Question

Prior to their phaseout in the 1980s, chemicals containing lead were commonly added to gasoline as anti-knocking agents. A 4.351 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 4.736 g of CO2(g) and 2.423 g of H2O(g). Insert subscripts below to appropriately display the empirical formula of the gasoline additive:

CHPb

Answer 1

Given that combustion of 4.351 g sample produced 4.736 g of CO2(g) and 2.423 g of H2O(g).

Mass of carbon present in 4.736 g of CO2 =4.736 g x molar mass of carbon/ molar mass of CO2

                                                                         = 4.736 g x 12 (g/mol) / 44 (g/mol)

                                                                         = 1.291 g

Mass of hydrogen present in 2.423 g of H2O = 2.423 g x molar mass of hydrogen/ molar mass of H2O

                                                                         = 2.423 g x2 (g/mol) / 18 (g/mol)

                                                                         = 0.269 g

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

1.291 g + 0.269 g + Mass of Pb = 4.351 g

Mass of Pb = 4.351 g – ( 1.291 g + 0.269 g )

                   = 2.791 g

Therefore, Mass of carbon = 1.291 g

                     Mass of hydrogen = 0.269 g

                    Mass of Pb = 2.791 g

Dividing each mass by elemental molar mass tells us how many moles of each element present in the sample of butyric acid.

For C, 1.291 g/ 12 (g/mol) = 0.107 mol C

For H, 0.269 g/ 1 (g/mol) = 0.269 mol H

For Pb, 2.791 g/ 207.2 (g/mol) = 0.0134 mol Pb

Divide each by smallest among them i.e. 0.0134

For C, 0.107/ 0.0134 = 7.98 mol C

For H, 0.269/ 0.0134 = 20.0 mol H

For Pb, 0.0134/ 0.0134 = Pb mol O

Round each value to the nearest integer,

C = 8

H = 20

Pb =1

Therefore, empirical formula of gasoline additive = C₈H₂₀Pb