A compound is 52.17 % C, 13.04 % H and 34.78 % O. What is its empirical formula?
assume you have 100 grams of the stuff.
then you have
52.17gC, convert to moles of C
13.04gH, convert to moles of H
34.78gO, conver to moles of O
Now look at the mole ratio. It is easiest if yo take the smaller number, and divide it into the others. In my head, I see something like this
C 4, H12, O4
then dividing it in my head
CH3O or CH2OH
do it with your calculator.
To determine the empirical formula of a compound, we need to find the simplest whole number ratio of the atoms present in the compound.
Step 1: Assume we have 100 grams of the compound so that the percentages become grams. This allows us to work with tangible values.
Step 2: Convert the given percentages into grams.
For Carbon (C): 52.17% of 100g = 52.17g
For Hydrogen (H): 13.04% of 100g = 13.04g
For Oxygen (O): 34.78% of 100g = 34.78g
Step 3: Convert the grams of each element to moles by dividing by their atomic mass.
The atomic mass of Carbon (C) = 12.01 g/mole
The atomic mass of Hydrogen (H) = 1.008 g/mole
The atomic mass of Oxygen (O) = 16.00 g/mole
For Carbon (C): moles = 52.17g / 12.01 g/mole = 4.34 moles
For Hydrogen (H): moles = 13.04g / 1.008 g/mole = 12.92 moles
For Oxygen (O): moles = 34.78g / 16.00 g/mole = 2.17 moles
Step 4: Divide each mole value by the smallest mole value to obtain the simplest whole number ratio. Round to the nearest whole number if needed.
The ratio becomes approximately: C4.34 : H12.92 : O2.17, which can be simplified to C2H6O.
Therefore, the empirical formula of the compound is C2H6O.