Types of Chemistry Problems

Each type of problem is worked out only once. Several other examples are also given with answers in parentheses for you to practice. I will use these in lecture as an instructional tool.

Naming Problems: See chart of ion names and activity for practice problems.
Generally, the main tool for ensuring that your formula or name is correct is to be able to identify ions (monatomic and polyatomic) and their charges. Charges must balance so that the ionic compound as a whole has a charge of 0. Remember that metal ions often can have more than one charge (but not at the same time!) so that you can have, for example, Fe2+ [iron (II)] and Fe3+ [iron (III)]
Molar Mass: The molar mass of a compound is the total of the molar masses of its constituent atoms. The molar mass of an element is the mass given in the periodic table expressed in grams. This is the average atomic mass and takes into account the mass of each elements’ isotopes and their abundance.

The molar mass of NaCl is 58.44 g/mol because 22.99 g/mol (Na) + 35.45 g/mol (Ca) = 58.44 g/mol.
CaCl2: Ca, 40.08 g/mol; Cl, 35.45 g/mol
(1 × 40.08 g/mol) + (2 × 35.45 g/mol) = 110.98 g/mol

When a compound contains a polyatomic ion or a repeated unit, multiply the subscripts inside the parentheses by the subscript outside the parentheses:
Ca3(PO4)2: Ca, 40.08 g/mol; P, 30.97 g/mol; O, 15.99 g/mol
(3 × 40.08 g/mol) + (2 × 1 × 30.97 g/mol) + (2 × 4 × 15.99 g/mol) =

CH3(CH2)3Cl: C, 12.01 g/mol; H, 1.01 g/mol; Cl, 35.45 g/mol
(1 × 12.01 g/mol) + (3 × 1.01 g/mol) + (3 × 12.01 g/mol) + (3 × 2 × 12.01 g/mol) + (1 × 35.45 g/mol) = 92.58 g/mol

Others to try: XeF2 (169.29 g/mol), SnCl4 (260.51 g/mol), SnCl2 (189.61 g/mol), CO2 (44.01 g/mol), NH3 (17.04 g/mol)
Percentage Composition: Each element in a compound forms a percentage of the total molar mass of that compound. The percentage composition of a compound gives the percent of the total mass of the compound represented by a particular element. The percentage by mass of an element can be calculated by dividing the molar mass of the element, times its subscript in the formula, by the total molar mass of the compound. To express the answer as a percent instead of a decimal fraction, multiply the answer by 100. The procedure can be repeated for each element in the compound and the resulting percentages should add up to 100%.

percentage hydrogen: mass of 2 mol H/molar mass of H2O = 2 × (1.01 g/mol)/(18.02 g/mol) = 0.1121
multiply by 100 to get the answer as a percent: 0.1121 x 100% = 11.21% H
percentage oxygen: mass of 1 mol H/molar mass of H2O = (16.00 g/mol)/(18.02 g/mol) = 0.8879
multiply by 100 to get the answer as a percent: 0.8879 x 100% = 88.79% O
88.79% O + 11.21% H = 100% H2O

Note: once you figure out the percentage for all but one element in a compound, just subtract the total percentages so far from 100% to find the last.

Others to try: CH4 (74.83% C; 25.17% H), LiAlH4 (18.28% Li; 71.07% Al; 10.65% H), NaOH (57.48% Na; 40.00% O; 2.52% H)
Empirical Formulas: You can use percentage composition data to find the formula of a compound. Rather, to be more specific, you can find the empirical formula of the compound. If you also have the total molar mass of the compound, you can usually find the molecular formula, too. The simplest way to find the formula is to assume you have 100 g of the compound and use the percentages given as amounts in grams. Convert grams to moles for each element and find the smallest whole number ratio of the moles of each element.

Given: a compound consisting of 63% Mn and 37% O

Assuming 100 g of the compound, you have 63 g Mn and 37 g O
63 g Mn × (1 mol Mn)/(54.94 g Mn) = 1.1 mol Mn
37 g O × (1 mol O)/(16.00 g O) = 2.3 mol O

To find the smallest whole number ratio divide the number of moles of each element by the number of moles found for the element present in the smallest number of moles:
1.1 mol Mn/1.1 = 1 mol Mn      2.3 mol O/1.1 = 2.1 mol O
The best ratio is Mn:O::1:2 and the formula is MnO2

Others to try: 39.32% Na, 60.66% Cl (NaCl); 30.45% N, 69.55% O (NO2); 34.97% Fe, 65.03% O (Fe2O3); 40.00% C, 53.28% O, 6.72% H (CH2O)
Molecular Formulas: Using an empirical formula you can find the molecular formula of a compound if you know its molar mass. All you need to do is find the formula weight of the empirical formula and find out how many times this mass can be divided into the given molar mass.

Given: empirical formula: CH2O; molar mass: 60.06 g/mol
Find the molar mass of the empirical formula (this is called the formula weight): it is 30.03 g/mol (see above for method)
Divide the given molar mass by the formula weight: 60.06/30.03 = 2
The factor you are looking for is 2: multiply this factor by the subscripts in the empirical formula to find the molecular formula: C2H4O2

Others to try: CH2O, 180.18 g/mol (C6H12O6); P2O5, 283.89 g/mol (P4O10); HgCl, 472.08 g/mol (Hg2Cl2)
Finding the Oxidation Number: Use the following rules to assign oxidation numbers:
1) Free elements have an oxidation number of 0 (H2, Br2, Na, Be, K, P, Fe, etc. have an oxidation number of 0).
2) For ions made of only one atom, the oxidation number is the same as the charge. All alkaline metals have an oxidation number of +1, alkaline earth metals have an oxidation number of +2 and aluminum always has an oxidation number of +3 in compounds.
3) Oxygen almost always has an oxidation number of –2 (exceptions are in H2O2 and O22– where it is –1)
4) Hydrogen has an oxidation number of +1 unless bonded to metals in binary compounds when it is –1.
5) Fluorine always has an oxidation number of –1. Other halogens generally have negative oxidation numbers except when they appear with O in acids when they have positive oxidation numbers.
6) The sum of the oxidation numbers of the elements in a neutral molecule must add up to zero. In a polyatomic ion the oxidation numbers must add up to the charge on the ion.
7) Oxidation numbers are usually but do not have to be integers: oxygen in O2 has an oxidation number of -½

Examples: PBr5 (P: +5; Br: –1), PBr3 (P: +3; Br –1), H2SO4 (H: +1; S: +6; O: –2), CCl4 (C: +4; Cl –1)
Balancing Chemical Equations: The following steps can be used to balance a chemical equation:
  1. Count the atoms of each element in the equation in its unbalanced form. Make a table for the reactant side (left) and one for the product side (right) showing the quantity of each element or polyatomic ion.
  2. Balance the atom or polyatomic ion that appears only once on each side by inserting a coefficient in front of the whole formula. Note: you cannot change the subscripts in the formulas since this would change the substance in the reaction! Sometimes an element or polyatomic ion will appear more than once on one side. Save these instances for later.
  3. Balance the other elements and polyatomic ions one by one, leaving O and H for last.
  4. Update your table for each change you make to the equation.
  5. When you are done, each element or polyatomic ion will have the same number next to it in both the reactant table and the product table.
H2 + O2 → H2O     2H2 + O2 → 2H2O
Reactants Products Reactants Products
H = 2
O = 2
H = 2
O = 1
H = 4
O = 2
H = 4
O = 2
P4 + KClO3 → KCl + P2O5     3P4 + 10KClO3 → 10KCl + 6P2O5
Reactants Products Reactants Products
P = 4
K = 1
Cl = 1
O = 3
P = 2
K = 1
Cl = 1
O = 5
P = 12
K = 10
Cl = 1
O = 30
P = 12
K = 10
Cl = 1
O = 30
Others to try: C3H8 + O2 → CO2 + H2O      |      (C3H8 + 5O2 → 3CO2 + 4H2O)
HCl + Cu → CuCl2 + 2      |      (2HCl + Cu → CuCl2 + 2)
NaClO3 → NaCl + O2      |      (2NaClO3 → 2NaCl + 3O2)
Cr2(SO4)3 + K3PO4 → K2SO4 + CrPO4      |      (Cr2(SO4)3 + 2K3PO4 → 3K2SO4 + 2CrPO4)
Stoichiometry: The key to stoichiometry is to always convert everything to moles before you even begin. When you have found the answer you can always convert to mass or volume.
Here are some sequences with which to work in order to solve different types of problems:
1. Starting with the number of moles of one of the substances in the balanced chemical equation you can easily find the number of moles of one of the other substances using the relevant molar ratio. Make sure to put the substance you know on the bottom of the ratio and the substance you need to find on the top of the ratio before you multiply.
2. Starting with a mass requires a further step: convert that mass into a number of moles using the molar mass of the substance. Then simply follow the same steps used above.
3. If the answer required is a mass, simply convert the number of moles you find to the mass using the molar mass of the substance in question.
4. If you are given a volume and a density, simply find the mass using D = m/V and then calculate moles and complete the problem. Or simply use this five-step method:
  1. Write correct formulas for all reactants and products, and balance the resulting equation.
  2. Convert the quantities of some or all given or known substances (usually reactants) into moles.
  3. Use the coefficients in the balanced equation to calculate the number of moles of the sought or unknown quantities (usually products) in the problem.
  4. Using the calculated numbers of moles and the molar masses, convert the unknown quantities are required (usually grams).
  5. Check that your answer is reasonable in physical terms.
C3H8 + 5O2 → 3CO2 + 4H2O
Moles Example: You are told that a bottle contains 40 mol C3H8 (propane) and asked how many moles of O2 you would need to burn it all.
The molar ratio is 5 mol O2 to 1 mol C3H8.
40 mol C3H8 × 5 mol O2/1 mol C3H8 = 200 mol O2
Mass Example: Given 32 g of O2 and an excess of C3H8 how much water would the reaction produce, in grams?
Convert mass to moles: 32 g O2 × 1 mol/16 g = 2 mol O2
Use molar ratio (4 mol H2O to 5 mol O2) to find moles of water: 2 mol O2 × 4 mol H2O/5 mol O2 = 1.6 mol H2O
Convert moles back to mass: 1.6 mol H2O × 18.02 g/1 mol = 28.83 g
Others to try: S + 6HNO3 → H2SO4 + 6NO2 2H2O
starting w/14 g S; how many grams of HNO3 are needed? (165.07 g)
3 mol H2O produced; how many grams of S reacted? (48.098 g)
10.0 L NO2 produced (D = 2.05 g/L); how many moles of S reacted? (3.74 ×10-1 mol)
Limiting Reagents: In nearly any reaction it will be the case that the physical quantities of the reactants will not be present in a perfectly stoichiometric amount. One of the reactants will limit the total output of the reaction because it will be used up first. These examples will show how to figure out which reactant is the limiting reactant and which is the excess reactant.
Generally, this is simply a special case of a stoichiometric calculation. Given an amount for both/all reactants you need to find out which one will run out first. The good thing about this is that it does not matter which of two reactants you choose, you can find the limiting reactant either way.
S + 3F2 → SF6 Four moles of S are added to 20 moles of F2
The number of moles of F2 needed to completely react with four moles of S is: 4 mol S × 3 mol F2/1 mol S = 12 mol F2
Thus, F2 is the excess reactant since we have more than 12 moles of fluorine gas and S the limiting reactant
But say we started with F2:
The number of moles of S needed to react with 20 moles of F2 is: 20 mol F2 × 1 mol S/3 mol F2 = 6.7 mol S
Thus, S is the limiting reactant since we do not have this much sulfur and F2 is the excess reactant
Once you have found which is the limiting reactant you can use that molar amount to find the theoretical yield of the reaction. Remember the advice from the general stoichiometry problems: convert masses to moles before trying to solve the problem!
Others to try: 2Al + Fe2O3 → Al2O3 + 2Fe
100 g Al/150 g Fe2O3 (1.85 mol Fe2O3 needed: this is the limiting reactant since only 0.94 mol are present; 0.94 mol Al2O3 produced; 95.84 g)
124 g Al/601 g Fe2O3 (2.3 mol Fe2O3 needed: this is the excess reactant; 2.3 mol Al2O3 produced; 235 g)
Percent Yield: Limiting reagents are not the only thing that limits the yield of a given product. The actual yield of a reaction is always less than 100% of what the stoichiometry predicts. Sometimes how much less is very difficult to measure, so that it appears that 100% product formation has occurred. Generally, however, much less than 100% of the theoretical yield is achieved. Therefore it is useful to know how to find the percent yield.
5Ca + V2O5 → 5 CaO + 2V
Theoretical Yield
Calculate the theoretical yield when 1.54 × 103 g of V2O5 react with 1.96 × 103 g of Ca.
First, we find the limiting reactant:
1.54 × 103 g · 1 mol/181.881 g = 8.47 mol V2O5
1.96 × 103 g · 1 mol/40.078 g = 48.9 mol
The number of moles of Ca to react completely with 8.47 mol V2O5 is 8.47 mol V2O5 · 5 mol Ca/1 mol V2O5 = 42.4 mol
So Ca is in excess and we should use the molar amount of V2O5 to determine the theoretical yield of V.
8.47 mol V2O5 · 2 mol V/1 mol V2O5 = 16.9 mol V
Percent Yield What is the percent yield if 803 g of V are obtained?
First, find out how many grams the theoretical yield of V represents: 16.9 mol · 50.941 g/mol = 863 g
Then just divide the actual yield (given) by the theoretical yield: 803g/863g = 0.930 × 100% = 93.0%
(notice how the units cancel out completely to give a unit-free ratio)
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