A compound isolated from oil of mint has a characteristic cool, refreshing odour. It is frequently used in cough drops to ease the pain of a sore throat. This compound contains only C, H, and O. If 95.6 mg of this compound burns completely in O2, and yields 269 mg of CO2 and 111 mg of H2O, find the empirical formula of the compound.
Compound + O2(g) --> CO2(g) + H2O(g)
mg C = 269 x (12/44) = ?
mg H = 111 x (2/18) = ?
mg O = 95.6 - g C - g o = ?
Then convert mg to g and to mols.
Find the ratio of the mols to one another with the smallest being 1.00. The easy way to do that is to divide the smallest number by itself, then divide the other numbers by the same small number. That will give yhou the empirical formula.
To determine the empirical formula of the compound, we need to find the ratio of the elements present in the compound.
First, let's find the amount of carbon (C), hydrogen (H), and oxygen (O) in the compound.
From the given information, we know that 95.6 mg of the compound produces 269 mg of CO2 and 111 mg of H2O when burned in oxygen.
To find the amount of carbon in the compound, we need to convert the mass of CO2 to the amount of carbon. Since there is only one carbon atom in each molecule of CO2, we can directly convert the mass of CO2 to the amount of carbon:
269 mg CO2 x (1 mol CO2 / 44.01 g CO2) x (1 mol C / 1 mol CO2) x (12.01 g C / 1 mol C) = X g C
Simplifying, we find:
(269 mg x 1 mol / 44.01 g) x (12.01 g / 1 mol) = 74.06 mg C
Similarly, we can find the amount of hydrogen in the compound by converting the mass of H2O to the amount of hydrogen:
111 mg H2O x (1 mol H2O / 18.02 g H2O) x (2 mol H / 1 mol H2O) x (1.01 g H / 1 mol H) = Y g H
Simplifying, we find:
(111 mg x 1 mol / 18.02 g) x (2.02 g / 1 mol) = 12.36 mg H
Now, to find the amount of oxygen, we simply subtract the total mass of carbon and hydrogen from the mass of the compound:
95.6 mg - (74.06 mg + 12.36 mg) = Z mg O
Simplifying:
9.18 mg O = Z mg O
Now that we have the amounts of carbon, hydrogen, and oxygen in the compound, we can calculate the simplest whole-number ratio. To do this, we divide each amount by the smallest amount:
Carbon: 74.06 mg / 9.18 mg = 8.07 ≈ 8
Hydrogen: 12.36 mg / 9.18 mg = 1.35 ≈ 1
Oxygen: 9.18 mg / 9.18 mg = 1
Therefore, the empirical formula of the compound is C8H10O.