What is the empirical formula of a hydride prepared from #4*g# of hydrogen, and #28*g# of oxygen?

You know that you have 4 grams of Hydrogen and 32 grams of Oxygen in your molecule so the first step is converting both weights given into moles. You use each molecules molecular weight (1.0079 g/mol for Hydrogen and 15.999 g/mol for Oxygen) to convert between moles and grams. These values are found on a periodic table!

#("4g H")/1 * ("1mol H")/("1.0079g H") = "4mol H"#

#("32g O")/1 * ("1mol O")/("15.999g O") = "2mol O"#

After you know how many moles of each compound you have, look at the ratio and see if you can reduce like you would a fraction.

#("4mol H")/("2mol O") = ("2mol H")/("1mol O") = H_2O#

This gives you your final answer!

The empirical formula is the simplest, whole number ratio defining constituent atoms in a species. And how do we get this? Well, we divide the elemental masses thru by the atomic masses of each element...........

#"Moles of hydrogen"=(4.0*g)/(1.0*g*mol^-1)=4*mol#

#"Moles of oxygen"=(32.0*g)/(16.0*g*mol^-1)=2*mol#

If we divide thru by the smallest molar quantity (that of oxygen) we get #H_2O# as the empirical formula.