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Stoichiometry and composition

Module by: Free High School Science Texts Project. E-mail the author

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 1: Empirical formula

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

Definition 2: 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 1: 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
    (1)
  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 %
    (2)

    Sulphur

    32 , 07 98 , 09 × 100 % = 32 , 69 % 32 , 07 98 , 09 × 100 % = 32 , 69 %
    (3)

    Oxygen

    64 98 , 09 × 100 % = 65 , 25 % 64 98 , 09 × 100 % = 65 , 25 %
    (4)

    (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 2: 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
    (5)

    Therefore,

    n ( Carbon ) = 52 , 2 12 , 01 = 4 , 35 mol n ( Carbon ) = 52 , 2 12 , 01 = 4 , 35 mol
    (6)
    n ( Hydrogen ) = 13 , 0 1 , 008 = 12 , 90 mol n ( Hydrogen ) = 13 , 0 1 , 008 = 12 , 90 mol
    (7)
    n ( Oxygen ) = 34 , 8 16 = 2 , 18 mol n ( Oxygen ) = 34 , 8 16 = 2 , 18 mol
    (8)
  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
    (9)

    Hydrogen

    12 , 90 2 , 18 = 6 12 , 90 2 , 18 = 6
    (10)

    Oxygen

    2 , 18 2 , 18 = 1 2 , 18 2 , 18 = 1
    (11)

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

Exercise 3: 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
    (12)
  2. Step 2. Calculate the number of moles of lead and oxygen in the reactants :
    n = m M n = m M
    (13)

    Lead

    207 207 = 1 mol 207 207 = 1 mol
    (14)

    Oxygen

    32 16 = 2 mol 32 16 = 2 mol
    (15)
  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 4: 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
    (16)
  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 5: 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
    (17)
    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
    (18)
    Next we do the following:
    0,02110,021=10,02110,021=1
    (19)
    and
    0,120,021=60,120,021=6
    (20)
    So the mole ratio of aluminium trichloride to water is:
    1:61:6
    (21)

  3. Step 3. Write down the final answer:

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

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

Figure 2
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

Definition 3: Molar volume of gases
1 mole of gas occupies 22,4dm322,4dm3 at S.T.P.

This applies to any gas that is at standard temperature and pressure. In grade 11 you will learn more about this and the gas laws.

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
(22)

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 4: 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 6: 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
    (23)
  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
    (24)

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

Exercise 7: 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
    (25)

    therefore

    n = C × V = 0 , 2 × 1 = 0 , 2 mol n = C × V = 0 , 2 × 1 = 0 , 2 mol
    (26)
  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
    (27)

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

Exercise 8: 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
    (28)
  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
    (29)
  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
    (30)

    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 9: 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 10: 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
    (31)
  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
    (32)

    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 11: 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
    (33)
  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
    (34)

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

Figure 3
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 4

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
    (35)
  • 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
    (36)
    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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