Calculate Grams Of Glucose In 1.5 M Solution

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Hey guys! Ever found yourself scratching your head over a molarity problem? Don't worry, it happens to the best of us. Today, we're going to break down a common chemistry calculation: figuring out how many grams of glucose are in a 1.5 M solution with a total volume of 1500 mL. We'll walk through each step, making sure you understand the logic behind it all. So, grab your calculators, and let's dive in!

Understanding the Problem

First, let's make sure we understand the problem. We need to determine the mass of glucose (in grams) present in a solution. This solution has a molar concentration of 1.5 M (molar) and a volume of 1500 mL. The key to solving this problem lies in understanding the relationship between molarity, moles, and mass, and that is what we will cover in this guide. The formula is your friend in these situations. Don't be intimidated! Let's break down these concepts first.

Molarity: The Concentration Key

Molarity, symbolized by 'M', is a measure of the concentration of a solution. It tells us how many moles of solute (in this case, glucose) are dissolved in one liter of solution. A 1.5 M solution means there are 1.5 moles of glucose per liter of solution. Thinking about molarity as a key to unlocking the relationship between moles and volume is helpful. Remember, 1 M is equal to 1 mole of solute per liter of solution (1 mol/L).

Moles: Counting Molecules

A mole is a unit of measurement used in chemistry to express amounts of a chemical substance. It's like a chemist's dozen! One mole contains Avogadro's number (approximately 6.022 x 10^23) of particles (atoms, molecules, ions, etc.). The mole is the bridge between the microscopic world of atoms and molecules and the macroscopic world we can measure in the lab. Converting between grams and moles is a fundamental skill in chemistry.

Grams: Mass in the Real World

Grams (g) are a unit of mass, which is a measure of how much matter an object contains. In this problem, we want to find the mass of glucose in grams. We'll use the molar mass of glucose to convert from moles to grams. The molar mass is the mass of one mole of a substance, and it's usually expressed in grams per mole (g/mol).

Step 1: Finding the Molar Mass of Glucose (C6H12O6)

The first step in solving our problem is to determine the molar mass of glucose. Glucose has the chemical formula C6H12O6, which tells us it has 6 carbon atoms, 12 hydrogen atoms, and 6 oxygen atoms. To calculate the molar mass, we need the atomic masses of each element from the periodic table:

  • Carbon (C): 12.01 g/mol
  • Hydrogen (H): 1.01 g/mol
  • Oxygen (O): 16.00 g/mol

Now, we multiply the atomic mass of each element by the number of atoms in the glucose molecule and add them together:

(6 C atoms * 12.01 g/mol) + (12 H atoms * 1.01 g/mol) + (6 O atoms * 16.00 g/mol) = 72.06 g/mol + 12.12 g/mol + 96.00 g/mol = 180.18 g/mol

So, the molar mass of glucose is approximately 180.18 g/mol. This means that one mole of glucose weighs 180.18 grams. Knowing the molar mass is crucial for converting between moles and grams.

Step 2: Converting Volume from Milliliters (mL) to Liters (L)

Molarity is defined as moles per liter (mol/L), so we need to convert the volume of our solution from milliliters (mL) to liters (L). There are 1000 mL in 1 L, so we can use the following conversion:

1500 mL * (1 L / 1000 mL) = 1.5 L

Therefore, our solution has a volume of 1.5 liters. It's essential to use consistent units in your calculations to avoid errors.

Step 3: Calculating Moles of Glucose

Now that we have the molarity (1.5 M) and the volume in liters (1.5 L), we can calculate the number of moles of glucose in the solution. We'll use the definition of molarity:

Molarity (M) = Moles of solute / Liters of solution

We can rearrange this formula to solve for moles:

Moles of solute = Molarity (M) * Liters of solution

Plugging in our values:

Moles of glucose = 1.5 M * 1.5 L = 2.25 moles

So, there are 2.25 moles of glucose in the solution. We're getting closer to our final answer! This step is a direct application of the definition of molarity.

Step 4: Converting Moles of Glucose to Grams

Finally, we can convert the moles of glucose to grams using the molar mass we calculated earlier (180.18 g/mol). We'll use the following formula:

Mass (grams) = Moles * Molar mass

Plugging in our values:

Mass of glucose = 2.25 moles * 180.18 g/mol = 405.405 grams

Rounding to a reasonable number of significant figures (let's say three), we get approximately 405 grams of glucose.

Final Answer

Therefore, there are approximately 405 grams of glucose in a 1.5 M solution with a total volume of 1500 mL. We've done it! We successfully calculated the mass of glucose using molarity, volume, and molar mass.

Key Takeaways

  • Understanding the definitions of molarity, moles, and molar mass is crucial for solving these types of problems.
  • Converting units to be consistent (e.g., mL to L) is essential for accurate calculations.
  • The formula Molarity = Moles / Liters is a cornerstone of solution chemistry.
  • The molar mass acts as a conversion factor between moles and grams.

Practice Makes Perfect

The best way to master these calculations is to practice! Try working through similar problems with different solutes, molarities, and volumes. You can even make up your own problems to challenge yourself. Remember, chemistry is like a puzzle, and each piece of information helps you solve it.

So there you have it, guys! We've successfully tackled a molarity problem and learned how to calculate the mass of glucose in a solution. I hope this step-by-step guide has been helpful. Keep practicing, and you'll become a chemistry whiz in no time!

Further Exploration

If you're interested in learning more about solution chemistry, here are some topics you might want to explore:

  • Dilution calculations
  • Preparing solutions of specific concentrations
  • Titration
  • Solution stoichiometry

These topics build upon the concepts we've discussed here and will further enhance your understanding of chemistry.

Remember, chemistry can be challenging, but it's also fascinating. Keep asking questions, keep exploring, and most importantly, keep having fun with it!