Question

Quinone, an oxygenated hydrocarbon, is a chemical used in the photography industry. When a 0.144-g sample is analyzed by combustion analysis, 0.352 g of carbon dioxide and 0.0480 g of water are produced. Find the empirical formula for quinone.

Answer 1

C3H2O1

Step-by-step explanation:

The following data were obtained from the question:

Mass of compound = 0.144 g

Mass of CO2 = 0.352 g

Mass of H2O = 0.0480 g

Empirical formula of compound =?

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):

Molar mass of CO2 = 12 + (2×16)

Molar mass of CO2 = 44 g/mol

Mass of CO2 = 0.352 g

Mass of C in the compound

= 12/44 × 0.352

= 0.096 g

For Hydrogen (H):

Molar mass of H2O = (2×1) + 16

Molar mass of H2O = 18 g/mol

Mass of H2O = 0.0480 g

Mass of H in the compound

= 2/18 × 0.0480

= 0.0053 g

For oxygen (O):

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

Mass of compound = 0.144 g

Mass of C = 0.096 g

Mass of H = 0.0053 g

Mass of O = 0.144 – (0.096 + 0.0053)

Mass of O = 0.144 – 0.1013

Mass of O = 0.0427 g

Finally, we shall determine the empirical formula of the compound as follow:

C = 0.096 g

H = 0.0053 g

O = 0.0427 g

Divide both side by their molar mass

C = 0.096 / 12 = 0.008

H = 0.0053 / 1 = 0.0053

O = 0.0427 / 16 = 0.0027

Divide by the smallest

C = 0.008 / 0.0027 = 3

H = 0.0053 / 0.0027 = 2

O = 0.0027 / 0.0027 = 1

Therefore, the empirical formula of the compound is C3H2O1