EL6 - Ar, Mr, Percentage Yield and Balancing Equations

Question 1

What do formula units include?

Answer 1

Single atoms, molecules, or groups of ions.

In most elements the formula unit is a single atom so their rfm = ram.
There are some exceptions – oxygen gas molecules are diatomic so the rfm is twice the ram of oxygen.

In many covalent compounds the formula unit is a molecule, e.g. in methane the formula unit is the CH4 molecule.

However in ionic compounds the formula unit is a group of ions. In calcium nitrate the formula (or repeating unit) is Ca(NO3)2 and contains a group of three ions – one Ca2+ and two NO3-. These groups of ions are not labelled with any special name so you just use the general name formula unit when referring to ionic compounds.

Question 2

rfm in g = molar mass

How do you calculate the amount in moles of formula units?

Answer 2

amount of moles in formula units = mass / molar mass

NOTE / EXAMPLE:

So 8g of methane contains 0.5 mol of CH4 formula units.
And 41g of calcium nitrate contains 0.25 mol of Ca(NO3)2 formula units.

As well as describing 16g of methane as consisting of one mole of formula units (or molecules) of CH4, you can also say that it contains one mole of carbon atoms and (1 x 4 =) four moles of hydrogen atoms.
Similarly, 164g of calcium nitrate contains one mole of formula units of Ca(NO3)2 – it also contains one mole of Ca2+ two moles 2 mol of NO3^- ions.

Question 3

What does Avogadro’s mean? So what actually is it and what’s its value?

Answer 3

The number of formula units in one mole of a substance.

6.02 x 10^23 formula units per mole.

One mole of atoms, molecules and electrons will all contain 6.02 x 10^23 mol-1.

Question 4

NOTE:

CHEMICAL FORMULA USING MOLES

Moles can be used to work out chemical formulae.
If a known mass of magnesium is reacted with oxygen to form magnesium oxide, you can find out the mass of oxygen that combined with the magnesium.

Worked example:

In an experiment, 0.84g of magnesium was burnt and combined with 0.56g of oxygen. Using these results, work out the chemical formula of the magnesium oxide produced.

Answer 4

Step 1:
Calculate the amount in moles of magnesium atoms in the reaction.

0.84g / 24 g mol-1 = 0.035 mol-1

Step 2:
Calculate the amount in moles of oxygen atoms in the reaction.

0.56g / 16 g mol-1 = 0.035 mol

Step 3:
Calculate the ratio of moles of atoms of Mg:O in magnesium oxide.

1:1

Step 4:
Write the formula of magnesium oxide.

MgO.

Question 5

NOTE:

CHEMICAL FORMULA USING PERCENTAGE MASS

Percentage mass can be used to work out chemical formulae.

Worked example:

The results from an experiment indicate that methane contains 75% mass of carbon and 25% mass of hydrogen.
Work out the formula of methane.

Answer 5

Step 1:
Calculate the mass of carbon in 100g of methane.

100g of methane contains 75g carbon.

Step 2:
Calculate the amount of carbon atoms in 75g.

75 / 12gmol-1 = 6.25 mol

Step 3:
Calculate the mass of hydrogen in 100g of methane.

100g methane contains 25g hydrogen.

Step 4: Calculate the amount of hydrogen atoms in 35g.

25 / 1gmol-1 = 25 mol

Step 5:
Calculate the ratio of moles of atoms of C:H in methane.

6.25 : 25 = 1:4

Step 6:
Write out formula of methane.

CH4

Question 6

NOTE:

In a methane molecule a central carbon atom is surrounded by four hydrogen atoms – the simple ratio of atoms is the same as the formula of the molecule.

This isn’t the case for all substances.

The molecular formula tells you the actual numbers of different types of atom.

The molecular formula of ethane is C2H6, but the simplest ratio for the moles of atoms of C:H is 1:3 – a calculation from percentage masses would give a formula CH3. This is the empirical formula.

Answer 6

SUBSTANCE:
ethene
benzene
butane
phosphorus(V) oxide
oxygen
bromine

MOLECULAR FORMULA:
C2H4
C6H6
C4H10
P4O10
O2
Br2

EMPIRICAL FORMULA:
CH2
CH
C2H5
P2O5
O
Br

Question 7

What is water of crystallisation?

Answer 7

The crystals of some ionic lattices include molecules of water.

These water molecules are fitted within the lattice in a regular manner, and are called water of crystallisation.

The crystals are said to be hydrated, for example,
CuSO4 * 5H20 is hydrated copper(ii) sulfate and is blue coloured.

When hydrated crystals are heated, the water of crystallisation is removed as steam leaving the anhydrous solid, for example,
CuSO4 is the anhydrous form of copper(ii) sulfate and is white.

Question 8

NOTE:

CALCULATING THE FORMULA OF A HYDRATED IONIC COMPOUNDS

The formula of hydrated compounds can be determined using a simple experimental technique.
The mass of a sample of the hydrated salt is measured.

The salt is then heated, with the mass measured at regular intervals.

When no more mass is lost, the formula can be determined.

WORKED EXAMPLE:

2.53g of hydrated magnesium chloride
(MgCl2 * xH2O) was heated to constant mass.

1.17g of solid remained.

What is the formula of the hydrated compound?

(Mr MgCl2 = 95.9)

Answer 8

Step 1:
Calculate the mass of water in 2.53g MgCl2.

2.53 - 1.17 = 1.36g

Step 2:
Calculate the number of moles of water.

1.36 / 18 = 0.076 moles

Step 3:
Calculate the number of moles of MgCl2 in the anhydrous solid. (Mr MgCl2) = 95.9)

1.17 / 95.9 = 0.012 moles

Step 4:
Calculate the ratio of moles of magnesium chloride to moles of water by dividing both values by the number of moles of magnesium chloride.

0. 012 / 0.012 = 1
1. 076 / 0.012 = 6.3

1 : 6.3

Step 5:
Use the ratio from step 4 to determine the formula of the hydrated magnesium chloride.

MgCl2 * 6H2O

Question 9

What factors can reduce the amount of products produced?

Answer 9

Loss of products from reaction vessels, particularly if they are several stages into the reaction.

Side-reactions occuring, producing unwanted by-products

Impurities in the reactants

Changes in temperature and pressure

If the reaction is an equilibrium system

Question 10

How do you work out percentage yield?

Answer 10

percentage yield = (experimental yield / theoretical yield) x 100

Question 11

NOTE:

CALCULATING PERCENTAGE YIELD

The theoretical yield can be calculated from the balanced equation for the particular reaction.

Worked example:

0.84g of magnesium was burnt in excess oxygen, producing 1.1g of magnesium oxide. Using these results, work out the percentage yield of the reaction.

Answer 11

Step 1:
Write a balanced equation for the reaction.

2Mg(s) + O2(g) –> 2MgO(s)

Step 2:
Calculate the amount of substance, in moles, of magnesium.

0.84g / 24gmol-1 = 0.035 mol

Step 3:
The balanced equation tells us that the ratio of Mg to MgO produced is 2:2 or as the lowest whole number ratio 1:1.
This means that the maximum amount of magnesium oxide that can be produced from this amount of magnesium is also 0.035 mol.

Step 4:
Calculate the formula mass of MgO.

Ar(Mg) = 24; Ar(O) = 16
Mr(MgO) = 24 + 16 = 40

Step 5:
Calculate the maximum mass of MgO you can expect to produce.

0.035 mol of MgO
so, 40 x 0.035 = 1.4g

Step 6:
The experiment produced 1.1g. Calculate the percentage yield.

(1.1g / 1.4g) x 100 = 79% (2.s.f)