Quantities in Chemical Reactions: Test Review

Everything for the Unit 3 test on one page: the key facts from all 11 lessons, the mistakes that cost the most marks, and a practice test that points you back to the lessons you need.

Key facts

Significant Figures & Unit Analysis

Multiplication/division: round the answer to the same number of sig figs as the measurement with the fewest sig figs.
Addition/subtraction: round the answer to the same number of decimal places as the measurement with the fewest decimal places.
Zeros between nonzero digits are always significant; leading zeros never are.
Trailing zeros are significant only when a decimal point is present — use scientific notation to remove ambiguity, e.g. $1.5 \times 1 0^{3}$ has 2 sig figs.
Exact numbers (counted items, defined conversions such as $1 m = 100 cm$ ) have unlimited sig figs and never limit an answer.
Carry guard digits through every intermediate step and round only the final answer.

The Mole Concept

$1 mol = 6.022 \times 1 0^{23}$ particles (atoms, molecules, ions, or formula units) — Avogadro's number, $N_{A}$ .
$n = \frac{N}{N _{A}}$ converts a particle count to moles.
$N = n \times N_{A}$ converts moles to a particle count.
The mole is the SI base unit for amount of substance — conceptually identical to "a dozen," just for an enormously larger group size.
$N_{A} \times 1 u = 1 g$ , so a particle's mass in u equals the mass of one mole of those particles in grams.
Atoms or ions inside a compound: moles of that atom = moles of compound × its subscript (e.g. $1 mol CaCl_{2}$ has $2 mol Cl^{-}$ ).

Molar Mass

Molar mass $M$ (g/mol) numerically equals the average atomic mass (amu) from the periodic table, summed for every atom in the formula.
$M (compound) = \sum (number of atoms \times atomic mass)$
A subscript outside a bracket multiplies every atom inside it: $Mg (OH)_{2}$ has 1 Mg, 2 O, and 2 H.
Molar mass always has units of g/mol.
Periodic-table atomic masses are weighted averages of an element's isotopes, which is why they are rarely whole numbers.
Diatomic elements ( $H_{2}, N_{2}, O_{2}, F_{2}, Cl_{2}, Br_{2}, I_{2})$ have double the atomic molar mass, e.g. $M (O_{2}) = 32.00 g/mol$ .

Mass, Moles & Number of Particles

$n = \frac{m}{M}$ — moles from mass, using molar mass.
$m = n \times M$ — mass from moles.
$N = n \times N_{A}$ — particles from moles.
$n = \frac{N}{N _{A}}$ — moles from particles.
Atoms of one element in a compound: moles of that atom = moles of compound × its subscript.
Mass of one particle $= \frac{M}{N _{A}}$ , e.g. one $H_{2} O$ molecule is $2.992 \times 1 0^{- 23} g$ .

Percentage Composition

$% element = \frac{mass of element in 1 mol of compound}{molar mass of compound} \times 100%$
The percentages of every element in a compound always sum to $100%$ (within rounding).
$mass of element in sample = % element \times mass of sample$
Percent composition is fixed for a given compound — it doesn't depend on sample size.
From lab data: $% element = \frac{measured mass of element}{measured mass of compound} \times 100% .$
Law of definite proportions: a pure compound always has the same composition by mass, so percent composition can identify it.

Empirical & Molecular Formulas

Assume a $100 g$ sample so each percent becomes grams directly.
Convert grams of each element to moles using $n = m / M$ .
Divide every mole value by the smallest mole value to get a ratio close to whole numbers.
A ratio ending in $.5$ needs every subscript doubled; $.33$ or $.67$ needs every subscript tripled.
Molecular formula = empirical formula $\times n$ , where $n = \frac{molar mass given}{empirical formula mass} .$
Combustion analysis: $m (C) = m (CO_{2}) \times \frac{12.01}{44.01}$ , $m (H) = m (H_{2} O) \times \frac{2.02}{18.02}$ ; oxygen is found by difference.

Hydrates

A hydrate's formula is $salt \cdot x H_{2} O$ ; the dot means "combined with," not multiplication.
Mass of water lost on heating = mass of hydrate $-$ mass of anhydrous salt remaining.
$x = \frac{mol H _{2} O lost}{mol anhydrous salt},$ rounded to the nearest whole number.
$% H_{2} O = \frac{mass H _{2} O}{mass hydrate} \times 100%$
$M (hydrate) = M (anhydrous salt) + x (18.02 g/mol)$
Water left behind makes $x$ too low; any other mass lost (spattering, decomposition) makes $x$ too high.

Mole Ratios from Balanced Equations

Coefficients in a balanced equation give the mole ratio between any two substances directly.
$mol wanted = mol given \times \frac{coefficient wanted}{coefficient given}$
The equation must be balanced before its coefficients can be trusted as a mole ratio.
Stoichiometry roadmap: given substance $\to$ mol given $\to$ mol wanted $\to$ wanted substance.
Mass is always conserved in a reaction, but the total number of moles is not (e.g. $2 H_{2} + O_{2} \to 2 H_{2} O$ : 3 mol become 2 mol).
Mole ratios apply equally to numbers of particles, since particles are proportional to moles.

Mass-to-Mass Stoichiometry

Roadmap: mass given $\to$ mol given ( $\div M_{given}$ ) $\to$ mol wanted (mole ratio) $\to$ mass wanted ( $\times M_{wanted}$ ).
$n = m / M$ and $m = n \times M$ are used once on each side of the mole-ratio step.
Masses of two different substances can never be compared or scaled directly — only their moles are related by the equation's coefficients.
The chemical equation must be balanced before any of this works.
Check with conservation of mass: total mass of reactants used = total mass of products formed.
Keep extra digits in intermediate values; round only the final answer to the sig figs of the given data.

Limiting & Excess Reagents

Divide each reactant's moles by its own coefficient; the smaller result identifies the limiting reagent.
All product-yield calculations must be based on the limiting reagent, never the excess reagent.
Excess remaining = starting mol of excess reagent $-$ mol of excess reagent consumed (via the mole ratio with the limiting reagent).
The limiting reagent is determined by the required ratio, not simply by which reactant has less starting mass or fewer moles.
After the reaction, the limiting reagent is at zero; mass is conserved, so mass of reactants used = mass of products formed.
Stoichiometric amounts: reactants mixed in exactly the equation's mole ratio, leaving no excess.

Percentage Yield

$% yield = \frac{actual yield}{theoretical yield} \times 100%$
Theoretical yield is calculated from stoichiometry, always based on the limiting reagent.
Actual yield is measured experimentally and is normally less than theoretical due to side reactions, incomplete reactions, or loss during transfer/purification.
A percent yield over 100% signals a measurement or purity error, not an unusually efficient reaction.
Planning: $theoretical needed = \frac{actual wanted}{% yield /100%};$ predicting: $actual = \frac{% yield}{100%} \times theoretical .$
For a multi-step process, overall yield = product of the step yields (as decimals).

Where marks get lost

Rounding at every intermediate step instead of only the final answer, which compounds small errors into a wrong last digit. (Significant Figures & Unit Analysis)
Applying the multiplication/division rule (sig-fig count) to an addition/subtraction problem, or the reverse. (Significant Figures & Unit Analysis)
Counting trailing zeros in a whole number like $500$ as significant without checking whether a decimal point is present. (Significant Figures & Unit Analysis)
Forgetting that exact counting numbers and defined conversion factors (e.g., $100 cm = 1 m$ ) have infinite sig figs and never limit the final answer. (Significant Figures & Unit Analysis)
Setting up a conversion factor upside-down so the unwanted units don't actually cancel, then not noticing because the final unit label was never checked against the arithmetic. (Significant Figures & Unit Analysis)
Treating "mole" as if it were a unit of mass (grams) instead of a count of particles. (The Mole Concept)
Rounding $N_{A}$ down to $6 \times 1 0^{23}$ too early in a multi-step problem and carrying that error through. (The Mole Concept)
Forgetting to multiply by the number of atoms per molecule when the question asks for atoms rather than molecules, e.g. for O atoms in $O_{2}$ . (The Mole Concept)
Reporting a particle-count answer with the wrong number of sig figs by ignoring how many sig figs were in the given mole amount. (The Mole Concept)
Dividing by $N_{A}$ when converting moles to particles, producing an impossible answer like $4 \times 1 0^{- 24}$ molecules; a particle count for any real sample is an enormous number. (The Mole Concept)
Forgetting that a subscript outside brackets multiplies every atom inside, e.g. treating $Mg (OH)_{2}$ as having only 1 O and 1 H. (Molar Mass)
Reading the atomic number instead of the average atomic mass off the periodic table. (Molar Mass)
Adding atomic masses without multiplying by the subscript first, e.g. treating $H_{2} O$ as $M (H) + M (O)$ . (Molar Mass)
Rounding each atomic mass to a whole number before adding, which shifts the final total off by a noticeable amount. (Molar Mass)
Using $16.00 g/mol$ for oxygen gas (or $35.45$ for chlorine gas) instead of the diatomic molar mass $32.00$ (or $70.90$ ). (Molar Mass)
Dividing by $M$ when the question calls for multiplying (going the wrong direction on the mole map), or the reverse. (Mass, Moles & Number of Particles)
Using the molar mass of a single element inside a compound instead of the molar mass of the whole compound. (Mass, Moles & Number of Particles)
Dropping units at intermediate steps, then reporting a final answer in the wrong quantity (e.g. giving a mass when particles were requested). (Mass, Moles & Number of Particles)
Calling formula units "molecules" for an ionic compound, or vice versa — the vocabulary should match the substance. (Mass, Moles & Number of Particles)
Stopping at the number of molecules when the question asked for atoms of one element, forgetting to multiply by that element's subscript. (Mass, Moles & Number of Particles)
Dividing by the atomic mass of just the element instead of by the molar mass of the entire compound. (Percentage Composition)
Forgetting to multiply the element's atomic mass by its subscript before dividing. (Percentage Composition)
Not checking that the percentages sum close to $100%$ , missing an arithmetic error. (Percentage Composition)
Confusing a percent-composition question with an empirical-formula question and trying to "solve for a formula" when only percentages were asked for. (Percentage Composition)
Multiplying a sample mass by the percentage as a whole number (e.g. $35.00 \times 25.0$ ) instead of as a decimal ( $0.3500 \times 25.0$ ), giving an answer 100 times too big. (Percentage Composition)
Dividing by the largest mole value instead of the smallest, which doesn't produce whole numbers. (Empirical & Molecular Formulas)
Rounding a ratio like $1.5$ to $2$ instead of recognizing it needs the whole formula multiplied by 2. (Empirical & Molecular Formulas)
Stopping at the empirical formula and reporting it as the final answer even when a molar mass was given, skipping the molecular-formula step. (Empirical & Molecular Formulas)
Rounding mole values to 1 or 2 significant figures before dividing, so a true ratio like $1.33$ turns into $1.3$ or $1$ and the formula comes out wrong. (Empirical & Molecular Formulas)
Forgetting to divide the given molar mass by the empirical formula's own mass, and instead just guessing the multiplier $n$ . (Empirical & Molecular Formulas)
Forgetting to subtract the two masses first, and treating the total hydrate mass as if it were the mass of water lost. (Hydrates)
Using the molar mass of the hydrate (salt + water) instead of the molar mass of the anhydrous salt alone when finding moles of salt. (Hydrates)
Getting a value like $x = 4.98$ and reporting it as-is instead of recognizing it should round to the nearest whole number, 5. (Hydrates)
Writing the hydrate formula without the centred dot, or placing the coefficient $x$ incorrectly, e.g. writing $x CuSO_{4} H_{2} O$ . (Hydrates)
Leaving out the water when calculating a hydrate's molar mass (using $159.62$ instead of $249.72 g/mol$ for $CuSO_{4} \cdot 5 H_{2} O$ ), which throws off percent water and any mass-to-mole conversion. (Hydrates)
Using the mole ratio upside-down — putting the given substance's coefficient on top instead of the wanted substance's. (Mole Ratios from Balanced Equations)
Reading coefficients from an equation that isn't actually balanced yet. (Mole Ratios from Balanced Equations)
Misreading a formula subscript as if it were the equation's coefficient (subscripts describe one molecule; coefficients describe how many molecules react). (Mole Ratios from Balanced Equations)
Applying a mole ratio directly to a mass instead of converting to moles first — mole ratios only ever apply to moles. (Mole Ratios from Balanced Equations)
Assuming moles are conserved like mass, e.g. expecting 3 mol of reactants in $2 H_{2} + O_{2} \to 2 H_{2} O$ to give 3 mol of water instead of 2 mol. (Mole Ratios from Balanced Equations)
Trying to scale masses directly by the coefficient ratio, e.g. assuming twice the coefficient means twice the mass (only moles scale that way, not mass, since molar masses differ). (Mass-to-Mass Stoichiometry)
Using the wrong molar mass at the wrong step — e.g. dividing by the wanted substance's molar mass instead of the given substance's. (Mass-to-Mass Stoichiometry)
Stopping at moles of the wanted substance and reporting that number as if it were already the mass answer. (Mass-to-Mass Stoichiometry)
Skipping the balancing step and using coefficients from an unbalanced (or incorrectly balanced) equation. (Mass-to-Mass Stoichiometry)
Rounding intermediate values too early, e.g. rounding $0.2468 mol$ to $0.25 mol$ and getting $10.1 g$ of MgO instead of $9.95 g$ . (Mass-to-Mass Stoichiometry)
Assuming the reactant with the smaller starting mass (or fewer starting moles) is automatically the limiting reagent, without dividing by coefficients first. (Limiting & Excess Reagents)
Comparing raw mole counts directly instead of mol ÷ coefficient for each reactant. (Limiting & Excess Reagents)
Basing the product-yield calculation on the excess reagent instead of the limiting reagent, after correctly identifying which one is limiting. (Limiting & Excess Reagents)
Calculating excess remaining in moles but forgetting to convert it to grams (or vice versa) when the question asks for a specific unit. (Limiting & Excess Reagents)
Finding the excess by subtracting the limiting reagent's moles straight from the excess reagent's moles, without first using the mole ratio to find how much of the excess reagent was actually consumed. (Limiting & Excess Reagents)
Dividing theoretical yield by actual yield instead of actual by theoretical, which gives a nonsensical result. (Percentage Yield)
Forgetting to first calculate the theoretical yield from the limiting reagent, and instead using the full mass of a reactant that wasn't actually limiting. (Percentage Yield)
Reporting percent yield as a decimal (e.g. "0.886") and forgetting to multiply by 100%. (Percentage Yield)
Treating a percent yield above 100% as evidence of an efficient reaction instead of recognizing it as a sign of a measurement or purity error. (Percentage Yield)
When planning, multiplying by the percent yield instead of dividing (e.g. $50.0 \times 0.850 = 42.5 g$ instead of $50.0 \div 0.850 = 58.8 g$ ), so too little reactant is used. (Percentage Yield)

Practice test

22 questions from across the unit. Each one shows the lesson it comes from.

Significant Figures & Unit Analysis
1
How many significant figures does $0.02050$ have?
Significant Figures & Unit Analysis
2
A student measures the mass of 12 identical beads as $3.46 g$ and divides by 12 to get the mass of one bead. How many sig figs should the answer have?
The Mole Concept
3
Which best explains why chemists use the mole instead of counting atoms one at a time?
The Mole Concept
4
Which sample contains the greatest total number of atoms?
Molar Mass
5
Calculate the molar mass of $NaCl$ .
g/mol
Molar Mass
6
A problem asks for the mass of chlorine gas that reacts. Which molar mass should you use for chlorine?
Mass, Moles & Number of Particles
7
How many moles are in $100.0 g$ of $NaCl$ ( $M = 58.44 g/mol$ )?
mol
Mass, Moles & Number of Particles
8
Which $10.0 g$ sample contains the most atoms in total?
Percentage Composition
9
Calculate the percent nitrogen by mass in $NH_{3}$ ( $M = 17.04 g/mol$ ).
%
Percentage Composition
10
Burning $0.486 g$ of magnesium produces $0.806 g$ of magnesium oxide. What is the experimental percent Mg by mass? Give 3 sig figs.
%
Empirical & Molecular Formulas
11
What information, beyond percent composition, is needed to go from an empirical formula to a molecular formula?
Empirical & Molecular Formulas
12
A hydrocarbon has the empirical formula $CH_{3}$ and a molar mass of $30.08 g/mol$ . How many hydrogen atoms are in its molecular formula?
Hydrates
13
A $2.86 g$ sample of $Na_{2} CO_{3} \cdot x H_{2} O$ is heated to constant mass, leaving $1.06 g$ of anhydrous $Na_{2} CO_{3}$ ( $M = 105.99 g/mol$ ). Find $x$ .
Hydrates
14
A $10.0 g$ sample of $BaCl_{2} \cdot 2 H_{2} O$ ( $M = 244.27 g/mol$ ) is heated to constant mass. What mass of anhydrous $BaCl_{2}$ ( $M = 208.23 g/mol$ ) remains? Give 3 sig figs.
g
Mole Ratios from Balanced Equations
15
For $2 C_{2} H_{2} + 5 O_{2} \to 4 CO_{2} + 2 H_{2} O$ , how many moles of $CO_{2}$ form from $1.5 mol C_{2} H_{2}$ ?
mol
Mole Ratios from Balanced Equations
16
Balance $Al + HCl \to AlCl_{3} + H_{2}$ , then find the moles of $H_{2}$ produced when $0.40 mol$ of Al reacts completely.
mol
Mass-to-Mass Stoichiometry
17
For $2 Na + Cl_{2} \to 2 NaCl$ , what mass of NaCl forms from $4.60 g$ of Na ( $M (Na) = 22.99$ , $M (NaCl) = 58.44 g/mol$ )?
g
Mass-to-Mass Stoichiometry
18
Methane burns by $CH_{4} + 2 O_{2} \to CO_{2} + 2 H_{2} O$ . What mass of $O_{2}$ is needed to burn $10.0 g$ of $CH_{4}$ completely? Give 3 sig figs.
g
Limiting & Excess Reagents
19
For $2 H_{2} + O_{2} \to 2 H_{2} O$ , $6.0 g$ of $H_{2}$ ( $M = 2.02$ ) reacts with $40.0 g$ of $O_{2}$ ( $M = 32.00 g/mol$ ). What mass of $H_{2} O$ ( $M = 18.02 g/mol$ ) forms?
g
Limiting & Excess Reagents
20
A furnace burns natural gas with plenty of excess air. Which substance is the limiting reagent?
Percentage Yield
21
For $2 Na + Cl_{2} \to 2 NaCl$ , the theoretical yield of NaCl from $4.60 g$ Na was $11.7 g$ . If $10.2 g$ is actually collected, what is the percent yield?
%
Percentage Yield
22
A medicine is made in two steps with yields of $90%$ and $70%$ . What is the overall percent yield?

Endless drill

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Quantities in Chemical Reactions: Test Review

Key facts

Where marks get lost

Practice test

Endless drill

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