Chemical formulas and the mole concept

Elements

Element Name (Symbol) - Atomic Number - Molar Mass, Hydrogen (H) - 1 - 1.01g, Helium (He) - 2 - 4.00g, Lithium (Li) - 3 - 6.94g, and so on.

Compounds

A compound contains more than one element. For example, a molecule of water contains two hydrogen atoms and one oxygen atom.

Formulas of compounds

A compound containing 40.00% carbon, 6.73% hydrogen and 53.27% oxygen by mass, determine the empirical formula.

C 40.00/12.01 = 3.33  1
H 6.73/1.01 = 6.66    2
O 53.27/16.00         1      = CH2O

Empirical formula: CH2

Molecular formula: C2H4

Avogadro’s constant

A mole of hydrogen gas contains 6.02 x 10^23 hydrogen atoms

Chemical reactions and equations

Chemical equations

Reactant + Reactant → Product + Product

State symbols

(s) - solid (l) - liquid (g) - gas (aq) - aqueous

Ionic equations

Ag+(aq) + NO3-(aq) + Na+(aq) + Cl-(aq) → AgCl(s) + Na+(aq) + NO3-(aq)

Na+(aq) and NO3-(aq) are spectator ions and do not take part in the reaction.

Mass and gaseous volume relationships

Measurement of molar quantities

Gases - Mass or volume may be used for gases.

Liquids - may be weighed or the volume. Density is usually expressed in g cm-3.

Solids - Weighted to obtain mass. 1.000kg = 1000g.

Solutions - 1.000 liter = 1.000 dm3 = 1000 cm

• Molar mass of a solid dissolved in distilled water to get a total volume of 1.00dm-3 would give you a 1.00 mol dm-3 solution.

Ideal gas equation

PV = nRT

• P = pressure in Pa (Nm-2) (1 atm = 1.013 x 10^5 Pa)

• T = absolute temperature in K

• V = volume in m^3 (1cm^3 = 1x10^-6 m^3)

• n = number of moles

• R = gas constant = 8.314 J K^-1 mol^-1

At a constant temperature: as the volume decreases the concentration of the particles increases and pressure increases.

Molar volume of a gas and calculations

Molar volume of a gas

At 273K and 1.013 x 10^5 Pa (1 atm) pressure this volume is 2.24 x 10^-2 m^3 (22.4dm^3 or 22400 cm^3)

(P1*V1)/T1 = (P2*V2)/T2

• P = pressure

• V = volume

• T = temperature

As long as the units are the same on both sides, it will work.

Calculations from equations

1. Write down the correct formulas for all the reactants and products.

2. Balance the equation to obtain the correct stoichiometry of the reaction.

3. If the amounts of all reactants are known work out which are in excess and which one is the limiting reagent. By knowing the limiting reagent the maximum yield of any of the products can be determined.

4. Work out the number of moles of the substance required.

5. Convert the number of moles into the mass or volume.

6. Express the answer to the correct number of significant figures and include the appropriate units.