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, H 2 O  H = (2) x 1.001, O = (1) x  H 2 O =

 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: NaCl  Na = (1) x , 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  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 Na 2 CO 3 there are:  2 x 6.02 x Na + ions  1 x 6.02 x CO 3 2- ions  Specifying the formula unit is extremely important.  One mole of oxygen atoms = (1) x 6.02 x  One mole of oxygen molecules = (2) x 6.02 x  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 = C 2 H 5 OH C = (2) x = H = (6) x = O = (1) x =  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 grams.  So what this means:  There are 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 C 2 H 5 OH x 1 mol C 2 H 5 OH = x 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, C 6 H 12 O 6 has an empirical formula of CH 2 O.  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, NH 3. N 2 (g) + 3H 2 (g)  2NH 3 (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. N 2 (g) + 3H 2 (g)  2NH 3 (g)  Thus, now we can say one molecule of N 2 gas reacts with 3 molecules of H 2 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 N 2 (g) g H 2 (g)  34.0g NH 3 (g)

 Another type of problem we can solve using balanced chemical equations is this:  Consider the Haber process again.  N 2 (g) + 3H 2 (g)  2NH 3 (g)  Suppose we start with 4.8 moles of H 2, 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 H 2 x 2 moles NH 3  3.2 moles NH 3 d 3 moles H 2

 We can put individual conversions together to solve the original problem: 907 kg NH 3 requires how much hydrogen? 9.07 x 10 5 g NH 3 x 1 mol NH 3 x 3 mol H 2 x 2.02g H 2 = 1.62 x 10 5 g H g NH 3 2 moles NH 3 1 mole H 2

 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:  2H 2 (g) + O 2 (g)  2H 2 O(g)  If we put 1 mole of O 2 and 1 mole of H 2 into the reaction vessel, how many moles of H 2 O will be produced?  1 mol O 2 x 2 mol H 2 O = 2 moles H 2 O 1 mol O 2  1 mol H 2 x 2 mol H 2 O = 1 mol H 2 O 2 mol H 2  H 2 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)  ZnCl 2 (s) + H 2 (g)  How many moles of H2 are produced?

Zn(s) + 2HCl(aq)  ZnCl 2 (s) + H 2 (g)  0.30 mol Zn x 1 mol H 2 = 0.30 mol H 2 1 mol Zn  0.52 mol HCl x 1 mol H 2 = 0.26 mol H 2 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