5.82 g isopropyl alcohol yield 7.001 g of water and 12.6 g of carbon dioxide upon combustion. The empirical formula is ________.

Question

Quizplus · Accepted Answer

The empirical formula can be determined by analyzing the combustion products. From the given data, we can calculate the moles of carbon dioxide and water produced, which directly relate to the moles of carbon and hydrogen in the original compound. The oxygen in the compound can be found by difference from the original mass. - From 12.6 g of CO2, we get 12.6 g / 44.01 g/mol (molar mass of CO2) = 0.286 moles of CO2, which equals 0.286 moles of C.- From 7.001 g of H2O, we get 7.001 g / 18.015 g/mol (molar mass of H2O) = 0.388 moles of H2O, which equals 0.776 moles of H (since each molecule of water has 2 hydrogen atoms).- The mass of oxygen in the isopropyl alcohol can be found by subtracting the mass of carbon and hydrogen from the total mass of isopropyl alcohol. The mass of carbon is 0.286 moles * 12.01 g/mol = 3.434 g. The mass of hydrogen is 0.776 moles * 1.008 g/mol = 0.782 g. So, the mass of oxygen is 5.82 g - (3.434 g + 0.782 g) = 1.604 g. This gives us 1.604 g / 16.00 g/mol = 0.100 moles of O.The simplest whole number ratio of C:H:O is approximately 3:8:1, which corresponds to the empirical formula C3H8O. However, due to the encoding error in the options, the correct choice is B (C₃H₈O), which seems to be an attempt to represent C3H8O but with formatting issues.

582 G Isopropyl Alcohol Yield 7