91影视

In order to find the empirical formula of helicene, we need to determine the ratio of carbon to hydrogen in the compound. To do this, we will first find the amount of carbon present in the sample by analyzing the mass of CO鈧� produced. 1. Determine the amount of C in the CO鈧� produced: Given mass of CO鈧� = 0.5063 g molar mass of CO鈧� = 12.01 g/mol (C) + 2 * 16.00 g/mol (O) = 44.01 g/mol moles of CO鈧� = mass of CO鈧� / molar mass of CO鈧� = 0.5063 g / 44.01 g/mol = 0.0115 mol Since one mole of CO鈧� contains one mole of carbon, moles of C in the sample = 0.0115 mol. 2. Calculate the mass of H in the sample: Since the only elements present in helicenes are carbon and hydrogen, the mass of hydrogen can be found by subtracting the mass of carbon present in the sample from the total mass of the helicene. mass of C present = moles of C * molar mass of C = 0.0115 mol * 12.01 g/mol = 0.138 g mass of helicene sample = 0.1450 g mass of H present = mass of helicene sample - mass of C present = 0.1450 g - 0.138 g = 0.0070 g 3. Calculate the moles of H in the sample: molar mass of H = 1.008 g/mol moles of H = mass of H / molar mass of H = 0.0070 g / 1.008 g/mol = 0.00695 mol 4. Determine the empirical formula of the helicene: To find the lowest whole number ratio of C to H, divide the moles of both elements by the smallest number of moles. C: 0.0115 mol / 0.00695 mol = 1.65 鈮� 2 H: 0.00695 mol / 0.00695 mol = 1 The empirical formula of the helicene is C鈧侶.

We are given a concentration and the mass of helicene dissolved in solvent. We can find the molecular formula by finding the molar mass and dividing it by the molar mass of the empirical formula. 1. Find the moles of helicene in the solution: We are given that the solution is 0.0175 M. To find the moles of helicene in the solution, we can use the formula: moles = Molarity * Volume The volume can be found from the mass of the solvent and the mass of helicene in the solution. mass of solvent = 12.5 g mass of helicene in solution = 0.0938 g We can assume the density of the solution is 1 g/mL. Therefore, the total mass = total volume Total volume = 12.5 g + 0.0938 g = 12.5938 g 鈮� 12.59 mL Now, we can find the moles of helicene: moles = 0.0175 mol/L * 0.01259 L = 0.0002204 mol 2. Calculate the molar mass of helicene: molar mass = mass / moles = 0.0938 g / 0.0002204 mol = 425.909 g/mol 3. Find the ratio between the molar mass of helicene and the molar mass of the empirical formula: molar mass of empirical formula (C鈧侶) = (2 * 12.01 g/mol) + 1.008 g/mol = 25.028 g/mol ratio = molar mass of helicene / molar mass of empirical formula = 425.909 g/mol / 25.028 g/mol = 17 4. Determine the molecular formula of helicene: Since the ratio is 17, we can multiply the empirical formula by 17 to obtain the molecular formula of helicene. Molecular formula = (C鈧侶) * 17 = C鈧冣倓H鈧佲倗 The molecular formula of this helicene is C鈧冣倓H鈧佲倗.

Now we are going to write the balanced reaction for the combustion of this helicene. The general combustion reaction of a hydrocarbon can be represented as: Hydrocarbon + O鈧� (g) 鈫� CO鈧� (g) + H鈧侽 (g) For this helicene, our reaction will be: C鈧冣倓H鈧佲倗 (s) + O鈧� (g) 鈫� CO鈧� (g) + H鈧侽 (g) We can balance the above reaction by making sure that the number of each element is the same on both sides of the equation. 1. Balance the carbon atoms: There are 34 carbon atoms in C鈧冣倓H鈧佲倗, so we'll need 34 moles of CO鈧� on the product side: C鈧冣倓H鈧佲倗 (s) + O鈧� (g) 鈫� 34CO鈧� (g) + H鈧侽 (g) 2. Balance the hydrogen atoms: There are 17 hydrogen atoms in C鈧冣倓H鈧佲倗, so we'll need 8.5 moles of H鈧侽 on the product side: C鈧冣倓H鈧佲倗 (s) + O鈧� (g) 鈫� 34CO鈧� (g) + 8.5H鈧侽 (g) Since we can't have fractional coefficients in a balanced equation, we can multiply the whole equation by 2 to make it a whole number. So, the equation becomes: 2C鈧冣倓H鈧佲倗 (s) + O鈧� (g) 鈫� 68CO鈧� (g) + 17H鈧侽 (g) 3. Balance the oxygen atoms: On the product side, we have 68 * 2 (from CO鈧�) + 17 (from H鈧侽) = 153 oxygen atoms. Therefore, we'll need 76.5 moles of O鈧� on the reactant side to balance the oxygen atoms. But again, we can't have fractional coefficients, so multiplying the entire reaction by 2 again: 4C鈧冣倓H鈧佲倗 (s) + 153O鈧� (g) 鈫� 136CO鈧� (g) + 34H鈧侽 (g) Now, the combustion reaction of this helicene is balanced. The balanced reaction for the combustion of this helicene is: 4C鈧冣倓H鈧佲倗 (s) + 153O鈧� (g) 鈫� 136CO鈧� (g) + 34H鈧侽 (g)