Cyclopropane, a substance used with oxygen as a general anesthetic, contains only two elements, carbon and hydrogen. When 1.00 g of this substance is completely combusted, 3.14 g of CO2 and 1.29 g of H2O are produced. What is the empirical formula of cyclopropane

CH₂

Explanation:

From the question given above, the following data were obtained:

Mass of compound = 1 g

Mass of CO₂ = 3.14 g

Mass of H₂O = 1.29 g

Empirical formula =?

Next, we shall determine the mass of Carbon and hydrogen present in the compound. This can be obtained as follow:

For Carbon, C:

Mass of CO₂ = 3.14 g

Molar mass of CO₂ = 12 + (2×16)

= 12 + 32

= 44 g/mol

Molar mass of C = 12 g/mol

Mass of C =?

Mass of C = molar mass of C/ Molar mass of CO₂ × Mass of CO₂

Mass of C = 12/44 × 3.14

Mass of C = 0.86 g

For hydrogen, H:

Mass of C = 0.86 g

Mass of compound = 1 g

Mass of H =?

Mass of H = (Mass of compound) – (mass of C)

Mass of H = 1 – 0.86

Mass of H = 0.14 g

Finally, we shall determine the empirical formula of the cyclopropane. This can be obtained as follow:

Mass of C = 0.86 g

Mass of H = 0.14 g

Divide by their molar mass

C = 0.86 / 12 = 0.07

H = 0.14 / 1 = 0.14

Divide by the smallest

C = 0.07 / 0.07 = 1

H = 0.14 / 0.07 = 2

Thus, the empirical formula of cyclopropane is CH₂

a 20-cm-long, cylindrical graphite (pure carbon) rod is inserted into an oxidizing atmosphere at 1145 k and 1.013 x 105 pa pressure. the oxidizing process is limited by the diffu- sion of oxygen counterflow to the carbon monoxide that is formed on the cylindrical surface.