Question

You perform a combustion analysis of 0.255g of a compound that contains C, H and O. You produce 0.561 g of CO₂ and 0.306g of H₂O.

A. How many grams of carbon and hydrogen did you produce?

a)0.154g C & 0.034g H b)0.225g C & 0.068g H c)0.0128g C & 0.017g H d)0.306g C & 0.278g H

B. How many grams of oxygen were there in the original compound?

a)0.081   b)0.091  c)0.055 d)0.067

C. What is the empirical formula of the compound?

a)C2H602 b)C3H8O2   c)C3H8O d)C3H6O

How many grams of oxygen were there in the original compound?

Answer 1

A)
let in compound number of moles of C, H and O be x, y and z respectively

Number of moles of CO2 = mass of CO2 / molar mass CO2
= 0.561/44
= 1.275*10^-2

Number of moles of H2O = mass of H2O / molar mass H2O
= 0.306/18
= 1.7*10^-2

Since 1 mol of CO2 has 1 mol of C
Number of moles of C in CO2= 1.275*10^-2
so, x = 1.275*10^-2
Mass of C = mol of C * molar mass of C
= 1.275*10^-2 mol * 12 g/mol
= 0.154 g

Since 1 mol of H2O has 2 mol of H
Number of moles of H = 2*1.7*10^-2 = 3.4*10^-2

Mass of H = mol of H * molar mass of H
= 3.4*10^-2 mol * 1 g/mol
= 0.034 g

Answer: a

B)
Molar mass of O = 16 g/mol


mass O = total mass - mass of C and H
= 0.255 - (0.154 + 0.034 )
= 0.067 g
Answer: d

C)

number of mol of O = mass of O / molar mass of O
= 6.7*10^-2/16.0
= 4.25*10^-3
so, z = 4.25*10^-3
Divide by smallest to get simplest whole number ratio:
C: 1.275*10^-2/4.25*10^-3 = 3
H: 3.4*10^-2/4.25*10^-3 = 8
O: 4.25*10^-3/4.25*10^-3 = 1


So empirical formula is:C3H8O
Answer: c