What is the empirical formula of the compound? ?

Question

A compound contains only C, H, and N. Combustion of 28.0 mg of the compound produces 26.8 mg CO2 and 32.9 mg H2O. What is the empirical formula of the compound?

Answer 1

The first step here is to convert the masses of the products you get into moles of each of the products.

CO2: The molar mass is 12.01 + 2(16.00) = 44.01g/mol

26.8mg CO2 = 0.0268g CO2

0.0268g CO2 (1 mol CO2/44.01g CO2) = 6.09x10^-4 mol CO2

H2O: The molar mass is 2(1.01) + 16.00 = 18.02g/mol

32.9mg H2O = 0.0329g H2O

0.0329g H2O (1 mol H2O/18.02g H2O) = 1.83x10^-3 mol H2O

Upon combustion, the number of moles of H2O produced is three times that of CO2.

Now, since matter cannot be created or destroyed, the mass of the reactants must equal the mass of the products. In this case, the reactants are the unknown compound and oxygen (combustion). Solve for the mass of oxygen gas, since we know its molecular formula (O2):

mass O2 = 26.8mg CO2 + 32.9mg H2O - 28.0mg unknown

mass O2 = 31.7mg O2

Since we know the molar mass of oxygen gas is 32g/mol, we can convert mass of reacted O2 to moles of O2:

31.7mg O2 = 0.0317g O2 (1 mol O2/32g O2) = 9.91x10^-4 mol O2.

1.83x10^-3/9.91x10^-4 = 2

This tells us that for every two moles of H2O produced, 1 moles of O2 gas is required.

Now we have unknown + 1.5O2 ----> 3H2O + CO2 and can solve for the empirical formula. First, multiply the whole reaction by two to get whole numbers to work with (unknown + 3O2 ----> 6H2O + 2CO2).  We have 12 H on the right side, so the unknown compound has 12 H in its empirical formula, and we have 2 C on the right side, so it also has 2 C in the empirical formula. For oxygen, we have 6 O on the left side and 10 O on the right, so ethyl butyrate must contribute 4 O. This gives us a molecular formula C2H12O4.  Recall that we multiplied by two, so now we divide by two to get the empirical formula:

CH6O2

Hope this helps.