2.6.1 Relative formula mass and percentage composition

Relative formula mass adds up every atom the formula shows

1. Read the formula and note how many atoms of each element it contains.
2. Multiply each element's $A_r$ by the number of its atoms, then add the results together.
3. A symbol with no subscript stands for one atom.
4. A number outside a bracket multiplies every atom inside that bracket.
5. $M_r$ carries no unit, because it compares one mass with another.
6. Take the $A_r$ values from the periodic table, decimals included, such as $35.5$ for chlorine.

Percentage by mass compares one element with the whole compound

1. The percentage by mass of an element is the share of a compound's total mass that the element contributes.
2. Find the total $A_r$ of that element in the formula, which is its $A_r$ multiplied by its number of atoms.
3. Divide that by the $M_r$ of the whole compound, then multiply by $100$: $$percentage\; by\; mass=\frac{total{A}_{r}of\; the\; element}{M_{r}of\; the\; compound}\times 100$$
4. Repeating the calculation for another element changes only the numerator.
5. The percentages of every element in a compound add to $100$, apart from small rounding differences.

A worked percentage: oxygen in calcium carbonate

1. The formula is ${CaCO}_3$, with $A_r$ values of $40$ for calcium, $12$ for carbon and $16$ for oxygen.
2. First find the relative formula mass of the compound: $$M_r=40+12+(3\times 16)=100$$
3. Then find the total contribution of oxygen in the formula: $$3\times 16=48$$
4. Divide by the $M_r$ and multiply by $100$ to reach the percentage: $$\frac{48}{100}\times 100=48\%$$
5. Oxygen makes up $48\%$ of the mass of calcium carbonate.

Where these calculations usually go wrong

1. Using the atomic number instead of the relative atomic mass changes every figure in the answer.
2. Ignoring a bracket subscript undercounts the atoms inside it.
3. Leaving a unit on $M_r$ is wrong, and so is leaving the $\%$ sign off a percentage.
4. Checking that the formula matches the compound named catches an error before the arithmetic starts.
5. Adding the percentages of all the elements gives a final check close to $100$.

2.6.2 Empirical and molecular formulae

An empirical formula gives a ratio, a molecular formula gives a count

1. The two formulae describe the same compound in different ways.
2. Glucose has the molecular formula $C_6{H}_{12}{O}_6$ and the empirical formula ${CH}_{2}O$.
3. Dividing every subscript by their highest common factor turns a molecular formula into an empirical one.
4. Ionic compounds are always written as empirical formulae, because a lattice contains no molecules.
5. Some molecular formulae are already in their simplest ratio, such as $H_{2}O$.

From reacting masses to an empirical formula

1. Write down the mass of each element present in the sample.
2. Divide each mass by that element's relative atomic mass: $$ratio\; number=\frac{mass}{A_r}$$
3. Divide every ratio number by the smallest of them.
4. Turn the results into whole numbers, multiplying them all by the same factor if one comes out as $1.5$ or $1.33$.
5. Use those whole numbers as the subscripts.
6. $2.40g$ of magnesium gives $2.40÷24=0.100$, and $1.60g$ of oxygen gives $1.60÷16=0.100$.
7. Dividing both by $0.100$ gives $1$ and $1$, so the empirical formula is $MgO$.

From percentage composition to an empirical formula

1. Treat the percentages as masses in a $100g$ sample, so $40.0\%$ carbon becomes $40.0g$ of carbon.
2. That works because only the ratio between the masses matters.
3. The rest of the method is identical: divide by $A_r$, divide by the smallest, then simplify.
4. For $40.0\%$ carbon, $6.7\%$ hydrogen and $53.3\%$ oxygen the ratio numbers are $3.33$, $6.7$ and $3.33$.
5. Dividing by $3.33$ gives $1$, $2$ and $1$, so the empirical formula is ${CH}_{2}O$.

From an empirical formula to a molecular formula

1. Add the $A_r$ values in the empirical formula to get its empirical formula mass.
2. Divide the compound's $M_r$ by that empirical formula mass.
3. The answer is a whole number, because a molecule contains whole atoms.
4. Multiply every subscript in the empirical formula by that number.
5. For glucose, ${CH}_{2}O$ has an empirical formula mass of $12+2+16=30$.
6. With $M_r=180$, the multiplier is $180÷30=6$.
7. Multiplying every subscript by $6$ gives $C_6{H}_{12}{O}_6$.

Finding the formula of magnesium oxide by experiment

1. Wear eye protection throughout, and move the hot crucible only with tongs.
2. Clean a length of magnesium ribbon with emery paper, which removes the oxide layer already on it.
3. Weigh an empty crucible with its lid, then weigh it again with the magnesium inside.
4. Heat strongly, lifting the lid a little at intervals so that air reaches the magnesium while the solid oxide stays inside.
5. Let the crucible cool, then weigh it again.
6. Reheat, cool and reweigh until the mass stops changing, which shows the reaction has reached constant mass.
7. The mass of magnesium is the second weighing minus the first.
8. The mass of oxygen is the final weighing minus the second, because that increase is the oxygen taken in.
9. With weighings of $20.00g$, $20.24g$ and $20.40g$, the masses are $0.24g$ of magnesium and $0.16g$ of oxygen.
10. Dividing by $24$ and by $16$ gives $0.010$ and $0.010$, a $1:1$ ratio, so the formula is $MgO$.

2.6.3 Conservation of mass

Atoms are rearranged, so the total mass cannot change

1. A reaction rearranges the atoms that are present, and it neither creates nor destroys any of them.
2. Every atom in the reactants ends up somewhere among the products.
3. The total mass of the products therefore equals the total mass of the reactants.
4. The mass of any one substance does change, because substances are used up and formed.
5. A balanced equation is that same statement written in symbols.

A closed system: a precipitate is a change of form, not of mass

1. A closed system is sealed, so nothing can enter it or leave it.
2. Mixing two solutions in a stoppered flask keeps every product inside.
3. In a precipitation reaction, two solutions produce an insoluble solid.
4. Lead nitrate solution and potassium iodide solution give a yellow solid: $${Pb(NO}_3{)}_2(aq)+2KI(aq)\to {PbI}_2(s)+2{KNO}_3(aq)$$
5. The flask looks cloudy afterwards, but the balance reads exactly the same.
6. The atoms now in the solid were already present in the two solutions.

A reaction that takes in a gas gains measured mass

1. A non-enclosed system lets matter cross its boundary, so a gas can enter or escape.
2. Magnesium burning in an open crucible takes in oxygen from the air: $$2Mg(s)+O_2(g)\to 2MgO(s)$$
3. Those oxygen atoms join the magnesium and stay in the product.
4. The magnesium oxide left behind therefore weighs more than the magnesium did.
5. Nothing has been created, because the extra mass came out of the air.

A reaction that gives off a gas loses measured mass

1. Calcium carbonate reacts with hydrochloric acid and releases carbon dioxide: $${CaCO}_3(s)+2HCl(aq)\to {CaCl}_2(aq)+H_{2}O(l)+{CO}_2(g)$$
2. The gas escapes from the open flask into the room.
3. The balance reading falls while the reaction runs, and steadies once it finishes.
4. The carbon dioxide still has mass, so nothing has been destroyed.
5. Running the same reaction in a sealed flask would show no change in mass at all.

Explaining a mass change

1. Decide first whether the system is closed or non-enclosed.
2. For a closed system, state that nothing enters or leaves, so the total mass is unchanged.
3. For a non-enclosed system, name the gas and say whether it entered or escaped.
4. Link that gas to the direction in which the balance reading moved.
5. Finish by stating that mass is still conserved once the surroundings are included.

2.6.4 Reacting masses and concentration in g dm⁻³

A balanced equation fixes the ratio in which substances react

1. The numbers written in front of the formulae are the coefficients, and they give the ratio in which the substances react.
2. The subscripts inside a formula count atoms, and are never used as the reacting ratio.
3. Turning that ratio into masses takes the relative formula mass of each substance.
4. Multiply each coefficient by that substance's $M_r$ to get the mass ratio the equation predicts.
5. For ${CaCO}_3\to CaO+{CO}_2$ the mass ratio is $100:56:44$.
6. Those figures balance, because $56+44=100$ and mass is conserved.

Scaling the mass ratio to the masses in the question

1. Write the balanced equation, and write the mass ratio underneath it.
2. Find the scale factor by dividing the mass given in the question by the ratio mass for that same substance.
3. Multiply every other mass in the ratio by that scale factor.
4. For $25.0g$ of ${CaCO}_3$ against a ratio mass of $100$: $$scale\; factor=\frac{25.0}{100}=0.250$$
5. The mass of $CaO$ is then the ratio mass of $56$ scaled by the same factor: $$m(CaO)=0.250\times 56=14.0g$$
6. The answer is smaller than the starting mass because carbon dioxide is given off as well.

Concentration in grams per cubic decimetre

1. A solution's concentration compares the mass of solute with the volume of solution it is dissolved in.
2. The calculation is: $$c=\frac{m}{V}$$
3. Here $c$ is in ${g\; dm}^{-3}$, $m$ is in $g$ and $V$ is in ${dm}^3$.
4. Volumes are usually measured in ${cm}^3$, and $1{dm}^3=1000{cm}^3$.
5. So $250{cm}^3$ becomes $250÷1000=0.250{dm}^3$.
6. Dissolving $5.0g$ of sodium chloride to make that volume of solution gives: $$c=\frac{5.0}{0.250}=20.0{g\; dm}^{-3}$$

Checking a mass or a concentration

1. Check the equation is balanced before taking any ratio from it.
2. Check every mass is in grams and every volume is in ${dm}^3$.
3. Check the total mass of the products does not exceed the total mass of the reactants.
4. Keep extra figures through the working and round once, at the end.
5. Check the unit on the final answer: $g$ for a mass and ${g\; dm}^{-3}$ for a concentration.

2.6.5 The mole and the Avogadro constant

One mole is both a count of particles and a mass

1. Counting atoms one at a time is impossible, so chemists count them in moles instead.
2. One mole of any substance contains $6.02\times 10^{23}$ particles.
3. The particles are whichever the formula represents: atoms for an element such as aluminium, molecules for $O_2$, formula units for $NaCl$, or ions when the question names an ion.
4. One mole of a substance has a mass in grams equal to its relative particle mass.
5. The relative particle mass is the $A_r$ for an element made of single atoms, and the $M_r$ for a compound.
6. One mole of water therefore has a mass of $18g$, and one mole of aluminium a mass of $27g$.

Converting between mass and amount

1. Divide a mass by the relative particle mass to get the amount: $$n=\frac{m}{M_r}$$
2. Multiply an amount by the relative particle mass to get the mass: $$m=n\times M_r$$
3. The mass is in $g$, the amount in $mol$, and the relative particle mass carries no unit.
4. So $36g$ of water is $36÷18=2.0mol$.
5. And $0.25mol$ of carbon dioxide has a mass of $0.25\times 44=11g$.

Converting between amount and number of particles

1. Multiply an amount by the Avogadro constant to get the number of particles: $$N=n\times 6.02\times 10^{23}$$
2. Divide a number of particles by the Avogadro constant to get the amount: $$n=\frac{N}{6.02\times 10^{23}}$$
3. So $0.50mol$ of oxygen contains $0.50\times 6.02\times 10^{23}=3.01\times 10^{23}$ molecules.
4. And $1.204\times 10^{24}$ chloride ions is $1.204\times 10^{24}÷(6.02\times 10^{23})=2.00mol$.
5. A number of particles carries no unit, although the particle itself has to be named.

Going from mass straight to particles

1. Mass and particle number are linked through the amount in moles.
2. From a mass, divide by the relative particle mass and then multiply by the Avogadro constant: $$N=\frac{m}{M_r}\times 6.02\times 10^{23}$$
3. From a number of particles, divide by the Avogadro constant and then multiply by the relative particle mass: $$m=\frac{N}{6.02\times 10^{23}}\times M_r$$
4. For $9.0g$ of aluminium, with $A_r=27$, the amount is $0.3333mol$ and the count is $2.0\times 10^{23}$ atoms.
5. For $3.01\times 10^{23}$ molecules of ${CO}_2$, the amount is $0.500mol$ and the mass is $0.500\times 44=22g$.

Choosing the right quantity

1. A mass in the question and a mass in the answer means the relative particle mass is used twice.
2. A particle count anywhere in the question means the Avogadro constant appears in the working.
3. The amount in moles sits in the middle of every one of these conversions.
4. Units settle most slips: $g$ for a mass, $mol$ for an amount, and no unit for a particle count.
5. The relative particle mass always comes from the substance named, never from a neighbouring one.

2.6.6 Limiting reactants and stoichiometry

The reactant that runs out first controls how much product forms

1. Reactants are rarely mixed in exactly the ratio the equation calls for.
2. One of them is used up before the other, and the reaction stops at that point.
3. Whatever is left over is described as being in excess.
4. The amount of product is set entirely by the reactant that ran out.
5. Adding more of the excess reactant changes nothing, because there is nothing left for it to react with.

Comparing amounts, not masses, identifies the limiting reactant

1. Equal masses of two substances are not equal amounts, because their particles have different masses.
2. Convert each reactant's mass into an amount first: $$n=\frac{m}{M_r}$$
3. Divide each amount by that substance's coefficient in the balanced equation.
4. The smallest of those results identifies the limiting reactant.
5. A reactant present in the larger mass can still be the limiting one.

A worked limiting reactant calculation

1. Burn $12.0g$ of magnesium with $10.0g$ of oxygen, where the equation is $2Mg+O_2\to 2MgO$.

2. Convert both masses, using $A_r(Mg)=24$ and $M_r(O_2)=32$:

   $$n(Mg)=\frac{12.0}{24}=0.500mol$$ $$n(O_2)=\frac{10.0}{32}=0.3125mol$$

3. Divide each amount by its coefficient: magnesium gives $0.500÷2=0.250$, and oxygen gives $0.3125÷1=0.3125$.

4. Magnesium gives the smaller result, so magnesium is the limiting reactant and the oxygen is in excess.

5. The reaction consumes only $0.250mol$ of oxygen, leaving $0.3125-0.250=0.0625mol$ unreacted.

6. The ratio $2Mg:2MgO$ means that $0.500mol$ of magnesium gives $0.500mol$ of magnesium oxide.

7. With $M_r(MgO)=40$, the mass of product is $0.500\times 40=20.0g$.

Deducing the stoichiometry from measured masses

1. Measured masses of the reactants and products can be turned back into the balancing numbers.
2. Divide each substance's mass by its relative formula mass, which gives a set of comparison numbers.
3. Divide every one of those by the smallest, then scale the results to whole numbers.
4. Those whole numbers are the coefficients in the balanced equation.
5. For $4.8g$ of magnesium, $3.2g$ of oxygen and $8.0g$ of magnesium oxide, the comparison numbers are $0.200$, $0.100$ and $0.200$.
6. Dividing by $0.100$ gives $2$, $1$ and $2$, so the equation is $2Mg+O_2\to 2MgO$.
7. The masses check out as well, because $4.8+3.2=8.0$.
8. At Higher tier those comparison numbers are the amounts in moles, and the method is identical.

Checking the finished equation

1. Count each kind of atom on both sides of the equation you have deduced.
2. Confirm the coefficients are whole numbers with no common factor left in them.
3. Confirm the total mass of the reactants matches the total mass of the products.
4. When a calculation follows, start from the limiting reactant rather than from whichever mass came first.
5. Check the final unit, since an amount is in $mol$ and a mass in $g$.