Formulae, Equations & Calculations

## Distinguish relative mass from molar mass - Relative atomic mass is the mean mass of an atom relative to one twelfth of the mass of a carbon-12 atom. Relative molecular mass uses the corresponding mean molecular mass. Both are dimensionless ratios. - A relative isotopic mass describes a specified isotope. Mass number instead counts protons and neutrons and is an integer; it is not an exact measurement of relative isotopic mass. - Molar mass is mass per amount, commonly expressed in g mol⁻¹. Its numerical correspondence with a relative mass in school calculations does not give the relative mass a unit. ## Calculate a weighted mean from isotope abundances - Multiply each supplied relative isotopic mass by its fractional abundance, then add the contributions. For percentage abundances, divide the total percentage-weighted sum by 100. - The more abundant isotope contributes more strongly. The result must lie between the supplied isotope masses and usually nearer the more abundant one. - Use the sample composition actually supplied. A simple unweighted mean is justified only when the abundances are equal. ## Use isotopic masses rather than integer mass numbers - When precise relative isotopic masses are supplied, use them in the weighted mean. Replacing each with its mass number changes the calculation. - For the supplied precise-isotope record, retain intermediate digits and apply the requested final precision. Do not round the isotope masses before weighting. - If a task explicitly uses integer mass numbers as an approximation, state that approximation. It does not redefine relative isotopic mass as an integer. ## Count every atom in a formula - Sum the supplied relative atomic masses for all atoms represented by a formula. A subscript outside brackets multiplies every atom within the brackets. - In (NH₄)₂SO₄, the bracket contains two nitrogen atoms and eight hydrogen atoms in total; the sulfate group contributes one sulfur and four oxygen atoms. - Use relative formula mass for an ionic solid rather than implying it contains discrete molecules. Relative molecular mass uses the actual molecular formula, not automatically the empirical formula. ## Infer an empirical formula from element masses - An empirical formula gives the simplest whole-number ratio of atoms of each element. Element masses must first be converted to amounts before ratios are compared. - Divide each element mass by the corresponding molar mass, then divide all resulting amounts by the smallest. Retain enough precision to identify the ratio. - The supplied oxide is assumed pure and contains only magnesium and oxygen. Such assumptions matter: unexplained impurities cannot simply be assigned to oxygen. ## Use percentage composition as a convenient mass basis - For mass percentages, imagine a 100 g sample so each percentage becomes the corresponding mass in grams. The chosen sample size cancels when amounts are normalised. - Divide each mass by its atomic molar mass and then compare the amounts. Equal percentages do not imply equal atom counts when elements have different atomic masses. - Small departures from integers may arise from rounded experimental data. Check whether one consistent small whole-number ratio fits all elements before writing the formula. ## Use a remainder only when composition is specified - Find an unlisted element by subtraction only when the problem states that the sample contains just the specified elements. A missing percentage alone does not identify an unknown element. - For the supplied C/H/O record, the oxygen percentage is the remainder after carbon and hydrogen. Convert that oxygen mass to amount before normalising. - Show the assumed composition and method. Working helps expose mistakes and can earn method credit where the mark scheme allows; no universal mark rule replaces the actual question. ## Resolve a half-integer amount ratio - Normalisation can produce a ratio such as 1:1.5. Multiply every part by the same factor to obtain the smallest whole-number ratio. - Do not round 1.5 to either 1 or 2. That changes the composition represented by the formula rather than expressing the same ratio. - For an iron oxide, compare the derived formula with the original element masses as a check. The experimental composition is evidence for the formula, not an instruction to assume a familiar oxide. ## Resolve a ratio involving thirds - A normalised ratio close to 1:1.333 can represent 3:4. Multiplying both entries by three preserves the ratio and gives whole numbers. - Keep intermediate digits so 1.333 is not prematurely reduced to 1. A simple fractional ratio should fit the precision of the composition data. - The empirical formula alone does not identify a molecular structure. Several molecular formulae or isomers can be compatible with one empirical ratio. ## Obtain a molecular formula from an empirical formula - A molecular formula gives the numbers of each atom in one molecule. It is a whole-number multiple of the empirical formula. - Calculate the relative mass of the empirical formula, divide the measured relative molecular mass by that value, then multiply every empirical subscript by the resulting integer. - This determines composition, not connectivity. Molecules with the same molecular formula may have different structures and properties. ## Check whether a claimed molecular formula is consistent - The molecular-to-empirical mass ratio should be a positive integer within the precision of the evidence. A strongly non-integer result signals inconsistent data or assumptions. - Do not round an exact ratio such as 2.5 to a convenient integer and silently change the relative molecular mass. Recheck composition, mass data and sample identity. - An experimental result with uncertainty may justify a nearby integer, but the uncertainty must support that inference. The supplied consistency record asks for exact values. ## Construct an electrically neutral ionic formula - Choose numbers of cations and anions so their total charges cancel. An ionic formula describes the constituent ion ratio rather than a discrete molecule. - Keep a polyatomic ion intact and use brackets when more than one is required. The overall subscript and each internal subscript have different jobs. - Simple-ion periodic trends are useful within their limits. Use a stated ion charge or oxidation state where needed rather than assigning every metal one fixed charge. ## Preserve polyatomic ions in formulae - Ionic compounds can contain polyatomic cations such as ammonium, so a metal is not required in every ionic compound. - For ammonium carbonate, neutrality determines how many intact ammonium and carbonate ions are present. Brackets prevent confusion about which atoms are multiplied. - Do not reduce subscripts in a way that destroys an ion’s identity. For example, peroxide contains O₂²⁻: the valid formula Na₂O₂ is not replaced by NaO. ## Balance coefficients without changing substances - A balanced chemical equation conserves the number of atoms of each element. Adjust coefficients in front of formulae while keeping each substance’s formula fixed. - An atom-count table can check each element separately. Use the smallest whole-number coefficients when requested, although a scaled coefficient set represents the same ratio. - Changing a subscript changes the substance. Changing an equation coefficient changes the amount represented without changing molecular or ionic identity. ## Balance a complete combustion equation - When fuel and complete-combustion products are specified, balance carbon and hydrogen first, then oxygen. A fractional oxygen coefficient can be cleared by scaling all coefficients. - This is a useful method rather than a chemical law about balancing order. The final atom totals determine whether the result is correct. - State symbols depend on the measurement conditions. Water may be vapour during combustion and liquid after cooling; phase does not change the atom balance. ## Interpret state symbols from the stated conditions - Use (s) for solid, (l) for liquid, (g) for gas and (aq) for a species dissolved in water. Aqueous does not mean every liquid substance. - An equation should represent the physical states relevant to its conditions. A soluble salt can be solid before dissolving and aqueous in its solution. - State information helps decide which species to write as ions in an aqueous ionic equation. It is not decorative notation. ## Remove spectators from an ionic equation - For an aqueous reaction, first identify the species actually present. Split the soluble strong electrolytes used in the model into ions, then cancel unchanged spectator ions. - Keep the precipitate as a solid formula. For the supplied mixture, the reacting ions form insoluble barium sulfate; sodium and chloride remain dissolved. - Check both atom balance and total charge in the net ionic equation. A spectator is unchanged across the reaction, not simply an ion that has a small coefficient. ## Keep weak acids in their appropriate molecular form - Do not split every aqueous substance into separate ions. Water, weak electrolytes and insoluble solids require their appropriate chemical forms. - Ethanoic acid is largely undissociated in the supplied model. Its reaction with hydroxide is CH₃COOH +OH⁻ →CH₃COO⁻ +H₂O, with sodium omitted as a spectator. - The dominant-species ionic equation still conserves atoms and charge. Use the stated reaction chemistry rather than mechanically cancelling formula fragments. ## Find the limiting reactant before calculating a maximum - Compare each reactant amount divided by its balanced-equation coefficient. The smallest value limits the possible reaction extent in a complete-conversion model. - Use that extent to calculate product and unreacted excess. Comparing raw masses or raw mole values alone can choose the wrong limiting species. - The theoretical maximum does not guarantee the isolated yield. Equilibrium or incomplete reaction can leave some limiting reactant unconverted in a real experiment. ## Compare actual and theoretical yield for the same product - Percentage yield is 100 times the actual product amount divided by its theoretical amount. Compare the same product in compatible units, using mass with mass or amount with amount. - The theoretical quantity comes from the balanced equation and the limiting available reactant. The actual quantity refers to the isolated product under the stated purity assumptions. - Incomplete conversion, competing reactions and losses during isolation can reduce yield. A lower measured yield alone does not identify which cause occurred. ## Calculate theoretical mass before percentage yield - A yield task may first require conversion of reactant mass to amount, application of the mole ratio and conversion of theoretical product amount to mass. - For the carbonate model, CaCO₃ and CaO have a 1:1 amount ratio but different molar masses. Comparing their masses directly as a yield would mix different substances. - Carry intermediate digits through the theoretical calculation. Only then compare the actual mass of the desired product with that theoretical mass. ## Interpret an apparent yield above 100 percent - A calculated mass recovery above 100% is a possible experimental observation, but it cannot represent a pure-product yield above the valid theoretical maximum. - Retained water, contamination, an incorrect reactant purity or the wrong reaction model can inflate the calculation. Identify which assumptions need checking. - Do not call a reaction more than perfectly efficient merely because the recovered solid is heavy. A mass measurement is not by itself a product-purity measurement. ## Calculate atom economy from a balanced equation - Percentage atom economy is the coefficient-weighted mass of the desired product divided by the coefficient-weighted total mass of the reactants, multiplied by 100. - It describes the ideal stoichiometric mass fraction reaching the desired product. It is not the percentage of the number of atoms, nor a measured experimental yield. - Use all net-consumed reactants in the balanced equation. Actual excess quantities, solvents and unchanged catalysts are considered in wider process metrics rather than inserted arbitrarily into this ratio. ## Include coefficients in both atom-economy totals - A formula mass alone is not enough when its stoichiometric coefficient exceeds one. Multiply each reactant and desired-product contribution by the coefficient in the same balanced equation. - The iron model forms two moles of Fe per equation event and consumes three moles of CO. Omitting either coefficient gives the wrong mass fraction. - Scaling every equation coefficient by the same factor leaves atom economy unchanged because the factor cancels between numerator and denominator. ## Compare routes that make the same desired product - A balanced ideal reaction with a single desired product has 100% atom economy because all net reactant atoms reach that product. This does not guarantee complete conversion or recovery. - A route that also forms a by-product allocates part of the reactant mass to that other substance. Specify the desired product before comparing atom economies. - For the two ethanol equations, include water in hydration and sodium hydroxide in substitution. Excluding a consumed reagent would inflate the apparent atom economy. ## Evaluate a process beyond atom economy - A high atom economy can reduce stoichiometric by-product waste, conserving raw materials and reducing separation or disposal demands. - An overall route judgement also needs yield, energy, solvent use, hazard, feedstock origin, cost and treatment of waste. One high percentage cannot settle every economic or environmental question. - The supplied process record contains trade-offs. Identify what each measurement supports and what further evidence is missing before claiming one route is best overall. ## Interpret water of crystallisation in a hydrate formula - Water of crystallisation is incorporated into a crystalline compound’s structure in a specified stoichiometric ratio. An anhydrous form contains no water of crystallisation. - The dot in MgSO₄·7H₂O associates seven water molecules with each salt formula unit. It does not multiply every atom of the salt by seven. - The copper sulfate comparison illustrates two solid forms, not a universal heating method. Suitable conditions can remove water, while excessive heating may decompose a salt. ## Calculate hydration using vessel-corrected masses - Subtract the empty vessel from both combined mass readings. The initial sample mass and final residue mass must refer to the same vessel and sample. - If only water is lost and the residue is pure anhydrous salt, the mass difference is water mass. Divide residue and water masses by their respective molar masses. - The water-to-salt amount ratio gives x in the hydrate formula. A mass ratio is not the same as this mole ratio, and the assumptions must match the actual chemical change. ## Use constant mass as evidence within a valid model - Repeated cooling and weighing can show that the measured mass no longer changes within the balance resolution. Compare like-for-like readings after suitable controlled heating cycles. - Constant mass alone does not prove complete dehydration. A stable partially hydrated or decomposed residue can also give repeated readings, so the chemical conditions matter. - Cool before weighing and limit rehydration during handling. Interpreting measurements requires both the observed mass pattern and a justified residue identity. ## Predict how deviations affect a calculated hydration number - If some water remains, the calculation overestimates anhydrous salt mass and underestimates water loss. Under that model the calculated water-to-salt ratio is too low. - If solid salt spatters out, its mass loss can be mistaken for water and the residue salt amount becomes too small. Both effects push the calculated hydration ratio upward. - These predictions depend on the stated isolated error. Several simultaneous errors can interact, so an unexplained result should prompt checks rather than an automatic correction.