If 0.100 g of a compound containing C, H, and O is combusted completed in an excess of oxygen and 0.281 g of CO2 and 0.0574 g of H2O are obtained, what is the e

Question

If 0.100 g of a compound containing C, H, and O is combusted completed in an excess of oxygen and 0.281 g of CO2 and 0.0574 g of H2O are obtained, what is the empirical formula of the compound?

In your answer list carbon first, followed by hydrogen and then oxygen. For example if the answer were C2H7O3, then give the answer as C2H7O3.

Answer 1

Hello Holly

 

We need to first determine the mass of C, H and O in the compound. Convert mass of CO2 and H2O to mass of C and H, respectively.

 

0.281 g CO2 x (1 mol CO2 / 44 g CO2) x (1 mol C / 1 mol CO2) x (12 g C / 1 mol C) = 0.0766 g C

 

0.0574 g H2O x (1 mol H2O / 18 g H2O) x (2 mol H / 1 mol H2O) x (1 g H / 1 mol H) = 0.00638 g H

 

Mass of O in 0.100 g of the compound = 0.100 - (mass of C + mass of H)

 

= 0.100 - (0.0766 + 0.00638)

= 0.017 g O (rounded to 3 decimal places after subtraction)

 

Now that we have the masses of C, H and O, we need to convert them to their respective # of moles.

 

C = 0.0766 g x 1 mol / 12 g = 0.00638 mol C

 

H = 0.00638 g x 1 mol / 1 g = 0.00638 mol H

 

O = 0.017 g x 1 mol / 16 g = 0.00106 mol O

 

Divide the moles of each element by the lowest # of moles (0.00106) to obtain the mole ratio of each element in the compound.

 

C = 0.00638 / 0.00106 = 6.01 ~ 6 mol C

 

H = 0.00638 / 0.00106 = 6.01 ~ 6 mol H

 

O = 0.00106 / 0.00106 = 1.00 ~ 1 mol O

 

The empirical formula is C6H6O.