Complete combustion of 5.80 g of a hydrocarbon produced 18.6 g of co2 and 6.36 g of h2o. what is the empirical formula for the hydrocarbon?

We need to first figure out how many carbons and hydrogen are from the products since they came from the hydrocarbon.

18.6 g (1 mol CO2/ 44 grams)( 1 mol C/ 1 mol CO2)= 0.423 moles C

6.36 g (1 mol H2O/ 18.02 grams)(2 mol H/ 1 mole H2O)= 0.706 mol H

Divide the both values by the smallest and multiple by any number until you get the smallest whole number

Carbon- 0.423/0.423= 1 x 3= 3

Hydrogen- 0.706/ 0.423= 1.7 x 3= 5

The formula is C3H5

The correct answer is c. orbital shape. "the angular momentum quantum number l describes the shape of an orbital. it corresponds to each orbital type, i.e. (0,1,2,−1)=(s,p,d,f,g,"