A compound that contains only Fe and O is 69.9% Fe by mass. What is the empirical formula of theis compound?
Attempt: 69.9g(1/56): 1
30.1g(1/16): 2
I got FeO2
but the answers say that it is Fe2O3
I got it since O comes out to be 1.5 * 2 = 3
To determine the empirical formula of the compound that contains only Fe (iron) and O (oxygen), you need to find the ratio of their atoms in the compound.
You correctly calculated the molar mass of the compound based on the percentage composition of Fe. The molar mass is given by:
(69.9g Fe / 56g/mol) + (30.1g O / 16g/mol) = 1.25 mol Fe + 1.88 mol O = 3.13 mol total
Now, divide the molar amounts of each element by the smallest value obtained. In this case, Fe has the smallest molar amount of 1.25 mol.
Fe: 1.25 mol / 1.25 mol = 1
O: 1.88 mol / 1.25 mol = 1.5
The resulting ratios indicate that there is one Fe atom and 1.5 O atoms in the empirical formula. To obtain whole numbers, we can multiply the ratio by 2 to eliminate the decimal:
Fe: 1 x 2 = 2
O: 1.5 x 2 = 3
Therefore, the empirical formula of the compound is Fe2O3, which indicates there are 2 Fe atoms and 3 O atoms in each molecule of the compound.
It seems like there might have been a calculation mistake in your attempt, leading to the incorrect empirical formula. Remember to double-check your calculations and consider rounding errors when working with decimal values.