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

Apr 29, 2017

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.

$\left(\text{4mol H")/("2mol O") = ("2mol H")/("1mol O}\right) = {H}_{2} O$

Apr 30, 2017

The empirical formula is ${H}_{2} O$.

#### 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...........

$\text{Moles of hydrogen} = \frac{4.0 \cdot g}{1.0 \cdot g \cdot m o {l}^{-} 1} = 4 \cdot m o l$

$\text{Moles of oxygen} = \frac{32.0 \cdot g}{16.0 \cdot g \cdot m o {l}^{-} 1} = 2 \cdot m o l$

If we divide thru by the smallest molar quantity (that of oxygen) we get ${H}_{2} O$ as the empirical formula.