Question

An unknown compound contains only CC , HH , and OO . Combustion of 5.30 g5.30 g of this compound produced 12.0 g12.0 g CO2CO2 and 4.93 g4.93 g H2OH2O . What is the empirical formula of the unknown compound? Insert subscripts as needed.

2.)

Prior to their phaseout in the 1980s, chemicals containing lead were commonly added to gasoline as anti‑knocking agents. A 6.763 g sample of one such additive containing only lead, carbon, and hydrogen was burned in an oxygen‑rich environment. The products of the combustion were 7.361 g CO2(g)7.361 g CO2(g) and 3.767 g H2O(g)3.767 g H2O(g) .

Determine the empirical formula of the gasoline additive?

Answer 1

1) Let the formula of the compound be

Number of moles of CO2 = Mass/Molar mass = 12.0/44.0 = 0.2727 moles = Number of moles of C in the sample

Number of moles of H2O = Mass/Molar mass = 4.93/18.0 = 0.2738 moles = 1/2 * Number of moles of H in the sample

Mass of sample = 5.30 = Mass of C + Mass of H + Mass of O

5.30 = 0.2727 * 12 + 0.2738 * 2 * 1 + Mass of O

Mass of O = 1.4798 g

Number of moles of O = Mass/molar mass = 1.498/16 = 0.093625 moles

Hence the ratio will be C:H:O is 3:6:1

Hence the empirical formula will be C3H6O

2)

Number of moles of CO2 = Mass/Molar mass = 7.361/44.0 = 0.1673 moles = Number of moles of C in the sample

Number of moles of H2O = Mass/Molar mass = 3.767/18.0 = 0.2092 moles = 1/2 * Number of moles of H in the sample

Mass of sample = 6.763 = Mass of C + Mass of H + Mass of Pb

6.763 = 0.1673 * 12 + 0.2092 * 2 * 1 + Mass of Pb

Mass of Pb = 6.763 - 0.1673 * 12 - 0.2092 * 2 * 1 = 4.337 g

Number of moles of Pb = 4.337/207.2 = 0.02093

Hence the ratio will be 0.1673:0.4184:0.02093 or 8:20:1

Hence the formula will be C8H20Pb

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