Unit 3

Calculations with Chemical Formulas

Types of Reactions

Synthesis reaction - two substances (generally elements) combine and form a compound.

Ionic compounds as products are always solids.

Decomposition reaction - a compound breaks up into the elements or in a few to simpler compounds.

  • carbonates produce carbon dioxide gas
  • nitrates produce nitrogen oxide gases
  • sulfates produce sulfur oxide gases
  • chlorates produce oxygen gas
  • oxide produces oxygen gas
  • hydroxides produce water vapor
  • peroxides (e.g. H2O2) produce oxygen gas and oxides (e.g. O2+ H2O)
  • metal oxide + water = base
  • non metal oxide + water = acid

Combustion reaction - occur when a hydrocarbon reacts with oxygen gas (also known as burning).

Fire triangle: a fuel (hydrocarbon), oxygen to burn it with, spark to ignite it.

Products in combustion are always carbon dioxide and water (although incomplete burning does cause some by-products like carbon monoxide).

Balancing Equations

  • balance only one element at a time
  • balance single elements last
  • leave oxygen and hydrogen to balance last
  • polyatomic ions that remain unchanged from reactant side to product side, keep as one balance these first

Don't forget to re-check the equation at the end, as amounts may have shifted.

Weights and Masses

Formula weight (FW) - sum of the atomic weights for the atoms in a chemical formula.

Molecular weight (MW) - the sum of the atomic weights of the atoms in a molecule.

Ionic compounds form three-dimensional crystal structures, so they only have empirical formulas and formula weights, not molecular formulas or molecular weights.

Molar mass - mass of 1 mol of a substance.

This is the mass listed on the periodic table (e.g. carbon has a mass of 12.01 g/mol, also 12.01 amu).

Avogadro's Number

Avogadro's Number is ~6.022 * 10^23

It's the number of particles in 1 mol.

A mol is a unit like a dozen, it's a measure of items of a type.

1 mol H2O contains 2 mol hydrogen and 1 mol oxygen

Moles are used to convert between grams and number of atoms or molecules of a substance — they're the connector unit.

Determining Empirical Formula from Percent Composition

Learn by example:


A compound of B and H is 81.10% B. What is its empirical formula?


When given percentages, use 100g to calculate

81.10g B * (1 mol / 10.81g) = 7.502 mol B

100 - 81.10g H * (1 mol / 1.01g) = 18.713 mol H

Divide each calculated result by lower number of moles:

7.502 / 7.502 = 1

18.713 / 7.502 = about 2.5 (you'll have to round, there's always experimental error)

Then, multiply to get a whole-number empirical formula from that ratio


Determining Molecular Formula from Empirical Formula

For this you need the molecular mass, which is found through experimentation.

Then, take the mass of the empirical formula (e.g. C2H3 = 27 g/mol) and compare with molecular mass (e.g. 81 g/mol), to find how much you need to multiply the empirical formula by to get the molecular formula (e.g. 81/27 = 3, so molecular formula is C6H9).


A hydrate is a salt that when crystallized forms an aqueous solution that incorporates a fixed amount of water in its crystalline matrix, even while appearing dry.

The number of moles of water present per one more of anhydrous salt is usually a simple whole number.


  1. calculate the mass lost during the heating, which is the mass of H2O lost
  2. calculate moles of H2O lost
  3. calculate moles of substance not including water
  4. similarly to finding empirical formulas, divide and find the ratio of substance : H2O

Combustion Analysis

in these problems, you are given the mass of the hydrocarbon, and resulting masses of CO2 and H2O.

C is determined by calculating moles of CO2 produced.

H is determined by calculating moles of H2O produced, then multiplying by two (because it's H2, there's 2 mol H per 1 mol H2O).

If the hydrocarbon is of the form CH_O or CH_N, you can calculate the mass of the O or N by subtracting the masses of C and H from the mass of the hydrocarbon.

Then, find the moles of O or N, and find the empirical formula by dividing moles by the lowest number of moles (like usual) and finding the ratio.

Limiting Reactant

The limiting reactant is the one that will be used up first in the reaction, and restricts the amount of product that can be created.

  • to calculate which reactant is the limiting reactant, convert the amount of each reactant given to moles
  • then, compare these amounts with the moles of reactant needed (remember, the coefficient = the # of moles needed)
  • just thinking about these ratios should make clear which reactant is limiting