The compound known as diethyl ether, commonly referred to as ether, contains carbon, hydrogen, and oxygen. A 1.751 g sample of ether was combusted in an oxygen rich environment to produce 4.159 g of CO 2 ( g ) and 2.128 g of H 2 O ( g ) .
Insert subscripts to complete the empirical formula of ether.

The empirical formula for the given compound is

Explanation:

The chemical equation for the combustion of ether follows:

where, 'x', 'y' and 'z' are the subscripts of carbon, hydrogen and oxygen respectively.

We are given:

Mass of

Mass of

Mass of sample = 1.751 g

We know that:

Molar mass of carbon dioxide = 44 g/mol

Molar mass of water = 18 g/mol

For calculating the mass of carbon:

In 44 g of carbon dioxide, 12 g of carbon is contained.

So, in 4.159 g of carbon dioxide, of carbon will be contained.

For calculating the mass of hydrogen:

In 18 g of water, 2 g of hydrogen is contained.

So, in 2.128 g of water, of hydrogen will be contained.

Mass of oxygen in the compound = (1.751) - (1.134 + 0.236) = 0.381 g

To formulate the empirical formula, we need to follow some steps:

Moles of Carbon =

Moles of Hydrogen =

Moles of Oxygen =

For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is 0.0238 moles.

For Carbon =

For Hydrogen =

For Oxygen =

The ratio of C : H : O = 4 : 10 : 1

Hence, the empirical formula for the given compound is

1. for the first question, we must find the mass of the anhydrous salt, mgso₄. the molar mass for mgso₄·7h₂o is 246.47 g/mol, while that of mgso₄ is 120.37 g/mol.7.834 g mgso₄·7h₂o * 1 mol mgso₄·7h₂o/246.47 g * 1 mol mgso₄/1 mol mgso₄·7h₂o * 20.37 g mgso₄/mol = 0.647 g mgso₄

2. mass of water = mass of sample - mass of anhydrous salemass of water = 7.834 - 0.647 = 7.19 gpercent water in hydrate = 7.19/7.834 * 100 = 91.74%