Question

Give the empirical formula of each of the following compounds if a sample contains (a) \(0.0130 \mathrm{~mol} \mathrm{C}, 0.0390\) mol \(\mathrm{H}\), and \(0.0065 \mathrm{~mol} \mathrm{O}\); (b) \(11.66 \mathrm{~g}\) iron and \(5.01 \mathrm{~g}\) oxygen; (c) \(40.0 \% \mathrm{C}, 67 \% \mathrm{H}\), and \(53.3 \% \mathrm{O}\) by mass

Short answer

The empirical formulas for each compound are: (a) \(C_2H_6O\), (b) \(Fe_2O_3\), and (c) \(CH_2O\).

Step-by-step solution

Exercise (a):

We have 0.0130 mol C, 0.0390 mol H, and 0.0065 mol O. First, let's find the smallest number of moles: \(Minimum~number~of~moles = min\{0.0130, 0.0390,0.0065\} = 0.0065\) Now, we'll divide the mole ratio by the smallest mole number: \(Ratio~of~C = 0.0130 / 0.0065 = 2\) \(Ratio~of~H = 0.0390 / 0.0065 = 6\) \(Ratio~of~O = 0.0065 / 0.0065 = 1\) The empirical formula is: \(C_2H_6O\)

Exercise (b):

We have 11.66 g of iron (Fe) and 5.01 g of oxygen (O). First, let's convert these amounts to moles: \(mol~Fe = \frac{11.66~g}{55.85~g/mol} \approx 0.2089~mol\) \(mol~O = \frac{5.01~g}{16.00~g/mol} \approx 0.3131~mol\) Now, we'll find the smallest number of moles and divide the mole ratio by it: \(Minimum~number~of~moles = min\{0.2089, 0.3131\} = 0.2089\) \(Ratio~of~Fe = 0.2089 / 0.2089 = 1\) \(Ratio~of~O = 0.3131 / 0.2089 \approx 1.5\) To get a whole number ratio for the oxygen, we'll multiply both ratios by 2: \(Ratio~of~Fe = 1 \times 2 = 2\) \(Ratio~of~O = 1.5 \times 2 = 3\) The empirical formula is: \(Fe_2O_3\)

Exercise (c):

We are given 40.0 % C, 6.7% H, and 53.3% O by mass. To find the empirical formula, assume that we have 100 g of the compound. Thus, we'll have 40.0 g of C, 6.7 g of H, and 53.3 g of O. Now, convert them into moles: \(mol~C = \frac{40.0~g}{12.01~g/mol} \approx 3.331\) \(mol~H = \frac{6.7}{1.008~g/mol} \approx 6.651\) \(mol~O = \frac{53.3~g}{16.00~g/mol} \approx 3.331\) Next, we will find the smallest mole number and divide the mole ratio by it: \(Minimum~number~of~moles = min\{3.331, 6.651, 3.331\} = 3.331\) \(Ratio~of~C = 3.331/3.331 = 1\) \(Ratio~of~H = 6.651/3.331 \approx 2\) \(Ratio~of~O = 3.331/3.331 = 1\) The empirical formula is: \(CH_2O\)

Key concepts

Understanding the Mole Concept

The mole concept is at the heart of chemistry, serving as the bridge between the microscopic world of atoms and molecules and the macroscopic world of grams and kilograms. It is defined as the amount of substance that contains as many entities (atoms, ions, molecules, or other particles) as there are atoms in exactly 12 grams of carbon-12, which is approximately \(6.022 \times 10^{23}\) entities, the Avogadro number.

When working with the mole concept, one key skill is converting between moles and grams. This is done using the molar mass, the weight of one mole of a substance, typically expressed in grams per mole (\text{g/mol}). In a textbook exercise, when given amounts in grams, one would convert these to moles by dividing the mass of the substance by its molar mass. For instance, to find the number of moles of carbon in 40 grams, you divide by carbon's molar mass (approximately \(12.01 \text{g/mol}\)).

In practical terms, when you're given the mass of an element and asked to determine the empirical formula, you convert the mass to moles to see how many moles of each element are present. By comparing these mole ratios, you can deduce the simplest whole number ratio of atoms within a compound, which constitutes the empirical formula.

Deciphering Chemical Composition

Chemical composition refers to the types and amounts of elements present in a substance. When you're deciphering chemical composition, such as finding empirical formulas, you're essentially figuring out the building blocks that make up a compound.

To illustrate, let's consider a sample situation where we have specific mole quantities of carbon (C), hydrogen (H), and oxygen (O). The strategy is to compare the number of moles of each element relative to the others. To achieve this, it's customary to divide all the moles by the smallest amount present; this helps us find the most reduced, or empirical, formula.

Example of Chemical Composition Analysis

In our first example from the textbook exercise, the mole ratios of the elements were simplified, leading to the formula \(C_2H_6O\). This is a representation of the chemical composition in its simplest form and gives us valuable insight into the structure and properties of the substance. Understanding the chemical composition is crucial not only for resolving textbook exercises but also for grasping the nature of chemical reactions and the substances involved.

Mastering Stoichiometry

Stoichiometry is where the mole concept and chemical composition come to life. It involves the calculation of reactants and products in chemical reactions. The stoichiometry is based on the conservation of mass and the principle that the amounts of reactants and products in a chemical reaction are proportional to their relative molecular weights and the coefficients in the balanced chemical equation.

In exercises where you're given different amounts of each element, as in the case of our second example with iron and oxygen, stoichiometry helps you determine the simplest ratio of elements, which directly relates to the empirical formula of the compound formed. After calculating the moles of each element (via the mole concept), we use the method of continuous division to obtain a whole number ratio that respects the stoichiometry of the chemical compound.

Applying Stoichiometry to Real Problems

For example, in exercise (b), we find that iron and oxygen combine in a whole number ratio to form iron(III) oxide with the empirical formula \(Fe_2O_3\). This stoichiometric balance is key to predicting the outcomes of chemical reactions and understanding the quantitative relationships within them.