Use the relations , molar mass , and

Phosphoric acid combines a single phosphate ion with three hydrogen ions. The correct molecular formula is . is the molecular formula for hydrophosphoric acid, is the molecular formula for phosphorous acid, and is just a charged phosphate ion, which has biochemical significance with regards to DNA structure and energy molecules like ATP. Phosphorus is not a diatomic atom.

Start by finding the molar mass of of  by adding up the molar masses of its constituent atoms.

Now, divide the molar mass of the molecule by the molar mass of its empirical formula.

In order to find the molecular formula, you will need to multiply the empirical formula by . Thus, the molecular formula is .

We're given the percent composition of the elements carbon and hydrogen in a given compound, and we're asked to determine the empirical formula.

The first thing we have to do is determine the relative molar ratios of each element. In this case, because we are given the percent composition, we can assume that we're starting off with  of the starting material. That way, we can convert each percentage directly into grams.

Following that, we need to use the molar mass of each element in order to convert it from grams to moles.

In other words, what we have found is that in our compound, for every  of carbon, there will also be  of hydrogen.

But remember, the empirical formula for any given compound tells us the lowest whole number ratio of the compound's constituent elements. Therefore, we'll need to divide each value by the lowest number out of all the values available.

What this tells us is that for every  of carbon in this compound, there will be  of hydrogen as well. This is equivalent to the previous ratio, but now it has been reduced to the lowest possible whole numbers.

Recall how to find the expression of the equilibrium constant for the simplified equation:

Since the given equation has gases, we will only consider the partial pressures of each gas in the expression for the equilibrium constant. Remember that only molecules in aqueous and gas forms are included in this expression. Pure solids and pure liquids are excluded.

Thus, we can then write the following equilibrium constant for the given equation:

Start by writing the equilibrium expression:

Now, create a chart like the following to keep track of the changes in concentration.

Initial	0.750	2.00	0.00
Change	-0.100	-0.025	0.200
Equilibrium	0.650	19.75	0.200

Since we know that the concentration of HCl decreased by , we can use the stoichiometric ratios to deduce the amount of change for the oxygen gas and the chlorine gas.

Plug in the equilibrium concentrations into the expression for the equilibrium constant.

Start by using the free energies of formation to find .

 , 

Recall the equation that links together  with the equilibrium constant, .

Plug in the given information and solve for .

Use algebra to solve for the partial pressure of .

Since  the reaction is not at equilibrium. This means that at equilibrium, the ratio of products to reactants is greater than at the given conditions. Thus, the reaction will move right towards the products to reach equilibrium.

Use an ice table and the  equation to solve.                      

Initial                                                       

Change                               

Equilibrium