91影视

From the combustion reaction, we will get water and carbon dioxide as products. Given that a 0.165 g sample of the compound produces 0.166 g of water and 0.403 g of carbon dioxide, we need to find how many moles of carbon, hydrogen, and oxygen are present in this sample. First, let's convert the mass of water and carbon dioxide to moles using their molar masses: For water: \(2\,H + 1\,O\) Molar mass of water (H2O) = \(2(1.008\,g/mol) + 15.999\,g/mol = 18.015\,g/mol\) Moles of water = \(\frac{0.166\,g}{18.015\,g/mol} = 0.009212\,mol\) For carbon dioxide: \(1\,C + 2\,O\) Molar mass of carbon dioxide (CO2) = \(12.01\,g/mol + 2(15.999\,g/mol)\) = \(44.009\,g/mol\) Moles of carbon dioxide = \(\frac{0.403\,g}{44.009\,g/mol} = 0.009158\,mol\) Now, we can find the moles of each element in the sample: Moles of carbon = Moles of carbon dioxide (each molecule of CO2 has 1 carbon atom) Moles of carbon = \(0.009158\,mol\) Moles of hydrogen = 2 * Moles of water (each molecule of H2O has 2 hydrogen atoms) Moles of hydrogen = \(2 \times 0.009212\,mol = 0.018424\,mol\) Lastly, we will find moles of oxygen. Since there is only carbon and hydrogen in water and carbon dioxide, the remaining mass of the sample must be oxygen. We know the mass of the sample (0.165 g) and the mass of carbon and hydrogen from the moles calculated above. Mass of carbon = Moles of carbon * Molar mass of carbon Mass of carbon = \(0.009158\,mol \times 12.01\,g/mol = 0.1099\,g\) Mass of hydrogen = Moles of hydrogen * Molar mass of hydrogen Mass of hydrogen = \(0.018424\,mol \times 1.008\,g/mol = 0.018571\,g\) Mass of oxygen = 0.165 g (sample mass) - 0.1099 g (carbon mass) - 0.018571 g (hydrogen mass) Mass of oxygen = 0.036529 g Moles of oxygen = \(\frac{0.036529\,g}{15.999\,g/mol} = 0.002285\,mol\)

Now we have moles of each element in the sample. To get the empirical formula, we will divide the moles of each element (C, H, and O) by the smallest mole value: Smallest mole value = 0.002285 mol (for oxygen) Mole ratio of carbon = \(\frac{0.009158}{0.002285} = 4\) Mole ratio of hydrogen = \(\frac{0.018424}{0.002285} = 8\) Mole ratio of oxygen = \(\frac{0.002285}{0.002285} = 1\) So, the empirical formula is C4H8O1, or simply C4H8O.

We're given the molar mass of valproic acid as 144 g/mol. First, we need to find the molar mass of the empirical formula: Molar mass of empirical formula (C4H8O) = \(4(12.01\,g/mol) + 8(1.008\,g/mol) + 15.999\,g/mol = 72\,g/mol\) Now we can find the ratio of the molar mass of the molecular formula to the molar mass of the empirical formula: Ratio = \(\frac{144\,g/mol}{72\,g/mol} = 2\) Therefore, the molecular formula is twice the empirical formula, which is C8H16O2.