Calculations with Chemical Formulas

and Equations

The molecular mass is the average mass of a molecule of a substance expressed in amu.

For example: Water, H2O H = (2) x 1.001, O = (1) x 15.999 H2O = 18.001

The formula mass is essentially the same thing as the molecular mass except it applies to ionic compounds.

It is the mass of the smallest formula unit.For example: NaClNa = (1) x 22.999, Cl = (1) x 35.45

Chemists use the concept of the mole to deal with quantities at the molecular level that contain such large numbers.

The mole is the quantity of a substance that contains as many molecules/formula units as the number of atoms in exactly 12 grams of carbon-12. It’s called Avogadro’s number. It equals 6.02 x 1023.

The mole in ionic terms means formula units.

For example: Consider 1 mole of ethanol.

One mole of ethanol contains as many ethanol molecules as there are carbon atoms in 1 mole of carbon-12.

Or,

One mole of ethanol contains as many ethanol molecules as there are carbon atoms in 12 grams of carbon-12.

For example: In a mole of Na2CO3 there are:2 x 6.02 x 1023 Na+ ions1 x 6.02 x 1023 CO3

2- ions

Specifying the formula unit is extremely important.One mole of oxygen atoms = (1) x 6.02 x 1023

One mole of oxygen molecules = (2) x 6.02 x 1023

The molecular mass, then, is the mass of one mole of a substance.Thus, 1 mole of carbon-12 = 12g/mol

The molar mass in amu is equal to the formula mass in g/mol.EtOH = C2H5OH

C = (2) x 12.011 = H = (6) x 1.001 = O = (1) x 15.999 =

How/why do we do this? We find the molar mass of compounds because it helps us in the preparation of solutions.

To calculate these things, we need dimensional analysis.

We just calculated the mass of 1 mole of EtOH to be 46.07 grams.

So what this means:There are 46.07 grams of EtOH in one mole

of EtOH, 46.07g EtOH/mol EtOH

How many moles of ethanol is 10.0 grams of ethanol?

10.0g C2H5OH x 1 mol C2H5OH = 2.170 x 10-1 mol EtOH

46.07g EtOH

When chemists discover a new compound, they like to know the formula.

This is expressed as the percent composition which is the mass percentage of each element in the compound.

Once we know the %-comp. we can calculate the formula of the substance with the smallest whole number subscripts.

If the compound is a molecular substance, the molar mass of the compound also has to be determined in order to obtain the molecular molecular formula.Mass % of x = mass of x in the whole/mass

of the whole x 100% Break the compound down to “100

grams.” In other words, how many grams of x are there in 100 grams of the whole?

For example: Glucose, C6H12O6 has an empirical formula of CH2O.Percent composition gives us the ratios of

the numbers of atoms in a compound.This can create some confusion because

many compounds have the same empirical formla and will have the same percent composition.

To get the molecular formula (which tells us what the compound is), we need two things:1. The percent composition.2. The molecular mass.

To determine the empirical formula, convert the masses of the elements into moles.

The molecular formula of any compound is just a multiple of its empirical formula. Thus, the molecular mass is some multiple of the empirical formula mass.

For any molecular compound,

Molecular mass = n x empirical formula mass.

n is the number of empirical formula units in the compound.

n = molecular mass/empirical formula mass.

The molecular formula is obtained by multiplying the subscripts of the empirical formula by n obtained in the above equation.

Stoichiometry is the calculations of reactants and products involved in chemical reactions.

It is based on the chemical equation and the relationship between mass and moles.

To illustrate stoichiometry, we’ll look at the Haber process—the process of producing ammonia, NH3.

N2(g) + 3H2(g) 2NH3(g)

For example, a legitimate question would be, how much hydrogen do we need to produce 907 kg of ammonia? First, we need to balance the chemical

equation.

N2(g) + 3H2(g) 2NH3(g)

Thus, now we can say one molecule of N2 gas reacts with 3 molecules of H2 gas, to give 2 molecules of ammonia gas.

We could also say moles instead of molecules.

Another thing we could do would be to convert the numbers of moles in the eqution into molar masses:

28.0g N2(g) + 6.06g H2(g) 34.0g NH3(g)

Another type of problem we can solve using balanced chemical equations is this:

Consider the Haber process again.

N2(g) + 3H2(g) 2NH3(g)

Suppose we start with 4.8 moles of H2, how much ammonia can we get?

We set up a single conversion keeping the following in mind:

Always keep what we are converting FROM on the bottom of the conversion and what we are converting TO on the top.

4.8 moles H2 x 2 moles NH3 3.2 moles NH3 d 3 moles H2

We can put individual conversions together to solve the original problem:

907 kg NH3 requires how much hydrogen?

9.07 x 105g NH3 x 1 mol NH3 x 3 mol H2 x 2.02g H2 = 1.62 x 105g H2

17.0g NH3 2 moles NH3 1 mole H2

Often times we add reactants to vessels in different amounts. Some of the reactants get used up, while others remain unused.

The question often arises as to what the limiting reagent is—that is, which chemical will be used up in the chemical reaction.

The chemical not used up is called the “excess reagent.”

Let’s consider the following: 2H2(g) + O2(g) 2H2O(g)

If we put 1 mole of O2 and 1 mole of H2 into the reaction vessel, how many moles of H2O will be produced?

1 mol O2 x 2 mol H2O = 2 moles H2O

1 mol O2

1 mol H2 x 2 mol H2O = 1 mol H2O

2 mol H2

H2 is the limiting reagent in this case.

To calculate the amount of product obtained from a particular chemical reaction, you need to know which reactant is the limiting reagent, andonly after you know this can you calculate the amount of product obtained.

Consider the following:

Zn(s) + 2HCl(aq) ZnCl2(s) + H2(g)

How many moles of H2 are produced?

Zn(s) + 2HCl(aq) ZnCl2(s) + H2(g)

0.30 mol Zn x 1 mol H2 = 0.30 mol H2

1 mol Zn

0.52 mol HCl x 1 mol H2 = 0.26 mol H2

2 mol HCl

HCl is the limiting reagent.

The numbers we usually obtain during these stoichiometric calculations are called the theoretical yields.

For many different reasons, the actual yield is often much less.

Thus, we often want to know the percentage yield of a particular reaction. To do this, we use the following equation:

% yield = actual yield x 100%theoretical yield