Calcula El Volumen De Acetato De Etilo: ¡Fácil!
Hey guys! Ever wondered how much space a certain number of molecules of ethyl acetate would take up? Well, buckle up, because today we're diving deep into the world of chemistry to figure out exactly that. We're talking about calculating the volume occupied by 5.32 x 10^22 molecules of ethyl acetate (C5H10O2), and trust me, it's not as intimidating as it sounds. We'll break it down step-by-step, making sure you understand every bit of it. So, grab your notebooks, maybe a cup of coffee, and let's get this chemical party started!
Understanding the Basics: Moles and Avogadro's Number
Before we jump into calculating the volume, we need to get a handle on a couple of fundamental chemistry concepts. First up, the mole. You've probably heard of it, right? Think of a mole as a chemist's dozen, but way, way bigger. It's a specific quantity of a substance that contains exactly 6.022 x 10^23 elementary entities (like atoms, molecules, or ions). This magical number is known as Avogadro's number. So, one mole of anything contains 6.022 x 10^23 of those things. It's a cornerstone of quantitative chemistry, allowing us to relate the microscopic world of atoms and molecules to the macroscopic world we can measure and observe.
Why is this so important for our problem? Well, we're given a number of molecules, not a mass or a volume directly. To bridge this gap and eventually find the volume, we first need to convert our number of molecules into moles. This is where Avogadro's number comes into play as our conversion factor. It's literally the bridge between counting individual molecules and thinking about substances in measurable units like grams or liters.
So, to recap, when you see a number of molecules and need to figure out its properties in terms of mass or volume, your first stop is usually converting those molecules into moles using Avogadro's number. It’s the universal language of chemistry for quantifying amounts of substances. And don't worry if the number 6.022 x 10^23 seems mind-bogglingly huge – it is! But it's also incredibly useful for making sense of the vast numbers of particles we're dealing with in chemical reactions and calculations. We'll be using this concept extensively, so make sure it’s clear in your mind. Ready for the next step? Let's move on!
Step 1: From Molecules to Moles
Alright guys, the first crucial step in our journey to find the volume occupied by ethyl acetate molecules is to convert the given number of molecules into moles. We've got 5.32 x 10^22 molecules of ethyl acetate (C5H10O2). Remember our superstar, Avogadro's number? It tells us that 1 mole of any substance contains 6.022 x 10^23 entities (in our case, molecules). So, to find out how many moles our given number of molecules represents, we'll use this relationship. It’s a simple division problem:
Number of Moles = (Number of Molecules) / (Avogadro's Number)
Plugging in our values:
Number of Moles = (5.32 x 10^22 molecules) / (6.022 x 10^23 molecules/mol)
Now, let's crunch those numbers. When you divide 5.32 by 6.022, you get approximately 0.8834. For the exponents, we have 10^22 divided by 10^23, which means we subtract the exponents: 22 - 23 = -1. So, 10^22 / 10^23 = 10^-1.
Putting it all together, we get:
Number of Moles ≈ 0.8834 x 10^-1 moles
Or, to express this in a more standard scientific notation:
Number of Moles ≈ 0.08834 moles
So, our starting quantity of 5.32 x 10^22 molecules of ethyl acetate is equivalent to about 0.08834 moles. This is a pretty sweet achievement because moles are a much more convenient unit to work with when we need to calculate mass or, ultimately, volume. See? We're already making progress! This conversion is key because it allows us to connect the number of particles to a measurable quantity that chemists use every day. It's like translating from a secret molecular language to a more practical one. Keep this number handy, as it's going to be our stepping stone for the next part of the calculation!
Step 2: Finding the Molar Mass of Ethyl Acetate
Alright team, we've successfully converted our molecules into moles. The next logical step is to figure out the mass of these moles. To do that, we need the molar mass of ethyl acetate (C5H10O2). The molar mass is essentially the mass of one mole of a substance, and it's expressed in grams per mole (g/mol). It's derived by summing up the atomic masses of all the atoms present in the chemical formula.
Ethyl acetate has the chemical formula C5H10O2. This means each molecule contains:
- 5 Carbon (C) atoms
- 10 Hydrogen (H) atoms
- 2 Oxygen (O) atoms
We'll need the atomic masses of these elements from the periodic table. Let's use approximate values for simplicity:
- Atomic mass of Carbon (C) ≈ 12.01 g/mol
- Atomic mass of Hydrogen (H) ≈ 1.01 g/mol
- Atomic mass of Oxygen (O) ≈ 16.00 g/mol
Now, let's calculate the molar mass of ethyl acetate:
Molar Mass (C5H10O2) = (Number of C atoms * Atomic mass of C) + (Number of H atoms * Atomic mass of H) + (Number of O atoms * Atomic mass of O)
Molar Mass = (5 * 12.01 g/mol) + (10 * 1.01 g/mol) + (2 * 16.00 g/mol)
Let's do the multiplication:
Molar Mass = (60.05 g/mol) + (10.10 g/mol) + (32.00 g/mol)
And now, the addition:
Molar Mass = 102.15 g/mol
So, the molar mass of ethyl acetate is approximately 102.15 grams per mole. This tells us that if we had a full mole (6.022 x 10^23 molecules) of ethyl acetate, it would weigh 102.15 grams. This is super handy because it gives us a direct way to convert between moles and mass. Knowing the molar mass is crucial for so many calculations in chemistry, including figuring out how much stuff you actually have in terms of weight. We're building momentum here, guys! This value will be key in our next calculation to find the total mass of ethyl acetate we're dealing with.
Step 3: Calculating the Total Mass of Ethyl Acetate
Fantastic work, everyone! We've figured out how many moles of ethyl acetate we have (approximately 0.08834 moles) and we've determined its molar mass (102.15 g/mol). Now, it's time to combine these two pieces of information to calculate the total mass of our ethyl acetate. This step is straightforward and uses the relationship between moles, molar mass, and mass:
Mass = Number of Moles × Molar Mass
Let's plug in the values we've found:
Mass = (0.08834 moles) × (102.15 g/mol)
Get ready for some calculation!
Mass ≈ 9.025 grams
So, the 5.32 x 10^22 molecules of ethyl acetate that we started with have a total mass of approximately 9.025 grams. It might not sound like a lot, but remember how incredibly tiny individual molecules are! This mass is comprised of billions upon billions of these little guys. This calculation is super important because it gives us a tangible weight for the amount of substance we're working with. It’s another key conversion that allows us to move from the abstract world of moles to something we can potentially measure on a scale. This mass value is the bridge to our final goal: calculating the volume. We're almost there, guys!
Step 4: Using Density to Find the Volume
We're at the final stretch, people! We’ve calculated the mass of our ethyl acetate, which is approximately 9.025 grams. Now, to find the volume it occupies, we need one more crucial piece of information: the density of ethyl acetate. The problem statement kindly provides this for us: the density of ethyl acetate is 902 kg/m³.
Density is defined as mass per unit volume. The formula is:
Density = Mass / Volume
We want to find the volume, so we can rearrange this formula to:
Volume = Mass / Density
Here’s a slight hitch: our mass is in grams (g), but our density is in kilograms per cubic meter (kg/m³). We need to make sure our units are consistent before we plug in the numbers. It's usually easiest to convert everything to a common set of units. Let's convert the mass from grams to kilograms:
- 1 kilogram (kg) = 1000 grams (g)
- So, 1 gram (g) = 0.001 kilograms (kg)
Our mass is 9.025 grams. Converting this to kilograms:
Mass in kg = 9.025 g × (1 kg / 1000 g)
Mass in kg = 0.009025 kg
Now that our mass is in kilograms, we can use the density as given:
Density = 902 kg/m³
Now we can calculate the volume:
Volume = Mass / Density
Volume = 0.009025 kg / 902 kg/m³
Let's do the division:
Volume ≈ 0.0000100055 m³
That's a very small number, and it's in cubic meters (m³). It's often more intuitive to express such small volumes in cubic centimeters (cm³) or milliliters (mL). Let's do that conversion:
- 1 meter (m) = 100 centimeters (cm)
- Therefore, 1 cubic meter (m³) = (100 cm)³ = 1,000,000 cm³
So, to convert our volume from m³ to cm³:
Volume in cm³ = 0.0000100055 m³ × (1,000,000 cm³/m³)
Volume in cm³ ≈ 10.0055 cm³
And since 1 cubic centimeter (cm³) is equal to 1 milliliter (mL), the volume is also approximately 10.0055 mL.
There you have it, guys! The 5.32 x 10^22 molecules of ethyl acetate occupy a volume of about 10.01 cubic centimeters (or milliliters). Pretty cool, right? We went from a count of molecules to a measurable volume using fundamental chemistry principles. This demonstrates the power of stoichiometry and density in relating the microscopic world to the macroscopic one we experience.
Conclusion: A Small Volume, a Big Concept
So, what have we learned today, amazing chemists? We’ve successfully calculated that 5.32 x 10^22 molecules of ethyl acetate will occupy approximately 10.01 cubic centimeters (or milliliters). This journey involved understanding key concepts like moles and Avogadro's number, calculating the molar mass of ethyl acetate, determining the total mass of the given molecules, and finally, using the density to convert that mass into a volume. It’s a fantastic example of how chemists piece together information to solve problems about the physical world around us.
Remember, even though individual molecules are incredibly tiny, a significant number of them can add up to a measurable quantity. The principles we used – stoichiometry, molar mass, and density – are foundational in chemistry and apply to countless other substances and calculations. Whether you're working in a lab, studying for an exam, or just curious about the world, these skills are invaluable. Keep practicing, keep asking questions, and never stop exploring the fascinating realm of chemistry. You guys totally rocked this!
