Calculations in Chemistry | Chemistry for Grade 10 PDF Download

Relative Formula Mass

Atoms have very little mass so their relative atomic masses are used. The relative atomic mass of an element, symbol Ar, is the relative mass of its atoms compared to the mass of a carbon-12 atom. The Ar values for elements are given in the periodic table. Since Ar is a measure of relative mass, it has no units.

Calculating relative formula mass

The relative formula mass of a substance made up of molecules is the sum of the relative atomic masses of the atoms in the numbers shown in the formula.
Relative formula mass has the symbol, Mr. To calculate the M_r for a substance:

• Work out how many atoms of each element there are in the chemical formula.
• Add together the A_r values for all the atoms of each element present.

For example, the formula for carbon dioxide is CO₂. It consists of one carbon atom (A_r = 12) and two oxygen atoms (A_r = 16):
M_r of CO₂ = 12 + 16 + 16 = 44
It could also be calculated this way:
M_r of CO₂ = (1 × 12) + (2 × 16) = 12 + 32 = 44
Like Ar values, Mr values are just numbers. They have no units because they are relative masses.

Relative formula masses of ionic compounds

Ionic compounds such as sodium chloride do not exist as molecules. However, their relative formula masses are calculated in the same way, from the numbers shown in the formula.

Solved Example
Example 1: Calculate the relative formula mass, M_r, of calcium hydroxide, Ca(OH)₂.
(A_r of Ca = 40, A_r of O = 16, A_r of H = 1)

M_r = 40 + (2 × 16) + (2 × 1)
= 40 + 32 + 2
= 74
It could also be calculated this way:
M_r = 40 + 2 × (16 + 1)
40 + 34 = 74

Example 2: Calculate the relative formula mass, M_r, of magnesium nitrate, Mg(NO₃)₂.
(A_r of Mg = 24, A_r of N = 14, A_r of O = 16)

M_r = 24 + (2 × 14) + (2 × 3 × 16)
= 24 + 28 + 96
= 148
It could also be calculated this way:
M_r = 24 + 2 × [14 + (16 × 3)]
= 24 + 124
= 148

Law of Conservation of Mass

The law of conservation of mass states that no atoms are lost or made in a chemical reaction. Instead, the atoms join together in different ways to form products. This is why, in a balanced symbol equation, the number of atoms of each element is the same on both sides of the equation.

Since atoms are not lost or made in a chemical reaction, the total mass of the products is equal to the total mass of the reactants. The sum of the relative formula masses of the reactants is equal to the sum of the relative formula masses of the products.

No atoms are created or destroyed when copper reacts with oxygen to form copper oxide

Calculations using the law

The mass of one substance in a reaction can be calculated if the masses of the other substances are known. For example:

1. Calcium carbonate is made up of 28 grams of calcium oxide and 22 grams of carbon dioxide

2. To find out the amount of calcium carbonate, add the 28 grams of calcium oxide and 22 grams of carbon dioxide

3. This makes a total of 50 grams of calcium carbonate

4. 48 grams of magnesium and an unknown amount of oxygen make 80 grams of magnesium oxide

5. Take the 80 grams of magnesium oxide and take away the 48 grams of magnesium

6. This leaves 32 grams of oxygen

Reactions in closed systems

No substances can enter or leave a closed system, such as a stoppered flask. Sometimes, reactions that happen in open beakers are closed systems, for example acid-alkali neutralisation reactions. Since all the reactants and products stay in the beaker, the total mass of the beaker and the substances in it stay the same during the reaction.

Reactions in non-enclosed systems

Substances can enter or leave a non-enclosed system. These systems include open flasks, boiling tubes or crucibles that let gases enter or leave. For example:

• copper carbonate decomposes on heating to make solid copper oxide, which stays in the boiling tube, and carbon dioxide gas, which escapes
• magnesium reacts with oxygen, gained from the air, to produce magnesium oxide

If a gas escapes, the total mass will look as if it has decreased. If a gas is gained, the total mass will look as if it has increased. However, the total mass stays the same if the mass of the gas is included.

Solved Example

10.0 g of calcium carbonate, CaCO₃, was heated in a thermal decomposition reaction. 5.60 g of solid remained after heating. The equation below represents the reaction:
CaCO₃(s) → CaO(s) + CO₂(g)
Explain the change in mass.

The mass appears to have decreased because one of the products, carbon dioxide, escapes to the air. The total mass of the solid and gas products will be 10.0 g.

Chemical Measurements

Whenever a measurement is made in chemistry, there is always some uncertainty in the result obtained. There are many causes of uncertainty in chemical measurements. For example it may be difficult to judge:

• whether a thermometer is showing a temperature of 24.0°C, 24.5°C or 25.0°C
• exactly when a chemical reaction has finished

There are two ways of estimating uncertainty:

• by considering the resolution of measuring instruments
• from the range of a set of repeat measurements

Estimating uncertainty from measuring instruments

• The resolution of a measuring instrument is the smallest change in a quantity that gives a change in the reading that can be seen. A thermometer with a mark at every 1.0°C has a resolution of 1.0°C. It has a higher resolution than a thermometer with a mark at every 2.0°C.
• The uncertainty of a measuring instrument is estimated as plus or minus (±) half the smallest scale division. For a thermometer with a mark at every 1.0°C, the uncertainty is ± 0.5°C. This means that if a student reads a value from this thermometer as 24.0°C, they could give the result as 24.0°C ± 0.5°C.
• For a digital measuring instrument, the uncertainty is half the last digit shown on its display. For a timer reading to 0.1 s, the uncertainty is ± 0.05 s.

Estimating uncertainty from sets of repeat measurements

For a set of repeat measurements, the uncertainty is ± half the range. This means that the value can be given as the mean value ± half the range.

Solved Example

The table shows five measurements for the volume of acid required in a neutralisation reaction.
Calculate the mean volume and estimate the uncertainty.

mean = = 24.0 cm³
range = (biggest value - smallest value)
= 25.0 - 23.0
= 2.0 cm³
uncertainty = ± half the range
= 2.0/2 cm³
= ± 1.0 cm³
So the volume is 24.0 cm3 ± 1.0 cm³.

Showing uncertainty on a graph

Uncertainty can also be shown on a graph. All the repeat readings for each value of the independent variable are plotted. Vertical lines joining these values represent the uncertainty.

A graph showing the repeat readings for each value of the independent variable. The short vertical lines represent uncertainty.

Concentration of Solutions

A solution forms when a solute dissolves in a solvent. The concentration of a solution is a measure of how 'crowded' the solute particles are. The more concentrated the solution, the more particles it contains in a given volume.

Calculating concentration

The concentration of a solution can be calculated using:

• the mass of dissolved solute in grams, g
• the volume of solution (or solvent) in cubic decimetres, dm³

The units for concentration can also be shown as g dm^-3, but this means the same as g/dm³.

Worked example

8 g of sodium hydroxide is dissolved in 2 dm3 of water. Calculate the concentration of the sodium hydroxide solution formed.

concentration =
concentration =
concentration = 4 g/dm³

Volume units

Apparatus used to measure volumes is usually marked in cm³ or ml. Although these are different units, they describe the same volume. For example, 250 ml = 250 cm³.

Volumes used in concentration calculations must be in dm3, not in cm3 or ml. It is useful to know that 1 dm³ = 1000 cm³. This means:

• divide by 1000 to convert from cm³ to dm³
• multiply by 1000 to convert from dm³ to cm³

For example, 250 cm³ is 0.25 dm3 (250 ÷ 1000). It is often easiest to convert from cm³ to dm³ before continuing with a concentration calculation.
Example: 100 cm³ of dilute hydrochloric acid contains 0.5 g of dissolved hydrogen chloride. Calculate the concentration of the acid in g/dm³.

volume of acid = 100 ÷ 1000 = 0.1 dm³
concentration of acid = 0.5/0.1
= 5 g/dm³

Calculating the mass of solute

Rearranging the equation for concentration allows the mass of solute to be calculated:

mass of solute in g = concentration in g/dm³ × volume in dm³

Solved Example: A solution of sodium chloride has a concentration of 10 g/dm3. What mass of sodium chloride is dissolved in 2 dm³ of the solution?

mass of solute in g = concentration in g/dm³ × volume in dm³
= 10 g/dm³ × 2 dm³
= 20 g

Higher tier

The concentration of a solution can be changed:

• concentration can be increased by dissolving more solute in a given volume of solution - this increases the mass of the solute
• concentration can be increased by allowing some of the solvent to evaporate - this decreases the volume of the solution