Isopropyl alcohol, sold as rubbing alcohol, is composed of C, H, and 0. Combustion of 0.255 g of isopropyl alcohol produces 0.561 g of CO2 and 0.306 g
of H20. Determine the empirical formula of isopropyl alcohol.

C₃H₈O

Explanation:

The following data were obtained from the question:

Mass of compound = 0.255 g

Mass of CO2 = 0.561 g

Mass of H2O = 0.306 g

Empirical formula =?

Next, we shall determine the mass of carbon (C), hydrogen (H) and oxygen (O) present in the compound. This can be obtained as follow:

For Carbon (C):

Mass of CO2 = 0.561 g

Molar mass of CO2 = 12 + (2×16) = 12 + 32 = 44 g/mol

Mass of C = 12/44 × 0.561

Mass of C = 0.153 g

For Hydrogen (H):

Mass of H2O = 0.306 g

Molar mass of H2O = (2×1) + 16 = 2 + 16 = 18 g/mol

Mass of H = 2/18 × 0.306

Mass of H = 0.034 g

For Oxygen (O):

Mass of compound = 0.255 g

Mass of C = 0.153 g

Mass of H = 0.034 g

Mass of O = Mass of compound – (Mass of C + Mass of H)

Mass of O = 0.255 – ( 0.153 + 0.034)

Mass of O = 0.255 – 0.187

Mass of O = 0.068 g

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

C = 0.153 g

H = 0.034 g

O = 0.068 g

Divide by their molar mass

C = 0.153 / 12 = 0.0128

H = 0.034 / 1 = 0.034

O = 0.068 / 16 = 0.0043

Divide by the smallest

C = 0.0128 / 0.0043 = 3

H = 0.034 /0.0043 = 8

O = 0.0043 / 0.0043 = 1

Thus, the empirical formula of the compound is C₃H₈O

b. accuracy of the data matters alot more than its number of digits, organization or amount of detail. quantitative observation is more accepted than the qualitative observation.