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Leaded gasoline contains an additive to prevent engine knocking. On analysis, the additive compound is found to contain carbon, hydrogen, and lead (Pb) (hence, leaded gasoline). When 51.36 g of this compound is burned in a combustion analysis apparatus, 55.90 g of CO2 and 28.61 g of H2O are produced. Determine the empirical formula of the gasoline additive.

Respuesta :

Answer:

PbC(8)H(20)

Explanation:

51.36g of the leaded gasoline when combusted yielded 55.90 CO2 and 28.61 H2O,

Fraction of carbon in the sample = molecular mass of carbon / molecular mass of CO2* mass of CO2 produced = 12.0107/ 44.01 * 55.90 = 15.26 g

Similarly,

Mass of hydrogen present in the sample = molecular mass of hydrogen/ molecular mass of water * Mass of water produced = 2(1.00784)/18.0153 *28.61 = 3.201

The mass of the compound was 51.36, mass of lead present in the sample = total mass of the sample - (mass of carbon + mass of hydrogen) = 51.36 - (15.26+3.201) = 32.899

To find the mole ratio of the substances in the compound, lead = mass of lead / molecular mass of lead = 32.899/ 207.2 = 0.159

Carbon = 15.26/12.0107 = 1.271

Hydrogen = 3.201/ 1.00784 = 3.1761

Divide each mole by the smallest mole to find the mole ratio

Lead = 0.159/0.159 =1

Carbon = 1.271/0.159 = 7.994 approx 8

Hydrogen = 3.1761/0.159 = 19.975approx 20

Therefore the empirical formula which is different from the molecular because it is the simplest form of the substance is equal to PbC8H20

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