1.25-1.30 State Symbols & Calculations

State symbols are always written in ______________ and ____________ letters ___________ the chemical formula.

brackets, lowercase, after

state symbol for solids

(s)

state symbol for liquids

(l)

state symbol for gases

(g)

state symbol for aqueous solutions

(aq)

Aqueous solution

Solution formed when a substance is dissolved in water.

How to balance chemical equations

1. Write the word equation first.
2. Write chemical formulae for reactants and products underneath.
3. Write the number of atoms there are for each element on the left hand side and right hand side.
4. Do NOT change subscript numbers as they determine the chemical structure and properties of a substance because they indicate the number of chemically bonded atoms.
5. Instead, change the coefficients (represent number of molecules) to the smallest suitable integer so the number of atoms for each element is equal on both the left hand side and right hand side. Ensure this is done for all elements to fully balance the equation.
6. Add state symbols after each reactant and product.

Relative molecular mass (Mr)

The sum of all Ar of each atom involved in a molecule or chemical compound, multiplied by the number of that atom.

Mr = Σ (Ar of atom \* number of atoms)

Mole (mol)

A unit for measuring the mass (“amount”) of a substance containing the same number of fundamental units as there are atoms in exactly 12.000g of carbon-12. 1 mole = RFM (Mr) of atom, molecule, or formulae, in grams.

1 mole contains … particles (whether these particles are atoms, molecules, or formulae) of a substance

6\.022 × 10²³

Avogadro’s number (a.k.a. Avogadro’s constant)

The number of particles in 1 mol of a substance, it is approximately equal to 6.022 × 10²³

Mass of substance (g) =

Number of moles of substance x Mr (or Ar) of substance

Number of moles of substance =

Mass of substance (g) / Mr (or Ar) of substance

Mr (or Ar) of substance =

Mass of substance (g) / Number of moles of substance

Ratio of coefficients for reactants and products in a balanced chemical equation =

Ratio of number of moles of the reactants and products

Number of … … by … gives you the number of moles of a substance

particles, divided, Avogadro’s constant

Number of … … by … gives you the number of particles of a substance

moles, multiplied, Avogadro’s constant

Number of … … by … gives you the … in grams of a substance

moles, multiplied, Mr or Ar (molar mass), mass

… in grams of a substance, … by … gives you the number of … of a substance

Mass, divided, Mr or Ar (molar mass), moles

Atomic mass unit (amu)

A unit of mass used to express atomic and molecular masses, equal to 1/12 of the mass of a carbon-12 atom (approximately the mass of a single proton or a single neutron).

Reacting masses

Masses of required reactants in a chemical reaction

Product masses

Masses of produced products in a chemical reaction

How can you calculate reacting masses and product masses?

1. Balanced chemical equations’ coefficients show the ratio of moles for involved reactants and products. This ratio can be used to calculate the reacting as well as product masses.
2. Write the balanced chemical equation.
3. Write the Ar or Mr of the reactants and products.
4. Write the ratio of the reactants and products (equal to ratio of coefficients in the balanced chemical equation).
5. Multiply each element’s Ar or each compound’s Mr by their corresponding ratio value (or by a relevant scale factor if the mass does not fulfill the ratio).
6. The ratio of the values found in step 5 is the ratio of the reacting and product masses.

Example: 48g of Mg, how much O2 needed and how much MgO produced?

Magnesium + Oxygen → Magnesium Oxide

2Mg + O2 → 2MgO

Ratio of Mg : O2 : MgO = 2 : 1 : 2

Mg’s Ar = 24

O2’s Mr = 2 x 16 = 32

MgO’s Mr = 1 x 24 + 1 x 16 = 40

Ratio of reacting masses (for 1 mole each) = 24 : 32 : 40

Ratio of reacting masses and product masses for

Mg: O2 : MgO (for the ratio number of moles) =

(24 x 2) : (32 x 1) : (40 x 2) = 48 : 32 : 80

Hence 32g of O2 is needed and 80g of MgO will be produced. Notice how the sum of the reactant masses (48g + 32g = 80g) should be equal to the sum of the product masses (80g) due to the law of conservation of mass.

Percentage by mass (percentage composition)

Amount of an element in a compound written as a percentage of the relative formula mass.

% by mass = \[(Ar of element x Number of atoms)/(Mr of Compound)\] \* 100%

Percentage yield

Ratio showing how much of a product was actually made in an experiment compared with the amount of product that was expected. It’s the actual yield of a chemical reaction written as a percentage of the theoretical yield.

% yield = (Actual yield / Theoretical yield) x 100%

Theoretical yield

Maximum mass of a product that could theoretically be made or expected to be made in a chemical reaction.

Actual yield

Mass of product that is actually obtained from the real chemical reaction.

Atom economy

Amount of starting materials that end up as useful products

Why is actual yield always … than theoretical yield?

less, incomplete reactions, inefficiency, loss of some of the product in the reaction, etc. → actual yield always < theoretical yield → % yield always < 100%

What does high % yield indicate?

high % yield → high atom economy → high efficiency:


1. Minimizes waste of non-renewable reactants.
2. Maximizes mass of useful products produced.
3. Reduces pollution from waste products.
4. Minimizes energy used in heating chemical reactions.
5. Minimizes energy used in running factories.
6. Reduces water usage for cooling chemical reactions.