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.

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

chemical digestion occurs in the stomach were hcl and pepsin forms chyme, the mouth in which salivary amylase breaks down starch to form bolus, and the small intestine specifically the duodenum.