Problem 108 - Introductory Chemistry Atoms First

Question

How many milliliters of a \(0.250 \mathrm{M}\) solution of glucose, \(\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}\), are required to obtain \(100.0 \mathrm{~g}\) of glucose?

Step-by-Step Solution

Answer

To obtain 100.0 g of glucose, 2220 mL of a 0.250 M glucose solution is required.

Step 1: Calculate the molecular weight of glucose

First, we need to find the molecular weight of glucose. The formula of glucose is C6H12O6, and the atomic weights of carbon, hydrogen, and oxygen are 12.01 g/mol, 1.01 g/mol, and 16.00 g/mol, respectively. Molecular weight of glucose = (6 × 12.01) + (12 × 1.01) + (6 × 16.00) = 72.06 + 12.12 + 96.00 = 180.18 g/mol

Step 2: Convert grams of glucose into moles

Now, we need to convert the given mass of glucose (100.0 g) into moles. To do this, we will use the molecular weight of glucose (180.18 g/mol) that we calculated in the previous step. Moles of glucose = 100.0 g / 180.18 g/mol = 0.555 moles

Step 3: Find the volume of the solution needed

Now that we have the moles of glucose, we can use the given molarity and the formula for molarity to find the volume of the solution in liters. Molarity formula: M = moles of solute / volume of solution in liters 0.250 M = 0.555 moles / volume of solution in liters Volume of solution in liters = 0.555 moles / 0.250 M = 2.22 L

Step 4: Convert the volume from liters to milliliters

Finally, we need to convert the volume of the solution from liters to milliliters. 1 L = 1000 mL Volume of solution in milliliters = 2.22 L × 1000 mL/L = 2220 mL So, 2220 mL of a 0.250 M solution of glucose is required to obtain 100.0 g of glucose.

Key Concepts

Molecular Weight

The molecular weight of a compound tells us the weight of one mole of that compound. It is essential for converting between grams and moles, a fundamental concept in chemistry. Glucose, with the formula \( \text{C}_6\text{H}_{12}\text{O}_6 \), is composed of carbon, hydrogen, and oxygen atoms.
To find the molecular weight of glucose:

• Multiply the number of carbon atoms (6) by the atomic weight of carbon (12.01 g/mol).
• Multiply the number of hydrogen atoms (12) by the atomic weight of hydrogen (1.01 g/mol).
• Multiply the number of oxygen atoms (6) by the atomic weight of oxygen (16.00 g/mol).

Add these values together to get the molecular weight:
\((6 \times 12.01) + (12 \times 1.01) + (6 \times 16.00) = 180.18 \, \text{g/mol}. \)
This value is the bridge that lets us move from the mass of glucose to its quantity in moles.

Moles Conversion

Moles are a way of expressing the amount of a chemical substance. When dealing with chemical reactions, we often convert between mass and moles to know how much of each substance is involved. Using the molecular weight, we can convert grams into moles through division.
To convert 100.0 grams of glucose into moles:

• Use the formula: \( \text{Moles of glucose} = \frac{\text{mass in grams}}{\text{molecular weight}} \).
• Substitute the values: \( \text{Moles of glucose} = \frac{100.0 \, \text{g}}{180.18 \, \text{g/mol}} \).
• The result is approximately \( 0.555 \) moles.

This conversion allows us to work with the quantity of the substance in a way that is independent of its molecular structure.

Molarity Formula

Molarity is the concentration of a solution expressed as the number of moles of solute per liter of solution. It's a key concept in preparing solutions and analyzing reactions. The formula for calculating molarity is:

• \( M = \frac{\text{moles of solute}}{\text{volume of solution in liters}} \).

In the glucose solution question, we need to find how much glucose is dissolved in a given volume. We establish a direct relationship using:

• The moles of glucose (0.555 moles).
• The molarity of the solution (0.250 M).

Rearranging the formula gives:\[\text{Volume of solution in liters} = \frac{\text{moles of solute}}{\text{Molarity}} = \frac{0.555}{0.250} = 2.22 \, \text{liters}\]
This approach directly provides the volume needed in the simplest manner possible.

Volume Conversion

Volume conversion is often necessary when dealing with liquid solutions, as we typically measure liquids in milliliters rather than liters. Knowing the conversion factor is crucial: 1 liter is equivalent to 1000 milliliters.
In the exercise, once the solution's volume has been calculated in liters (2.22 L), convert it to milliliters using a simple multiplication:

• \( \text{Volume in milliliters} = 2.22 \, \text{liters} \times 1000 \, \text{mL/L} \).
• The result is 2220 mL.

By applying this conversion, we translate the volume into a unit that is often more practical for laboratory and real-life applications.