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Quantitative Aspects of Chemical Change

An equation for a chemical reaction can provide us with a lot of useful information. It tells us what the reactants and the products are in the reaction, and it also tells us the ratio in which the reactants combine to form products. Look at the equation below:

Fe + S → FeS Fe + S → FeS

In this reaction, every atom of iron (FeFe) will react with a single atom of sulphur (SS) to form one molecule of iron sulphide (FeSFeS). However, what the equation doesn't tell us, is the quantities or the amount of each substance that is involved. You may for example be given a small sample of iron for the reaction. How will you know how many atoms of iron are in this sample? And how many atoms of sulphur will you need for the reaction to use up all the iron you have? Is there a way of knowing what mass of iron sulphide will be produced at the end of the reaction? These are all very important questions, especially when the reaction is an industrial one, where it is important to know the quantities of reactants that are needed, and the quantity of product that will be formed. This chapter will look at how to quantify the changes that take place in chemical reactions.

The Mole

Sometimes it is important to know exactly how many particles (e.g. atoms or molecules) are in a sample of a substance, or what quantity of a substance is needed for a chemical reaction to take place.

You will remember from Relative atomic mass that the relative atomic mass of an element, describes the mass of an atom of that element relative to the mass of an atom of carbon-12. So the mass of an atom of carbon (relative atomic mass is 12u12u) for example, is twelve times greater than the mass of an atom of hydrogen, which has a relative atomic mass of 1u1u. How can this information be used to help us to know what mass of each element will be needed if we want to end up with the same number of atoms of carbon and hydrogen?

Let's say for example, that we have a sample of 12g12g carbon. What mass of hydrogen will contain the same number of atoms as 12g12g carbon? We know that each atom of carbon weighs twelve times more than an atom of hydrogen. Surely then, we will only need 1g1g of hydrogen for the number of atoms in the two samples to be the same? You will notice that the number of particles (in this case, atoms) in the two substances is the same when the ratio of their sample masses (12 g carbon: 1g hydrogen = 12:1) is the same as the ratio of their relative atomic masses (12 u: 1 u = 12:1).

To take this a step further, if you were to weigh out samples of a number of elements so that the mass of the sample was the same as the relative atomic mass of that element, you would find that the number of particles in each sample is 6,022× 10236,022×1023. These results are shown in Table 1 below for a number of different elements. So, 24,31g24,31g of magnesium (relative atomic mass=24,31urelative atomic mass=24,31u) for example, has the same number of atoms as 40,08g40,08g of calcium (relative atomic mass=40,08urelative atomic mass=40,08u).

Table 1: Table showing the relationship between the sample mass, the relative atomic mass and the number of atoms in a sample, for a number of elements.
Element Relative atomic mass (u) Sample mass (g) Atoms in sample
Hydrogen (HH) 1 1 6,022× 10236,022×1023
Carbon (CC) 12 12 6,022× 10226,022×1022
Magnesium (MgMg) 24.31 24.31 6,022× 10236,022×1023
Sulphur (SS) 32.07 32.07 6,022× 10236,022×1023
Calcium (CaCa) 40.08 40.08 6,022× 10236,022×1023

This result is so important that scientists decided to use a special unit of measurement to define this quantity: the mole or 'mol'. A mole is defined as being an amount of a substance which contains the same number of particles as there are atoms in 12 g of carbon. In the examples that were used earlier, 24,31g24,31g magnesium is one mole of magnesium, while 40,08g40,08g of calcium is one mole of calcium. A mole of any substance always contains the same number of particles.

Definition 1: Mole

The mole (abbreviation 'n') is the SI (Standard International) unit for 'amount of substance'. It is defined as an amount of substance that contains the same number of particles (atoms, molecules or other particle units) as there are atoms in 12g12g carbon.

In one mole of any substance, there are 6,022× 10236,022×1023 particles.

Definition 2: Avogadro's number

The number of particles in a mole, equal to 6,022× 10236,022×1023. It is also sometimes referred to as the number of atoms in 12g12g of carbon-12.

If we were to write out Avogadro's number then it would look like: 602200000000000000000000602200000000000000000000. This is a very large number. If we had this number of cold drink cans, then we could cover the surface of the earth to a depth of over 300km300km! If you could count atoms at a rate of 10 million per second, then it would take you 2 billion years to count the atoms in one mole!

We can build up to the idea of Avogadro's number. For example, if you have 12 eggs then you have a dozen eggs. After this number we get a gross of eggs, which is 144 eggs. Finally if we wanted one mole of eggs this would be 6,022× 10236,022×1023. That is a lot of eggs!

Note: Interesting Fact :

The original hypothesis that was proposed by Amadeo Avogadro was that 'equal volumes of gases, at the same temperature and pressure, contain the same number of molecules'. His ideas were not accepted by the scientific community and it was only four years after his death, that his original hypothesis was accepted and that it became known as 'Avogadro's Law'. In honour of his contribution to science, the number of particles in one mole was named Avogadro's number.

Moles and mass

  1. Complete the following table:
    Table 2
    ElementRelative atomic mass (u)Sample mass (g)Number of moles in the sample
    Hydrogen1.011.01 
    Magnesium24.3124.31 
    Carbon12.0124.02 
    Chlorine35.4570.9 
    Nitrogen 42.08 
    Click here for the solution
  2. How many atoms are there in...
    1. 1 mole of a substance
    2. 2 moles of calcium
    3. 5 moles of phosphorus
    4. 24,31g24,31g of magnesium
    5. 24,02g24,02g of carbon
    Click here for the solution

Molar Mass

Definition 3: Molar mass

Molar mass (M) is the mass of 1 mole of a chemical substance. The unit for molar mass is grams per mole or g·mol-1g·mol-1.

Refer to Table 1. You will remember that when the mass, in grams, of an element is equal to its relative atomic mass, the sample contains one mole of that element. This mass is called the molar mass of that element.

You may sometimes see the molar mass written as MmMm. We will use MM in this book, but you should be aware of the alternate notation.

It is worth remembering the following: On the periodic table, the relative atomic mass that is shown can be interpreted in two ways.

  1. The mass of a single, average atom of that element relative to the mass of an atom of carbon.
  2. The mass of one mole of the element. This second use is the molar mass of the element.
Table 3: The relationship between relative atomic mass, molar mass and the mass of one mole for a number of elements.
Element Relative atomic mass (u) Molar mass (g·mol-1g·mol-1) Mass of one mole of the element (g)
Magnesium 24,31 24,31 24,31
Lithium 6,94 6,94 6,94
Oxygen 16 16 16
Nitrogen 14,01 14,01 14,01
Iron 55,85 55,85 55,85

Exercise 1: Calculating the number of moles from mass

Calculate the number of moles of iron (FeFe) in a 11,7g11,7g sample.

Solution

  1. Step 1. Find the molar mass of iron :

    If we look at the periodic table, we see that the molar mass of iron is 55,85g·mol-155,85g·mol-1. This means that 1 mole of iron will have a mass of 55,85g55,85g.

  2. Step 2. Use the molar mass and sample mass to calculate the number of moles of iron :

    If 1 mole of iron has a mass of 55,85g55,85g, then: the number of moles of iron in 111,7g111,7g must be:

    111 , 7 g 55 , 85 g · mol - 1 = 2 mol 111 , 7 g 55 , 85 g · mol - 1 = 2 mol
    (1)

    There are 2 moles of iron in the sample.

Exercise 2: Calculating mass from moles

You have a sample that contains 5 moles of zinc.

  1. What is the mass of the zinc in the sample?
  2. How many atoms of zinc are in the sample?

Solution

  1. Step 1. Find the molar mass of zinc :

    Molar mass of zinc is 65,38g·mol-165,38g·mol-1, meaning that 1 mole of zinc has a mass of 65,38g65,38g.

  2. Step 2. Calculate the mass of zinc, using moles and molar mass. :

    If 1 mole of zinc has a mass of 65,38g65,38g, then 5 moles of zinc has a mass of: 65,38g×5mol=326,9g65,38g×5mol=326,9g (answer to a)

  3. Step 3. Use the number of moles of zinc and Avogadro's number to calculate the number of zinc atoms in the sample. :
    5 × 6 , 022 × 10 23 = 30 , 115 × 10 23 5 × 6 , 022 × 10 23 = 30 , 115 × 10 23
    (2)

    (answer to b)

Moles and molar mass

  1. Give the molar mass of each of the following elements:
    1. hydrogen
    2. nitrogen
    3. bromine
    Click here for the solution
  2. Calculate the number of moles in each of the following samples:
    1. 21,62g21,62g of boron (BB)
    2. 54,94g54,94g of manganese (MnMn)
    3. 100,3g100,3g of mercury (HgHg)
    4. 50g50g of barium (BaBa)
    5. 40g40g of lead (PbPb)
    Click here for the solution

An equation to calculate moles and mass in chemical reactions

The calculations that have been used so far, can be made much simpler by using the following equation:

n ( number of moles ) = m ( mass of substance in g ) M ( molar mass of substance in g · mol - 1 ) n ( number of moles ) = m ( mass of substance in g ) M ( molar mass of substance in g · mol - 1 )
(3)

Tip:

Remember that when you use the equation n=mMn=mM, the mass is always in grams (gg) and molar mass is in grams per mol (g·mol-1g·mol-1).

The equation can also be used to calculate mass and molar mass, using the following equations:

m = n × M m = n × M
(4)

and

M = m n M = m n
(5)

The following diagram may help to remember the relationship between these three variables. You need to imagine that the horizontal line is like a 'division' sign and that the vertical line is like a 'multiplication' sign. So, for example, if you want to calculate 'M', then the remaining two letters in the triangle are 'm' and 'n' and 'm' is above 'n' with a division sign between them. In your calculation then, 'm' will be the numerator and 'n' will be the denominator.

Figure 1
Figure 1 (CG11C6_001.png)

Exercise 3: Calculating moles from mass

Calculate the number of moles of copper there are in a sample that weighs 127g127g.

Solution

  1. Step 1. Write the equation to calculate the number of moles :
    n = m M n = m M
    (6)
  2. Step 2. Substitute numbers into the equation :
    n = 127 63 , 55 = 2 n = 127 63 , 55 = 2
    (7)

    There are 2 moles of copper in the sample.

Exercise 4: Calculating mass from moles

You are given a 5 mol sample of sodium. What mass of sodium is in the sample?

Solution

  1. Step 1. Write the equation to calculate the sample mass :
    m = n × M m = n × M
    (8)
  2. Step 2. Substitute values into the equation :

    MNa=22,99g·mol-1Na=22,99g·mol-1

    Therefore,

    m = 5 × 22 , 99 = 114 , 95 g m = 5 × 22 , 99 = 114 , 95 g
    (9)

    The sample of sodium has a mass of 114 , 95 g114,95g.

Exercise 5: Calculating atoms from mass

Calculate the number of atoms there are in a sample of aluminium that weighs 80 , 94 g80,94g.

Solution

  1. Step 1. Calculate the number of moles of aluminium in the sample :
    n = m M = 80 , 94 26 , 98 = 3 moles n = m M = 80 , 94 26 , 98 = 3 moles
    (10)
  2. Step 2. Use Avogadro's number to calculate the number of atoms in the sample:

    Number of atoms in 3 mol aluminium =3 ×6,022×1023=3×6,022×1023

    There are 18,069×102318,069×1023 aluminium atoms in a sample of 80,94g80,94g.

Some simple calculations

  1. Calculate the number of moles in each of the following samples:
    1. 5,6g5,6g of calcium
    2. 0,02g0,02g of manganese
    3. 40g40g of aluminium
    Click here for the solution
  2. A lead sinker has a mass of 5g5g.
    1. Calculate the number of moles of lead the sinker contains.
    2. How many lead atoms are in the sinker?
    Click here for the solution
  3. Calculate the mass of each of the following samples:
    1. 2,5mol2,5mol magnesium
    2. 12mol12mol lithium
    3. 4,5× 10254,5× 1025 atoms of silicon
    Click here for the solution

Molecules and compounds

So far, we have only discussed moles, mass and molar mass in relation to elements. But what happens if we are dealing with a molecule or some other chemical compound? Do the same concepts and rules apply? The answer is 'yes'. However, you need to remember that all your calculations will apply to the whole molecule. So, when you calculate the molar mass of a molecule, you will need to add the molar mass of each atom in that compound. Also, the number of moles will also apply to the whole molecule. For example, if you have one mole of nitric acid (HNO3HNO3), it means you have 6,022× 10236,022×1023 molecules of nitric acid in the sample. This also means that there are 6,022× 10236,022×1023 atoms of hydrogen, 6,022× 10236,022×1023 atoms of nitrogen and (3× 6,022× 10233×6,022×1023) atoms of oxygen in the sample.

In a balanced chemical equation, the number that is written in front of the element or compound, shows the mole ratio in which the reactants combine to form a product. If there are no numbers in front of the element symbol, this means the number is '1'.

e.g. N2+3H2→2NH3N2+3H2→2NH3

In this reaction, 1 mole of nitrogen reacts with 3 moles of hydrogen to produce 2 moles of ammonia.

Exercise 6: Calculating molar mass

Calculate the molar mass of H2SO4H2SO4.

Solution

  1. Step 1. Use the periodic table to find the molar mass for each element in the molecule:

    Hydrogen =1,008 g·mol-1=1,008g·mol-1; Sulphur = =32,07 g·mol-1=32,07g·mol-1; Oxygen =16 g·mol-1=16g·mol-1

  2. Step 2. Add the molar masses of each atom in the molecule :
    M ( H 2 S O 4 ) = ( 2 × 1 , 008 ) + ( 32 , 07 ) + ( 4 × 16 ) = 98 , 09 g · mol - 1 M ( H 2 S O 4 ) = ( 2 × 1 , 008 ) + ( 32 , 07 ) + ( 4 × 16 ) = 98 , 09 g · mol - 1
    (11)

Exercise 7: Calculating moles from mass

Calculate the number of moles there are in 1kg1kg of MgCl2MgCl2.

Solution

  1. Step 1. Write the equation for calculating the number of moles in the sample :
    n = m M n = m M
    (12)
  2. Step 2. Calculate the values that you will need, to substitute into the equation :
    1. Convert mass into grams
      m=1kg×1 000=1 000gm=1kg×1 000=1 000g
      (13)
    2. Calculate the molar mass of MgCl2MgCl2.
      M(MgCl2)=24,31+(2×35,45)=95,21g·mol-1M(MgCl2)=24,31+(2×35,45)=95,21g·mol-1
      (14)
  3. Step 3. Substitute values into the equation :
    n = 1000 95 , 21 = 10 , 5 mol n = 1000 95 , 21 = 10 , 5 mol
    (15)

    There are 10 , 5 moles10,5moles of magnesium chloride in a 1kg1kg sample.

Exercise 8: Calculating the mass of reactants and products

Barium chloride and sulphuric acid react according to the following equation to produce barium sulphate and hydrochloric acid.

BaCl 2 + H 2 SO 4 → BaSO 4 + 2 HCl BaCl 2 + H 2 SO 4 → BaSO 4 + 2 HCl

If you have 2g2g of BaCl2BaCl2...

  1. What quantity (in g) of H2SO4H2SO4 will you need for the reaction so that all the barium chloride is used up?
  2. What mass of HClHCl is produced during the reaction?

Solution

  1. Step 1. Calculate the number of moles of BaCl2BaCl2 that react. :
    n = m M = 2 208 , 24 = 0 , 0096 mol n = m M = 2 208 , 24 = 0 , 0096 mol
    (16)
  2. Step 2. Determine how many moles of H2SO4H2SO4 are needed for the reaction :

    According to the balanced equation, 1 mole of BaCl2BaCl2 will react with 1 mole of H2SO4H2SO4. Therefore, if 0 , 0096 mol0,0096mol of BaCl2BaCl2 react, then there must be the same number of moles of H2SO4H2SO4 that react because their mole ratio is 1:1.

  3. Step 3. Calculate the mass of H2SO4H2SO4 that is needed. :
    m = n × M = 0 , 0096 × 98 , 086 = 0 , 94 g m = n × M = 0 , 0096 × 98 , 086 = 0 , 94 g
    (17)

    (answer to 1)

  4. Step 4. Determine the number of moles of HClHCl produced. :

    According to the balanced equation, 2 moles of HClHCl are produced for every 1 mole of the two reactants. Therefore the number of moles of HClHCl produced is (2 ×0,00962×0,0096), which equals 0,0096moles0,0096moles.

  5. Step 5. Calculate the mass of HClHCl. :
    m = n × M = 0 , 0192 × 35 , 73 = 0 , 69 g m = n × M = 0 , 0192 × 35 , 73 = 0 , 69 g
    (18)

    (answer to 2)

Group work : Understanding moles, molecules and Avogadro's number

Divide into groups of three and spend about 20 minutes answering the following questions together:

  1. What are the units of the mole? Hint: Check the definition of the mole.
  2. You have a 56g56g sample of iron sulphide (FeSFeS)
    1. How many moles of FeSFeS are there in the sample?
    2. How many molecules of FeSFeS are there in the sample?
    3. What is the difference between a mole and a molecule?
  3. The exact size of Avogadro's number is sometimes difficult to imagine.
    1. Write down Avogadro's number without using scientific notation.
    2. How long would it take to count to Avogadro's number? You can assume that you can count two numbers in each second.

Figure 2
Khan academy video on the mole - 1

More advanced calculations

  1. Calculate the molar mass of the following chemical compounds:
    1. KOHKOH
    2. FeCl3FeCl3
    3. Mg(OH)2Mg(OH)2
    Click here for the solution
  2. How many moles are present in:
    1. 10g10g of Na2Na2SO44
    2. 34g34g of Ca(OH)2Ca(OH)2
    3. 2,45×10232,45×1023 molecules of CH4CH4?
    Click here for the solution
  3. For a sample of 0,2moles0,2moles of potassium bromide (KBrKBr), calculate...
    1. the number of moles of K+K+ ions
    2. the number of moles of Br-Br- ions
    Click here for the solution
  4. You have a sample containing 3moles3moles of calcium chloride.
    1. What is the chemical formula of calcium chloride?
    2. How many calcium atoms are in the sample?
    Click here for the solution
  5. Calculate the mass of:
    1. 3moles3moles of NH4OHNH4OH
    2. 4,2moles4,2moles of Ca(NO3)2Ca(NO3)2
    Click here for the solution
  6. 96,2g96,2g sulphur reacts with an unknown quantity of zinc according to the following equation: Zn+S→ZnSZn+S→ZnS
    1. What mass of zinc will you need for the reaction, if all the sulphur is to be used up?
    2. What mass of zinc sulphide will this reaction produce?
    Click here for the solution
  7. Calcium chloride reacts with carbonic acid to produce calcium carbonate and hydrochloric acid according to the following equation: CaCl2+H2CO3→CaCO3+2HClCaCl2+H2CO3→CaCO3+2HCl If you want to produce 10g10g of calcium carbonate through this chemical reaction, what quantity (in g) of calcium chloride will you need at the start of the reaction?
    Click here for the solution

The Composition of Substances

The empirical formula of a chemical compound is a simple expression of the relative number of each type of atom in that compound. In contrast, the molecular formula of a chemical compound gives the actual number of atoms of each element found in a molecule of that compound.

Definition 4: Empirical formula

The empirical formula of a chemical compound gives the relative number of each type of atom in that compound.

Definition 5: Molecular formula

The molecular formula of a chemical compound gives the exact number of atoms of each element in one molecule of that compound.

The compound ethanoic acid for example, has the molecular formula CH3COOHCH3COOH or simply C2H4O2C2H4O2. In one molecule of this acid, there are two carbon atoms, four hydrogen atoms and two oxygen atoms. The ratio of atoms in the compound is 2:4:2, which can be simplified to 1:2:1. Therefore, the empirical formula for this compound is CH2OCH2O. The empirical formula contains the smallest whole number ratio of the elements that make up a compound.

Knowing either the empirical or molecular formula of a compound, can help to determine its composition in more detail. The opposite is also true. Knowing the composition of a substance can help you to determine its formula. There are four different types of composition problems that you might come across:

  1. Problems where you will be given the formula of the substance and asked to calculate the percentage by mass of each element in the substance.
  2. Problems where you will be given the percentage composition and asked to calculate the formula.
  3. Problems where you will be given the products of a chemical reaction and asked to calculate the formula of one of the reactants. These are often referred to as combustion analysis problems.
  4. Problems where you will be asked to find number of moles of waters of crystallisation.

Exercise 9: Calculating the percentage by mass of elements in a compound

Calculate the percentage that each element contributes to the overall mass of sulphuric acid (H2SO4H2SO4).

Solution

  1. Step 1. Write down the relative atomic mass of each element in the compound:

    Hydrogen=1,008× 2=2,016uHydrogen=1,008×2=2,016u

    Sulphur=32,07uSulphur=32,07u

    Oxygen=4× 16=64uOxygen=4×16=64u

  2. Step 2. Calculate the molecular mass of sulphuric acid :

    Use the calculations in the previous step to calculate the molecular mass of sulphuric acid.

    Mass = 2 , 016 + 32 , 07 + 64 = 98 , 09 u Mass = 2 , 016 + 32 , 07 + 64 = 98 , 09 u
    (19)
  3. Step 3. Convert the mass of each element to a percentage of the total mass of the compound :

    Use the equation:

    Percentage by mass=atomic massmolecular mass of H2SO4×100%Percentage by mass=atomic massmolecular mass of H2SO4×100%

    Hydrogen

    2 , 016 98 , 09 × 100 % = 2 , 06 % 2 , 016 98 , 09 × 100 % = 2 , 06 %
    (20)

    Sulphur

    32 , 07 98 , 09 × 100 % = 32 , 69 % 32 , 07 98 , 09 × 100 % = 32 , 69 %
    (21)

    Oxygen

    64 98 , 09 × 100 % = 65 , 25 % 64 98 , 09 × 100 % = 65 , 25 %
    (22)

    (You should check at the end that these percentages add up to 100%!)

    In other words, in one molecule of sulphuric acid, hydrogen makes up 2,06% of the mass of the compound, sulphur makes up 32,69% and oxygen makes up 65,25%.

Exercise 10: Determining the empirical formula of a compound

A compound contains 52.2% carbon (CC), 13.0% hydrogen (HH) and 34.8% oxygen (OO). Determine its empirical formula.

Solution

  1. Step 1. If we assume that we have 100g100g of this substance, then we can convert each element percentage into a mass in grams :

    Carbon =52,2g=52,2g, hydrogen =13,0g=13,0g and oxygen =34,8g=34,8g

  2. Step 2. Convert the mass of each element into number of moles :
    n = m M n = m M
    (23)

    Therefore,

    n ( Carbon ) = 52 , 2 12 , 01 = 4 , 35 mol n ( Carbon ) = 52 , 2 12 , 01 = 4 , 35 mol
    (24)
    n ( Hydrogen ) = 13 , 0 1 , 008 = 12 , 90 mol n ( Hydrogen ) = 13 , 0 1 , 008 = 12 , 90 mol
    (25)
    n ( Oxygen ) = 34 , 8 16 = 2 , 18 mol n ( Oxygen ) = 34 , 8 16 = 2 , 18 mol
    (26)
  3. Step 3. Convert these numbers to the simplest mole ratio by dividing by the smallest number of moles :

    In this case, the smallest number of moles is 2.18. Therefore...

    Carbon

    4 , 35 2 , 18 = 2 4 , 35 2 , 18 = 2
    (27)

    Hydrogen

    12 , 90 2 , 18 = 6 12 , 90 2 , 18 = 6
    (28)

    Oxygen

    2 , 18 2 , 18 = 1 2 , 18 2 , 18 = 1
    (29)

    Therefore the empirical formula of this substance is: C2H6OC2H6O. Do you recognise this compound?

Exercise 11: Determining the formula of a compound

207g207g of lead combines with oxygen to form 239g239g of a lead oxide. Use this information to work out the formula of the lead oxide (Relative atomic masses: Pb=207uPb=207u and O=16uO=16u).

Solution

  1. Step 1. Calculate the mass of oxygen in the reactants :
    239 - 207 = 32 g 239 - 207 = 32 g
    (30)
  2. Step 2. Calculate the number of moles of lead and oxygen in the reactants :
    n = m M n = m M
    (31)

    Lead

    207 207 = 1 mol 207 207 = 1 mol
    (32)

    Oxygen

    32 16 = 2 mol 32 16 = 2 mol
    (33)
  3. Step 3. Deduce the formula of the compound :

    The mole ratio of Pb:OPb:O in the product is 1:2, which means that for every atom of lead, there will be two atoms of oxygen. The formula of the compound is PbO2PbO2.

Exercise 12: Empirical and molecular formula

Vinegar, which is used in our homes, is a dilute form of acetic acid. A sample of acetic acid has the following percentage composition: 39,9% carbon, 6,7% hyrogen and 53,4% oxygen.

  1. Determine the empirical formula of acetic acid.
  2. Determine the molecular formula of acetic acid if the molar mass of acetic acid is 60g·mol-160g·mol-1.

Solution

  1. Step 1. Calculate the mass of each element in 100g100g of acetic acid. :

    In 100g100g of acetic acid, there is 39,9gC39,9gC, 6,7gH6,7gH and 53,4gO53,4gO

  2. Step 2. Calculate the number of moles of each element in 100g100g of acetic acid. :

    n = m M n = m M

    n C = 39 , 9 12 = 3 , 33 mol n H = 6 , 7 1 = 6 , 7 mol n O = 53 , 4 16 = 3 , 34 mol n C = 39 , 9 12 = 3 , 33 mol n H = 6 , 7 1 = 6 , 7 mol n O = 53 , 4 16 = 3 , 34 mol
    (34)
  3. Step 3. Divide the number of moles of each element by the lowest number to get the simplest mole ratio of the elements (i.e. the empirical formula) in acetic acid. :

    Empirical formula is CH2OCH2O

  4. Step 4. Calculate the molecular formula, using the molar mass of acetic acid. :

    The molar mass of acetic acid using the empirical formula is 30g·mol-130g·mol-1. Therefore the actual number of moles of each element must be double what it is in the empirical formula.

    The molecular formula is therefore C2H4O2C2H4O2 or CH3COOHCH3COOH

Exercise 13: Waters of crystallisation

Aluminium trichloride (AlCl3AlCl3) is an ionic substance that forms crystals in the solid phase. Water molecules may be trapped inside the crystal lattice. We represent this as: AlCl3·nH2OAlCl3·nH2O. A learner heated some aluminium trichloride crystals until all the water had evaporated and found that the mass after heating was 2,8g2,8g. The mass before heating was 5g5g. What is the number of moles of water molecules in the aluminium trichloride?

Solution

  1. Step 1. Work out the mass of water molecules lost:

    We first need to find n, the number of water molecules that are present in the crystal. To do this we first note that the mass of water lost is 5-2,8=2,25-2,8=2,2.

  2. Step 2. Work out the mass ratio and mole ratio:

    The next step is to work out the mass ratio of aluminium trichloride to water and the mole ratio. The mass ratio is:

    2,8:2,22,8:2,2
    (35)
    To work out the mole ratio we divide the mass ratio by the molecular mass of each species:
    2,8133:2,218=0,021:0,122,8133:2,218=0,021:0,12
    (36)
    Next we do the following:
    0,02110,021=10,02110,021=1
    (37)
    and
    0,120,021=60,120,021=6
    (38)
    So the mole ratio of aluminium trichloride to water is:
    1:61:6
    (39)

  3. Step 3. Write down the final answer:

    And now we know that there are 6 moles of water molecules in the crystal.

Figure 3
Khan academy video on molecular and empirical formulae - 1

Figure 4
Khan academy video on mass composition - 1

Moles and empirical formulae

  1. Calcium chloride is produced as the product of a chemical reaction.
    1. What is the formula of calcium chloride?
    2. What percentage does each of the elements contribute to the mass of a molecule of calcium chloride?
    3. If the sample contains 5g5g of calcium chloride, what is the mass of calcium in the sample?
    4. How many moles of calcium chloride are in the sample?
    Click here for the solution
  2. 13g13g of zinc combines with 6,4g6,4g of sulphur. What is the empirical formula of zinc sulphide?
    1. What mass of zinc sulphide will be produced?
    2. What percentage does each of the elements in zinc sulphide contribute to its mass?
    3. Determine the formula of zinc sulphide.
    Click here for the solution
  3. A calcium mineral consisted of 29,4% calcium, 23,5% sulphur and 47,1% oxygen by mass. Calculate the empirical formula of the mineral.
    Click here for the solution
  4. A chlorinated hydrocarbon compound was analysed and found to consist of 24,24% carbon, 4,04% hydrogen and 71,72% chlorine. From another experiment the molecular mass was found to be 99g·mol-199g·mol-1. Deduce the empirical and molecular formula.
    Click here for the solution

Molar Volumes of Gases

It is possible to calculate the volume of one mole of gas at STP using what we know about gases.

  1. Write down the ideal gas equation pV=nRTpV=nRT, therefore V=nRTpV=nRTp
  2. Record the values that you know, making sure that they are in SI units You know that the gas is under STP conditions. These are as follows: p=101,3kPa=101300Pap=101,3kPa=101300Pa n=1moln=1mol R=8,31J·K-1·mol-1R=8,31J·K-1·mol-1 T=273KT=273K
  3. Substitute these values into the original equation.
    V = n R T p V = n R T p
    (40)
    V = 1 mol × 8 , 31 J · K - 1 · mol - 1 × 273 K 101300 Pa V = 1 mol × 8 , 31 J · K - 1 · mol - 1 × 273 K 101300 Pa
    (41)
  4. Calculate the volume of 1mole1mole of gas under these conditions The volume of 1mole1mole of gas at STP is 22,4×10-3m3=22,4dm322,4×10-3m3=22,4dm3.

Tip:

The standard units used for this equation are PP in PaPa, VV in m3m3 and TT in KK. Remember also that 1000cm3=1dm31000cm3=1dm3 and 1000dm3=1m31000dm3=1m3.

Exercise 14: Ideal Gas

A sample of gas occupies a volume of 20dm320dm3, has a temperature of 200K200K and has a pressure of 105Pa105Pa. Calculate the number of moles of gas that are present in the sample.

Solution

  1. Step 1. Convert all values into SI units :

    The only value that is not in SI units is volume. V=0,02m3V=0,02m3.

  2. Step 2. Write the equation for calculating the number of moles in a gas :

    We know that pV=nRTpV=nRT

    Therefore,

    n = p V R T n = p V R T
    (42)
  3. Step 3. Substitute values into the equation to calculate the number of moles of the gas :
    n = 105 × 0 , 02 8 , 31 × 280 = 2 , 1 2326 , 8 = 0 , 0009 moles n = 105 × 0 , 02 8 , 31 × 280 = 2 , 1 2326 , 8 = 0 , 0009 moles
    (43)

Using the combined gas law

  1. An enclosed gas(i.e. one in a sealed container) has a volume of 300cm3300cm3 and a temperature of 300K300K. The pressure of the gas is 50kPa50kPa. Calculate the number of moles of gas that are present in the container.
    Click here for the solution
  2. What pressure will 3mol3mol of gaseous nitrogen exert if it is pumped into a container that has a volume of 25dm325dm3 at a temperature of 29 0C290C?
    Click here for the solution
  3. The volume of air inside a tyre is 19 litres and the temperature is 290K290K. You check the pressure of your tyres and find that the pressure is 190kPa190kPa. How many moles of air are present in the tyre?
    Click here for the solution
  4. Compressed carbon dioxide is contained within a gas cylinder at a pressure of 700 kPa. The temperature of the gas in the cylinder is 310K310K and the number of moles of gas is 13mols13mols of carbon dioxide. What is the volume of the gas inside the cylinder?
    Click here for the solution

Molar concentrations of liquids

A typical solution is made by dissolving some solid substance in a liquid. The amount of substance that is dissolved in a given volume of liquid is known as the concentration of the liquid. Mathematically, concentration (C) is defined as moles of solute (n) per unit volume (V) of solution.

C = n V C = n V
(44)

For this equation, the units for volume are dm3dm3. Therefore, the unit of concentration is mol· dm -3mol· dm -3. When concentration is expressed in mol· dm -3mol· dm -3 it is known as the molarity (M) of the solution. Molarity is the most common expression for concentration.

Tip:

Do not confuse molarity (M) with molar mass (M). Look carefully at the question in which the M appears to determine whether it is concentration or molar mass.
Definition 6: Concentration

Concentration is a measure of the amount of solute that is dissolved in a given volume of liquid. It is measured in mol· dm -3mol· dm -3. Another term that is used for concentration is molarity (M)

Exercise 15: Concentration Calculations 1

If 3,5g3,5g of sodium hydroxide (NaOH) is dissolved in 2,5 dm 32,5 dm 3 of water, what is the concentration of the solution in mol· dm -3mol· dm -3?

Solution

  1. Step 1. Convert the mass of NaOHNaOH into moles :
    n = m M = 3 , 5 40 = 0 , 0875 mol n = m M = 3 , 5 40 = 0 , 0875 mol
    (45)
  2. Step 2. Calculate the concentration of the solution. :
    C = n V = 0 , 0875 2 , 5 = 0 , 035 C = n V = 0 , 0875 2 , 5 = 0 , 035
    (46)

    The concentration of the solution is 0,035mol· dm -30,035mol· dm -3 or 0,035M0,035M

Exercise 16: Concentration Calculations 2

You have a 1 dm 31 dm 3 container in which to prepare a solution of potassium permanganate (KMnO4KMnO4). What mass of KMnO4KMnO4 is needed to make a solution with a concentration of 0,2M0,2M?

Solution

  1. Step 1. Calculate the number of moles of KMnO4KMnO4 needed :
    C = n V C = n V
    (47)

    therefore

    n = C × V = 0 , 2 × 1 = 0 , 2 mol n = C × V = 0 , 2 × 1 = 0 , 2 mol
    (48)
  2. Step 2. Convert the number of moles of KMnO4KMnO4 to mass. :
    m = n × M = 0 , 2 × 158 , 04 = 31 , 61 g m = n × M = 0 , 2 × 158 , 04 = 31 , 61 g
    (49)

    The mass of KMnO4KMnO4 that is needed is 31,61g31,61g.

Exercise 17: Concentration Calculations 3

How much sodium chloride (in g) will one need to prepare 500 cm 3500 cm 3 of solution with a concentration of 0,01M0,01M?

Solution

  1. Step 1. Convert all quantities into the correct units for this equation :
    V = 500 1 000 = 0 , 5 dm 3 V = 500 1 000 = 0 , 5 dm 3
    (50)
  2. Step 2. Calculate the number of moles of sodium chloride needed :
    n = C × V = 0 , 01 × 0 , 5 = 0 , 005 mol n = C × V = 0 , 01 × 0 , 5 = 0 , 005 mol
    (51)
  3. Step 3. Convert moles of sodium chloride to mass :
    m = n × M = 0 , 005 × 58 , 45 = 0 , 29 g m = n × M = 0 , 005 × 58 , 45 = 0 , 29 g
    (52)

    The mass of sodium chloride needed is 0,29g0,29g

Molarity and the concentration of solutions

  1. 5,95g5,95g of potassium bromide was dissolved in 400 cm 3400 cm 3 of water. Calculate its molarity.
    Click here for the solution
  2. 100g100g of sodium chloride (NaCl) is dissolved in 450 cm 3450 cm 3 of water.
    1. How many moles of NaCl are present in solution?
    2. What is the volume of water (in dm 3 dm 3)?
    3. Calculate the concentration of the solution.
    4. What mass of sodium chloride would need to be added for the concentration to become 5,7mol· dm -35,7mol· dm -3?
    Click here for the solution
  3. What is the molarity of the solution formed by dissolving 80g80g of sodium hydroxide (NaOHNaOH) in 500 cm 3500 cm 3 of water?
    Click here for the solution
  4. What mass (g) of hydrogen chloride (HClHCl) is needed to make up 1000 cm 31000 cm 3 of a solution of concentration 1mol· dm -31mol· dm -3?
    Click here for the solution
  5. How many moles of H2SO4H2SO4 are there in 250 cm 3250 cm 3 of a 0,8M0,8M sulphuric acid solution? What mass of acid is in this solution?
    Click here for the solution

Stoichiometric calculations

Stoichiometry is the calculation of the quantities of reactants and products in chemical reactions. It is also the numerical relationship between reactants and products. In representing chemical change showed how to write balanced chemical equations. By knowing the ratios of substances in a reaction, it is possible to use stoichiometry to calculate the amount of either reactants or products that are involved in the reaction. The examples shown below will make this concept clearer.

Exercise 18: Stoichiometric calculation 1

What volume of oxygen at S.T.P. is needed for the complete combustion of 2 dm 32 dm 3 of propane (C3H8C3H8)? (Hint: CO2CO2 and H2OH2O are the products in this reaction (and in all combustion reactions))

Solution

  1. Step 1. Write a balanced equation for the reaction. :

    C 3 H 8 ( g ) + 5 O 2 ( g ) → 3 C O 2 ( g ) + 4 H 2 O ( g ) C 3 H 8 ( g ) + 5 O 2 ( g ) → 3 C O 2 ( g ) + 4 H 2 O ( g )

  2. Step 2. Determine the ratio of oxygen to propane that is needed for the reaction. :

    From the balanced equation, the ratio of oxygen to propane in the reactants is 5:1.

  3. Step 3. Determine the volume of oxygen needed for the reaction. :

    1 volume of propane needs 5 volumes of oxygen, therefore 2 dm 32 dm 3 of propane will need 10 dm 310 dm 3 of oxygen for the reaction to proceed to completion.

Exercise 19: Stoichiometric calculation 2

What mass of iron (II) sulphide is formed when 5,6g5,6g of iron is completely reacted with sulphur?

Solution

  1. Step 1. Write a balanced chemical equation for the reaction. :

    Fe ( s ) + S ( s ) → FeS ( s ) Fe ( s ) + S ( s ) → FeS ( s )

  2. Step 2. Calculate the number of moles of iron that react. :
    n = m M = 5 , 6 55 , 85 = 0 , 1 mol n = m M = 5 , 6 55 , 85 = 0 , 1 mol
    (53)
  3. Step 3. Determine the number of moles of FeSFeS produced. :

    From the equation 1mole1mole of FeFe gives 1mole1mole of FeSFeS. Therefore, 0,1moles0,1moles of iron in the reactants will give 0,1moles0,1moles of iron sulphide in the product.

  4. Step 4. Calculate the mass of iron sulphide formed :
    m = n × M = 0 , 1 × 87 , 911 = 8 , 79 g m = n × M = 0 , 1 × 87 , 911 = 8 , 79 g
    (54)

    The mass of iron (II) sulphide that is produced during this reaction is 8,79g8,79g.

When we are given a known mass of a reactant and are asked to work out how much product is formed, we are working out the theoretical yield of the reaction. In the laboratory chemists never get this amount of product. In each step of a reaction a small amount of product and reactants is 'lost' either because a reactant did not completely react or some of the product was left behind in the original container. Think about this. When you make your lunch or supper, you might be a bit hungry, so you eat some of the food that you are preparing. So instead of getting the full amount of food out (theoretical yield) that you started preparing, you lose some along the way.

Exercise 20: Industrial reaction to produce fertiliser

Sulphuric acid (H2SO4H2SO4) reacts with ammonia (NH3NH3) to produce the fertiliser ammonium sulphate ((NH4NH4)2SO42SO4) according to the following equation:

H 2 SO 4 ( aq ) + 2 NH 3 ( g ) → ( NH 4 ) 2 SO 4 ( aq ) H 2 SO 4 ( aq ) + 2 NH 3 ( g ) → ( NH 4 ) 2 SO 4 ( aq )

What is the maximum mass of ammonium sulphate that can be obtained from 2,0kg2,0kg of sulphuric acid?

Solution

  1. Step 1. Convert the mass of sulphuric acid into moles :
    n ( H 2 SO 4 ) = m M = 2 000 g 98 , 078 g · mols - 1 = 20 , 39 mols n ( H 2 SO 4 ) = m M = 2 000 g 98 , 078 g · mols - 1 = 20 , 39 mols
    (55)
  2. Step 2. Calculate the maximum amount of ammonium sulphate that can be produced :

    From the balanced equation, the mole ratio of H2SO4H2SO4 in the reactants to (NH4)2SO4(NH4)2SO4 in the product is 1:1. Therefore, 20,39mols20,39mols of H2SO4H2SO4 of (NH4)2SO4(NH4)2SO4.

    The maximum mass of ammonium sulphate that can be produced is calculated as follows:

    m = n × M = 20 , 41 mol × 132 g · mol - 1 = 2694 g m = n × M = 20 , 41 mol × 132 g · mol - 1 = 2694 g
    (56)

    The maximum amount of ammonium sulphate that can be produced is 2,694kg2,694kg.

Figure 5
Khan academy video on stoichiometry - 1

Stoichiometry

  1. Diborane, B2H6B2H6, was once considered for use as a rocket fuel. The combustion reaction for diborane is: B2H6(g)+3O2(g)→2HBO2(g)+2H2O(l)B2H6(g)+3O2(g)→2HBO2(g)+2H2O(l) If we react 2,37grams2,37grams of diborane, how many grams of water would we expect to produce?
    Click here for the solution
  2. Sodium azide is a commonly used compound in airbags. When triggered, it has the following reaction: 2NaN3(s)→2Na(s)+3N2(g)2NaN3(s)→2Na(s)+3N2(g) If 23,4grams23,4grams of sodium azide is used, how many moles of nitrogen gas would we expect to produce?
    Click here for the solution
  3. Photosynthesis is a chemical reaction that is vital to the existence of life on Earth. During photosynthesis, plants and bacteria convert carbon dioxide gas, liquid water, and light into glucose (C6H12O6C6H12O6) and oxygen gas.
    1. Write down the equation for the photosynthesis reaction.
    2. Balance the equation.
    3. If 3moles3moles of carbon dioxide are used up in the photosynthesis reaction, what mass of glucose will be produced?
    Click here for the solution

Figure 6

Summary

  • It is important to be able to quantify the changes that take place during a chemical reaction.
  • The mole (n) is a SI unit that is used to describe an amount of substance that contains the same number of particles as there are atoms in 12g12g of carbon.
  • The number of particles in a mole is called the Avogadro constant and its value is 6,022× 10236,022×1023. These particles could be atoms, molecules or other particle units, depending on the substance.
  • The molar mass (M) is the mass of one mole of a substance and is measured in grams per mole or g·mol-1g·mol-1. The numerical value of an element's molar mass is the same as its relative atomic mass. For a compound, the molar mass has the same numerical value as the molecular mass of that compound.
  • The relationship between moles (n), mass in grams (m) and molar mass (M) is defined by the following equation:
    n=mMn=mM
    (57)
  • In a balanced chemical equation, the number in front of the chemical symbols describes the mole ratio of the reactants and products.
  • The empirical formula of a compound is an expression of the relative number of each type of atom in the compound.
  • The molecular formula of a compound describes the actual number of atoms of each element in a molecule of the compound.
  • The formula of a substance can be used to calculate the percentage by mass that each element contributes to the compound.
  • The percentage composition of a substance can be used to deduce its chemical formula.
  • One mole of gas occupies a volume of 22,4 dm 322,4 dm 3.
  • The concentration of a solution can be calculated using the following equation,
    C=nVC=nV
    (58)
    where C is the concentration (in mol· dm -3mol· dm -3), n is the number of moles of solute dissolved in the solution and V is the volume of the solution (in dm -3 dm -3).
  • Molarity is a measure of the concentration of a solution, and its units are mol· dm -3mol· dm -3.
  • Stoichiometry is the calculation of the quantities of reactants and products in chemical reactions. It is also the numerical relationship between reactants and products.
  • The theoretical yield of a reaction is the maximum amount of product that we expect to get out of a reaction

End of chapter exercises

  1. Write only the word/term for each of the following descriptions:
    1. the mass of one mole of a substance
    2. the number of particles in one mole of a substance
    Click here for the solution
  2. Multiple choice: Choose the one correct answer from those given.
    1. 5g5g of magnesium chloride is formed as the product of a chemical reaction. Select the true statement from the answers below:
      1. 0.08 moles of magnesium chloride are formed in the reaction
      2. the number of atoms of ClCl in the product is 0,6022× 10230,6022×1023
      3. the number of atoms of MgMg is 0,05
      4. the atomic ratio of MgMg atoms to ClCl atoms in the product is 1:1
      Click here for the solution
    2. 2 moles of oxygen gas react with hydrogen. What is the mass of oxygen in the reactants?
      1. 32 g
      2. 0,125 g
      3. 64 g
      4. 0,063 g
      Click here for the solution
    3. In the compound potassium sulphate (K2SO4K2SO4), oxygen makes up x% of the mass of the compound. x = ...
      1. 36.8
      2. 9,2
      3. 4
      4. 18,3
      Click here for the solution
    4. The molarity of a 150 cm 3150 cm 3 solution, containing 5 g of NaClNaCl is...
      1. 0,09M0,09M
      2. 5,7×10-4M5,7×10-4M
      3. 0,57M0,57M
      4. 0,03M0,03M
      Click here for the solution
  3. Calculate the number of moles in:
    1. 5 g of methane (CH4CH4)
    2. 3,4 g of hydrochloric acid
    3. 6,2 g of potassium permanganate (KMnO4KMnO4)
    4. 4 g of neon
    5. 9,6 kg of titanium tetrachloride (TiCl4TiCl4)
    Click here for the solution
  4. Calculate the mass of:
    1. 0,2 mols of potassium hydroxide (KOHKOH)
    2. 0,47 mols of nitrogen dioxide
    3. 5,2 mols of helium
    4. 0,05 mols of copper (II) chloride (CuCl2CuCl2)
    5. 31,31×102331,31×1023 molecules of carbon monoxide (COCO)
    Click here for the solution
  5. Calculate the percentage that each element contributes to the overall mass of:
    1. Chloro-benzene (C6H5ClC6H5Cl)
    2. Lithium hydroxide (LiOHLiOH)
    Click here for the solution
  6. CFC's (chlorofluorocarbons) are one of the gases that contribute to the depletion of the ozone layer. A chemist analysed a CFC and found that it contained 58,64% chlorine, 31,43% fluorine and 9,93% carbon. What is the empirical formula?
    Click here for the solution
  7. 14 g of nitrogen combines with oxygen to form 46 g of a nitrogen oxide. Use this information to work out the formula of the oxide.
    Click here for the solution
  8. Iodine can exist as one of three oxides (I2O4;I2O5;I4O9I2O4;I2O5;I4O9). A chemist has produced one of these oxides and wishes to know which one they have. If he started with 508 g of iodine and formed 652 g of the oxide, which form has he produced?
    Click here for the solution
  9. A fluorinated hydrocarbon (a hydrocarbon is a chemical compound containing hydrogen and carbon.) was analysed and found to contain 8,57% H, 51,05% C and 40,38% F.
    1. What is its empirical formula?
    2. What is the molecular formula if the molar mass is 94,1g⋅mol−194,1g⋅mol−1?
    Click here for the solution
  10. Copper sulphate crystals often include water. A chemist is trying to determine the number of moles of water in the copper sulphate crystals. She weighs out 3 g of copper sulphate and heats this. After heating, she finds that the mass is 1,9 g. What is the number of moles of water in the crystals? (Copper sulphate is represented by CuSO4⋅xH2OCuSO4⋅xH2O).Click here for the solution
  11. 300 cm 3300 cm 3 of a 0,1mol· dm -30,1mol· dm -3 solution of sulphuric acid is added to 200 cm 3200 cm 3 of a 0,5mol· dm -30,5mol· dm -3 solution of sodium hydroxide.
    1. Write down a balanced equation for the reaction which takes place when these two solutions are mixed.
    2. Calculate the number of moles of sulphuric acid which were added to the sodium hydroxide solution.
    3. Is the number of moles of sulphuric acid enough to fully neutralise the sodium hydroxide solution? Support your answer by showing all relevant calculations. (IEB Paper 2 2004)
    Click here for the solution
  12. A learner is asked to make 200 cm 3200 cm 3 of sodium hydroxide (NaOHNaOH) solution of concentration 0,5mol· dm -30,5mol· dm -3.
    1. Determine the mass of sodium hydroxide pellets he needs to use to do this.
    2. Using an accurate balance the learner accurately measures the correct mass of the NaOHNaOH pellets. To the pellets he now adds exactly 200 cm 3200 cm 3 of pure water. Will his solution have the correct concentration? Explain your answer.
    3. The learner then takes 300 cm 3300 cm 3 of a 0,1mol· dm -30,1mol· dm -3 solution of sulphuric acid (H2SO4H2SO4) and adds it to 200 cm 3200 cm 3 of a 0,5mol· dm -30,5mol· dm -3 solution of NaOHNaOH at 250C250C.
    4. Write down a balanced equation for the reaction which takes place when these two solutions are mixed.
    5. Calculate the number of moles of H2SO4H2SO4 which were added to the NaOH solution.
    6. Is the number of moles of H2SO4H2SO4 calculated in the previous question enough to fully neutralise the NaOHNaOH solution? Support your answer by showing all the relevant calculations. (IEB Paper 2, 2004)
    Click here for the solution

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