Stoichiometry and Reactions — IMAT Chemistry Formulas and Traps

The Mole, Molar Mass and Avogadro's Number

Relative atomic mass, relative formula mass and molar mass

Relative formula mass

M_r = \sum (\text{number of atoms} \times A_r)

$M_r$ relative formula mass (no unit)
$A_r$ relative atomic mass of each element

Amount of substance: moles from a mass

Moles from mass

n = \frac{m}{M}

$n$ amount of substance, in mol
$m$ mass, in g
$M$ molar mass, in g/mol

Avogadro's number: counting molecules and atoms

Number of particles

N = n \times N_A = \frac{m}{M} \times N_A

$N$ number of particles (molecules or formula units)
$N_A$ Avogadro's number, $6.02 \times 10^{23}\ \text{mol}^{-1}$

Common traps

A bracket multiplies everything inside it

In

\mathrm{Ca(NO_3)_2}

there are 2 nitrogen atoms and 6 oxygen atoms, not 1 and 3. Forgetting the number outside the bracket gives a formula mass that is too small, and that wrong value is usually among the options.

Kilograms must become grams before you divide

Molar masses are in grams per mole. Dividing a mass in kilograms by a molar mass gives an answer 1000 times too small. Convert first: 3.2 kg is 3200 g.

Nitrogen gas is made of two-atom molecules

A mass of

\mathrm{N_2}

divided by 28 gives moles of molecules, and the number of nitrogen atoms is twice that. Dividing by 28 and stopping gives half the right number of atoms, and that answer is usually an option. (Dividing the mass by 14 gives moles of atoms directly.) Decide first whether the question counts atoms or molecules.

Percentage Composition, Empirical and Molecular Formulas

Learn this subtopic in the notes

Percentage by mass of an element in a compound

Percentage by mass

\%\,\text{element} = \frac{x \times A_r}{M_r} \times 100

$x$ number of atoms of the element in the formula
$A_r$ relative atomic mass of the element
$M_r$ relative formula mass of the compound

Empirical formula from masses or percentages

Mole ratio of elements

\text{ratio of atoms} = \frac{m_1}{A_{r,1}} : \frac{m_2}{A_{r,2}} : \dots

$m_1, m_2$ masses of each element in the sample, in g
$A_{r,1}, A_{r,2}$ their relative atomic masses

Molecular formula from the empirical formula and molar mass

Molecular formula multiplier

n = \frac{M_r(\text{compound})}{M_r(\text{empirical formula})}

$n$ whole number that multiplies every subscript

Common traps

Share of atoms is not share of mass

Two of the five atoms in

\mathrm{Fe_2O_3}

are iron, but iron is 70% of the mass, not 40%, because an iron atom is much heavier than an oxygen atom. Always weight each atom by its

A_r

.

Do not round 1.5 or 1.33 to the nearest whole number

A ratio of 1 : 1.5 is 2 : 3, and 1 : 1.33 is 3 : 4. Rounding them to 1 : 1 or 1 : 2 gives a wrong formula, and IMAT lists those wrong formulas as options. Only values within about 0.1 of a whole number may be rounded.

Different compounds can share one empirical formula

Ethyne

\mathrm{C_2H_2}

and benzene

\mathrm{C_6H_6}

both have the empirical formula CH. Analysis by mass cannot tell them apart; only the molar mass can. And sometimes the two formulas are the same, as in methane,

\mathrm{CH_4}

.

Conservation of Mass, Balancing and Ionic Equations

Learn this subtopic in the notes

Conservation of mass and balancing an equation

Conservation of mass

\sum m_{\text{reactants}} = \sum m_{\text{products}}

$m$ mass of each substance taking part, in g

Balancing the combustion of organic compounds

Complete combustion

\mathrm{C_xH_yO_z} + \left(x + \tfrac{y}{4} - \tfrac{z}{2}\right)\mathrm{O_2} \rightarrow x\,\mathrm{CO_2} + \tfrac{y}{2}\,\mathrm{H_2O}

$x, y, z$ numbers of C, H and O atoms in one molecule of the fuel

Common traps

Never balance by changing a subscript

Turning

\mathrm{H_2O}

into

\mathrm{H_2O_2}

to fix the oxygen count makes hydrogen peroxide, a different substance. Options that balance the atoms with an impossible formula (MgOH,

\mathrm{Mg_2O}

) are wrong even when the atoms add up.

Count the oxygen already inside the fuel

Alcohols, sugars and other oxygen-containing fuels bring some oxygen atoms with them, so they need less

\mathrm{O_2}

than the hydrocarbon with the same carbon and hydrogen. Forgetting this gives a coefficient that is too large.

Simplest ionic equation means no spectators

An equation that lists every ion on both sides is balanced, but it is not the simplest ionic equation. Remove every ion that appears unchanged on both sides; what is left must still balance in atoms and in charge.

Types of Reaction, Displacement and Precipitates

Learn this subtopic in the notes

The main types of chemical reaction

Type	Pattern	Example	How to spot it
Synthesis	A + B → AB	$2\mathrm{Mg} + \mathrm{O_2} \rightarrow 2\mathrm{MgO}$	Several reactants, one product
Decomposition	AB → A + B	$\mathrm{CaCO_3} \rightarrow \mathrm{CaO} + \mathrm{CO_2}$ (heated)	One reactant, several products
Single displacement	A + BC → AC + B	$\mathrm{Fe} + \mathrm{CuSO_4} \rightarrow \mathrm{FeSO_4} + \mathrm{Cu}$	An element in, a different element out
Double displacement	AB + CD → AD + CB	$\mathrm{AgNO_3} + \mathrm{NaCl} \rightarrow \mathrm{AgCl} + \mathrm{NaNO_3}$	Two compounds swap ions
Precipitation	ions (aq) → insoluble solid	$\mathrm{Fe^{3+}} + 3\mathrm{OH^-} \rightarrow \mathrm{Fe(OH)_3(s)}$	A solid appears when two solutions mix
Neutralisation	acid + base → salt + water	$\mathrm{HCl} + \mathrm{NaOH} \rightarrow \mathrm{NaCl} + \mathrm{H_2O}$	Net change is $\mathrm{H^+} + \mathrm{OH^-} \rightarrow \mathrm{H_2O}$
Combustion	fuel + oxygen → oxides	$\mathrm{CH_4} + 2\mathrm{O_2} \rightarrow \mathrm{CO_2} + 2\mathrm{H_2O}$	Oxygen is a reactant; heat is given out

Displacement reactions and the reactivity series

Reaction	General result	Example
Group 1 metal or calcium + cold water	hydroxide + hydrogen	$2\mathrm{Na} + 2\mathrm{H_2O} \rightarrow 2\mathrm{NaOH} + \mathrm{H_2}$ Magnesium reacts the same way with cold water, but only very slowly.
Magnesium, zinc or iron + steam	oxide + hydrogen	$\mathrm{Mg} + \mathrm{H_2O(g)} \rightarrow \mathrm{MgO} + \mathrm{H_2}$
Metal above hydrogen + dilute acid	salt + hydrogen	$\mathrm{Zn} + 2\mathrm{HCl} \rightarrow \mathrm{ZnCl_2} + \mathrm{H_2}$
Metal + salt solution of a less reactive metal	the less reactive metal is deposited	$\mathrm{Fe} + \mathrm{CuSO_4} \rightarrow \mathrm{FeSO_4} + \mathrm{Cu}$
Halogen + salt of a less reactive halogen	the less reactive halogen is released	$\mathrm{Cl_2} + 2\mathrm{NaBr} \rightarrow 2\mathrm{NaCl} + \mathrm{Br_2}$

The formula of each product follows the ion charges:

\mathrm{Mg^{2+}}

with

\mathrm{OH^-}

gives

\mathrm{Mg(OH)_2}

, never MgOH.

Solubility rules for predicting a precipitate

Compounds of	Usually	Main exceptions
Sodium, potassium, ammonium	Soluble	None at this level
Nitrates	Soluble	None at this level
Chlorides, bromides, iodides	Soluble	Silver and lead(II) salts are insoluble
Sulfates	Soluble	Barium and lead(II) insoluble; calcium only slightly soluble
Carbonates	Insoluble	Sodium, potassium and ammonium carbonates dissolve
Hydroxides	Insoluble	Group 1 hydroxides dissolve; calcium hydroxide slightly

Common traps

A reaction can have more than one label

Burning magnesium is both synthesis and combustion; neutralisation is also a double displacement; every single displacement is also a redox reaction. An option is not wrong just because another label also fits.

Cold water gives a hydroxide, steam gives an oxide

A reactive metal in cold water makes the metal hydroxide and hydrogen. The oxide forms when a less reactive metal is heated in steam. An option pairing cold water with an oxide is wrong.

The precipitate never contains sodium, potassium or nitrate

Their compounds all dissolve, so they stay in solution as spectator ions. An option showing sodium chloride or potassium nitrate as the solid formed is wrong.

Reacting Masses, Limiting Reagent and Percentage Yield

Learn this subtopic in the notes

Reacting masses through the mole ratio

Mole ratio from the equation

\frac{n_B}{n_A} = \frac{b}{a}

$n_A, n_B$ moles of substances A and B
$a, b$ their coefficients in the balanced equation

Limiting reagent and the reactant in excess

Finding the limiting reagent

\text{limiting} = \text{smallest of } \frac{n_A}{a},\ \frac{n_B}{b}

$n_A, n_B$ moles of each reactant at the start
$a, b$ their coefficients in the balanced equation

Percentage yield

\%\,\text{yield} = \frac{\text{actual yield}}{\text{theoretical yield}} \times 100

actual yieldmass of product obtained, in g
theoretical yieldmaximum mass of product from the limiting reagent, in g

Common traps

Coefficients are a ratio of moles, not of grams

In

2\mathrm{Mg} + \mathrm{O_2} \rightarrow 2\mathrm{MgO}

, 2 g of magnesium does not react with 1 g of oxygen. Convert to moles before using the coefficients, and back to grams afterwards.

The limiting reagent is not simply the smaller mass or the fewer moles

Compare moles divided by coefficients. 0.10 mol of magnesium is limiting against 0.075 mol of oxygen, because the equation needs twice as much magnesium as oxygen. Then calculate the product from the limiting reagent, never from the one in excess.

Compare product with product, not product with reactant

Percentage yield is actual product mass over the theoretical mass of the same product. Dividing the product mass by the mass of reactant used is a common wrong option, because the two substances have different molar masses.

Stoichiometry and Reactions formulas

The Mole, Molar Mass and Avogadro's Number

Relative atomic mass, relative formula mass and molar mass

Amount of substance: moles from a mass

Avogadro's number: counting molecules and atoms

Common traps

Percentage Composition, Empirical and Molecular Formulas

Percentage by mass of an element in a compound

Empirical formula from masses or percentages

Molecular formula from the empirical formula and molar mass

Common traps

Conservation of Mass, Balancing and Ionic Equations

Conservation of mass and balancing an equation

Balancing the combustion of organic compounds

Common traps

Types of Reaction, Displacement and Precipitates

The main types of chemical reaction

Displacement reactions and the reactivity series

Solubility rules for predicting a precipitate

Common traps

Reacting Masses, Limiting Reagent and Percentage Yield

Reacting masses through the mole ratio

Limiting reagent and the reactant in excess

Common traps

More IMAT Chemistry formula sheets