Problem 45 - Chemistry: The Central Science

Question

Determine the empirical formulas of the compounds with the following compositions by mass: (a) \(10.4 \% \mathrm{C}, 27.8 \% \mathrm{~S}\), and \(61.7 \% \mathrm{Cl}\) (b) \(21.7 \%\) C, \(9.6 \% \mathrm{O}\), and \(68.7 \% \mathrm{~F}\) (c) \(32.79 \% \mathrm{Na}, 13.02 \% \mathrm{Al}\), and \(54.19 \% \mathrm{~F}\)

Step-by-Step Solution

Answer

The empirical formulas for the given compounds are: (a) \(CS_2Cl_2\), (b) \(C_2OF_6\), and (c) \(Na_3AlF_6\).

Step 1: Convert mass percentages to grams

Assume a 100-gram sample of the compound. This means we have 10.4 grams of C, 27.8 grams of S, and 61.7 grams of Cl.

Step 2: Convert grams to moles

Using the molar masses of each element (C: 12.01 g/mol, S: 32.07 g/mol, Cl: 35.45 g/mol), convert the grams to moles: moles of C = \(\frac{10.4\text{ g}}{12.01\text{ g/mol}}\) moles of S = \(\frac{27.8\text{ g}}{32.07\text{ g/mol}}\) moles of Cl = \(\frac{61.7\text{ g}}{35.45\text{ g/mol}}\)

Step 3: Find the mole ratios

Divide each value by the smallest number of moles calculated to get the mole ratios: mole ratio of C: \(\frac{\text{moles of C}}{\text{smallest moles}}\) mole ratio of S: \(\frac{\text{moles of S}}{\text{smallest moles}}\) mole ratio of Cl: \(\frac{\text{moles of Cl}}{\text{smallest moles}}\)

Step 4: Write the empirical formula

If the mole ratios are whole numbers, write the empirical formula using the mole ratios as subscripts. If not, round the ratios to the nearest whole number or multiply them by a factor to obtain whole numbers. In this case, the empirical formula is: \(CS_2Cl_2\) (b) Compound with 21.7% C, 9.6% O, and 68.7% F

Step 1: Convert mass percentages to grams

Assume a 100-gram sample of the compound. This means we have 21.7 grams of C, 9.6 grams of O, and 68.7 grams of F.

Step 2: Convert grams to moles

Using the molar masses of each element (C: 12.01 g/mol, O: 16.00 g/mol, F: 19.00 g/mol), convert the grams to moles: moles of C = \(\frac{21.7\text{ g}}{12.01\text{ g/mol}}\) moles of O = \(\frac{9.6\text{ g}}{16.00\text{ g/mol}}\) moles of F = \(\frac{68.7\text{ g}}{19.00\text{ g/mol}}\)

Step 3: Find the mole ratios

Divide each value by the smallest number of moles calculated to get the mole ratios: mole ratio of C: \(\frac{\text{moles of C}}{\text{smallest moles}}\) mole ratio of O: \(\frac{\text{moles of O}}{\text{smallest moles}}\) mole ratio of F: \(\frac{\text{moles of F}}{\text{smallest moles}}\)

Step 4: Write the empirical formula

In this case, the empirical formula is: \(C_2OF_6\) (c) Compound with 32.79% Na, 13.02% Al, and 54.19% F

Step 1: Convert mass percentages to grams

Assume a 100-gram sample of the compound. This means we have 32.79 grams of Na, 13.02 grams of Al, and 54.19 grams of F.

Step 2: Convert grams to moles

Using the molar masses of each element (Na: 22.99 g/mol, Al: 26.98 g/mol, F: 19.00 g/mol), convert the grams to moles: moles of Na = \(\frac{32.79\text{ g}}{22.99\text{ g/mol}}\) moles of Al = \(\frac{13.02\text{ g}}{26.98\text{ g/mol}}\) moles of F = \(\frac{54.19\text{ g}}{19.00\text{ g/mol}}\)

Step 3: Find the mole ratios

Divide each value by the smallest number of moles calculated to get the mole ratios: mole ratio of Na: \(\frac{\text{moles of Na}}{\text{smallest moles}}\) mole ratio of Al: \(\frac{\text{moles of Al}}{\text{smallest moles}}\) mole ratio of F: \(\frac{\text{moles of F}}{\text{smallest moles}}\)

Step 4: Write the empirical formula

In this case, the empirical formula is: \(Na_3AlF_6\)

Key Concepts

Mole Concept

Understanding the mole concept is fundamental in chemistry. It's a unit of measurement that allows chemists to count atoms, molecules, and other chemical entities by weighing them. A mole is defined as exactly 6.022 x 10²³ particles, which is Avogadro's number. This means that one mole of any substance contains the same number of particles as there are atoms in exactly 12 grams of carbon-12.

To put this into perspective, consider that every task in stoichiometry starts with converting mass to moles, using the formula:
\[ \text{moles} = \frac{\text{mass (g)}}{\text{molar mass (g/mol)}} \].
This equation is central to empirical formula calculations and enables the comparison of the amount of different substances through a common measure, moles.

Molar Mass

Molar mass is essentially the mass of one mole of a substance, typically expressed in grams per mole (g/mol). It's a property specific to each chemical element and compound, based on the atomic weights of its constituent atoms. For instance, the molar mass of carbon is 12.01 g/mol, which accounts for the mass of one mole of carbon atoms.

To calculate the molar mass of a compound, add up the atomic masses of all atoms in the molecular formula. This step is crucial in converting grams to moles, as we need to know how much one mole of a substance weighs to determine how many moles we have in a given sample. The concept of molar mass bridges the gap between the microscale (atoms and molecules) and the macroscale (grams and liters), enabling stoichiometric calculations in empirical formula determination.

Stoichiometry

Stoichiometry involves the quantification of reactants and products in a chemical reaction. It's based on the conservation of mass and the mole concept, with balanced chemical equations showing the mole ratios of reactants that combine and products that form. Therefore, stoichiometry allows us to predict the amounts of substances consumed and produced in a reaction.

When determining empirical formulas, stoichiometry comes into play after converting the mass percentage of each element to moles. Next, we find the simplest whole-number mole ratio of the elements to write the empirical formula, reflecting the basic proportion in which the atoms combine. This simplification represents the most reduced form of the compound's composition, which is a cornerstone in understanding chemical formulas.

Percent Composition

Percent composition is another vital concept in empirical formula calculation. It represents the percentage by mass of each element in a compound. We often start by assuming a 100-gram sample to simplify calculations, which directly translates percentages to grams. For instance, a compound with 30% carbon means that there are 30 grams of carbon per 100 grams of the compound.

The percent composition is crucial because it helps to determine the relative abundance of each element in a compound. Knowing the mass percentage, we can calculate the number of moles and thereafter, the empirical formula by dividing the individual elements' mole quantities by the smallest number of moles obtained within the compound. Accurately determining the percent composition of elements is the beginning point for empirical formula calculations and plays an integral role in quantitative analysis.