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Quantitative Chemistry

Relative Atomic Mass (A_r)

The relative atomic mass (Ar) of an element is its relative average mass of an atom compared to the mass of an atom of carbon-12.

Image: Pixabay

 

Relative Formula Mass (Mr)

The relative formula mass of a compound is the sum of all the relative atomic masses of the atoms in the compound. The numbers in the formula will state how many atoms are in that compound.

e.g. HCl consists of one hydrogen atom and one chlorine atom

Ar:  H=1      Cl=35.5

Mr: HCl = 1 + 35.5 = 36.5

e.g. CO₂ consists of one carbon atom and two oxygen atoms

Ar: C=12     O=16

Mr: CO₂ = (1 x 12) + (2 x 16) = 40

Ar and Mr values do not have any units.

 

 

The law of conservation of mass

The law of conservation of mass states that during a chemical reaction, no atoms are created or lost they are just rearranged.

2Mg + O₂    —->   2MgO

This means mass is never lost or gained. The total mass of the reactants equals the total mass of the products.

Image: Pixabay

 

Some reactions may show a change in mass. This is often because a reactant or product is a gas. The gas is taken in if it is a reactant or can be lost if it is a product.

e.g. Mg + 2HCl   —–>   MgCl₂  + H₂

The mass of the products would reduce as hydrogen gas is lost to the surroundings.

 

Closed systems

If a reaction occurs in a closed system, nothing can enter or leave the system and  the mass in a chemical reaction will remain constant.

Image: Pixabay

 

If it is a non-enclosed system, changes in mass can occur, such as when a gas is released. However the total mass of the reactants and products will always remain the same and the mass of the gas should be included in the reaction calculation.

 

 

The mole

 

A mole is a quantity that is used in chemistry.

One mole of any substance contains 6×10²³ atoms or particles. The number is known as Avogadro’s Constant.

• One mole of carbon contains 6×10²³ atoms
• One mole of oxygen contains 6×10²³ atoms
• One mole of hydrogen contains 6×10²³ atoms
• One mole of carbon dioxide contains 6×10²³

The mass of one mole of a substance in grams is equal to its relative atomic/ formula mass.

• One mole of carbon-12 has a mass of 12g
• One mole of oxygen-16 has a mass of 16g
• One mole of hydrogen-1 has a mass of 1g
• One mole of carbon dioxide has a mass of 40g

 

Each of these quantities would have 6×10²³ atoms or molecules. One mole of carbon atoms has the same number of atoms as you would have molecules in 1 mole of carbon dioxide.

 

Calculating the number of particles

Avogadro’s constant is given the symbol N_A.

To calculate the number of particles in a substance, the following equation can be used:

N = n × N_A

N = the number of particles in a substance
n = the number of moles (mol)
N_A = Avogadro’s constant 6.02 × 10²³

 

Balancing equations and Moles

The law of conservation of mass states that no atoms are created or lost in a reaction. This means there must be the same number of atoms of each element on both sides of the equation.

2Mg + O₂    —–>   2MgO

 

Reactants	Products
Mg=2	Mg=2
O=2	O=2

 

• This equation is balanced as there are the same number of each atom on both sides of the equation.
• Using moles in a balanced symbol equation shows the ratio of reactants to products.
• The equation also shows that two moles of magnesium react with one mole of oxygen to produce two moles of magnesium oxide.

 

 

How to work out the balancing numbers in an equation

E.g.

12g of Mg reacts with 8g of O to produce MgO. Determine the balanced symbol equation for the reaction.

Ar: Mg= 24, O= 16

Step 1: Find out the amounts of moles of each of the reactants.

Mg = 12 ÷ 24 = 0.5 mol

O₂ = 8 ÷ 32 = 0.25 mol

Step 2: Divide both amounts by the smaller value.

Mg = 0.5 ÷ 0.25 = 2

O₂ = 0.25 ÷ 0.25 = 1

2 mol of Mg reacts with 1 mol of O_2.  Using this information, it is then possible to balance the rest of the equation in the normal way.

2 Mg + O₂    —–>   2 MgO

 

Calculating moles, mass and relative formula mass

 

The number of moles of a substance can be calculated by dividing its mass by its relative formula mass.

Image: BBC bitesize

 

• Moles= Mass ÷ RFM
• Mass=Moles x RFM
• RFM= Mass ÷ Moles

 

E.g.

Calculate the number of moles in 220g of K₂S.

• K₂S consists of 2 potassium atoms (Ar 39) and 1 sulphur atom (Ar 32).
• Calculate the Mr of K₂S = (39 × 2) + 32 = 110
• moles = mass ÷ Mr
• moles = 220 ÷ 110 = 2 moles

 

What is the mass of 0.9 moles of Fe(NO₃)₃(H₂O)₉?

(Ar): Fe = 56, O = 16, N = 14, H = 1

• Calculate the Mr of Fe(NO₃)₃(H₂O)_{9 =} 404
• mass = moles × Mr
  mass = 0.9 × 404 = 6g

 

 

Working out masses in equations

 

To work out the masses involved in an equation:

Step 1: Write out a balanced symbol equation.

Step 2: Write in the relative atomic and relative formula masses for the reactants and products involved in the chemical reaction.

1. 2Mg + O₂ 2MgO
2. (2 × 24) + (2 × 16) 2x(24 + 16)

           80g —–> 80g

 

Reacting masses

In order to work out the mass of a substance that is reacting you must work out the ratio of moles of each substance.

 

e.g. Calculate the mass of water made when burning 300g of methane.

Step 1: Write a balanced equation.

CH₄ + 2O₂    —–>  2H₂O + CO₂

 

Step 2: Work out the relative formula mass of each compound.

Ar: C = 12   O = 16  H = 1

CH₄: 12 + (1 × 4) =16

O₂: 16 × 2 =32

H₂O: (2 x1) + 16 = 18

CO₂: 12 +(16×2) =44

 

Step 3: Work out the masses involved in the equation.

CH₄ + 2O₂    —–>   2H₂O + CO₂

16g  + (2 x 32)   —–>   (2x 18) + 44

 

Step 4: From the equation it can be seen that 16g of methane reacts to produce 36g of water.

 

Step 5: Work out the unknown mass of water.

• Work out the number of moles of methane by using the given mass
• Moles= Mass ÷ RFM
• Moles : 300 ÷ 16 = 18.75 mol

By looking at the equation you can see that 1 mol of methane makes 2 mol of water. The ratio is 1:2.

• Work out the number of moles of water using the ratio
• 75 mol of methane makes ( 2 x 18.75) = 37.5 moles of water
• Work out the mass of water using : Mass = Moles x RFM
• Mass = 37.5 x 18 =675g

Or use: (known mass ÷ M_r) x M_r of unknown mass

       (300 ÷ 16) x (18×2) = 675g

 

Limiting reactants

• The limiting reactant is the reactant that is completely used up in a chemical reaction.
• The limiting reactant determines how much product is made.
• The reaction will end once the reactant is used up.
• There will be some of the other reactant left over. This reactant is said to be in
• The maximum mass of a product can be calculated by working out the mass of the limiting reactant and then using that to calculate the mass of the product.

 

 

Calculating concentration using moles

The concentration of a solution tells you the amount of substance in a certain volume of liquid. The higher the concentration the more particles of a substance present in that volume.

The pressure of a gas also states how many particles are present in a specific volume.

Concentration (g/dm³) = Mass (g) ÷ volume (dm³)

 

• Concentration can be given in mol/dm³ or g/dm³
• The volume must always be given in dm³
• cm³    —–>   dm³ = ÷ 1000
• dm³    —–>   cm³ = x 1000 

Concentration in mol/dm³ of a solution can be calculated by using the equation:

• concentration = moles ÷ volume (dm³)
• number of moles (of solute) = concentration (mol/dm³) x volume (dm³)
• volume (dm³) = moles ÷ concentration (mol/dm³)

 

 

• To covert the concentration of a substance from g/dm³ to mol/dm³ divide by the M_r
• To covert the concentration of a substance from mol/dm³ to g/dm³ multiply by the M_r

 

 

Percentage Yield

• The theoretical yield is the mass of product that should be made during a chemical reaction.
• The actual yield is how much product is actually made.

 

Percentage yield =

 (actual mass of product ÷ theoretical maximum mass of product) x100

 

The loss of product can be due to:

• Product being lost when filtered.
• Some of the reactants may not react as expected and produce by-products.
• The reaction may be a reversible one and therefore the reaction may not go to completion.

 

Atom Economy

Atom economy will tell you how useful a reaction is. It is a measure of the percentage of starting materials that end up as useful products.

Atom economy =

Relative formula mass of desired product ÷ Sum of relative formula masses of all reactants

 

Gas Volumes

The volume of one mole of any gas at room temperature and pressure is 24dm³ (24000 cm³).

• Volume of gas = Amount in mol × Molar volume
• Amount in mol = Volume of gas ÷ Molar volume
• Molar volume = 24dm³

e.g.  Determine the volume of 0.55 mol of nitrogen monoxide at room temperature and pressure.

       volume = 0.55 × 24
                    = 13.2dm³

 

Avogadro’s Law

When the temperature and pressure remain constant, Avogadro’s law states that different gases that have the same volume contain equal numbers of molecules.

So for example 1 mol of argon gas occupies the same volume as 1 mol of methane gas.

e.g. H₂  (g)+ Cl₂ (g)   —–>   2HCl (g)

It can be seen from the equation that the ratio between hydrogen and chlorine is 1:1.

This means that if there is 10cm³ of hydrogen then it needs 10cm³ of chlorine to react completely. This will produce 20cm³ of hydrogen chloride gas (as the ratio is 1:2).

 

Reaction pathways

A reaction pathway describes the route taken to make a product. There is often more than one way something can be made.

Choosing a particular pathway will depend on:

• Raw materials needed (e.g. are they renewable?)
• Percentage yield
• Atom economy
• Position of equilibrium
• Rate of reaction
• Usefulness of any by-products

 

Uncertainty

All measurements have some degree of uncertainty. Uncertainty is a measure of the variability in scientific data. It is measured by finding the resolution of the scientific equipment being used or from the range of a set of scientific data.

 

Random errors

Random errors can be caused by human error such as a poor technique when taking measurements or by equipment that is faulty. They are random and cannot be predicted.

e.g. multiple balances could show different mass values for the same object.

Systematic errors

Systematic errors are produced consistently: if the experiment is repeated, the same error will occur.

Uncertainty = range of results  ÷ 2

The range is the difference between the largest and smallest value.

 

 

Questions 

1. Calculate the concentration of a solution with a mass of 2.05g and a volume of 5dm³.
2. What is the M_r of sulphuric acid?
3. Calculate the mass of water produced when burning 600g of methane.
4. Student A wanted to find out how much dilute sulfuric acid is needed to react with exactly 50.0cm3 of a sodium hydroxide solution. The results are given in the table. What is the range and the uncertainty?

Repeat	1	2	3	Mean
Volume of acid cm3	23.13	24.00	23.56	23.56

 

Answers

1. Concentration = mass ÷ volume
   Concentration = 2.05g ÷ 5dm³
   = 0.41g/dm³

 

2. M_r: H₂SO₄ = 98

 

3. 1350g

 

4. Range = 24.00 – 23.13 = 0.87cm³

          Uncertainty = 0.87 ÷ 2 = 0.44 cm³