Percent Composition assignment 1.20.02 g of sulfur reacts with 29.98 g of oxygen to form a compound. What is the percent composition? 2.Determine the percent composition of Al(OH)_(3) 3. Determine the empirical formula for each compound. a) ethylene (C_(2)H_(4)) b) ascorbic acid (C_(6)H_(8)O_(6)) c 4. Which of these molecular formulas are also empirical formulas: ethyl ether (C_(4)H_(10)O) aspirin (C_(9)H_(8)O_(4)) ethyl acetate (C_(4)H_(8)O_(2)) glucose (C_(6)H_(12)O_(6)) 5. Determine the empirical formula of a compound that consists of 12.00 g Ni, 4.91 g C,and 13.09 g O. 6. A compound is composed of 50.70% carbon, 4.27% hydrogen, and 45.03% oxygen. Determine the empirical formula for this compound. 7. A compound is composed of 57.14% carbon, 4.80% hydrogen, and 38.06% oxygen. Its molar mass is 252.24g/mol Determine the empirical and molecular formulas for this compound.

1. To find the percent composition of sulfur and oxygen in the compound, we need to calculate the total mass of the compound and then determine the percentage of each element.Total mass of the compound = mass of sulfur + mass of oxygenTotal mass of the compound = 20.02 g + 29.98 g = 49.00 gPercent composition of sulfur = (mass of sulfur / total mass of the compound) * 100Percent composition of sulfur = (20.02 g / 49.00 g) * 100 ≈ 40.82%Percent composition of oxygen = (mass of oxygen / total mass of the compound) * 100Percent composition of oxygen = (29.98 g / 49.00 g) * 100 ≈ 59.18%Answer: The percent composition of the compound is approximately 40.82% sulfur and 59.18% oxygen.2. To determine the percent composition of , we need to calculate the molar mass of the compound and then find the percentage of each element.Molar mass of = (1 * 26.98) + (3 * 15.999) + (3 * 1.008) = 78.00 g/molPercent composition of aluminum = (molar mass of aluminum / molar mass of ) * 100Percent composition of aluminum = (26.98 g/mol / 78.00 g/mol) * 100 ≈ 34.72%Percent composition of oxygen = (molar mass of oxygen / molar mass of ) * 100Percent composition of oxygen = (3 * 15.999 g/mol / 78.00 g/mol) * 100 ≈ 62.68%Percent composition of hydrogen = (molar mass of hydrogen / molar mass of ) * 100Percent composition of hydrogen = (3 * 1.008 g/mol / 78.00 g/mol) * 100 ≈ 2.56%Answer: The percent composition of is approximately 34.72% aluminum, 62.68% oxygen, and 2.56% hydrogen.3. a) The empirical formula for ethylene is already given as .b) The empirical formula for ascorbic acid is already given as .c) The empirical formula for ethyl ether is already given as .Answer: The empirical formulas for the compounds are:a) b) c) 4. To determine which molecular formulas are also empirical formulas, we need to check if the subscripts in the molecular formulas are the smallest possible whole numbers.Ethyl ether : The subscripts are already in the smallest possible whole numbers, so it is an empirical formula.Aspirin : The subscripts are already in the smallest possible whole numbers, so it is an empirical formula.Ethyl acetate : The subscripts are already in the smallest possible whole numbers, so it is an empirical formula.Glucose : The subscripts are already in the smallest possible whole numbers, so it is an empirical formula.Answer: All the given molecular formulas are empirical formulas.5. To determine the empirical formula of the compound, we need to convert the given masses of each element to moles and then find the simplest whole number ratio.Moles of Ni = 12.00 g / 58.69 g/mol ≈ 0.205 molMoles of C = 4.91 g / 12.01 g/mol ≈ 0.410 molMoles of O = 13.09 g / 16.00 g/mol ≈ 0.815 molTo find the simplest whole number ratio, we divide each mole value by the smallest mole value (0.205 mol).Ratio of Ni = 0.205 mol / 0.205 mol = 1Ratio of C = 0.410 mol / 0.205 mol ≈ 2Ratio of O = 0.815 mol / 0.205 mol ≈ 4Answer: The empirical