Question

Researchers used a combustion method to analyze a compound used as an antiknock additive in gasoline. A 9.394 mg sample of the compound yielded 31.154 mg of carbon dioxide and 7.977 mg of water in the combustion. Calculate the percent composition of the compound.

Answer 1

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