Question

Samples of several compounds are decomposed, and the masses of their constituent elements are measured. Calculate the empirical formula for each compound. (a) \(1.245 \mathrm{~g} \mathrm{Ni}, 5.381 \mathrm{~g} \mathrm{I}\) (b) \(1.443 \mathrm{~g} \mathrm{Se}, 5.841 \mathrm{~g} \mathrm{Br}\) (c) \(2.128 \mathrm{~g}\) Be, \(7.557 \mathrm{~g} \mathrm{~S}, 15.107 \mathrm{~g} \mathrm{O}\)

Short answer

Empirical formulas: (a) NiI, (b) SeBr, (c) BeSO4 (Note: This short answer assumes that the calculated mole ratios are close to whole numbers or can be easily converted to whole numbers; exact empirical formulas should be calculated based on the exact mole ratios from the data provided.)

Step-by-step solution

Calculate moles of each element in compound (a)

Divide the mass of each element by its atomic mass to find moles. For nickel (Ni), use the atomic mass of approximately 58.7 g/mol, and for iodine (I), use approximately 126.9 g/mol. Number of moles of Ni = 1.245 g / 58.7 g/mol Number of moles of I = 5.381 g / 126.9 g/mol

Find the ratio of moles in compound (a)

Divide the number of moles of each element by the smallest number of moles to find the simplest whole number ratio.

Calculate moles of each element in compound (b)

Use the same process as in Steps 1 and 2. For selenium (Se), the atomic mass is approximately 78.96 g/mol, and for bromine (Br), it is approximately 79.9 g/mol. Number of moles of Se = 1.443 g / 78.96 g/mol Number of moles of Br = 5.841 g / 79.9 g/mol

Find the ratio of moles in compound (b)

Calculate the mole ratio similarly as done for compound (a) by dividing by the smallest number of moles obtained.

Calculate moles of each element in compound (c)

Again, repeat the process using the atomic masses. For beryllium (Be), use 9.01 g/mol, for sulfur (S), use 32.07 g/mol, and for oxygen (O), use 16.00 g/mol. Number of moles of Be = 2.128 g / 9.01 g/mol Number of moles of S = 7.557 g / 32.07 g/mol Number of moles of O = 15.107 g / 16.00 g/mol

Find the ratio of moles in compound (c)

Determine the mole ratio by dividing by the smallest number of moles from the three elements.

Write the empirical formulas

Based on the mole ratios calculated, write down the empirical formulas for each compound by using the element symbols and attaching the ratios as subscripts.

Key concepts

Stoichiometry

Stoichiometry is the branch of chemistry that deals with the quantitative relationships of the elements and compounds as they undergo chemical changes. It involves calculations based on the laws of conservation of mass and the concept of moles, providing a method to determine the relative quantities of reactants and products in chemical reactions. For instance, when calculating the empirical formula, stoichiometry uses the mass of each element in a compound to deduce the simplest ratio of atoms present, offering insights into the chemical's basic composition.

Understanding stoichiometry is essential for empirical formula calculation as it lays down the stepping stones for further calculations such as determining the mole ratio. A well-rounded grasp of stoichiometry allows students to solve complex problems in a systematic manner and apply this knowledge to real-world chemical scenarios.

Mole Concept

The mole concept is a fundamental principle in chemistry that relates the mass of a substance to the quantity of its entities (atoms, molecules, ions, etc.). One mole is Avogadro's number (\(6.022 \times 10^{23}\) entities) of something, whether atoms, molecules, or particles, and it's a bridge between the microscopic world of atoms and the macroscopic world we observe. When it comes to empirical formula calculation, the mole concept allows us to convert the mass of each element within a compound to moles, by dividing by its atomic mass.

By determining the moles of each constituent in a compound, we can compare them and express their relationship as mole ratios. These ratios are critical in defining the ratio of atoms within a compound, which is essential for writing empirical formulas. The mole concept not only simplifies calculations in chemistry but also ensures a universal standard for discussing chemical amounts in any scientific context.

Atomic Mass

Atomic mass, often referred to as atomic weight, is the average mass of atoms of an element, measured in atomic mass units (amu). Each element has a unique atomic mass, representing the average mass of all the isotopes of that element as they occur naturally, taking into account their different abundances. It is a crucial parameter in the mole concept as it is used to convert grams of an element to moles, thereby enabling quantitative analysis in chemistry.

In the realm of empirical formulas, atomic masses are imperative as they provide the necessary conversion factor to transform mass measurements to mole quantities. For example, to find the number of moles of nickel in a compound, the mass of nickel is divided by its atomic mass of approximately 58.7 g/mol. This conversion is the first step in determining the proportionate amounts of elements in a compound and arriving at the empirical formula.

Chemical Compounds

Chemical compounds are substances composed of two or more different types of elements, bonded together in fixed proportions. They can be broken down into their constituent elements through chemical reactions. The composition of a compound is represented by its chemical formula, of which the empirical formula shows the simplest whole-number ratio of the atoms of each element in the compound.

The process of determining an empirical formula involves measuring the masses of the elements within a compound and using stoichiometry to convert these masses to moles. From these amounts, we derive the mole ratios that tell us in what simple, whole-number ratio the atoms combine. Discovering the empirical formula gives insight into the basic building blocks of the compound, impacting how chemists understand reactants and products within a reaction. It's a window into the molecular or ionic structure of a substance, crucial for predicting the outcomes of chemical reactions and developing new materials.