Answer:
The percent composition of the compound is 90.5 % C and 9.5 % H
Explanation:
Step 1: Data given
Mass of compound = 9.394 mg
Mass of CO2 yielded = 31.154 mg
Mass of H2O yielded = 7.977 mg
Molar mass of CO2 = 44.01 g/mol
Molar mass of H2O = 18.02 g/mol
Step 2: Calculate moles CO2
moles of CO2 = (0.031154 g / 44.01 g/mol) = 7.08 * 10^-4 mol CO2
Step 3: Calculate moles C
moles of C = moles of CO2 * (1 mol C / 1 mol CO2)
moles of C = 7.08 * 10^-4 mol
Step 4: Calculate moles H2O
moles of H2O = (0.007977 g / 18.02 g/mol) = 4.43 * 10^-4 mol H2O
Step 5: Calculate moles of H
moles of H = moles of H2O * (2 mol H / 1 mol H2O)
moles of H = 4.43* 10^-4 *2 = 8.86 * 10^-4 mol H
Step 6: Calculate mass of C
mass C = moles C * molar mass C
mass C = 7.08 * 10^-4 mol*12.01 g/mol
mass C = 0.0085 grams
Step 7: Calculate mass of H
mass H = moles H * molar mass H
mass H = 8.86 * 10^-4 mol*1.01 g/mol
mass H = 0.000894 grams
Step 8: Calculate total mass of compound =
0.0085 grams + 0.000894 grams = 0.009394 grams = 9.394 mg
Step 9: Calculate the percent composition:
% C = (8.50 mg / 9.394 mg) x 100 = 90.5%
% H = (0.894 mg / 9.394 mg) x 100 = 9.5%
The percent composition of the compound is 90.5 % C and 9.5 % H
