Question

Propenoic acid, C3H4O2, is a reactive organic liquid that is used in the manufacturing of plastics, coatings, and adhesives. An unlabeled container is thought to contain this liquid. A 0.442-g sample is combusted to produce 0.220 g of water and 0.636 g of carbon dioxide.Calculate the experimental mass % of C and H in the unknown liquid used in the experiment.

Answer 1

Answer:

C = 39%

H = 5.43%

Explanation:

Firstly, we calculate the number of moles of each.

For carbon, we use carbon iv oxide

Mass here = 0.636g

Number of moles of carbon iv oxide = mass of carbon iv oxide ÷ molar mass of carbon iv oxide. Molar mass = 44g/mol

Number of moles = 0.636 ÷ 44 = 0.014mole

Since 1 mole of carbon iv oxide contains 1 mole carbon, 0.014 mole of carbon iv oxide is also produced.

Mass of carbon = 0.014 × 12 = 0.173g, where 12 is the a.m.u of carbon.

For hydrogen, we use water .

Mass of water = 0.220g

No of moles of water = 0.22 ÷ 18 = 0.012 mole

1 mole of water has two moles of hydrogen, thus the amount of hydrogen produced = 0.012 × 2 = 0.024mole

Mass of hydrogen produced = 0.024 × 1 = 0.024g

The percentage composition are as follows:

Carbon = 0.173/0.442 × 100 = 39%

Hydrogen = 0.024/0.442 × 100 = 5.43%