3.1.1 Writing formulae and balanced chemical equations

Formulae are built from chemical symbols

1. Each element has a chemical symbol of one or two letters, such as H for hydrogen and Mg for magnesium.
2. The first letter of a symbol is always a capital letter, and a second letter is always lower case, as in Ca for calcium and Cl for chlorine.
3. The Periodic Table you are given shows the symbol, name, atomic number and relative atomic mass of every element, so you can look up any symbol you are unsure of.
4. Some symbols come from Latin names, so sodium is Na, potassium is K, iron is Fe, copper is Cu, silver is Ag and lead is Pb.
5. A chemical formula combines symbols to show which elements a substance contains.
   1. A small number written after and below a symbol, called a subscript, gives the number of atoms of that element, and no number means one atom.

Formulae of elements and simple covalent compounds

1. Metals, carbon and the noble gases are written as their symbol alone, such as Fe, C and He.
2. Seven elements exist as diatomic molecules, so their formulae are $H_2$, $N_2$, $O_2$, $F_2$, ${Cl}_2$, ${Br}_2$ and $I_2$.
   1. Hydrogen gas is therefore $H_2$, while H on its own stands for a single hydrogen atom.
3. A simple covalent compound is made of small molecules, and its formula counts every atom in one molecule.
   1. Water is $H_{2}O$, ammonia is ${NH}_3$, methane is ${CH}_4$ and hydrogen chloride is $HCl$.
4. In the names of covalent compounds, the prefixes mono, di and tri mean one, two and three atoms of the element they are attached to.
   1. Carbon monoxide is $CO$, carbon dioxide is ${CO}_2$ and nitrogen dioxide is ${NO}_2$.

Formulae and names of simple ionic compounds

1. An ionic compound usually forms when a metal reacts with a non-metal, and the metal is named first.
2. A name ending in -ide usually means the compound contains only two elements, as in sodium chloride, $NaCl$, and magnesium oxide, $MgO$.
   1. Hydroxide is the exception, because it contains oxygen and hydrogen, as in sodium hydroxide, $NaOH$.
3. A name ending in -ate means the compound also contains oxygen, as in calcium carbonate, ${CaCO}_3$, copper sulfate, ${CuSO}_4$, and sodium nitrate, ${NaNO}_3$.
4. The formula gives the simplest ratio of ions, because the ions form a giant lattice rather than separate molecules.
   1. In ${MgCl}_2$ there are two chloride ions for every magnesium ion.
5. The common acids have the formulae $HCl$ for hydrochloric acid, $H_2{SO}_4$ for sulfuric acid and ${HNO}_3$ for nitric acid.

Balanced equations conserve atoms on both sides

1. In a reaction, atoms are rearranged into new substances, but no atoms are made or destroyed.
   1. The total mass of the products therefore equals the total mass of the reactants, which is the conservation of mass.
2. An equation shows the reactants on the left of the arrow and the products on the right, with a plus sign between substances.
3. To balance an equation, write a balancing number in front of a formula, which multiplies every atom in that formula.
   1. In $2H_{2}O$ there are four hydrogen atoms and two oxygen atoms.
4. Never change the subscripts in a formula, because that would change the substance.
   1. Changing $H_{2}O$ to $H_2{O}_2$ would turn water into hydrogen peroxide.

Four steps from names to a balanced equation

1. Start with a word equation, such as magnesium + oxygen $\to$ magnesium oxide.
2. Replace each name with its correct formula, using the Periodic Table for any symbol you need: $$Mg+O_2\to MgO$$
   1. Oxygen is written as $O_2$, because it is one of the diatomic elements.
3. Balance one element at a time, leaving any element that appears in several formulae until last.
   1. Putting $2$ in front of $MgO$ balances oxygen, and then $2$ in front of $Mg$ balances magnesium: $$2Mg+O_2\to 2MgO$$
4. Finish by recounting every atom on both sides, and use the smallest whole-number balancing numbers that work.

3.1.2 Formulae of compounds from ions, and state symbols

Ionic compounds have no overall charge

1. Metal atoms lose electrons to form positive ions, and non-metal atoms gain electrons to form negative ions.
2. The charge on a simple ion can be predicted from the group number of the element.
   1. Group 1 metals form $1+$ ions, Group 2 metals form $2+$ ions and aluminium, in Group 3, forms ${Al}^{3+}$.
   2. Group 6 non-metals form $2-$ ions and Group 7 non-metals form $1-$ ions, such as $O^{2-}$ and ${Cl}^{-}$.
3. A negative ion made from one element takes the ending -ide, so chlorine forms chloride ions and oxygen forms oxide ions.
4. Transition metals can form ions with different charges, shown by a Roman numeral in the name, so iron(II) is ${Fe}^{2+}$ and iron(III) is ${Fe}^{3+}$.
5. In the compound, the total positive charge must equal the total negative charge.

Polyatomic ions stay together with one overall charge

1. Some ions contain several atoms covalently bonded together, and the charge belongs to the whole group.
2. The common negative ones are hydroxide, ${OH}^{-}$, nitrate, ${NO}_3^{-}$, carbonate, ${CO}_3^{2-}$, and sulfate, ${SO}_4^{2-}$.
   1. An ion name ending in -ate shows that the ion contains oxygen.
3. The ammonium ion, ${NH}_4^{+}$, is a common positive polyatomic ion.
4. When a formula needs more than one polyatomic ion, the ion goes inside brackets with the number after them, as in ${Ca(OH)}_2$.
   1. The 2 after the bracket doubles every atom inside it, so ${Ca(OH)}_2$ contains one calcium, two oxygen and two hydrogen atoms.
5. The atoms inside a polyatomic ion never change, so sulfate is always ${SO}_4$ in a formula.

Writing ionic formulae so the charges cancel

1. Write down the formula and charge of each ion.
2. Find the smallest numbers of each ion that give equal and opposite total charges.
   1. For ${Mg}^{2+}$ and ${Cl}^{-}$, one magnesium ion balances two chloride ions, giving ${MgCl}_2$.
3. Write the positive ion first and show the number of each ion as a subscript, leaving out any 1.
4. Leave the charges out of the final formula, because the compound has no overall charge.
5. When the charges are the same size, one of each ion is needed, so ${Mg}^{2+}$ and $O^{2-}$ give $MgO$.
6. Ammonium sulfate needs two ${NH}_4^{+}$ ions for each ${SO}_4^{2-}$ ion, giving $({NH}_4{)}_2{SO}_4$.

State symbols show solid, liquid, gas or aqueous

1. A state symbol is written in brackets straight after each formula in an equation.
2. The four state symbols are $(s)$ for solid, $(l)$ for liquid, $(g)$ for gas and $(aq)$ for aqueous.
   1. Aqueous means dissolved in water, so a solution of sodium chloride is $NaCl(aq)$.
3. Pure water at room temperature is $H_{2}O(l)$, but steam is $H_{2}O(g)$.
4. Acids and alkalis in the laboratory are solutions, so they are written as $HCl(aq)$ or $NaOH(aq)$.
5. A precipitate forms as a solid when two solutions are mixed, so it takes the symbol $(s)$.

Choosing state symbols from the reaction conditions

1. Metals and solid compounds added to a reaction, such as magnesium ribbon or marble chips, are $(s)$.
2. Gases given off as bubbles, such as hydrogen and carbon dioxide, are $(g)$.
3. Magnesium reacting with hydrochloric acid is written as: $$Mg(s)+2HCl(aq)\to {MgCl}_2(aq)+H_2(g)$$
4. Marble chips reacting with hydrochloric acid are written as: $${CaCO}_3(s)+2HCl(aq)\to {CaCl}_2(aq)+H_{2}O(l)+{CO}_2(g)$$
5. Mixing silver nitrate and sodium chloride solutions gives a white precipitate of silver chloride: $${AgNO}_3(aq)+NaCl(aq)\to AgCl(s)+{NaNO}_3(aq)$$
   1. The sodium nitrate stays dissolved, so it is $(aq)$.

3.1.3 Conservation of mass and observed changes in mass

Mass is conserved because atoms are only rearranged

1. During a reaction, bonds between atoms break and new bonds form, so the atoms rearrange into new substances.
2. The same atoms are present at the end as at the start, so the total mass cannot change.
   1. The total mass of the products therefore equals the total mass of the reactants.
3. A balanced equation shows this, because it has the same number of each type of atom on both sides.
   1. In $2Mg+O_2\to 2MgO$, there are two magnesium atoms and two oxygen atoms on each side.
4. The law applies to every reaction, including reactions that give off a gas or form a precipitate.

Conservation of mass lets you calculate a missing mass

1. Add the masses of all the reactants to find the total mass of the products.
2. If one product mass is unknown, subtract the known masses from the total.
   1. Heating $10.0g$ of calcium carbonate gives $5.6g$ of calcium oxide, so the mass of carbon dioxide made is $10.0-5.6=4.4g$.
3. The masses in a reaction are always in the same proportion, so using twice the mass of reactant gives twice the mass of each product.
   1. Heating $20.0g$ of calcium carbonate therefore gives $11.2g$ of calcium oxide and $8.8g$ of carbon dioxide.
4. Check that the product masses add up to the reactant total, since any difference points to an arithmetic slip or a substance that was left out.

Mass falls in an open container as gas escapes

1. Marble chips react with hydrochloric acid in an open flask to make calcium chloride, water and carbon dioxide gas.
2. The balance reading falls as the reaction goes on.
3. Gas particles are far apart and move quickly in random directions, so they spread out of the open flask into the room.
   1. The balance no longer weighs these escaped particles, so the measured mass drops by the mass of carbon dioxide lost.
4. Heating copper carbonate in an open tube also loses mass, because carbon dioxide leaves: $${CuCO}_3\to CuO+{CO}_2$$
5. If the escaped gas were collected and weighed, the total mass of all the products would equal the mass of the reactants.

Mass rises when a product gains gas from air

1. When magnesium ribbon is heated in a crucible, it reacts with oxygen from the air to form magnesium oxide.
2. The magnesium oxide has a greater mass than the magnesium that was heated.
   1. Oxygen particles from the air bond to the magnesium atoms and stay in the solid product, so the balance now weighs them.
3. The mass gained equals the mass of oxygen that reacted.
4. Iron wool heated in air also gains mass, because the iron combines with oxygen to form iron oxide.
5. Before the reaction, the oxygen was part of the air and was not on the balance, so no mass was created.

In an enclosed system, mass stays the same

1. In a sealed container, no particles can leave or enter.
2. Marble chips and acid in a flask closed with a bung give the same balance reading before and after the reaction.
   1. The carbon dioxide particles stay inside the flask, so they are still weighed.
3. A reaction with no gas involved keeps the same mass, even in an open beaker.
   1. Mixing silver nitrate and sodium chloride solutions forms a precipitate, but the total mass of the beaker and its contents does not change.
4. An observed change in mass therefore always means that a substance has crossed the boundary of the container.

3.1.4 Half equations and balanced ionic equations

Half equations show electrons lost or gained

1. Many reactions involve electron transfer, in which one species loses electrons and another gains them.
2. A half equation shows one of these changes on its own, with each electron written as $e^{-}$.
3. When a species loses electrons, the electrons appear on the right-hand side: $$Na\to {Na}^{+}+e^{-}$$
4. When a species gains electrons, the electrons appear on the left-hand side: $${Cu}^{2+}+2e^{-}\to Cu$$
5. A correct half equation balances both the atoms and the total charge on each side.
   1. In the copper half equation, the left side has a total charge of $2+$ plus $2-$, which is zero, matching the uncharged copper atom on the right.

Writing half equations by balancing atoms then charge

1. Write the starting species and the species it becomes, such as ${Cl}^{-}\to {Cl}_2$.
2. Balance the atoms with balancing numbers: $$2{Cl}^{-}\to {Cl}_2$$
3. Work out the total charge on each side, which here is $2-$ on the left and $0$ on the right.
4. Add electrons to the side that is more positive until the charges match: $$2{Cl}^{-}\to {Cl}_2+2e^{-}$$
5. The same method gives ${Al}^{3+}+3e^{-}\to Al$ and $2H^{+}+2e^{-}\to H_2$.
   1. Hydrogen forms diatomic molecules, so two hydrogen ions are needed to make each $H_2$ molecule.

Ionic equations show only the particles that change

1. In solution, an ionic compound is split into separate ions, so $NaCl(aq)$ is present as ${Na}^{+}(aq)$ and ${Cl}^{-}(aq)$.
2. Some of these ions take part in the reaction, while others stay unchanged in the solution.
   1. The unchanged ions appear on both sides of the full equation, so they are left out of the ionic equation.
3. Solids, liquids, gases and covalent molecules such as water are kept as whole formulae, because they are not present as separate ions.
4. Every acid and alkali neutralisation shares one ionic equation: $$H^{+}(aq)+{OH}^{-}(aq)\to H_{2}O(l)$$
   1. The metal ion from the alkali and the negative ion from the acid are both spectator ions.

Building ionic equations by cancelling spectator ions

1. Start from the full balanced equation with state symbols.
2. Split every aqueous ionic compound into its ions, keeping the balancing numbers.
3. Cross out any ion that appears unchanged on both sides.
4. Write what is left, then check that the atoms and charges balance.
5. For magnesium reacting with hydrochloric acid, the chloride ions are spectators, leaving $Mg(s)+2H^{+}(aq)\to {Mg}^{2+}(aq)+H_2(g)$.
   1. The total charge is $2+$ on each side, so this ionic equation is balanced.

Two half equations combine into an ionic equation

1. In an electron-transfer reaction, every electron lost by one species is gained by another.
2. Multiply one or both half equations so that the numbers of electrons match.
3. Add the half equations and cancel the electrons, which now appear equally on both sides.
4. For magnesium displacing copper from copper sulfate solution, $Mg\to {Mg}^{2+}+2e^{-}$ and ${Cu}^{2+}+2e^{-}\to Cu$ already have two electrons each.
   1. Adding them gives $Mg(s)+{Cu}^{2+}(aq)\to {Mg}^{2+}(aq)+Cu(s)$, with the sulfate ions as spectators.
5. For aluminium displacing copper, each aluminium atom loses three electrons and each copper ion gains two, so the aluminium half equation is multiplied by $2$ and the copper half equation by $3$, giving six electrons each.
   1. Adding and cancelling gives $2Al(s)+3{Cu}^{2+}(aq)\to 2{Al}^{3+}(aq)+3Cu(s)$, with a total charge of $6+$ on each side.

3.1.5 The mole and the Avogadro constant

Chemists count particles in moles

1. One drop of water contains more than $10^{21}$ molecules, so chemists count particles in very large, fixed groups.
2. One mole of any substance contains $6.02\times 10^{23}$ particles.
   1. This number is the Avogadro constant, written in standard form because it is so large.
3. The amount of a substance in moles is given the unit mol.
4. Always say which particles you are counting, because one mole of oxygen molecules, $O_2$, contains two moles of oxygen atoms.
5. For an ionic compound, one mole of formula units of $NaCl$ contains one mole of ${Na}^{+}$ ions and one mole of ${Cl}^{-}$ ions.

The Avogadro constant converts moles into particles

1. Multiply the amount in moles by the Avogadro constant to find the number of particles: $$number\; of\; particles=moles\times 6.02\times 10^{23}$$
2. Divide a number of particles by the Avogadro constant to find the amount in moles.
   1. $3.01\times 10^{23}$ water molecules is $\frac{3.01\times 10^{23}}{6.02\times 10^{23}}=0.500mol$.
3. To count the atoms in molecules, multiply by the number of atoms in each molecule.
   1. One mole of $H_{2}O$ contains two moles of hydrogen atoms and one mole of oxygen atoms.

Mass of one mole equals relative formula mass

1. The mass of one mole of an element is its relative atomic mass in grams, so one mole of carbon atoms has a mass of $12.0g$.
2. The mass of one mole of a compound is its relative formula mass in grams, so one mole of water, $M_r=18.0$, has a mass of $18.0g$.
3. This works because relative masses compare particles on the same scale, and one mole always contains the same number of particles.
   1. A magnesium atom is about twice as heavy as a carbon atom, so $24.3g$ of magnesium contains the same number of atoms as $12.0g$ of carbon.
4. The mass of one mole of a substance is called its molar mass, measured in grams per mole.
5. Equal masses of different substances therefore contain different numbers of particles.

Moles equal mass divided by relative formula mass

1. Calculate the amount in moles from a mass using: $$moles=\frac{mass\; (g)}{M_r}$$
2. Rearrange it to find the mass from an amount in moles: $$mass\; (g)=moles\times M_r$$
3. Always work out $M_r$ first, using the relative atomic masses from the Periodic Table.
   1. For calcium carbonate, $M_r({CaCO}_3)=40.1+12.0+(3\times 16.0)=100.1$.
4. So $0.300mol$ of calcium carbonate has a mass of $0.300\times 100.1=30.03g$, which is $30.0g$ to three significant figures.

Linking mass and number of particles through moles

1. To go from a mass to a number of particles, convert the mass into moles first and then multiply by the Avogadro constant.
   1. $4.86g$ of magnesium is $\frac{4.86}{24.3}=0.200mol$, which contains $0.200\times 6.02\times 10^{23}=1.20\times 10^{23}$ atoms.
2. To go from a number of particles to a mass, divide by the Avogadro constant and then multiply by $M_r$.
3. The mass of one atom in grams is its relative atomic mass divided by the Avogadro constant.
   1. One magnesium atom has a mass of $\frac{24.3}{6.02\times 10^{23}}=4.04\times 10^{-23}g$.
4. Because moles count particles, equal amounts in moles of two substances contain the same number of particles, whatever their masses.

3.1.6 Stoichiometry, limiting reactants and reacting masses

Balancing numbers give the reacting ratio in moles

1. In $2Mg+O_2\to 2MgO$, two moles of magnesium react with one mole of oxygen to make two moles of magnesium oxide.
2. This mole ratio, $2:1:2$, is the stoichiometry of the reaction.
3. The reacting masses come from multiplying each amount in moles by its relative formula mass.
   1. Two moles of magnesium have a mass of $2\times 24.3=48.6g$, and they react with $32.0g$ of oxygen to make $80.6g$ of magnesium oxide.
4. The mass ratio, $48.6:32.0:80.6$, is therefore not the same as the mole ratio.
5. The mole ratio holds at any scale, so $0.10mol$ of magnesium reacts with $0.050mol$ of oxygen.

Reacting masses can be turned into balancing numbers

1. Convert the mass of each reactant and product into moles by dividing by its $M_r$.
2. Divide every amount by the smallest amount to get a simple ratio.
   1. If a value is a half, such as $1.5$, multiply every value by $2$ to reach whole numbers.
3. Use the whole-number ratio as the balancing numbers in the equation.
4. Check that the equation also balances atom by atom, which confirms that the formulae and the ratio fit together.

Calculating reacting masses using the mole ratio

1. Write the balanced equation and pick out the ratio between the substance you know and the one you want.
2. Convert the known mass into moles using $moles=\frac{mass}{M_r}$.
3. Use the mole ratio to find the moles of the substance you want.
4. Convert those moles into a mass using $mass=moles\times M_r$.
5. The same chain works in reverse to find the mass of reactant needed for a given mass of product.
   1. To make $9.53g$ of magnesium chloride, $M_r=95.3$, by $Mg+2HCl\to {MgCl}_2+H_2$, you need $\frac{9.53}{95.3}=0.100mol$ of ${MgCl}_2$ and so $0.100mol$ of magnesium, which is $0.100\times 24.3=2.43g$.

The limiting reactant is used up first

1. A reaction stops when one reactant runs out, even if some of the others remain.
2. Any reactant left over at the end was present in excess.
3. The amount of product depends on the limiting reactant, so doubling the limiting reactant doubles the mass of product.
   1. Adding more of a reactant that is already in excess makes no more product.
4. Chemists often use a cheap reactant in excess, so that all of a more expensive reactant is used up.
5. When marble chips are left in the flask after the fizzing stops, the hydrochloric acid was the limiting reactant.

Find the limiting reactant before calculating product mass

1. Convert each reactant mass into moles.
2. Divide each amount by its balancing number in the equation.
   1. The reactant with the smaller result is the limiting reactant.
3. Use the moles of the limiting reactant, never the excess one, to calculate the mass of product.
4. The amount of excess reactant left over is the amount supplied minus the amount that reacts.