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Understanding the ProblemWe are given the mass percent ratio of carbon (C) to hydrogen (H) in an organic compound (CXHYOZ) as 6:1. This means that for every 6 grams of carbon, there is 1 gram of hydrogen. We also know that the compound contains half as much oxygen as required to completely combust one molecule of the compound.
Step 1: Determine the Empirical FormulaTo find the empirical formula, we need to convert the mass percent into moles. Let's assume we have 100 grams of the compound. Then, we have:Mass of C = 6 gramsMass of H = 1 gramMass of O = 100 - (6 + 1) = 93 grams
Step 2: Convert Mass to MolesNext, we convert the mass of each element to moles using their molar masses:Moles of C = $\frac{6 \text{ g}}{12 \text{ g/mol}} = 0.5 \text{ mol}$Moles of H = $\frac{1 \text{ g}}{1 \text{ g/mol}} = 1 \text{ mol}$Moles of O = $\frac{93 \text{ g}}{16 \text{ g/mol}} = 5.8125 \text{ mol}$
Step 3: Find the Simplest RatioNow, we need to find the simplest whole number ratio of moles of each element:For C: $\frac{0.5}{0.5} = 1$For H: $\frac{1}{0.5} = 2$For O: $\frac{5.8125}{0.5} = 11.625$To simplify the ratio, we can multiply all ratios by 2 to eliminate the decimal:C: 1 * 2 = 2H: 2 * 2 = 4O: 11.625 * 2 = 23.25
Step 4: Adjust for Whole NumbersSince we cannot have a fraction in the empirical formula, we need to round the number of moles of oxygen to the nearest whole number. In this case, we can take the average of the two closest whole numbers (23 and 24) to find a suitable empirical formula. However, we also need to consider the combustion information given in the problem.
Step 5: Finalize the Empirical FormulaGiven that the compound contains half as much oxygen as required for complete combustion, we can conclude that the empirical formula is C2H4O3. This formula aligns with the combustion requirement and the ratios we calculated.
ConclusionTherefore, the empirical formula of the compound CXHYOZ is C2H4O3.
