Question

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

Short answer

The molecular formulas are: a. \(S_4N_4H_4\) b. \(N_3P_3Cl_6\) c. \(Co_2C_8O_8\) d. \(S_4N_4\)

Step-by-step solution

a. SNH \( (188.35\: g/mol)\)

First, we need to find the molar mass of the empirical formula SNH. The atomic masses of S (sulfur), N (nitrogen) and H (hydrogen) are approximately 32, 14, and 1, respectively. Therefore, the molar mass of the empirical formula SNH is \(32 + 14 + 1 = 47\: g/mol\). Now, we will divide the molecular formula molar mass by the empirical formula molar mass: \(\frac{188.35}{47} \approx 4\) Then, multiply each element in the empirical formula by 4: \(SNH \times 4 = S_4N_4H_4\) The molecular formula is \(S_4N_4H_4\).

b. NPCl2 \( (347.64\: g/mol)\)

First, we need to find the molar mass of the empirical formula NPCl2. The atomic masses of N (nitrogen), P (phosphorus), and Cl (chlorine) are approximately 14, 31, and 35.5, respectively. Therefore, the molar mass of the empirical formula NPCl2 is: \(14 + 31 + 2(35.5) = 116\: g/mol\). Now, we will divide the molecular formula molar mass by the empirical formula molar mass: \(\frac{347.64}{116} \approx 3\) Then, multiply each element in the empirical formula by 3: \(NPCl_2 \times 3 = N_3P_3Cl_6\) The molecular formula is \(N_3P_3Cl_6\).

c. CoC4O4 \( (341.94\: g/mol)\)

First, we need to find the molar mass of the empirical formula CoC4O4. The atomic masses of Co (cobalt), C (carbon), and O (oxygen) are approximately 59, 12, and 16, respectively. Therefore, the molar mass of the empirical formula CoC4O4 is: \(59 + 4(12) + 4(16) = 171\: g/mol\). Now, we will divide the molecular formula molar mass by the empirical formula molar mass: \(\frac{341.94}{171} \approx 2\) Then, multiply each element in the empirical formula by 2: \(CoC_4O_4 \times 2 = Co_2C_8O_8\) The molecular formula is \(Co_2C_8O_8\).

d. SN \( (184.32\: g/mol)\)

First, we need to find the molar mass of the empirical formula SN. The atomic masses of S (sulfur) and N (nitrogen) are approximately 32 and 14, respectively. Therefore, the molar mass of the empirical formula SN is: \(32 + 14 = 46\: g/mol\). Now, we will divide the molecular formula molar mass by the empirical formula molar mass: \(\frac{184.32}{46} \approx 4\) Then, multiply each element in the empirical formula by 4: \(SN \times 4 = S_4N_4\) The molecular formula is \(S_4N_4\).

Key concepts

Empirical Formulas

Empirical formulas represent the simplest whole-number ratio of the elements present in a compound. They provide a basic structure, showing only the relative proportions of atoms rather than the actual number of atoms in a compound. For instance, the empirical formula for glucose is CH\(_2\)O, which indicates that the carbon, hydrogen, and oxygen are in a 1:2:1 ratio.
While empirical formulas do not provide detailed molecular information, they lay the groundwork for calculating molecular formulas when combined with molar mass data. Determining molecular formulas involves both empirical data and molar mass, helping us understand the true structural composition of a compound.

Molar Mass Calculation

Molar mass is an essential concept in chemistry to determine the weight of one mole of a substance. It is calculated by adding up the atomic masses of all atoms present in a formula. For example, if we look at the compound NPCl\(_2\), we would calculate the molar mass by summing the atomic masses of nitrogen (14 g/mol), phosphorus (31 g/mol), and chlorine (35.5 g/mol each).
In chemical conversions and problems involving empirical formulas, molar mass plays a crucial role. Knowing the empirical formula's molar mass allows you to find a factor to scale up to the molecular formula, revealing the exact number of atoms in each molecule.

Atomic Mass of Elements

Atomic mass provides the average mass of atoms of an element, measured in atomic mass units (amu). This value considers the different isotopes of an element and their respective abundances. For example, the atomic mass of chlorine is about 35.5 amu because of its isotopic composition.
Atomic mass is fundamental in molecular calculations, serving as the building block for computing both empirical and molar masses. Having precise atomic masses of elements helps in determining the correct molecular formula, giving insight into a compound's structure.

Chemical Conversion

Chemical conversion often involves transforming empirical data into usable molecular information. This means taking known empirical formulas and converting them to molecular formulas using the molecular mass. For example, given an empirical formula SNH with a molar mass of 47 g/mol, and a compound molar mass of 188.35 g/mol, we can calculate the factor for transformation.
By dividing the molar mass of the compound by the empirical formula's molar mass, we get a multiplication factor, like 4 in SNH's case. We then use this factor to multiply the subscripts of the empirical formula to determine the molecular formula, which provides a deeper understanding of the compound’s actual molecular structure.