#### Conservation of Mass

• ###### Explain the law of conservation of mass applied to: a) a closed system including a precipitation reaction in a closed flask b) a non-enclosed system including a reaction in an open flask that takes in or gives out a gas
• The Law of Conservation of Mass states that no matter is lost or gained during a chemical reaction.
• Mass is always conserved, therefore the total mass of the reactants is equal to the total mass of the products, which is why all chemical equations must be balanced.
• A precipitation reaction is one in which two solutions react to form an insoluble solid called a precipitate.
• If the reaction flask is closed and no other substance can enter or leave the system, then the total mass of the reaction flask will remain constant.
• For example, the reaction between calcium chloride and sodium sulfate produces a precipitate of calcium sulfate.
• If carried out in a closed system then the mass before and after the reaction will be the same.
• The balanced equation is:

#### CaCl2 + Na2SO4 → CaSO4 + 2NaCl

Diagram showing the conservation of mass in a precipitation reaction

• If the reaction flask is open and a gaseous product is allowed to escape, then the total mass of the reaction flask will change as product mass is lost when the gas leaves the system.
• For example, the reaction between hydrochloric acid and calcium carbonate produces carbon dioxide gas:

#### 2HCl + CaCO3 → CaCl2 + H2O + CO2

• Mass will be lost from the reaction flask unless it is closed.

#### Calculating Mass

• ###### Calculate the masses of reactants and products from balanced equations, given the mass of one substance.

Balancing Equations

Word Equations

• These show the reactants and products of a chemical reaction using their full chemical names.
• The arrow (which is spoken as “goes to” or “produces”) implies the conversion of reactants into products.
• Reaction conditions or the name of a catalyst can be written above the arrow.

Writing and Balancing Chemical Equations

• These use the chemical symbols of each reactant and product.
• When balancing equations, there needs to be the same number of atoms of each element on either side of the equation.
• The following nonmetals must be written as molecules: H2, N2, O2, F2, Cl2, Br2 and I2.
• Work across the equation from left to right, checking one element after another.
• If there is a group of atoms, for example a nitrate group (NO3) that has not changed from one side to the other, then count the whole group as one entity rather than counting the individual atoms.
• Examples:
• Acid-base neutralisation reaction:

#### NaOH + HCl → NaCl + H2O

• Redox reaction:

#### 2Fe2O3 + 3C → 4Fe + 3CO2

• In each equation there are equal numbers of each atom on either side of the reaction arrow so the equations are balanced.

Using state symbols:

State symbols are written after formulae in chemical equations to show which physical state each substance is in:

Example

###### Aluminium (s)  +  Copper (II) Oxide (s) → Aluminium Oxide (s)  +  Copper (s)

Unbalanced symbol equation:

#### Al     +     CuO     →     Al2O3     +     Cu

ALUMINIUM: There is 1 aluminium atom on the left and 2 on the right so if you end up with 2, you must start with 2. To achieve this, it must be 2Al

#### 2Al     +     CuO     →     Al2O3     +     Cu

OXYGEN: There is 1 oxygen atom on the left and 3 on the right so if you end up with 3, you must start with 3. To achieve this, it must be 3CuO.

#### 2Al     +     3CuO     →     Al2O3     +     Cu

COPPER: There are 3 copper atoms on the left and 1 on the right. The only way of achieving 3 on the right is to have 3Cu

#### 2Al     +     3CuO     →     Al2O3     +     3Cu

Calculating reacting masses

• Chemical equations can be used to calculate the moles or masses of reactants and products.

Calculate the Mass of Magnesium Oxide that can be made by completely burning 6g of Magnesium in Oxygen

• To do this use information from the question to find the amount in moles of the substances being considered.
• Then identify the ratio between the substances using the balanced chemical equation.
• For example for the following reaction:

#### 2ZnOH + Ca → Ca(OH)2 + 2Zn

• If there were 2 moles of Ca used and excess ZnOH, then you would produce 2 x 2 = 4 moles of Zn.

Example 1:

Calculate the Mass of Magnesium Oxide that can be made by completely burning 6g of Magnesium in Oxygen

#### Magnesium (s)  + Oxygen (g)  →   Magnesium Oxide (s)

Symbol Equation:

#### 2Mg     +     O2       →       2MgO

Relative Formula Mass:    Magnesium : 24             Magnesium Oxide : 40

Step 1 Calculate the moles of Magnesium Used in reaction

Moles = Mass ÷ Mr                                       Moles = 6 ÷ 24 = 0.25

Step 2 Find the Ratio of Magnesium to Magnesium Oxide using the balanced Chemical Equation

Magnesium               Magnesium Oxide

Mol                            2                                      2

Ratio                         1                                      1

Mol                          0.25                                 0.25

Moles of Magnesium Oxide = 0.25

Step 3 Find the Mass of Magnesium Oxide

Moles of Magnesium Oxide = 0.25

Mass = Moles x Mr                                       Mass = 0.25 x 40 = 10 g

Mass of Magnesium Oxide Produced = 10 g

Example 2:

Calculate the Mass, in Tonnes, of Aluminium that can be Produced from 51 Tonnes of Aluminium Oxide

#### Aluminium Oxide (s)  →   Aluminium (s)   +   Oxygen (g)

Symbol Equation:

#### 2Al2O3     →       4Al       +       3O2

Ar and Mr:                   Aluminium : 27    Oxygen : 16    Aluminium Oxide : 102

1 Tonne = 106 g

Step 1 Calculate the moles of Aluminium Oxide Used

Mass of Aluminium Oxide in Grams = 51 x 106 = 51,000,000  g

Moles = Mass ÷ Ar                             Moles = 51,000,000 ÷ 102 = 500,000

Step 2 Find the Ratio of Aluminium Oxide to Aluminium using the balanced Chemical Equation

Aluminium Oxide         Aluminium

Mol                            1                                     2

Ratio                         1                                     2

Mol                      500,000                         1,000,000

Moles of Aluminium = 1,000,000

Step 3 Find the Mass of Aluminium

Moles of Aluminium = 1,000,000

Mass in grams = Moles x Ar                Mass = 1,000,000 x 27 = 27,000,000

Mass in Tonnes = 27,000,000 ÷ 106 = 27 Tonnes

Mass of Aluminium Produced = 27 Tonnes

#### Calculating Concentration

• ###### Calculate the concentration of solutions in g dm–3
• Concentration is expressed as mass per given volume of solution, using the equation:
• Moles can be converted to grams by multiplying the moles by the relative atomic/formula mass of the substance.
• The equation can also be rearranged to solve for moles or volume:
• Volume = moles ÷ concentration
• Moles = concentration x volume

Example:

25.0 cm3 of 0.050 mol / dm3 sodium carbonate was completely neutralised by 20.00 cm3 of dilute hydrochloric acid. Calculate the concentration, in mol / dm3 of the hydrochloric acid.

Step 1 Calculate the amount, in moles, of sodium carbonate reacted by  rearranging the equation for amount of substance (mol) and dividing by 1000 to convert cm3 to dm3

Amount of Na2CO3  =  (25.0 x 0.050) ÷ 1000  =  0.00125 mol

Step 2 Calculate the amount, in moles, of hydrochloric acid reacted

#### Na2CO3  +  2HCl  →  2NaCl  +  H2O  +  CO2

1 mol of Na2CO3 reacts with 2 mol of HCl, so the Molar Ratio is 1 : 2

Therefore 0.00125 moles of Na2CO3 react with 0.00250 moles of HCl

Step 3 Calculate the concentration, in mol / dm3 of the Hydrochloric Acid

1 dm3 = 1000 cm3

Volume of HCl =  20 ÷ 1000  =   0.0200 dm3

Concentration HCl (mol / dm3) =  0.00250 ÷ 0.0200  =  0.125

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### Author: Morgan

Morgan’s passion for the Periodic Table begun on his 10th birthday when he received his first Chemistry set. After studying the subject at university he went on to become a fully fledged Chemistry teacher, and now works in an international school in Madrid! In his spare time he helps create our fantastic resources to help you ace your exams.