Molecular Formula Calculation: Organic Compound Analysis
Hey guys! Ever wondered how chemists figure out the exact recipe, or molecular formula, of a mysterious organic compound? Let's dive into a fascinating problem where we'll crack the code of an organic substance made of carbon, hydrogen, and oxygen. This isn't just about crunching numbers; it's about understanding the fundamental principles that govern the composition of molecules. We'll break down the steps, making it super clear and easy to follow, so you'll feel like a pro in no time! So, grab your thinking caps, and let's get started on this exciting chemical journey!
Understanding the Problem: The Organic Mystery
Okay, so here’s the puzzle: We have an organic compound that's 40% carbon, 6.6% hydrogen, and the rest is oxygen. Think of it like a pie chart – those percentages tell us the relative amounts of each element in the compound. Now, we also know that when we dissolve 15 grams of this stuff in 25 grams of water, the freezing point drops to -6.2 °C. That freezing point depression is a crucial clue! It's like a chemical detective giving us a wink. We’ve also got the atomic weights (Ar) of carbon (C = 12), hydrogen (H = 1), and oxygen (O = 16), which are our essential tools for converting percentages into actual element ratios. The ultimate goal? To figure out the molecular formula, which is like the compound's secret identity card – it tells us exactly how many atoms of each element are in one molecule. Is it , or something else entirely? That's what we're here to discover!
Breaking Down the Composition
The first step in solving this molecular mystery is to figure out the empirical formula. This is the simplest whole-number ratio of atoms in the compound. Think of it as the basic building block ratio before we figure out the full structure. We know the percentages of each element: 40% carbon, 6.6% hydrogen, and the rest is oxygen. To find the percentage of oxygen, we subtract the percentages of carbon and hydrogen from 100%: 100% - 40% - 6.6% = 53.4% oxygen. Now, we can pretend we have 100 grams of the compound, which makes these percentages easy to convert to grams: 40 grams of carbon, 6.6 grams of hydrogen, and 53.4 grams of oxygen. The magic now happens when we convert these masses into moles – the chemist's favorite unit for counting atoms and molecules. We divide the mass of each element by its atomic weight (Ar): For carbon: 40 g / 12 g/mol = 3.33 mol. For hydrogen: 6.6 g / 1 g/mol = 6.6 mol. For oxygen: 53.4 g / 16 g/mol = 3.34 mol. Now we have the molar ratios, but they're not quite whole numbers yet. To get there, we divide each mole value by the smallest one (3.33 mol): Carbon: 3.33 mol / 3.33 mol = 1. Hydrogen: 6.6 mol / 3.33 mol ≈ 2. Oxygen: 3.34 mol / 3.33 mol ≈ 1. So, the simplest ratio of carbon to hydrogen to oxygen is 1:2:1. That gives us our empirical formula: . We're one big step closer to cracking the code!
Freezing Point Depression: A Chilling Clue
Now comes the really cool part – using the freezing point depression to figure out the molecular weight of our compound! You see, when you dissolve something in water, it messes with the water's freezing point. It's like adding salt to icy roads in winter; it lowers the freezing point and helps melt the ice. The amount the freezing point drops depends on how much "stuff" (in terms of moles) you dissolve. This is described by the formula: ΔTf = Kf * m * i. Let’s break it down: ΔTf is the freezing point depression, which is the difference between the freezing point of pure water (0 °C) and the freezing point of the solution (-6.2 °C in our case). So, ΔTf = 0 °C - (-6.2 °C) = 6.2 °C. Kf is the freezing point depression constant for water, which is a known value: 1.86 °C kg/mol. m is the molality of the solution, which is the number of moles of solute (our organic compound) per kilogram of solvent (water). And i is the van't Hoff factor, which tells us how many particles one molecule of solute breaks into when dissolved. For organic compounds that don't ionize (break into ions) in water, i is usually 1. Now we can plug in the values and solve for molality (m): 6.2 °C = 1.86 °C kg/mol * m * 1. Solving for m gives us: m = 6.2 °C / (1.86 °C kg/mol) = 3.33 mol/kg. Molality is moles of solute per kilogram of solvent, we have 15 grams of solute in 25 grams of water, so that gives us the molality. And boom! With this freezing point wizardry, we’re unlocking the compound's secrets!
Calculating Molecular Weight and Finding the Formula
Alright, let's put those molality skills to work! We've figured out that the molality (m) of our solution is 3.33 mol/kg. Remember, molality is moles of solute per kilogram of solvent. We dissolved 15 grams of our mystery compound in 25 grams (0.025 kg) of water. We can use this information to find the number of moles of the compound: Moles = molality * kilograms of solvent = 3.33 mol/kg * 0.025 kg = 0.0833 moles. Now, we know the mass of the compound (15 grams) and the number of moles (0.0833 moles), so we can calculate the molecular weight (MW): MW = mass / moles = 15 g / 0.0833 mol = 180 g/mol (approximately). The molecular weight is like the total weight of all the atoms in one molecule of our compound. We're on the home stretch now! We already found the empirical formula to be . To find the molecular formula, we need to compare the empirical formula weight with the molecular weight. The empirical formula weight is the sum of the atomic weights of the atoms in the empirical formula: 1 carbon (12) + 2 hydrogens (2 * 1) + 1 oxygen (16) = 30 g/mol. Now, we divide the molecular weight by the empirical formula weight: 180 g/mol / 30 g/mol = 6. This tells us that the molecular formula is 6 times the empirical formula. So, we multiply the subscripts in the empirical formula by 6 to get the molecular formula: . And there we have it! The molecular formula of our organic compound is . Victory!
Conclusion: The Molecular Formula Unveiled
So, guys, we did it! We took a complex problem with percentages, freezing points, and atomic weights, and we cracked the code to reveal the molecular formula of our organic compound: . It's like a chemical detective story, where each piece of information is a clue that leads us to the final answer. We started by finding the empirical formula, the simplest ratio of atoms, using the percentage composition. Then, we used the freezing point depression to calculate the molecular weight, which gave us the total weight of all atoms in the molecule. Finally, comparing the empirical formula weight with the molecular weight allowed us to scale up the empirical formula to the molecular formula. This process isn't just about following steps; it's about understanding the relationships between different properties and how they all fit together. Remember, chemistry is all about connecting the dots – from percentages to freezing points to the grand finale of the molecular formula. Keep practicing, keep exploring, and who knows? Maybe you'll be the one to solve the next big molecular mystery! Keep rocking, future chemists!
