Problem 32 - Chemistry: Atoms First

Question

Calculate the mole fraction of each solute and solvent: (a) \(583 \mathrm{g}\) of \(\mathrm{H}_{2} \mathrm{SO}_{4}\) in \(1.50 \mathrm{kg}\) of water \(-\) the acid solution used in an automobile battery (b) \(0.86 \mathrm{g}\) of \(\mathrm{NaCl}\) in \(1.00 \times 10^{2} \mathrm{g}\) of water \(-\) a solution of sodium chloride for intravenous injection (c) \(46.85 \mathrm{g}\) of codeine, \(\mathrm{C}_{18} \mathrm{H}_{21} \mathrm{NO}_{3},\) in \(125.5 \mathrm{g}\) of ethanol, \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}\) (d) \(25 \mathrm{g}\) of \(\mathrm{I}_{2}\) in \(125 \mathrm{g}\) of ethanol, \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}\)

Step-by-Step Solution

Answer

For each solution, the mole fraction of each component is determined by dividing the number of moles of the solute by the total number of moles of both solute and solvent. The moles of each component are calculated using their respective molecular weights and the provided masses for each solution.

Step 1 - Calculate Moles of H2SO4

Firstly, calculate the number of moles of H2SO4 using its molecular weight. The molecular weight of H2SO4 is 98.079 g/mol. Use the formula: moles = mass (g) / molecular weight (g/mol). For H2SO4, it will be moles of H2SO4 = 583 g / 98.079 g/mol.

Step 2 - Calculate Moles of Water

Similarly, calculate the moles of water using its molecular weight (18.015 g/mol): moles of water = mass (g) / molecular weight (g/mol). With given mass 1.50 kg (or 1500 g), moles of water = 1500 g / 18.015 g/mol.

Step 3 - Calculate Mole Fraction of H2SO4

The mole fraction of H2SO4 is calculated by its moles divided by the total moles of the solution. Mole fraction of H2SO4 = moles H2SO4 / (moles H2SO4 + moles water).

Step 4 - Calculate Mole Fraction of Water

The mole fraction of water is similarly calculated by moles of water divided by the total moles of the solution. Mole fraction of water = moles water / (moles H2SO4 + moles water).

Step 5 - Repeat Calculations for NaCl Solution

Repeat steps 1 to 4 for NaCl solution using 0.86 g of NaCl and 100 g of water. Use the molecular weight of NaCl (58.44 g/mol) and of water (18.015 g/mol).

Step 6 - Repeat Calculations for Codeine Solution

Repeat steps 1 to 4 for the codeine solution using 46.85 g of codeine and 125.5 g of ethanol. The molecular weight of codeine (C18H21NO3) is 299.364 g/mol and ethanol (C2H5OH) is 46.068 g/mol.

Step 7 - Repeat Calculations for I2 Solution

Lastly, repeat steps 1 to 4 for the I2 solution using 25 g of I2 and 125 g of ethanol. Use the molecular weight of I2 (253.80894 g/mol) and ethanol (46.068 g/mol).

Key Concepts

Solution Concentration

When we talk about solution concentration, we are essentially discussing how much solute is present in a given amount of solvent. The concentration of a solution can be expressed in many ways, such as molarity, molality, weight percent, and mole fraction.

One particularly useful concentration measure is the mole fraction, which is the ratio of the number of moles of a component to the total number of moles of all components in the solution. It is a unitless quantity, making it versatile for various applications, such as predicting the boiling point elevation or freezing point depression.

To calculate the mole fraction, you need to know the number of moles of each constituent in the solution. This straightforward ratio can provide a quick insight into the proportion of the components within a mixture, which is essential for precisely preparing solutions in both laboratory and industrial settings.

Molecular Weight

The molecular weight, also known as the molecular mass, is the sum of the atomic weights of the atoms in a molecule. It's usually expressed in atomic mass units (amu) or grams per mole (g/mol).

Knowing the molecular weight is crucial because it allows us to convert between the mass of a substance and the number of moles, which is a fundamental step in stoichiometric calculations. For example, if we have a compound like sodium chloride (NaCl), which has a molecular weight of 58.44 g/mol, this tells us that one mole of NaCl weighs 58.44 grams.

For complex molecules like codeine, with the formula \( \mathrm{C}_{18} \mathrm{H}_{21} \mathrm{NO}_{3} \) , summing the atomic weights of each atom present gives its molecular weight. It is this value that allows scientists and students to carry out precise chemical quantifications and prepare solutions with specific concentrations.

Chemical Molarity

When scientists speak of chemical molarity, they're referring to a very common unit of concentration, useful for a wide range of chemical reactions and solutions. Molarity is defined as the number of moles of solute per liter of solution.

It's given the symbol \( M \) and is calculated using the formula \( M = \frac{\text{moles of solute}}{\text{liters of solution}} \). This measurement is significant because it allows chemists to predict how substances in a solution will react with one another, based on the number of molecules present rather than their mass.

For accurate experimentation and reaction predictions, understanding and being able to calculate molarity is key. It ensures the correct proportions of reactants in chemical reactions, which is essential for achieving the desired product yield and efficacy.

Stoichiometry

Stoichiometry is the field of chemistry that involves the quantitative relationships between the reactants and products in a chemical reaction. It's based on the law of conservation of mass and the concept that elements combine in fixed ratios of whole numbers.

This branch of chemistry uses balanced chemical equations to calculate the amounts of reactants needed or products formed. Getting familiar with stoichiometry is crucial for anyone in the field of chemistry as it's applied everywhere, from the lab bench to industrial processes.

In the context of solution preparation, stoichiometry is used to determine the precise amount of a substance needed to react completely with another substance, as seen in mole fraction calculations. For example, the stoichiometry of a reaction would dictate how much \( \mathrm{H}_{2} \mathrm{SO}_{4} \) is required to neutralize a specific amount of NaOH. These relationships are foundational for mastering chemistry and essential for any aspiring scientist or student.