Download 10 4.0 g of magnesium oxide was found to contain 2.4 g of

GCSE Revision (Higher)
Formulae
The chemical formula of a compound shows how many of each type of atom join together to make the units
which make up the compound.
For example, in iron sulfide every iron atom is joined to one sulfur atom, so we show its formula as FeS. In
sodium oxide, there are two sodium atoms for every oxygen atom, so we show its formula as Na2O. Notice that
the '2' is written as a subscript, so Na2O would be wrong.
This diagram shows that one carbon atom and two oxygen atoms combine to make up the units of carbon
dioxide. Its chemical formula is written as CO2.
Carbon dioxide units contain one carbon atom and two oxygen atoms
Sometimes you see more complex formulae such as Na2SO4 and Fe(OH)3 :


a unit of Na2SO4 contains two sodium atoms, one sulfur atom and four oxygen atoms joined together
a unit of Fe(OH)3 contains one iron atom, three oxygen atoms and three hydrogen atoms - the brackets
show that the 3 applies to O and H
Chemical equations
Word equations are useful to show which chemicals react together (the reactants) and which chemicals are
produced (the products). Symbol equations allow chemists to work out the masses that will react or be
produced.
Copper and oxygen reaction - getting a balanced equation
Balanced symbol equations show what happens to the different atoms in reactions. For example, copper and
oxygen react together to make copper oxide.
Take a look at this word equation for the reaction:
copper + oxygen → copper oxide


copper and oxygen are the reactants because they are on the left of the arrow
copper oxide is the product because it is on the right of the arrow
If we just replace the words shown above by the correct chemical formulas, we will get an unbalanced
equation, as shown here:
Cu + O2 → CuO
Notice that there are unequal numbers of each type of atom on the left-hand side compared with the right-hand
side. To make things equal, you need to adjust the number of units of some of the substances until you get equal
numbers of each type of atom on both sides.
Here is the balanced symbol equation:
2Cu + O2 → 2CuO
You can see that now there are two copper atoms and two oxygen atoms on each side. This matches what
happens in the reaction.
Two atoms of copper react with two atoms of oxygen to form two molecules of copper oxide
Remember: never change a formula to balance an equation
State symbols
Sometimes it is useful to know whether the reactants and products in a chemical reaction are solids, gases,
liquids or dissolved in water. We can add state symbols to a symbol equation to show this.
State symbol and meaning
Symbol
Meaning
(s)
Solid
(l)
Liquid
(g)
Gas
(aq)
Aqueous (dissolved in water)
For example, for the reaction between sodium and water, this is the symbol equation with state symbols:
2Na(s) + 2H2O(l) → 2NaOH(aq) + H2(g)
Conservation of mass
Mass is never lost or gained in chemical reactions. We say that mass is always conserved. In other words, the
total mass of products at the end of the reaction is equal to the total mass of the reactants at the beginning. This
is because no atoms are created or destroyed during chemical reactions.
The principle of conservation of mass allows you to work out the mass of one substance in a reaction if the
masses of the other substances are known. For example:
Percentage yield
For a given mass of reactants, it is possible to calculate the maximum mass that could be made of a certain product. This
mass is called the predicted yield. The actual yield is the mass of product made when the reaction is actually carried out.
Calculating percentage yield
The percentage yield is calculated using this formula:
Equation: Percentage yield = Actual yield obtained/(divided) by the Theoretical or predicted yield x 100
For example, if the predicted yield is 20 g but the actual yield is 15 g:
Percentage yield = (15 ÷ 20) × 100 = 75%
Losing product
A 100 per cent yield means that no product has been lost, while a 0 per cent yield means that no product has been
made. There are several reasons why the percentage yield of a product might be less than 100 per cent. These include:
loss when filtering
loss when evaporating
loss when transferring liquids
not all reactants reacting to make product
Percentage purity
Percentage purity of a substance can be calculated by dividing the mass of the pure chemical by the total mass of the
sample, and then multiplying this number by 100.
Calcium carbonate
?
Calcium oxide
28g
+
Carbon dioxide
22g
Calcium carbonate is made up of 28 grams of calcium oxide and 22 grams of carbon dioxide
In practice, it is not always possible to get the entire calculated amount of product from a reaction. This is
because:



reversible reactions may not be completed
some product may be lost when it is removed from the reaction mixture
some of the reactants may react in an unexpected way
Empirical formulae
The empirical formula of a compound is the simplest whole number ratio of atoms of each element in the
compound. It is determined using data from experiments and therefore empirical.
For example, the molecular formula of glucose is C6H12O6 but the empirical formula is CH2O. This is because
we can divide each number in C6H12O6 by 6 to make a simpler whole number ratio.
On the other hand, a compound which has the empirical formula of CH2 could have a molecular formula of
C2H4, C3H6, C4H8 or even C13H26.
You can use information about reacting masses to calculate the formula of a compound. Here is an example:
3.2 g of sulphur reacts with oxygen to produce 6.4 g of sulphur oxide. What is the formula of the oxide?
You need to use the fact that the Ar (relative atom mass) of sulphur is 32 and the Ar of oxygen is 16.
Step
Action
Result
1
Write the element symbols
S
O
2
Write the masses
6.4 g – 3.2 g = 3.2 g
3.2 g
3
Write the Ar values
32
16
4
Divide masses by Ar
3.2 ÷ 32 = 0.1 3.2 ÷ 16 = 0.2
5
Divide by the smallest number 0.1 ÷ 0.1 = 1 0.2 ÷ 0.1 = 2
6
Write the formula
SO2
The action at Step 5 usually gives you the simplest whole number ratio straightaway. Sometimes it does not,
so you might get 1 and 1.5. In this example, you would multiply both numbers by 2, giving 2 and 3 (instead of
rounding 1.5 up to 2).
Converting the empirical formula to a molecular formula
From the empirical formula, you can work out the molecular formula if you know the relative formula mass
(Mr) of the compound.
Add up the atomic masses of the atoms in the empirical formula.
For example, the empirical formula of a hydrocarbon is CH2 and its Mr is 42.



the mass of the atoms in the empirical formula is 14
42 ÷ 14 = 3
so you need to multiply the numbers in the empirical formula by 3
The molecular formula of the hydrocarbon is therefore C3H6.
Empirical formulae experiments
Finding the empirical formula of a metal oxide
The empirical formula of magnesium oxide can be calculated using the following experiment, which finds the
mass of the magnesium and oxygen atoms in a sample of the compound.
1. Weigh a crucible (with its lid).
2. Put a sample of clean magnesium ribbon into the crucible and weigh it with the lid. Calculate the mass
of magnesium by subtracting the mass of the empty crucible.
3. Strongly heat the crucible over a Bunsen burner for several minutes.
4. Carefully lift the lid from time to time to allow sufficient air into the crucible for the magnesium to fully
oxidise without letting any magnesium oxide escape.
5. Continue heating until the mass of the crucible reaches a constant (maximum) mass, indicating that the
reaction is complete.
6. Measure the mass of the crucible and contents again. Calculate the mass of the magnesium oxide by
subtracting the mass of the empty crucible.
7. To work out the empirical formula, you need the mass of the magnesium (from step 2) and the mass of
the oxygen atoms as well. To find the mass of the oxygen atoms, subtract the mass of magnesium used
from the mass of the magnesium oxide (from step 6).
8. Now divide each of the two masses by the relative atomic masses of the elements and simplify the ratio.
Finding the molecular formula - crystallisation
Some ionic compounds have a fixed number of water molecules trapped inside their crystal lattice. The formula
of a hydrated ionic compound can be calculated by dehydrating it:
1. Record the mass of an evaporating basin.
2. Add a known mass of hydrated copper sulfate.
3. Heat over a Bunsen burner, gently stirring, until the blue copper sulfate has turned completely white and
the mass has reached a constant (minimum) mass, indicating that all the water has been lost.
4. Record the mass of the evaporating basin and contents and then use this to calculate the mass of the
white anhydrous copper sulfate remaining, and also the mass of the water lost.
5. Divide the mass of the white copper sulfate by the Mr (relative formula mass) of CuSO4 (159.5) and
divide the mass of water by the Mr of water (18). Simplify the ratio of waters to copper sulfates to find
the whole number represented here by ‘x’: CuSO4.xH2O
Reacting mass calculations
If you know the mass of a reactant and a product, you can use simple ratios to calculate reacting masses and
product masses.
Example 1 – reactant mass
Question:
When 12 g of carbon is burned in air, 44 g of carbon dioxide is produced. What mass of carbon
is needed to produce 11 g of carbon dioxide?
Answer:
12 g of carbon makes 44 g of carbon dioxide.
12 ÷ 44 g of carbon will make 1 g of carbon dioxide.
You will need 11 × (12 ÷ 44 g) = 3 g of carbon to make 11 g of carbon dioxide.
Example 2 – product mass
Question:
When 5.0 g of calcium carbonate is decomposed by heating, it produces 2.2 g of carbon dioxide.
What mass of calcium carbonate is needed to produce 8.8 g of carbon dioxide?
Answer:
5 g of calcium carbonate makes 2.2 g of carbon dioxide.
5 ÷ 2.2 g of calcium carbonate will make 1 g of carbon dioxide.
You will need 8.8 × (5 ÷ 2.2 g) = 20 g of carbon to make 8.8 g of carbon dioxide.
More reacting mass calculations
The mass of a product or reactant can be calculated using the balanced equation. Follow these steps:
1. Write out the balanced symbol equation. Underline the two substances you are interested in.
2. Write the given mass of a substance under its formula.
3. Work out the total relative formula mass (Mr) for each substance (the one you know and the one you
are trying to find out). Write these under their formulae.
4. Calculate: unknown mass = known mass ÷ total Mr of known substance × total Mr of unknown
substance
Worked example
Question:
Calculate the mass of iron (Fe) that can be extracted from 8 g of Fe2O3 in the reaction with
carbon.
Answer:
Step
4Fe + 3CO2
1 2Fe2O3 + 3C →
?g
2 8g
4 × 56 = 224
3 2 × 160 = 320
4 mass of iron formed = (8 ÷ 320) × 224 = 5.6 g
Mole calculations in solutions
Concentration calculations
You can calculate the amount of a substance (in moles) in a solution if you know the volume and concentration.
You can also work out the concentration of an acid reacting with an alkali, or vice versa.
Concentration, amount of solute and volume of solution are linked by this equation:
Concentration in mol/dm3 = amount in mol ÷ volume in dm3
This equation can be rearranged to find the amount of solute or volume of solution.
Divide a volume in cm3 by 1000 to convert it to a volume in dm3.
Calculating concentration
Question:
0.5 mol of solute is dissolved in 250 cm3 of solution. Calculate its concentration.
Answer:
250 cm3 = 250 ÷ 1000 = 0.25 dm3
Concentration = 0.5 ÷ 0.25 = 2.0 mol/dm3
Calculating amount of solute
Rearranging the equation: amount = concentration × volume
Question:
Calculate the amount of solute dissolved in 2 dm3 of a 0.1 mol/dm3 solution.
Answer:
Amount = 0.1 × 2 = 0.2 mol/dm3
Calculating volume
Rearranging the equation: volume = amount ÷ concentration
Question:
Calculate the volume of a 2 mol/dm3 solution that contains 0.5 mol of solute.
Answer:
Volume = 0.5 ÷ 2 = 0.25 dm3
Note that 0.25 dm3 is the same as 250 cm3 (0.25 × 1000).
Concentrations can also be expressed in grams per decimetre cubed (g/dm3). To convert the two different units
of concentration, use the relative formula mass (Mr).
Percentage yield
For a given mass of reactants, it is possible to calculate the maximum mass that could be made of a certain
product. This mass is called the predicted yield. The actual yield is the mass of product made when the
reaction is actually carried out.
Chemical formulae and equations
1
What is the correct formula for carbon dioxide?
C2 O
CO2
CO2
2
The formula for sulfuric acid is H2SO4. How many of each type of atom are there?
Two hydrogens, one sulfur and four oxygens
One hydrogen, two sulfurs and four oxygens
Two hydrogens, two sulfurs and four oxygens
3
The formula for ammonium sulfate is (NH4)2SO4. How many hydrogen atoms are there in this formula?
4
2
8
4
What is used in the middle of a chemical equation?
=
→
Δ
In the reaction CH4 + O2 → CO2 + H2O, what is the formula of a reactant?
5
CH4
CO2
H2O
6
When hydrogen (H2) reacts with oxygen (O2), water is produced (H2O). Which of these shows a
balanced symbol equation for this reaction?
H2 + O2 → H2O
2H2 + O2 → 2H2O
2H2 + 2O2 → 2H2O
7
What is the state symbol for solid?
(s)
(l)
(g)
8
What is the meaning of the state symbol (aq)?
Liquid
Dissolved in water
Gaseous
9
What is the empirical formula of C4H8O2?
CH2O
C8H16O4
C2H4O
10
4.0 g of magnesium oxide was found to contain 2.4 g of magnesium. The rest was oxygen. What is the
empirical formula of magnesium oxide?
MgO2
Mg2O
MgO
Chemical calculations
1
If 24 g of magnesium reacts with 16 g of oxygen to produce 40 g of magnesium oxide, what mass of
magnesium would you need to start with to produce 10 g of magnesium oxide?
6 g of magnesium
12 g of magnesium
4 g of magnesium
2
What is the relative formula mass of calcium carbonate, CaCO3?
50
100
68
3
The relative formula mass of CaO is 56 and the relative formula mass of CO2 is 44. Work out the mass
of CaO that can be obtained from 200 g of CaCO3. CaCO3 → CaO + CO2.
56 g
224 g
112 g
4
What mass of CO2 will be produced when 50 g of CaCO3 decomposes?
22 g
44 g
11 g
5
What is 20 cm3 in dm3?
0.002 dm3
0.02 dm3
0.2 dm3
6
What is the concentration of a solution if 0.25 mol is dissolved in 500 cm3 of solution?
1 mol/dm3
2 mol/dm3
0.5 mol/dm3
7
How many moles of solute are there in 25 cm3 of a solution that has a concentration of 0.4 mol/dm3?
0.01 moles
10 moles
0.0625 moles
8
If a solution of NaOH is 0.1 mol/dm3, what is its concentration in g/dm3?
40 g/dm3
4 g/dm3
0.4 g/dm3
9
In a reaction, the predicted yield was 80 g, but the actual yield was 60 g. What was the percentage
yield?
75%
133%
50%
10
What is the percentage yield if the actual yield was 2.4 g and the predicted yield was 3.0 g?
75%
80%
125%
Answers
1
What is the correct formula for carbon dioxide?
CO2
2
The formula for sulfuric acid is H2SO4. How many of each type of atom are there?
Two hydrogens, one sulfur and four oxygens
3
The formula for ammonium sulfate is (NH4)2SO4. How many hydrogen atoms are there in this formula?
8
4
What is used in the middle of a chemical equation?
5
→
In the reaction CH4 + O2 → CO2 + H2O, what is the formula of a reactant?
CH4
6
When hydrogen (H2) reacts with oxygen (O2), water is produced (H2O). Which of these shows a
balanced symbol equation for this reaction?
2H2 + O2 → 2H2O
7
What is the state symbol for solid?
(s)
8
What is the meaning of the state symbol (aq)?
Dissolved in water
9
What is the empirical formula of C4H8O2?
C2H4O
10
4.0 g of magnesium oxide was found to contain 2.4 g of magnesium. The rest was oxygen. What is the
empirical formula of magnesium oxide?
MgO
1
If 24 g of magnesium reacts with 16 g of oxygen to produce 40 g of magnesium oxide, what mass of
magnesium would you need to start with to produce 10 g of magnesium oxide?
6g of magnesium
If 24 g of Mg produces 40 of MgO, and you want 10 g, you need to divide all the reacting quantities by 4. So 24
g of Mg divided by 4 equals 6 g. You divided the mass of Mg by 10.
2
What is the relative formula mass of calcium carbonate, CaCO3?
100
The relative formula mass is calculated by adding together the masses of the atoms in the compound: 40 + 12 +
16 + 16 + 16 = 100.
3
The relative formula mass of CaO is 56 and the relative formula mass of CO2 is 44. Work out the mass
of CaO that can be obtained from 200 g of CaCO3. CaCO3 → CaO + CO2.
112 g
The ratio of RFMs for this reaction is 100 → 56 + 44. But starting with 200 g of CaCO3 means we will end up
with twice as much CaO: 112 g.
4
What mass of CO2 will be produced when 50 g of CaCO3 decomposes?
22 g
Working from the ratio of RFMs for this reaction, we see that 100 g of CaCO3 will produce 44 g of CO2. So 50
g of CaCO3 will make 22 g of CO2.
5
What is 20 cm3 in dm3?
0.02 dm3
To convert cm3 to dm3, you divide by 1000. So 20 ÷ 1000 = 0.02 dm3.
6
What is the concentration of a solution if 0.25 mol is dissolved in 500 cm3 of solution?
0.5 mol/dm3
Convert the volume to dm3 (0.5) and then divide the moles by the volume: 0.25 ÷ 0.5 = 0.5 mol/dm3.
7
How many moles of solute are there in 25 cm3 of a solution that has a concentration of 0.4 mol/dm3?
0.01 moles
Convert the volume to dm3 (0.025) and then multiply this by the concentration: 0.025 x 0.4 = 0.01 moles.
8
If a solution of NaOH is 0.1 mol/dm3, what is its concentration in g/dm3?
4 g/dm3
To convert mol/dm3 to g/dm3, multiply by the RFM. So 0.1 x 40 = 4 g/dm3.
9
In a reaction, the predicted yield was 80 g, but the actual yield was 60 g. What was the percentage
yield?
75%
Percentage yield is calculated by dividing the actual yield by the predicted yield and then multiplying by 100.
So (60 ÷ 80) x 100 = 75%.
10
What is the percentage yield if the actual yield was 2.4 g and the predicted yield was 3.0 g?
80%
Percentage yield is calculated by dividing the actual yield by the predicted yield and then
multiplying by 100. So (2.4 ÷ 3.0) x 100 = 80%.