC}_{4} \mathrm{O}_{4}(341.94 \mathrm{g} / \mathrm{mol})\) b. \(\mathrm{NPCl}_{2}(347.64 \mathrm{g} / \mathrm{mol})\) d. \(\mathrm{SN}(184.32 \mathrm{g} / \mathrm{mol})\)

Understanding the empirical formula is the foundation of many chemistry problems. An empirical formula represents the simplest whole-number ratio of the elements within a compound. It doesn't provide the exact number of atoms, but rather the relative number of atoms of each element present. For instance, the empirical formula for glucose is CH₂O, which indicates that the ratio of carbon to hydrogen to oxygen is 1:2:1. However, glucose's molecular formula is C₆H₁₂O₆, revealing that the substance is comprised of six times the number of atoms represented in the empirical formula.

To determine an empirical formula, one must find the lowest common denominator that can be used to express the ratio of elements in a compound as whole numbers. This often involves the use of molar mass calculations and a deep understanding of the chemical composition of the substance.

A cornerstone of solving chemistry exercises is the ability to calculate molar masses. The molar mass of a substance corresponds to the weight of one mole of that substance, typically expressed in grams per mole (g/mol). One mole of a substance contains Avogadro's number (approximately 6.022 x 10²³) of molecules or atoms.

To calculate the molar mass of a compound, one should add the molar masses of all individual elements present, weighted by their respective number of atoms in the formula. Let's take water (H₂O) as an example: it has two hydrogen atoms and one oxygen atom. The molar mass of a hydrogen atom is approximately 1.008 g/mol, and for oxygen, it is about 16.00 g/mol. Thus, the molar mass of water equals (2 x 1.008) + (1 x 16.00) = 18.016 g/mol.

Practical Calculation

In the provided exercise, the molar mass of the empirical formula SNH was calculated by adding the molar masses of sulfur (32.06 g/mol), nitrogen (14.01 g/mol), and hydrogen (1.008 g/mol), leading to the sum of 47.078 g/mol. Such calculations are essential when converting between moles and grams, a common requirement in quantitative chemistry.

The chemical composition of a molecule is represented by its molecular formula, which provides information on the exact number of each type of atom present in the molecule. This distinguishes it from the empirical formula, which only reflects the simplest ratio. Chemical composition is central in learning about the characteristics and reactions of different substances.

Using the information about the molar mass and the empirical formula, one can deduce the molecular formula of a compound, which follows the steps highlighted in the exercise. It involves determining the ratio of the molar mass given to the molar mass of the empirical formula. This ratio is then used to multiply the subscripts in the empirical formula to obtain the molecular formula with actual counts of atoms. For example, if the empirical formula is CH₂O and the molar mass indicates that the actual substance is eight times the mass of the empirical formula, it would imply the molecular formula is C₈H₁₆O₈.

Application to Exercise

In the exercise, by calculating the ratio and multiplying it by the empirical formula, we find the molecular formula for each compound. The molecular formula not only gives us a clearer picture of the makeup of a compound but is also invaluable for stoichiometric calculations and understanding the physical and chemical properties of the substance.