Question

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

Answer 1

Yep!

Explanation:

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!

Answer 2

The empirical formula is `H_2O`.

Explanation:

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.