What is Mr in chemistry? iGCSE relative molecular mass
Mr stands for relative molecular mass. It is the sum of the relative atomic masses (Ar) of every atom in a molecule. For ionic compounds the equivalent quantity is called relative formula mass, but Cambridge iGCSE uses Mr for both. Mr has no units, because it is a ratio compared to one-twelfth of a carbon-12 atom.
This guide covers what Mr means at iGCSE, how to calculate it from a formula, the difference between Mr and molar mass, and the mole and percentage-by-mass calculations that flow from it. These are central to Cambridge iGCSE Chemistry Section 3 (stoichiometry) and turn up in every Paper 4.
Add up the atomic masses
Mr is the sum of the Ar values of every atom in the molecule. Use the values from the iGCSE periodic table.
No units
Mr is a relative number, compared to one-twelfth the mass of a carbon-12 atom. Do not write g or kg after it.
Same number as molar mass
Molar mass in grams per mole is numerically equal to Mr. So Mr of water is 18, and its molar mass is 18 g/mol.
What Mr means
Mr is defined as the relative mass of one molecule of a substance compared to one-twelfth of a carbon-12 atom. Because it is a ratio, Mr has no units. The number tells you how many times heavier a molecule is than a twelfth of a carbon-12 atom.
The formal Cambridge iGCSE definition is: Relative molecular mass is the sum of the relative atomic masses of all the atoms shown in the formula. For ionic compounds (which do not form discrete molecules) the same quantity is called relative formula mass, but the iGCSE specification still uses the symbol Mr.
Mr vs Ar Ar is the relative atomic mass of a single atom of an element (for example, Ar of oxygen is 16). Mr is the relative mass of a whole molecule (for example, Mr of O2 is 32). Mr is always built by adding up Ar values.
How to calculate Mr
To calculate Mr from a formula, multiply the Ar of each element by the number of atoms of that element in the formula, then add the totals. Always use the Ar values printed on the Cambridge iGCSE periodic table, not values you remember from elsewhere.
The four steps are: Write out each element and the number of atoms, look up the Ar of each element, multiply Ar by the number of atoms, then add the products together. Show every step in the exam, because mark schemes reward correct method even when the final number is wrong.
| Compound | Formula | Calculation | Mr |
|---|---|---|---|
| Water | H2O | (2 x 1) + (1 x 16) | 18 |
| Carbon dioxide | CO2 | (1 x 12) + (2 x 16) | 44 |
| Methane | CH4 | (1 x 12) + (4 x 1) | 16 |
| Sodium chloride | NaCl | (1 x 23) + (1 x 35.5) | 58.5 |
| Calcium carbonate | CaCO3 | (1 x 40) + (1 x 12) + (3 x 16) | 100 |
| Sulfuric acid | H2SO4 | (2 x 1) + (1 x 32) + (4 x 16) | 98 |
| Ammonium nitrate | NH4NO3 | (2 x 14) + (4 x 1) + (3 x 16) | 80 |
Worked example: Mr of magnesium hydroxide
Calculate the Mr of magnesium hydroxide, Mg(OH)2.
Step 1: Identify each element and the number of atoms. Mg = 1, O = 2, H = 2. Note that the small 2 outside the brackets multiplies both the O and the H inside.
Step 2: Look up Ar values. Mg = 24, O = 16, H = 1.
Step 3: Multiply and add. (1 x 24) + (2 x 16) + (2 x 1) = 24 + 32 + 2 = 58.
The Mr of Mg(OH)2 is 58. No units. The most common mistake here is forgetting that the 2 outside the bracket applies to both elements inside it.
Brackets in formulas A subscript outside a bracket multiplies every atom inside the bracket. For example, in Al2(SO4)3 there are 2 Al, 3 S, and 12 O. Cambridge iGCSE examiner reports flag bracket errors as a top reason for lost marks in this topic.
Mr and molar mass
Molar mass is the mass of one mole of a substance, measured in grams per mole (g/mol). Conveniently, the molar mass in g/mol is numerically equal to the Mr (or for elements, the Ar).
So if water has Mr 18, one mole of water has a mass of 18 g. If carbon dioxide has Mr 44, one mole of CO2 weighs 44 g. This is why Mr is the gateway to all mole calculations on Cambridge iGCSE Paper 4.
Using Mr in mole calculations
The mole formula is one you will use constantly: Number of moles = mass / Mr. Rearrange to find mass = moles x Mr, or Mr = mass / moles. These three forms appear directly on Paper 4.
Worked example: How many moles are in 9 g of water? Mr of H2O is 18, so moles = 9 / 18 = 0.5 mol. Going the other way: What is the mass of 2 moles of carbon dioxide? Mr of CO2 is 44, so mass = 2 x 44 = 88 g.
The mole triangle Draw a triangle with mass at the top and moles and Mr at the bottom. Cover the quantity you want to find. If you cover mass, you get moles x Mr. If you cover moles, you get mass / Mr. If you cover Mr, you get mass / moles. Many students find this faster than rearranging the algebra under exam pressure.
Percentage by mass calculations
Cambridge iGCSE Paper 4 often asks for the percentage by mass of a particular element in a compound. The formula is straightforward: Percentage by mass = (number of atoms x Ar of that element / Mr of the compound) x 100.
Worked example: What is the percentage by mass of nitrogen in ammonium nitrate (NH4NO3)? Mr is 80 (from the table above). There are 2 nitrogen atoms, each with Ar 14, so the total nitrogen mass is 28. Percentage by mass of N = (28 / 80) x 100 = 35%. This kind of calculation is important in fertiliser questions, because it shows how much usable nitrogen a fertiliser delivers per kilogram.
Where students lose marks
Cambridge iGCSE examiner reports flag the same problems every series. Most are not about chemistry knowledge, they are about arithmetic care and respecting the formula.
Mistakes that lose marks on Paper 4 Forgetting that subscripts outside brackets multiply every atom inside. Adding Ar values without multiplying by the number of atoms first. Writing units (g, kg) after Mr (it has none). Confusing Mr with Ar in the calculation. Using Ar values from memory instead of the values printed on the Cambridge iGCSE periodic table, which is the source of truth. Slipping a decimal place when an element like chlorine has Ar 35.5.
Worked example: Empirical formula from percentage composition
A compound has the composition by mass: 40% carbon, 6.7% hydrogen, and 53.3% oxygen. Find its empirical formula.
Step 1: Assume 100 g of the compound, so you have 40 g C, 6.7 g H, and 53.3 g O.
Step 2: Divide each mass by the element's Ar. C = 40 / 12 = 3.33 mol. H = 6.7 / 1 = 6.7 mol. O = 53.3 / 16 = 3.33 mol.
Step 3: Divide each result by the smallest. C = 1, H = 2, O = 1. The empirical formula is CH2O, which is the simplest ratio of atoms. This kind of question links Mr, moles, and percentages all in one. Cambridge iGCSE Paper 4 asks it most years.
iGCSE Mr revision checklist
- Mr is the sum of the Ar of every atom in a formula
- Mr has no units because it is a ratio
- Multiply Ar by the number of atoms, then add the totals
- Subscripts outside brackets multiply every atom inside
- Use the Ar values from the Cambridge iGCSE periodic table
- Molar mass in g/mol is numerically equal to Mr
- Number of moles = mass / Mr
- Percentage by mass = (n x Ar / Mr) x 100
