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) \(21.7 \% \mathrm{C}, 9.6 \% \mathrm{O}\), and \(68.7 \% \mathrm{~F}\) (c) \(32.79 \% \mathrm{Na}, 13.02 \% \mathrm{Al}\), and \(54.19 \% \mathrm{~F}\)

Short answer

The empirical formulas for the given compounds are: (a) CS2Cl2 (b) C3OF6 (c) Na3AlF6

Step-by-step solution

Convert mass percentages to grams

Since the mass percentages can be viewed as a direct measurement from 100 g of the compound, we can directly convert the mass percentages to grams. Assumptions: (a) 10.4 g of C, 27.8 g of S, and 61.7 g of Cl (b) 21.7 g of C, 9.6 g of O, and 68.7 g of F (c) 32.79 g of Na, 13.02 g of Al, and 54.19 g of F

Convert grams to moles

To convert grams into moles, we'll divide the amount of each element by their respective atomic weight. (a) Moles of C = 10.4 g / 12.01 g/mol = 0.866 moles Moles of S = 27.8 g / 32.07 g/mol = 0.867 moles Moles of Cl = 61.7 g / 35.45 g/mol = 1.74 moles (b) Moles of C = 21.7 g / 12.01 g/mol = 1.807 moles Moles of O = 9.6 g / 16.00 g/mol = 0.6 moles Moles of F = 68.7 g / 19.00 g/mol = 3.613 moles (c) Moles of Na = 32.79 g / 22.99 g/mol = 1.426 moles Moles of Al = 13.02 g / 26.98 g/mol = 0.4825 moles Moles of F = 54.19 g / 19.00 g/mol = 2.852 moles

Determine the ratio of moles

To find the simplest whole number ratio for each compound, we will divide each number of moles by the smallest number of moles in the compound and round to the nearest whole number. (a) Ratio of C:S:Cl = 0.866/0.866: 0.867/0.866: 1.74/0.866 = 1:1:2 (b) Ratio of C:O:F = 1.807/0.6 : 0.6/0.6 : 3.613/0.6 = 3:1:6 (c) Ratio of Na:Al:F = 1.426/0.4825 : 0.4825/0.4825 : 2.852/0.4825 = 3:1:6

Write the empirical formulas

Now that we have the ratios of the elements in each compound, we can write the empirical formulas. (a) CS_2Cl_2 (b) C_3O_1F_6 or C_3OF_6 (c) Na_3Al_1F_6 or Na_3AlF_6

Key concepts

Percent Composition by Mass

Understanding the percent composition by mass of a compound is essential for determining its empirical formula. The percent composition indicates the mass percentage of each element in the compound. It reflects how much of the compound's weight comes from each constituent element.

For instance, if a compound has a mass composition of 40% carbon and 60% oxygen, it suggests that 40 grams of carbon and 60 grams of oxygen would be present in a 100-gram sample of the compound. This knowledge allows chemists to calculate the number of moles of each element in a sample, which is the first step in determining the empirical formula.

Mole Concept

The mole is a fundamental concept in chemistry, serving as a bridge between the atomic world and the quantities we measure in the laboratory. One mole represents 6.022 x 10²³ entities (Avogadro's number) and is akin to a 'chemist's dozen'.

Using the mole concept, chemists convert the mass of an element to the number of moles by dividing the mass in grams by the element's atomic weight. This provides a way to compare different substances on a per-particle basis, enabling us to balance chemical equations and calculate the compositions of compounds.

Chemical Stoichiometry

Chemical stoichiometry deals with the relationships between the quantities of reactants and products in a chemical reaction. It is based on the conservation of mass and the mole concept.

To find the empirical formula of a compound, we apply stoichiometry by determining the mole ratio of the elements. Once the number of moles is known, we can find the simplest whole number ratio by dividing each mole value by the smallest number of moles present for those elements, creating a balanced and reduced ratio that reflects the empirical formula.

Atomic Weight

The atomic weight, or atomic mass, is the average mass of atoms of an element, measured in atomic mass units (amu). It accounts for the existence of isotopes and their relative abundances. The atomic weight is crucial in converting between grams and moles, as it tells us how many grams are in one mole of that element.

When determining the empirical formula, each element's mass percent is divided by its atomic weight to find the number of moles. This allows us to compare different elements on an equal footing, a pivotal step towards establishing the compound's empirical formula.

Molecular Formula

The molecular formula gives the actual number of atoms of each element in a molecule of a compound, which can be a multiple of the empirical formula. Sometimes, both formulas are the same, but if the compound's molar mass is known and it is higher than that calculated from the empirical formula, the molecular formula can be determined.

For example, if the empirical formula is CH₂O and the molar mass is known to be approximately 60 g/mol, the molecular formula would be C₂H₄O₂, which is twice the empirical formula mass. This gives chemists a complete understanding of the compound's composition.