Complete combustion of 7.40 g of a hydrocarbon produced 22.4 g of co2 and 11.5 g of h2o. what is the empirical formula for the hydrocarbon? show work ?

C2H5 First, you need to figure out the relative ratios of moles of carbon and hydrogen. You do this by first looking up the atomic weight of carbon, hydrogen, and oxygen. Then you use those atomic weights to calculate the molar masses of H2O and CO2. Carbon = 12.0107 Hydrogen = 1.00794 Oxygen = 15.999 Molar mass of H2O = 2 * 1.00794 + 15.999 = 18.01488 Molar mass of CO2 = 12.0107 + 2 * 15.999 = 44.0087 Now using the calculated molar masses, determine how many moles of each product was generated. You do this by dividing the given mass by the molar mass. moles H2O = 11.5 g / 18.01488 g/mole = 0.638361 moles moles CO2 = 22.4 g / 44.0087 g/mole = 0.50899 moles The number of moles of carbon is the same as the number of moles of CO2 since there's just 1 carbon atom per CO2 molecule. Since there's 2 hydrogen atoms per molecule of H2O, you need to multiply the number of moles of H2O by 2 to get the number of moles of hydrogen. moles C = 0.50899 moles H = 0.638361 * 2 = 1.276722 We can double check our math by multiplying the calculated number of moles of carbon and hydrogen by their respective atomic weights and see if we get the original mass of the hydrocarbon. total mass = 0.50899 * 12.0107 + 1.276722 * 1.00794 = 7.400185 7.400185 is more than close enough to 7.40 given rounding errors, so the double check worked. Now to find the empirical formula we need to find a ratio of small integers that comes close to the ratio of moles of carbon and hydrogen. 0.50899 / 1.276722 = 0.398669 0.398669 is extremely close to 4/10, so let's reduce that ratio by dividing both top and bottom by 2 giving 2/5. Since the number of moles of carbon was on top, that ratio implies that the empirical formula for this unknown hydrocarbon is C2H5

The answer would be c. fish

Ithink the answer is 56.