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Prior to their phaseout in the 1980s, chemicals containing lead were commonly added to gasoline as anti-knocking agents. a 2.475 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 2.694 g of co2(g) and 1.378 g of h2o(g). insert subscripts below to appropriately display the empirical formula of the gasoline additive:

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Final answer:

To determine the empirical formula of a gasoline additive made of lead, carbon, and hydrogen, we calculate the moles of carbon from the CO2 and hydrogen from the H2O produced in combustion. The leftover mass is attributed to lead. Using this data, the empirical formula can be obtained by finding the simplest whole number ratio of the elements.

Step-by-step explanation:

The student is asking us to determine the empirical formula of a gasoline additive that is composed only of lead, carbon, and hydrogen. This can be done by analyzing the amounts of CO2 and H2O produced from the combustion of the substance in an oxygen-rich environment. To find the empirical formula, the mass of carbon in CO2 and the mass of hydrogen in H2O need to be calculated first, and from there, we deduce the amount of lead in the original sample by subtracting the masses of carbon and hydrogen from the total mass of the sample.

From 2.694 g of CO2 produced, we can calculate the number of moles of carbon, as carbon has a molar mass of approximately 12 g/mol. Similarly, from 1.378 g of H2O produced, we can determine the number of moles of hydrogen, with hydrogen having a molar mass of approximately 1 g/mol. The remaining mass will be that of lead (Pb), which has a molar mass of roughly 207 g/mol. Once we have the moles of each element, we can convert these to the simplest whole number ratio to find the empirical formula of the additive.

User RobertoBr
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4.007-2.694 g of CO2 = 2.694 x 12 / 44 g of C = 0.7347 g of C = 0.0911 moles of C 2.050-1.378 g of H2O = 2.050 x 2 / 18 g of H = 0.1531 g and 0.1531 moles mass of Pb = 2.475 - ( 0.0911 + 0.1531) g = 2.23 g = 2.23 / 207 = 0.0108 moles
molar ratio of C:H:Pb = 0.0911: 0.1531: 0.0108divide by the smallest number and we get 8.0: 20.0: 1.0
empirical formula is C8H20Pb
tetraethyl lead
User Pavel Shastov
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