Problem 24 - Chemistry: Atoms First

Question

Consider this question: What is the mass of the solute in 0.500 L of 0.30 M glucose, \(C_{6} \mathrm{H}_{12} \mathrm{O}_{6}\), used for intravenous injection? (a) Outline the steps necessary to answer the question. (b) Answer the question.

Step-by-Step Solution

Answer

The mass of the solute (glucose) is 27.024 grams.

Step 1: Identify the Given Information

The exercise provides the volume of the solution, which is 0.500 liters, and the molarity of the glucose solution, which is 0.30 M. Molarity (M) is defined as the number of moles of solute per liter of solution.

Step 2: Calculate the Number of Moles of Glucose

To find the number of moles of glucose, use the molarity formula which is Molarity = Moles of solute / Volume of solution (in liters). Multiply the volume of the solution by the molarity to get the number of moles: Moles of glucose = 0.500 L * 0.30 M.

Step 3: Calculate the Mass of Glucose

Once we have the number of moles of glucose, we can convert it to grams using the molar mass of glucose. The molar mass of glucose (C_{6}H_{12}O_{6}) is approximately 180.16 g/mol. Mass of glucose = Moles of glucose * Molar mass of glucose.

Step 4: Write the Final Answer

Calculate the mass of glucose using the number of moles obtained in Step 2 and the molar mass from Step 3 to find the mass of glucose in grams.

Key Concepts

Molar Mass of Glucose

Understanding the molar mass of a compound is a fundamental concept in chemistry that connects the microscopic world of atoms and molecules to the macroscopic world we can measure. For glucose, which has the chemical formula \( C_{6}H_{12}O_{6} \), the molar mass is the sum of the atomic masses of all the atoms in one molecule of glucose.

Each carbon (\(C\)) atom has an atomic mass of approximately 12.01 g/mol, hydrogen (\(H\)) has about 1.01 g/mol, and oxygen (\(O\)) is roughly 16.00 g/mol. As glucose comprises six carbon atoms, twelve hydrogen atoms, and six oxygen atoms, calculating its molar mass involves multiplying the atomic mass of each element by the number of atoms it has in the molecule and adding all these values together: \[Molar\, mass\, of\, glucose = (6 \times 12.01 \, g/mol) + (12 \times 1.01 \, g/mol) + (6 \times 16.00 \, g/mol) = 180.16 \, g/mol.\]

The molar mass serves as a bridge in various calculations, such as converting between moles and grams, which is essential for molarity calculations and stoichiometry.

Concentration of Solutions

In chemistry, the concentration of a solution is a measure of the amount of solute that is dissolved in a given quantity of solvent. Molarity, often represented by the letter \(M\), is one of the most common units used to express concentration. Molarity is defined as the number of moles of solute per liter of solution.

For accurate and effective experiments or medical applications such as intravenous injections, it's critical to prepare solutions with the correct molarity. Ensuring the proper concentration impacts the chemical properties and reactions involved. The formula to calculate molarity is: \[Molarity = \frac{Moles\, of\, solute}{Volume\, of\, solution\, (in\, liters)}\]

Using the molarity of a solution, we can determine the mass of the solute that needs to be used, as demonstrated in the textbook exercise. When dealing with concentrations, it is also vital to ensure the proper units are used to avoid calculation errors and to achieve precise results.

Stoichiometry

Stoichiometry is the section of chemistry that relates the quantities of substances in chemical reactions. It is based on the conservation of mass and the concept of the mole. Stoichiometry involves calculations that use the relationships between reactants and or/products in a chemical reaction to determine the quantities of substances involved.

Molarity plays a key role in stoichiometry when dealing with solutions. In the exercise example, once we calculated the number of moles of glucose needed, we used stoichiometry to convert those moles to grams, using the molar mass of glucose.

Key Stoichiometric Steps

• Determine what information is provided and what is being solved for.
• Use molarity calculations to find the number of moles involved if the problem involves solutions.
• Apply the mole concept and the molar mass of substances to interchange between mass and moles, as necessary.
• Use balanced chemical equations to find the stoichiometry of the reaction, and calculate the mass or volume of reactants and products as required.

Overall, understanding molarity and molar masses are critical for stoichiometric calculations, which in turn, are essential for practical applications in both laboratory settings and industry.