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A sample of an unknown inorganic compound weighing 6.25 grams was analyzed to determine its empirical and molecular formulas. After combustion, the sample produced 9.375 grams of carbon dioxide and 3.125 grams of water. What is the empirical and molecular formula of the inorganic compound?

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To determine the empirical and molecular formulas of the unknown inorganic compound, we first need to find the moles of each element present in the compound.1. Calculate the moles of carbon  C  and hydrogen  H  from the products of combustion:From the given data, we have 9.375 grams of carbon dioxide  CO  and 3.125 grams of water  HO .Moles of C in CO:Molar mass of CO = 12.01 g/mol  C  + 2 * 16.00 g/mol  O  = 44.01 g/molMoles of C =  9.375 g CO  /  44.01 g/mol  = 0.213 mol CMoles of H in HO:Molar mass of HO = 2 * 1.01 g/mol  H  + 16.00 g/mol  O  = 18.02 g/molMoles of H =  3.125 g HO  /  18.02 g/mol  = 0.173 mol H2. Determine the moles of oxygen  O  in the unknown compound:Since the compound is inorganic, we can assume that the remaining mass is due to oxygen.Mass of O = Mass of the compound - Mass of C - Mass of HMass of C = 0.213 mol C * 12.01 g/mol = 2.56 gMass of H = 0.173 mol H * 1.01 g/mol = 0.175 gMass of O = 6.25 g  compound  - 2.56 g  C  - 0.175 g  H  = 3.515 gMoles of O =  3.515 g O  /  16.00 g/mol  = 0.220 mol O3. Determine the empirical formula:Divide the moles of each element by the smallest number of moles to get the mole ratio:Mole ratio of C:H:O = 0.213:0.173:0.220 / 0.173 = 1.23:1:1.27Since these ratios are close to whole numbers, we can round them to get the empirical formula:Empirical formula = CHO = CHO4. Determine the molecular formula:To find the molecular formula, we need the molar mass of the unknown compound. Unfortunately, the problem does not provide this information. Therefore, we cannot determine the molecular formula without additional information.In conclusion, the empirical formula of the unknown inorganic compound is CHO. The molecular formula cannot be determined without more information.

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