Question

Fluoxymesterone, \(\mathrm{C}_{20} \mathrm{H}_{29} \mathrm{FO}_{3}\), is an anabolic steroid. A \(500 \mathrm{~mL}\) solution is prepared by dissolving \(10.0 \mathrm{mg}\) of the steroid in water, \(1.0 \mathrm{~mL}\) portion of this solution is diluted to a final volume of \(1.00 \mathrm{~L}\). What is the resulting molarity ? (a) \(1.19 \times 10^{-10}\) (b) \(1.19 \times 10^{-7}\) (c) \(5.95 \times 10^{-8}\) (d) \(2.38 \times 10^{-11}\)

Short answer

The resulting molarity after dilution is \(5.54 \times 10^{-8} \text{ M}\), which corresponds to answer (c) \(5.95 \times 10^{-8}\).

Step-by-step solution

- Calculate the molarity of the original solution

First, convert the mass of Fluoxymesterone into moles using its molar mass. The molar mass of Fluoxymesterone (C20H29FO3) can be calculated by adding the molar masses of all the atoms in the compound: Molar mass of C20H29FO3 = (20 \times 12.01) + (29 \times 1.008) + (16.00) + (3 \times 16.00) = 360.44 g/mol.Next, convert the 10.0 mg of the steroid to grams (10 mg = 0.010 g) and then to moles: Number of moles = \(\frac{0.010 \text{ g}}{360.44 \text{ g/mol}} = 2.77 \times 10^{-5} \text{ moles}\).Now, calculate the molarity of the solution by dividing the number of moles by the volume in liters (500 mL = 0.500 L): Molarity = \(\frac{2.77 \times 10^{-5} \text{ moles}}{0.500 \text{ L}} = 5.54 \times 10^{-5} \text{ M}\).

- Calculate the molarity after dilution

In the dilution process, the moles of Fluoxymesterone remain the same, but the volume is increased. After taking a 1.0 mL portion from the original solution, we can calculate the moles contained in this portion: Moles in 1.0 mL = \(5.54 \times 10^{-5} \text{ M} \times 0.001 \text{ L} = 5.54 \times 10^{-8} \text{ moles}\).Now, calculate the molarity after this portion is diluted to a final volume of 1.00 L (1000 mL): Molarity after dilution = \(\frac{5.54 \times 10^{-8} \text{ moles}}{1.00 \text{ L}} = 5.54 \times 10^{-8} \text{ M}\), which simplifies to 5.54 \times 10^{-8} M.

Key concepts

Molar Mass

Understanding molar mass is essential for a variety of chemical calculations, especially when dealing with chemical reactions and solution preparation. The molar mass of a substance is the weight in grams of one mole of that substance. It is the sum of the atomic weights of all the atoms in the molecule, with the atomic weights provided in the periodic table in atomic mass units (amu).

For example, the molar mass of Fluoxymesterone, which has the chemical formula \(\mathrm{C}_{20} \mathrm{H}_{29} \mathrm{FO}_{3}\), is calculated by adding together the molar masses of all its atoms. This includes carbon (C), hydrogen (H), fluorine (F), and oxygen (O). By multiplying the number of each type of atom by its respective atomic weight and then summing them, we obtain the compound's molar mass. This is crucial since knowing the molar mass allows us to convert from grams to moles, a fundamental step in calculating molarity. Moreover, the molar mass serves as a conversion factor and helps to understand the quantitative aspects of chemical equations.

Solution Dilution

Solution dilution refers to the process of reducing the concentration of a solute in a solution, usually by adding more solvent. The principle governing dilution is that the number of moles of solute present remains unchanged during the dilution process. Therefore, when you have a solution and you increase the volume by adding solvent, the concentration decreases.

Dilution can be simply expressed using the equation \( M1 \times V1 = M2 \times V2 \), where \( M1 \) and \( M2 \) represent the molarity before and after dilution, and \( V1 \) and \( V2 \) represent the volume before and after dilution, respectively. When diluting a solution, it is vital to ensure that the solute is uniformly mixed to achieve an even concentration throughout. Accurate calculations and proper technique are key to effective solution dilution which is vital in laboratory settings, industrial processes, and even in pharmaceutical preparations.

Moles to Grams Conversion

Converting moles to grams and vice versa is a fundamental skill in chemistry known as stoichiometry. This conversion hinges on the principle that a mole of any substance contains the same number of entities, Avogadro's number (\(6.022 \times 10^{23}\)), and the mass of a mole is equal to the substance's molar mass.

To convert moles to grams, you multiply the number of moles by the molar mass of the substance: \(\text{mass in grams} = \text{moles} \times \text{molar mass}\). Conversely, to convert grams to moles, you divide the mass of the substance by its molar mass: \(\text{moles} = \frac{\text{mass in grams}}{\text{molar mass}}\). These conversions are indispensable for calculating molarity, which describes the concentration of a solution by expressing the number of moles of solute per liter of solution. Through this basic conversion, stoichiometry enables chemists to measure out precise amounts of substances for reactions and solutions.