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The Mole and Mass Conversions

Every time a baker measures flour, they use a scale, but a chemist needs to know the exact number of particles to make a reaction work. The unit used for the amount of substance is the mole (mol). One mole of any substance always contains $6.02 \times 10^{23}$ particles, a fixed value known as the Avogadro constant.

To convert between mass and moles, we use the relative formula mass ( $M_r$ ), which is the total sum of the values for all atoms in a chemical formula. The mass of one mole of a substance is its , which is numerically equal to the but is measured in grams per mole (g/mol).

Mass must always be in grams (g) before using this equation. If an exam question provides mass in other units, you must convert it first: multiply kilograms by $1000$ to get grams, or divide milligrams by $1000$ to get grams. For extremely large or small mole values, Edexcel examiners reward the correct use of standard form.

Calculate the number of moles in $4.5 \text{ kg}$ of water ( $H_2O$ ). (Relative atomic masses: $H = 1$ , $O = 16$ )

Convert the mass into grams: $4.5 \text{ kg} \times 1000 = 4500 \text{ g}$
Calculate the relative formula mass ( $M_r$ ): $M_{r} of H_{2} O = (2 \times$

Empirical Formulas

You can describe a bicycle by listing every single part it has, or by simply stating the basic ratio of two wheels to one frame. In chemistry, the molecular formula tells you the exact number and type of atoms in a single molecule. In contrast, the empirical formula represents the simplest whole-number ratio of atoms of each element present in a compound.

For ionic compounds, the standard chemical formula provided (such as $NaCl$ ) is always its empirical formula. If a question provides element percentages instead of masses, assume you have a $100 \text{ g}$ sample; the percentage value then directly equals the mass in grams.

A compound contains $2.4 \text{ g}$ of carbon and $0.8 \text{ g}$ of hydrogen. Determine its empirical formula. (Relative atomic masses: $C = 12$ , $H = 1$ )

Step	Carbon ( $C$ )	Hydrogen ( $H$ )
1. Mass (g)	$2.4$	$0.8$
2. $A_r$

Deducing Stoichiometry from Masses (Higher Tier Only)

How do industrial chemists know exactly how much of each reactant to mix without wasting any? They rely on stoichiometry, which is the relationship between the relative quantities of substances in a reaction. This is shown by the stoichiometric coefficients (the large numbers in front of formulas).

You can determine these coefficients by calculating the molar ratio of all substances involved. You must never alter the small subscript numbers in a chemical formula to balance an equation.

In a reaction, $46 \text{ g}$ of sodium ( $Na$ ) reacts with $71 \text{ g}$ of chlorine gas ( $Cl_2$ ) to produce of sodium chloride (). Write the balanced equation with state symbols.

Calculate $M_r$ for each: $Na = 23$ , $Cl_2 = 71$ ,

Yield and Reacting Masses

The theoretical yield is the maximum mass of product possible, assuming $100\%$ conversion. The actual yield obtained in practice is always less due to side reactions, reversible processes, or product lost during transfer. The maximum amount of product is restricted by the limiting reactant, which is completely used up first.

$\text{Percentage Yield} = \frac{\text{Actual Yield}}{\text{Theoretical Yield}} \times 100$

A student reacts $1.4 \text{ kg}$ of lithium ( $Li$ ) with excess oxygen to produce lithium oxide ( $Li_2O$ ). The actual yield is $2.5 \text{ kg}$ . Calculate the percentage yield. Equation: ()

Reactant Moles: $1400 \text{ g} \div 7 = 200 \text{ mol } Li$
Molar Ratio: $4:2$ ratio (from equation) means $100 \text{ mol } Li_2O$ should form.

Students often divide the relative atomic mass by the given mass when calculating moles; always divide the mass by the relative atomic mass.

If your percentage yield calculation results in a value greater than 100%, you have likely divided the theoretical yield by the actual yield instead of the other way around.

In multi-step calculation questions, examiners award separate marks for intermediate steps like finding the molar ratio, so always write out your working clearly even if you cannot finish the math.

When determining an empirical formula, if your final ratio ends in a decimal like 1.5, do not round it up to 2; instead, multiply all numbers in the ratio by 2 to achieve a whole-number ratio.

Always check the units of mass given in the question; failing to convert milligrams (mg) or kilograms (kg) into grams (g) before calculating moles is a very frequent reason for lost marks.

The unit for the amount of substance, where one mole contains exactly 6.02 × 10²³ particles.

The number of particles in one mole of a substance, equal to 6.02 × 10²³.

The sum of the relative atomic masses of all the atoms present in a chemical formula.

The average mass of an atom of an element compared to 1/12th of the mass of an atom of carbon-12.

The mass of one mole of a substance, expressed in grams per mole (g/mol).

A chemical formula that shows the actual number and type of each atom in a single molecule of a compound.

A chemical formula that shows the simplest whole-number ratio of atoms of each element present in a compound.

The mathematical relationship between the relative quantities of reacting substances and products in a chemical reaction.

The large numbers placed in front of chemical formulas in a balanced equation to show the molar proportions of substances.

The maximum possible mass of a product that can be formed from a given amount of reactants, assuming 100% conversion.

The physical mass of a product that is practically obtained at the end of a chemical reaction.

The reactant in a chemical reaction that is completely used up first, preventing any further product from being formed.

A measure of the efficiency of a chemical reaction, calculated by dividing the actual yield by the theoretical yield and multiplying by 100.