A \(0.25 \mathrm{M}\) solution of lithium carbonate \(\left(\mathrm{Li}_{2} \mathrm{CO}_{3}\right)\), a drug used to treat manic depression, is prepared. (a) How many moles of \(\mathrm{Li}_{2} \mathrm{CO}_{3}\) are present in \(45.8 \mathrm{~mL}\) of the solution? (b) How many grams of \(\mathrm{Li}_{2} \mathrm{CO}_{3}\) are in \(750 \mathrm{~mL}\) of the same solution? (c) How many milliliters of the solution would be needed to supply \(6.0 \mathrm{~g}\) of the solute? (d) If the solution has a density of \(1.22 \mathrm{~g} / \mathrm{mL}\), what is its mass percent?

Understanding molarity is essential for analyzing and preparing chemical solutions. Molarity, denoted as M, is defined as the number of moles of a solute per liter of solution. It allows chemists and students alike to calculate how concentrated a solution is.

To find the molarity of a given solution, one should use the formula: \[ \text{Molarity (M)} = \frac{\text{moles of solute}}{\text{volume of solution in liters}} \]. In practical terms, if you have a volume of a solution and its concentration, calculating the number of moles involved becomes straightforward, and this serves as the foundation for many other calculations in chemistry.

For example, if a solution has a molarity of 0.25 M and you have 45.8 mL of it, converting this volume to liters (since molarity requires volume in liters) allows you to calculate the moles of solute by multiplying the molarity by the volume in liters. This step is critical for any subsequent calculations involving that solution, such as finding out how much solute is present in grams or how much solution is needed for a specific amount of the solute.

Once the number of moles in a substance is determined through molarity calculations, the next step is often to convert this quantity into grams—the unit more commonly used in laboratory settings. To perform this conversion, the molar mass of the substance is needed, which is the weight of one mole of a substance and is expressed in grams per mole (g/mol).

The conversion formula is: \[ \text{Mass (g)} = \text{moles} \times \text{molar mass (g/mol)} \]. For lithium carbonate (\( \text{Li}_2\text{CO}_3 \)), if its molar mass is 73.89 g/mol, and you have calculated the moles of lithium carbonate in a particular volume of solution using the molarity, then simply multiply the number of moles by 73.89 g/mol to obtain the mass in grams.

The mass percent of a solution is a way to express the concentration of a solute in a solution. It tells you how much solute is present in a given mass of the entire solution and is calculated by the formula: \[ \text{Mass percent} = \frac{\text{mass of solute}}{\text{total mass of solution}} \times 100\text{%} \].

To find the mass percent, you need the mass of the solute (which you can calculate using the molarity and the molar mass) and the total mass of the solution. In the case of a lithium carbonate solution with a density of 1.22 g/mL, this density value can be used to find the mass of one liter of the solution (since density is mass per unit volume). Once you have the mass of the solute and the total mass of the solution, apply the formula to get the mass percent. This information can be particularly useful in medicine, where specific concentrations of substances are crucial.

Stoichiometry is the section of chemistry that focuses on the quantitative relationships between the reactants and products in a chemical reaction. It involves using balanced chemical equations to calculate moles, masses, volumes, and energy changes.

Stoichiometry requires a balanced equation as the basis for calculations and requires that all mole ratios in the reaction are understood. This extends to molarity calculations when solutions react together. For instance, knowing the stoichiometry of a reaction allows you to predict how much of a reactant you need to completely react with another, or how much product you can expect to produce from certain amounts of reactants.

Understanding the stoichiometric coefficients in the equations, one can set up mole-to-mole comparisons, which serve as conversion factors. These factors then allow for the direct translation of moles from one substance involved in the reaction to moles of another, and consequently to grams or liters of solution, integrating the knowledge of molarity, mass percent, and mole-to-gram conversions in practical applications.