Problem 46 - Chemistry: Principles and Reactions

Question

Determine the simplest formulas of the following compounds: (a) saccharin, the artificial sweetener, which has the composition \(45.90 \% \mathrm{C}, 2.75 \% \mathrm{H}, 26.20 \% \mathrm{O}, 17.50 \% \mathrm{~S}\), and \(7.65 \% \mathrm{~N}\) (b) allicin, the compound that gives garlic its characteristic odor, which has the composition \(6.21 \% \mathrm{H}, 44.4 \% \mathrm{C}, 9.86 \% \mathrm{O}\), and \(39.51 \% \mathrm{~S}\). (c) sodium thiosulfate, the fixer used in developing photographic film, which has the composition \(30.36 \% \mathrm{O}, 29.08 \% \mathrm{Na}\), and \(40.56 \% \mathrm{~S}\).

Step-by-Step Solution

Answer

Answer: The empirical formulas of the given compounds are (a) Saccharin: \(C_7H_5O_3SN\), (b) Allicin: \(C_6H_{10}O_2S_2\), and (c) Sodium thiosulfate: \(Na_2SO_3\).

Step 1: Convert percentages to moles

First, we need to calculate the moles of each element in saccharin. To do that, we can assume we have 100 g of the compound and then divide the mass of each element by their respective atomic masses: - Moles of C: \(\frac{45.90}{12.01} = 3.826\) - Moles of H: \(\frac{2.75}{1.008} = 2.731\) - Moles of O: \(\frac{26.20}{16.00} = 1.637\) - Moles of S: \(\frac{17.50}{32.07} = 0.545\) - Moles of N: \(\frac{7_65}{14.01} = 0.546\)

Step 2: Find the smallest whole number ratio of moles

Find the smallest whole number ratio by dividing each moles value by the smallest moles value obtained: - Ratio of C: \(\frac{3.826}{0.545} = 7.02 \approx 7\) - Ratio of H: \(\frac{2.731}{0.545} = 5.01 \approx 5\) - Ratio of O: \(\frac{1.637}{0.545} = 3.00 \approx 3\) - Ratio of S: \(\frac{0.545}{0.545} = 1.00 \approx 1\) - Ratio of N: \(\frac{0.546}{0.545} = 1.00 \approx 1\)

Step 3: Empirical Formula of Saccharin

The empirical formula of saccharin is \(C_7H_5O_3SN\). (b) Allicin:

Step 1: Convert percentages to moles

Calculate the moles of each element in allicin, assuming we have 100 g of the compound: - Moles of H: \(\frac{6.21}{1.008} = 6.162\) - Moles of C: \(\frac{44.4}{12.01} = 3.697\) - Moles of O: \(\frac{9.86}{16.00} = 0.616\) - Moles of S: \(\frac{39.51}{32.07} = 1.232\)

Step 2: Find the smallest whole number ratio of moles

Find the smallest whole number ratio by dividing each moles value by the smallest moles value obtained: - Ratio of H: \(\frac{6.162}{0.616} = 10.00 \approx 10\) - Ratio of C: \(\frac{3.697}{0.616} = 6.00 \approx 6\) - Ratio of O: \(\frac{0.616}{0.616} = 1.00 \approx 1\) - Ratio of S: \(\frac{1.232}{0.616} = 2.00 \approx 2\)

Step 6: Empirical Formula of Allicin

The empirical formula of allicin is \(C_6H_{10}O_2S_2\). (c) Sodium thiosulfate:

Step 1: Convert percentages to moles

Calculate the moles of each element in sodium thiosulfate, assuming we have 100 g of the compound: - Moles of O: \(\frac{30.36}{16.00} = 1.898\) - Moles of Na: \(\frac{29.08}{22.99} = 1.265\) - Moles of S: \(\frac{40.56}{32.07} = 1.265\)

Step 2: Find the smallest whole number ratio of moles

Find the smallest whole number ratio by dividing each moles value by the smallest moles value obtained: - Ratio of O: \(\frac{1.898}{1.265} = 1.50 \approx 2\) - Ratio of Na: \(\frac{1.265}{1.265} = 1.00 \approx 1\) - Ratio of S: \(\frac{1.265}{1.265} = 1.00 \approx 1\)

Step 9: Empirical Formula of Sodium Thiosulfate

The empirical formula of sodium thiosulfate is \(Na_2SO_3\).

Key Concepts

Mole-to-Atom Conversion

Understanding mole-to-atom conversion is crucial when encountering problems like the empirical formula determination. Since moles are a unit that measures the amount of substance, converting them to atoms aligns with the Avogadro's number (\(6.022 \times 10^{23}\) atoms per mole), which is the bridge between the microscopic world (atoms) and the macroscopic (grams and moles).

When calculating empirical formulas, you start with the percentage composition, then convert each percentage to moles using respective atomic masses. This is because chemical formulas are expressed in mole ratios, not grams or percentages. The mole concept allows us to count atoms by weighing them, thus providing a link between mass and number of particles.

Application in Empirical Formula Determination

In the given exercises, each percentage represents a mass portion in a 100 g sample. Dividing this mass by the atomic mass of the element provides you with the number of moles, acting as a stepping stone to the empirical formula. After calculating moles for each element, these values are compared to determine the smallest whole number ratio, which is itself a conversion from moles to the relative number of atoms—illustrating the essence of mole-to-atom conversion in empirical formula determination.

Percentage Composition

The percentage composition is a foundational concept in chemistry because it describes the fraction of each element within a compound by mass. To determine the empirical formula from percentage composition, the total composition of a compound is assumed to be 100%, and thus each percentage gives the mass of that particular element in 100 grams of the compound.

This method simplifies calculations since it is easier to work with a fixed sample size (100 g) and directly converts to moles. The percentages can be thought of as parts per hundred, directly translating to grams in this sample.

Significance in Empirical Formula Calculations

Knowing the percentage composition of each element leads directly to mole quantities once you divide by the atomic mass of each element, as seen in the exercise solutions. It is this transition from percentage to mole ratio that forms the empirical formula. This is because the empirical formula reflects the simplest whole number ratio of atoms in a compound, which is most effectively analyzed in terms of moles rather than mass percentage.

Chemical Stoichiometry

Chemical stoichiometry is the part of chemistry that refers to the quantitative relationships between reactants and products in a chemical reaction. It's based on the conservation of mass where the total mass of reactants equals the total mass of the products. In the context of determining empirical formulas, stoichiometry comes into play as it involves the calculation of the relative amounts of elements in a compound.

Moles are central to stoichiometry, bridging the gap between the mass of a substance and the chemical formulas. For empirical formula determination, stoichiometry dictates that once you have mole ratios, you can deduce the simplest whole number ratio of atoms, akin to a basic stoichiometric conversion from moles to atoms.

Role in Determining Formulas

The exercises provided illustrate stoichiometric principles by converting mass percentages to moles, then finding the ratio of moles that align with integer values, and subsequently arriving at the empirical formula. It’s an illustration of stoichiometry that does not involve a chemical reaction per se but uses the same principles to balance the relative numbers of atoms within a single compound.