What is the empirical formula of the compound with a mass percent composition of 70.0% Fe and 30.0% O? (Type your answer using the format CH4 for CH4
Fe2O3
Take a 100 g sample which gives you
70.0 g Fe and 30.0 g O.
mols Fe = 70.0/55.85 = about 1.25
mols O = 30.0/16 = 1.875
Find the ratio; the easy way to do that is to divide the smaller number by itself and do that to the other number too.
1.25/1.25 = 1
1.875/1.25 = 1.5
This obviously 2:3 so the formula is
Fe2O3.
To determine the empirical formula of a compound, we need to find the ratio of atoms present in the compound based on the mass percent composition of each element.
In this case, we have a compound with a mass percent composition of 70.0% Fe and 30.0% O.
Step 1: Assume we have a 100g sample of the compound. This means we have 70.0g of Fe and 30.0g of O.
Step 2: Calculate the number of moles for each element. To do this, divide the mass of each element by its molar mass. The molar mass of Fe is 55.85 g/mol, and the molar mass of O is 16.00 g/mol.
Number of moles of Fe = 70.0g Fe / 55.85 g/mol = 1.25 mol Fe
Number of moles of O = 30.0g O / 16.00 g/mol = 1.875 mol O
Step 3: Divide the number of moles of each element by the smallest number of moles to obtain the simplest, whole number ratio of atoms.
Smallest number of moles = 1.25 mol Fe (rounded to 1 decimal place)
Number of moles of Fe / Smallest number of moles = 1.25 mol Fe / 1.25 mol Fe = 1
Number of moles of O / Smallest number of moles = 1.875 mol O / 1.25 mol Fe = 1.5
Step 4: Multiply all the ratios by a whole number to convert them into whole numbers if necessary.
In this case, multiplying by 2 will give us the simplest whole number ratio:
Fe:O = 2:3
Step 5: Write the empirical formula using the ratio of atoms.
Fe2O3
Therefore, the empirical formula of the compound with a mass percent composition of 70.0% Fe and 30.0% O is Fe2O3.