Question

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

Short answer

The empirical formulas for the given compounds are: (a) \(\mathrm{CSCl_{3}}\) (b) \(\mathrm{CF_{2}O}\) (c) \(\mathrm{Na_{3}AlF_{6}}\)

Step-by-step solution

Conversion to grams

Assume a 100 g sample, so we have 10.4 g of \(\mathrm{C}\), 27.8 g of \(\mathrm{S}\), and 61.7 g of \(\mathrm{Cl}\).

Conversion to moles

With the molar masses of \(\mathrm{C}\), \(\mathrm{S}\), and \(\mathrm{Cl}\) are 12.01 g/mol, 32.07 g/mol, and 35.45 g/mol, respectively. We calculate the number of moles for each element: \(n_\mathrm{C} = \frac{10.4}{12.01}\) \(n_\mathrm{S} = \frac{27.8}{32.07}\) \(n_\mathrm{Cl} = \frac{61.7}{35.45}\)

Mole Ratios

Calculate the mole ratios by dividing each number of moles by the smallest one among them: r = \(\frac{n_\mathrm{C}}{\min(n_\mathrm{C}, n_\mathrm{S}, n_\mathrm{Cl})} : \frac{n_\mathrm{S}}{\min(n_\mathrm{C}, n_\mathrm{S}, n_\mathrm{Cl})} : \frac{n_\mathrm{Cl}}{\min(n_\mathrm{C}, n_\mathrm{S}, n_\mathrm{Cl})}\)

Empirical Formula

Find the simplest whole number ratio to determine the empirical formula. In this case, it is \(\mathrm{CSCl_{3}}\). (b) 21.7 % \(\mathrm{C}\), 9.6 % \(\mathrm{O}\), and 68.7 % \(\mathrm{F}\) Follow the same steps as in (a) to find the empirical formula:

Conversion to grams

Assume a 100 g sample, so we have 21.7 g of \(\mathrm{C}\), 9.6 g of \(\mathrm{O}\), and 68.7 g of \(\mathrm{F}\).

Conversion to moles

With the molar masses of \(\mathrm{C}\), \(\mathrm{O}\), and \(\mathrm{F}\) are 12.01 g/mol, 16.00 g/mol, and 19.00 g/mol, respectively. We calculate the number of moles for each element: \(n_\mathrm{C} = \frac{21.7}{12.01}\) \(n_\mathrm{O} = \frac{9.6}{16.00}\) \(n_\mathrm{F} = \frac{68.7}{19.00}\)

Mole Ratios

Calculate the mole ratios as in (a): r = \(\frac{n_\mathrm{C}}{\min(n_\mathrm{C}, n_\mathrm{O}, n_\mathrm{F})} : \frac{n_\mathrm{O}}{\min(n_\mathrm{C}, n_\mathrm{O}, n_\mathrm{F})} : \frac{n_\mathrm{F}}{\min(n_\mathrm{C}, n_\mathrm{O}, n_\mathrm{F})}\)

Empirical Formula

Find the simplest whole number ratio to determine the empirical formula. In this case, it is \(\mathrm{CF_{2}O}\). (c) 32.79 % \(\mathrm{Na}\), 13.02 % \(\mathrm{Al}\), and the remainder \(\mathrm{F}\) Follow the same steps as in (a) and (b) to find the empirical formula:

Conversion to grams

Assume a 100 g sample, so we have 32.79 g of \(\mathrm{Na}\), 13.02 g of \(\mathrm{Al}\), and the remainder 54.19 g of \(\mathrm{F}\).

Conversion to moles

With the molar masses of \(\mathrm{Na}\), \(\mathrm{Al}\), and \(\mathrm{F}\) are 22.99 g/mol, 26.98 g/mol, and 19.00 g/mol, respectively. We calculate the number of moles for each element: \(n_\mathrm{Na} = \frac{32.79}{22.99}\) \(n_\mathrm{Al} = \frac{13.02}{26.98}\) \(n_\mathrm{F} = \frac{54.19}{19.00}\)

Mole Ratios

Calculate the mole ratios as in (a) and (b): r = \(\frac{n_\mathrm{Na}}{\min(n_\mathrm{Na}, n_\mathrm{Al}, n_\mathrm{F})} : \frac{n_\mathrm{Al}}{\min(n_\mathrm{Na}, n_\mathrm{Al}, n_\mathrm{F})} : \frac{n_\mathrm{F}}{\min(n_\mathrm{Na}, n_\mathrm{Al}, n_\mathrm{F})}\)

Empirical Formula

Find the simplest whole number ratio to determine the empirical formula. In this case, it is \(\mathrm{Na_{3}AlF_{6}}\). The empirical formulas for the given compounds are: (a) \(\mathrm{CSCl_{3}}\) (b) \(\mathrm{CF_{2}O}\) (c) \(\mathrm{Na_{3}AlF_{6}}\)

Key concepts

Mole Concept

Understanding the mole concept is essential for unraveling the intricate world of chemistry. A mole is simply a unit used to express amounts of a chemical substance. Think of it as the equivalent of a 'dozen' for eggs, but instead for atoms and molecules. One mole contains exactly 6.022 x 10²³ particles (Avogadro's number) - be it atoms, molecules, or ions.

When dealing with empirical formulas, the mole concept allows us to count atoms in large quantities by weighing them. We take the percent composition by mass and convert these to grams as if we had a 100 g sample. This simplifies calculations because percentage becomes directly equal to the mass in grams. The next step, as shown in the exercise solutions, is to convert this mass to moles using the molar mass of each element. Mole ratios are then used to find the empirical formula of a compound, which reflects the simplest whole-number ratio of atoms in the molecule.

Molar Mass

The molar mass of a substance is the weight in grams of one mole of that substance. It's typically expressed in units of grams per mole (g/mol). To calculate molar mass, one must sum the masses of all the atoms in a molecule. The periodic table becomes a critical tool here as it provides the atomic weights of each element, which is essentially the molar mass of each atom.

For example, the atomic weight of Carbon (C) is approximately 12.01 g/mol and that of Chlorine (Cl) is approximately 35.45 g/mol. These values are pivotal in calculating the number of moles of each element in a substance when given the mass. From the step-by-step solution provided, it's clear that accurately determining molar mass is crucial for converting grams into moles, a necessary conversion for finding empirical formulas.

Stoichiometry

Stoichiometry is the calculation of reactants and products in chemical reactions in chemistry. It can be thought of as the 'mathematics of chemistry' and relies heavily on the mole concept and molar masses. Stoichiometry allows us to predict the quantities of substances consumed and produced in a given reaction.

In the context of determining empirical formulas, stoichiometry involves the mole ratios of the elements within a compound. These ratios are obtained by dividing the moles of each element by the smallest number of moles observed in the step-by-step solutions. The result gives us the simplest ratio or proportions in which the atoms combine to form the compound. This fundamental stoichiometric principle sheds light on the quantitative relationships within chemical formulas and reactions, making it a cornerstone of chemical equation balancing and compound analysis.