Answer key · Unit 3

The mole.

Each lesson’s practice questions, their correct answers and the mistakes they spot. Answers read off the Scene use the same setting the learner sees.

Misconception catalogue

Answer key 4

Lesson 1 of 4 · About 20 minutes

Relative masses

An atom weighs about 10⁻²³ grams, so chemists compare masses instead of weighing atoms: relative atomic mass, and a compound’s relative formula mass, added up from its formula.

Open the lesson

The 5 steps and their answers

Step 1 of 5
Open this step
Too light to weigh
Practice · Calculate
Question
How many times heavier is an iron atom than a carbon atom? Use the table of relative atomic masses.

Correct answer
4.65 ± 0.05
Setup, choices and feedback
The Scene at this step
Substance
Carbon (C)
Mass
12 g
Sample / Particles
Particles
Accepted within
± 0.05
A fixed value in the question
Wrong answers it recognises
- 0.215 ± 0.01
  Misconception: Turns a ratio or a division upside down, such as M ÷ m for the moles.
  Feedback from the catalogue
  Check with the units: g ÷ g/mol = mol, so $n = m / M$ ⁠. If you divided the other way, $M / m$ ⁠, the unit is 1/mol, which is not an amount. And in “how many times” questions, divide the thing asked about by the thing it is compared with.
Feedback for a right answer
$\dfrac{55.85}{12.01} \approx 4.65$ ⁠. So $12$ ⁠ grams of carbon and $55.85$ ⁠ grams of iron hold the same number of atoms: each iron atom is $4.65$ ⁠ times heavier, and exactly $4.65$ ⁠ times the grams makes up for it.

Hint
Divide iron’s $A_r$ ⁠ by carbon’s.

Worked solution
1. $A_r(\mathrm{Fe}) = 55.85$ ⁠, $A_r(\mathrm{C}) = 12.01$ ⁠.
2. The ratio: $\dfrac{55.85}{12.01} \approx 4.65$ ⁠.
3. An iron atom is about $4.65$ ⁠ times heavier than a carbon atom.
If you got $0.215$ ⁠, you divided the wrong way round: that is the fraction of an iron atom’s mass that a carbon atom weighs.
Step 2 of 5
Open this step
Formula mass
Practice · Calculate
Question
What is the relative formula mass of water, H₂O?

Correct answer
18.02 ± 0.1
Setup, choices and feedback
The Scene at this step
Substance
Water (H₂O)fixed
Mass
18 g
Sample / Particles
Particles
Other settings
Readings/nodes/bench/props/readings = "mass"
Accepted within
± 0.1
Computed from the Scene at this step’s setting
Relative formula mass Mr (relative atomic mass for an element)/nodes/bench/outputs/formulaMass
Wrong answers it recognises
- 17.01 ± 0.1
  Misconception: Counts each element in a formula once, leaving out the subscripts.
  Feedback from the catalogue
  The small number below a symbol says how many atoms of that element are in the formula unit, and it belongs only to the symbol before it. H₂O has two hydrogens, and CaCO₃ three oxygens. When working out a formula mass, multiply each $A_r$ ⁠ by its number of atoms before adding.
Feedback for a right answer
$M_r = 2 \times 1.008 + 16.00 = 18.02$ ⁠. Almost all of a water molecule’s mass is the oxygen.

Hint
Two hydrogens and one oxygen.

Worked solution
1. H₂O has $2$ ⁠ hydrogen atoms and $1$ ⁠ oxygen atom.
2. Hydrogen: $2 \times 1.008 = 2.016$ ⁠. Oxygen: $16.00$ ⁠.
3. $M_r = 2.016 + 16.00 \approx 18.02$ ⁠.
If you got $17.01$ ⁠, you counted one hydrogen: the small $2$ ⁠ says there are two.
Step 3 of 5
Open this step
How many atoms of each kind
Practice · Calculate
Question
What is the relative formula mass of calcium carbonate, CaCO₃?

Correct answer
100.1 ± 0.2
Setup, choices and feedback
The Scene at this step
Substance
Chalk (CaCO₃)fixed
Mass
50 g
Sample / Particles
Sample
Other settings
Readings/nodes/bench/props/readings = "mass"
Accepted within
± 0.2
Computed from the Scene at this step’s setting
Relative formula mass Mr (relative atomic mass for an element)/nodes/bench/outputs/formulaMass
Wrong answers it recognises
- 68.09 ± 0.2
  Misconception: Counts each element in a formula once, leaving out the subscripts.
  Feedback from the catalogue
  The small number below a symbol says how many atoms of that element are in the formula unit, and it belongs only to the symbol before it. H₂O has two hydrogens, and CaCO₃ three oxygens. When working out a formula mass, multiply each $A_r$ ⁠ by its number of atoms before adding.
Feedback for a right answer
$M_r = 40.08 + 12.01 + 3 \times 16.00 = 100.09$ ⁠. Almost half of chalk’s mass is oxygen.

Hint
One calcium, one carbon, three oxygens.

Worked solution
1. CaCO₃: $1$ ⁠ calcium, $1$ ⁠ carbon, $3$ ⁠ oxygen.
2. $40.08 + 12.01 + 3 \times 16.00$ ⁠.
3. $= 40.08 + 12.01 + 48.00 = 100.09$ ⁠.
If you got $68.09$ ⁠, you counted one oxygen. The $3$ ⁠ belongs to the oxygen, and says there are three.
Step 4 of 5
Open this step
Counting atoms
Practice · Calculate
Question
How many atoms are in one molecule of glucose, C₆H₁₂O₆?

Correct answer
24 ± 0.01
Setup, choices and feedback
The Scene at this step
Substance
Glucose (C₆H₁₂O₆)fixed
Mass
50 g
Sample / Particles
Particles
Other settings
Readings/nodes/bench/props/readings = "mass"
Accepted within
± 0.01
Computed from the Scene at this step’s setting
Atoms in one formula unit/nodes/bench/outputs/atomsPerParticle
Wrong answers it recognises
- 3 ± 0.01
  Misconception: Counts each element in a formula once, leaving out the subscripts.
  Feedback from the catalogue
  The small number below a symbol says how many atoms of that element are in the formula unit, and it belongs only to the symbol before it. H₂O has two hydrogens, and CaCO₃ three oxygens. When working out a formula mass, multiply each $A_r$ ⁠ by its number of atoms before adding.
Feedback for a right answer
$6 + 12 + 6 = 24$ ⁠ atoms, of three elements. In the Scene you can see the ring and the OH groups around it.

Hint
Add up the small numbers.

Worked solution
1. Carbon $6$ ⁠, hydrogen $12$ ⁠, oxygen $6$ ⁠.
2. $6 + 12 + 6 = 24$ ⁠.
If you answered $3$ ⁠, you counted the kinds of atom, the elements, not the atoms.
Step 5 of 5
Open this step
Find it
Practice · Reach the target
Question
Choose in Substance the substance whose relative formula mass is about $100$ ⁠, then press Check.

Goal
- 100.1 ± 0.5
  At the start of the step: 18.02
Setup, choices and feedback
The Scene at this step
Substance
Water (H₂O)
Mass
50 g
Sample / Particles
Sample
Other settings
Readings/nodes/bench/props/readings = "mass"
Read from the Scene at
Relative formula mass Mr (relative atomic mass for an element)/nodes/bench/outputs/formulaMass
One setting that reaches it
Substance
Chalk (CaCO₃)
Mass
0 g
Feedback for a right answer
Chalk, CaCO₃: $100.09$ ⁠. Ethanol is $46.07$ ⁠, salt $58.44$ ⁠ and glucose $180.16$ ⁠. Of the elements, copper and iron are nearer $60$ ⁠, and gold is far above.

Feedback for a wrong answer
Not yet. Try the compounds, and work out each one’s mass from its formula.

Hint
Work out $M_r$ ⁠ for the compounds in the list.

Worked solution
1. Water $18.02$ ⁠, ethanol $46.07$ ⁠, salt $58.44$ ⁠, glucose $180.16$ ⁠.
2. Chalk: $40.08 + 12.01 + 3 \times 16.00 = 100.09$ ⁠.
3. Choose chalk.

Lesson 2 of 4 · About 25 minutes

Counting by weighing

The mole: a fixed number of particles, 6.02 × 10²³, chosen so that a mole of a substance weighs its formula mass in grams. How to weigh out a mole, and why a mole of gold is heavier than a mole of aluminium.

Open the lesson

The 5 steps and their answers

Step 1 of 5
Open this step
Counting by weighing
Practice · Calculate
Question
The balance holds 12 g of carbon. How many atoms are in it? Type it in E notation: for example 3e22 for $3 \times 10^{22}$ ⁠.

Correct answer
6.017 × 10²³ ± 1%
Setup, choices and feedback
The Scene at this step
Substance
Carbon (C)
Mass
12 g
Sample / Particles
Sample
Other settings
Readings/nodes/bench/props/readings = "molar-mass"
Accepted within
± 1%
Computed from the Scene at this step’s setting
Atoms, molecules or formula units: N = n · Nₐ/nodes/bench/outputs/particles
Wrong answers it recognises
- 6.022 × 10⁻²³ ± 1%
  Misconception: Writes the Avogadro constant with a negative power of ten, 6.02 × 10⁻²³.
  Feedback from the catalogue
  The Avogadro constant is a huge number: $6.022 \times 10^{23}$ ⁠, six hundred thousand billion billion. $10^{-23}$ ⁠ is a tiny number, close to zero. What is tiny is the mass of one atom, about $10^{-23}$ ⁠ grams; the number of atoms in a few grams is huge.
- 7.23 × 10²⁴ ± 1%
  Misconception: Multiplies the grams by the Avogadro constant, as if the grams were moles.
  Feedback from the catalogue
  The Avogadro constant counts particles per mole, not per gram. First turn grams into moles, $n = m / M$ ⁠, and only then multiply: $N = n \times N_A$ ⁠. Multiplying the grams directly gives a number $M$ ⁠ times too big.
Feedback for a right answer
About $6.02 \times 10^{23}$ ⁠: one mole. Counting one atom a second, it would take about a million times the age of the universe.

Hint
$12$ ⁠ grams of carbon is about one mole.

Worked solution
1. Carbon’s molar mass: 12.01 g/mol.
2. $n = \dfrac{12}{12.01} \approx 1.00$ ⁠ mol.
3. $N = n \times N_A \approx 1.00 \times 6.022 \times 10^{23} = 6.02 \times 10^{23}$ ⁠.
If you typed 6.02e-23, the power has the wrong sign: $10^{-23}$ ⁠ is a tiny number, not a huge one.
Step 2 of 5
Open this step
The same number, a different mass
Practice · Predict, then look
Question
The balance holds one mole of aluminium, 27 g. Swap it for one mole of gold, exactly the same number of atoms. What will the balance read?

Correct answer
- Correct Much more than 27 g
Setup, choices and feedback
The Scene at this step
Substance
Aluminium (Al)
Mass
27 gfixed
Sample / Particles
Sample
Choices
- Correct Much more than 27 g
- Wrong 27 g again: a mole is a mole
  Misconception: Thinks a mole of any substance has the same mass.
  Feedback from the catalogue
  A mole is a fixed number of particles, $6.022 \times 10^{23}$ ⁠, just as a dozen is twelve. A mole of a substance weighs its formula mass in grams, so moles of different substances weigh differently: a mole of aluminium 27 g, a mole of gold 197 g. Just as a dozen eggs and a dozen watermelons do not weigh the same.
- Wrong Less than 27 g
  Feedback
  A gold atom, $A_r = 197$ ⁠, is much heavier than an aluminium atom, $A_r = 27$ ⁠. The same number of heavier atoms weighs more.
Once the prediction is locked in, the Scene changes
Substance
Gold (Au)
Mass
197 gfixed
The Scene then shows: 1 mol · Amount of substance n = m / M/nodes/bench/outputs/moles
A prediction is never marked as a failure: the learner sees the outcome and the explanation either way.
Feedback for a right answer
197 g, about seven times as much. A mole is a number, not a mass: a mole of any substance weighs its formula mass in grams, and gold is $7.3$ ⁠ times heavier than aluminium.

Worked solution
1. A mole of any substance weighs its $M_r$ ⁠ in grams.
2. Aluminium: $A_r = 26.98$ ⁠, so a mole weighs 27 g.
3. Gold: $A_r = 196.97$ ⁠, so a mole weighs 197 g.
If you predicted 27 g again, you thought of a mole as a fixed amount of mass. A mole is a fixed number of particles.
Step 3 of 5
Open this step
Molar mass
Practice · Calculate
Question
What is the molar mass of ethanol, C₂H₅OH?

Correct answer
46.07 g/mol ± 0.1 g/mol
Setup, choices and feedback
The Scene at this step
Substance
Ethanol (C₂H₅OH)fixed
Mass
50 g
Sample / Particles
Sample
Other settings
Readings/nodes/bench/props/readings = "mass"
Accepted within
± 0.1 g/mol
Computed from the Scene at this step’s setting
Mass of one mole: Mr grams/nodes/bench/outputs/molarMass
Any unit of the same quantity is accepted; it is converted before checking.
Wrong answers it recognises
- 45.06 g/mol ± 0.1 g/mol
  Misconception: Counts each element in a formula once, leaving out the subscripts.
  Feedback
  $45.06$ ⁠ counts five hydrogens. There is one more in OH: six in all.
- 29.02 g/mol ± 0.1 g/mol
  Misconception: Counts each element in a formula once, leaving out the subscripts.
  Feedback from the catalogue
  The small number below a symbol says how many atoms of that element are in the formula unit, and it belongs only to the symbol before it. H₂O has two hydrogens, and CaCO₃ three oxygens. When working out a formula mass, multiply each $A_r$ ⁠ by its number of atoms before adding.
Feedback for a right answer
46.07 g/mol: $2 \times 12.01 + 6 \times 1.008 + 16.00$ ⁠. A mole of ethanol, about 58 mL, weighs 46 g.

Hint
Two carbons, six hydrogens, one oxygen.

Worked solution
1. C₂H₅OH: carbon $2$ ⁠, hydrogen $5 + 1 = 6$ ⁠, oxygen $1$ ⁠.
2. $2 \times 12.01 = 24.02$ ⁠; $6 \times 1.008 = 6.048$ ⁠; $16.00$ ⁠.
3. $M = 24.02 + 6.048 + 16.00 \approx 46.07$ ⁠, in g/mol.
Step 4 of 5
Open this step
Weighing out a mole
Practice · Reach the target
Question
Change Mass until the balance holds exactly one mole of iron, then press Check.

Goal
- 1 mol ± 0.005 mol
  At the start of the step: 0.1791 mol
Setup, choices and feedback
The Scene at this step
Substance
Iron (Fe)fixed
Mass
10 g
Sample / Particles
Sample
Other settings
Readings/nodes/bench/props/readings = "mass"
Read from the Scene at
Amount of substance n = m / M/nodes/bench/outputs/moles
One setting that reaches it
Mass
55.6 g
Feedback for a right answer
55.8 g: iron’s $A_r$ ⁠ is $55.85$ ⁠, so a mole weighs 55.85 g. That small heap holds $6.02 \times 10^{23}$ ⁠ iron atoms.

Feedback for a wrong answer
Not yet. A mole of iron weighs 55.85 g.

Hint
A mole weighs its $A_r$ ⁠ in grams.

Worked solution
1. $A_r(\mathrm{Fe}) = 55.85$ ⁠, so $M = 55.85$ ⁠ g/mol.
2. $m = n \times M = 1 \times 55.85 = 55.85$ ⁠ grams.
3. Set Mass to 55.8 g or 55.9 g.
Step 5 of 5
Open this step
The same mass, a different number
Practice · Choose
Question
Which holds more atoms: 12 g of carbon or 12 g of iron?

Correct answer
- Correct The carbon
Setup, choices and feedback
The Scene at this step
Substance
Carbon (C)
Mass
12 g
Sample / Particles
Particles
Choices
Shown in a random order.
- Correct The carbon
- Wrong The iron
  Feedback
  An iron atom is $4.65$ ⁠ times heavier than a carbon atom, so the same mass holds fewer iron atoms.
- Wrong Equal masses hold equal numbers of atoms
  Misconception: Thinks equal masses of different substances hold the same number of particles.
  Feedback from the catalogue
  Equal masses hold equal numbers of particles only when the particles weigh the same. An iron atom is $4.65$ ⁠ times heavier than a carbon atom, so 12 g of iron holds $4.65$ ⁠ times fewer atoms than 12 g of carbon. To compare numbers, compare moles: $n = m / M$ ⁠.
Feedback for a right answer
The carbon: 12 g of carbon is a whole mole, $6.0 \times 10^{23}$ ⁠ atoms, and 12 g of iron only $0.21$ ⁠ mol, $1.3 \times 10^{23}$ ⁠ atoms. Change Substance to see.

Hint
Compare how many moles each one is.

Worked solution
1. Carbon: $n = \dfrac{12}{12.01} \approx 1.00$ ⁠ mol.
2. Iron: $n = \dfrac{12}{55.85} \approx 0.215$ ⁠ mol.
3. More moles means more atoms: the carbon has $4.65$ ⁠ times as many.

Lesson 3 of 4 · About 25 minutes

Grams, moles and particles

Two formulas link what you weigh to what you count: n = m ÷ M and N = n × Nₐ. Working both ways, molecules against atoms, and checking with units.

Open the lesson

The 5 steps and their answers

Step 1 of 5
Open this step
From grams to moles
Practice · Calculate
Question
How many moles of water are in 9 g?

Correct answer
0.4996 mol ± 1%
Setup, choices and feedback
The Scene at this step
Substance
Water (H₂O)fixed
Mass
9 gfixed
Sample / Particles
Sample
Other settings
Readings/nodes/bench/props/readings = "molar-mass"
Accepted within
± 1%
Computed from the Scene at this step’s setting
Amount of substance n = m / M/nodes/bench/outputs/moles
Any unit of the same quantity is accepted; it is converted before checking.
Wrong answers it recognises
- 162.1 mol ± 1%
  Misconception: Multiplies the mass by the molar mass to get moles.
  Feedback from the catalogue
  A mole weighs $M$ ⁠ grams, so to find how many moles are in $m$ ⁠ grams you divide: $n = \dfrac{m}{M}$ ⁠. 9 g of water is $9 / 18 = 0.5$ ⁠ mol, not $9 \times 18 = 162$ ⁠. The units show the slip: g × g/mol is not a mole.
- 2 mol ± 1%
  Misconception: Turns a ratio or a division upside down, such as M ÷ m for the moles.
  Feedback from the catalogue
  Check with the units: g ÷ g/mol = mol, so $n = m / M$ ⁠. If you divided the other way, $M / m$ ⁠, the unit is 1/mol, which is not an amount. And in “how many times” questions, divide the thing asked about by the thing it is compared with.
Feedback for a right answer
$n = \dfrac{9}{18.02} \approx 0.50$ ⁠ mol: half a mole, since 9 g is half the mass of a mole of water.

Hint
$n = \dfrac{m}{M}$ ⁠.

Worked solution
1. The givens: $m = 9$ ⁠ grams, $M = 18.02$ ⁠ g/mol.
2. $n = \dfrac{m}{M}$ ⁠.
3. $n = \dfrac{9}{18.02} \approx 0.50$ ⁠ mol.
If you got $162$ ⁠, you multiplied: g × g/mol is not a mole. If you got $2$ ⁠, you divided the wrong way round.
Step 2 of 5
Open this step
From moles to grams
Practice · Reach the target
Question
Change Mass until the balance holds a quarter of a mole, 0.25 mol, of table salt, then press Check.

Goal
- 0.25 mol ± 1%
  At the start of the step: 0.8556 mol
Setup, choices and feedback
The Scene at this step
Substance
Salt (NaCl)fixed
Mass
50 g
Sample / Particles
Sample
Other settings
Readings/nodes/bench/props/readings = "molar-mass"
Read from the Scene at
Amount of substance n = m / M/nodes/bench/outputs/moles
One setting that reaches it
Mass
14.5 g
Feedback for a right answer
14.6 g: $0.25 \times 58.44 = 14.61$ ⁠. A heaped spoonful of salt.

Feedback for a wrong answer
Not yet. Multiply the amount by the molar mass, 58.44 g/mol.

Hint
$m = n \times M$ ⁠, and $M$ ⁠ is in the readings.

Worked solution
1. $M(\mathrm{NaCl}) = 22.99 + 35.45 = 58.44$ ⁠ g/mol.
2. $m = n \times M = 0.25 \times 58.44 \approx 14.6$ ⁠ grams.
3. Set Mass to 14.6 g.
Step 3 of 5
Open this step
From moles to particles
Practice · Calculate
Question
How many water molecules are in 4.5 g of water? Type it in E notation: for example 3e22 for $3 \times 10^{22}$ ⁠.

Correct answer
1.504 × 10²³ ± 1%
Setup, choices and feedback
The Scene at this step
Substance
Water (H₂O)fixed
Mass
4.5 gfixed
Sample / Particles
Particles
Other settings
Readings/nodes/bench/props/readings = "molar-mass"
Accepted within
± 1%
Computed from the Scene at this step’s setting
Atoms, molecules or formula units: N = n · Nₐ/nodes/bench/outputs/particles
Wrong answers it recognises
- 4.513 × 10²³ ± 1%
  Misconception: Mixes up the number of molecules or formula units with the number of atoms in them.
  Feedback from the catalogue
  Mind what you are counting. A mole of water is a mole of molecules, but three moles of atoms, since each molecule has three. A mole of CaCO₃ holds three moles of oxygen atoms. Multiply the number of units by the number of atoms of the kind asked about in each one.
- 2.71 × 10²⁴ ± 1%
  Misconception: Multiplies the grams by the Avogadro constant, as if the grams were moles.
  Feedback from the catalogue
  The Avogadro constant counts particles per mole, not per gram. First turn grams into moles, $n = m / M$ ⁠, and only then multiply: $N = n \times N_A$ ⁠. Multiplying the grams directly gives a number $M$ ⁠ times too big.
Feedback for a right answer
$1.50 \times 10^{23}$ ⁠ molecules: $0.25$ ⁠ mol, a quarter of the Avogadro constant. The same drop holds three times as many atoms: $4.5 \times 10^{23}$ ⁠.

Hint
Moles first, $n = m / M$ ⁠, then $N = n \times N_A$ ⁠.

Worked solution
1. $n = \dfrac{4.5}{18.02} \approx 0.250$ ⁠ mol.
2. $N = 0.250 \times 6.022 \times 10^{23} \approx 1.50 \times 10^{23}$ ⁠.
3. In E notation: 1.5e23.
If you got $4.5 \times 10^{23}$ ⁠, you counted atoms: each molecule has three. If you got $2.7 \times 10^{24}$ ⁠, you multiplied the grams by $N_A$ ⁠ without dividing by the molar mass.
Step 4 of 5
Open this step
Mass from moles
Practice · Calculate
Question
What is the mass of 2.00 mol of chalk, CaCO₃?

Correct answer
200.2 g ± 0.5%
Setup, choices and feedback
The Scene at this step
Substance
Chalk (CaCO₃)
Mass
100 g
Sample / Particles
Sample
Accepted within
± 0.5%
A fixed value in the question
Any unit of the same quantity is accepted; it is converted before checking.
Wrong answers it recognises
- 50.04 g ± 0.5%
  Misconception: Turns a ratio or a division upside down, such as M ÷ m for the moles.
  Feedback from the catalogue
  Check with the units: g ÷ g/mol = mol, so $n = m / M$ ⁠. If you divided the other way, $M / m$ ⁠, the unit is 1/mol, which is not an amount. And in “how many times” questions, divide the thing asked about by the thing it is compared with.
Feedback for a right answer
$m = 2.00 \times 100.09 = 200.2$ ⁠ grams. Check it: set Mass to 200.2 g, and the readings show 2.00 mol.

Hint
$m = n \times M$ ⁠, and chalk’s $M$ ⁠ is about 100 g/mol.

Worked solution
1. $M(\mathrm{CaCO_3}) = 100.09$ ⁠ g/mol.
2. $m = n \times M = 2.00 \times 100.09$ ⁠.
3. $m \approx 200.2$ ⁠ grams.
If you got $50$ ⁠, you divided the molar mass by the number of moles. The units warn you: (g/mol) ÷ mol is not grams.
Step 5 of 5
Open this step
Which holds the most
Practice · Choose
Question
Which of these substances holds the most particles in one gram?

Correct answer
- Correct Water, M = 18 g/mol
Setup, choices and feedback
The Scene at this step
Substance
Water (H₂O)
Mass
1 gfixed
Sample / Particles
Sample
Choices
Shown in a random order.
- Correct Water, M = 18 g/mol
- Wrong Glucose, M = 180 g/mol
  Feedback
  A glucose molecule is ten times heavier than a water molecule, so one gram holds ten times fewer of them.
- Wrong Gold, M = 197 g/mol
  Feedback
  Gold atoms are very heavy: one gram holds only $0.005$ ⁠ mol.
- Wrong All the same, since the mass is the same
  Misconception: Thinks equal masses of different substances hold the same number of particles.
  Feedback from the catalogue
  Equal masses hold equal numbers of particles only when the particles weigh the same. An iron atom is $4.65$ ⁠ times heavier than a carbon atom, so 12 g of iron holds $4.65$ ⁠ times fewer atoms than 12 g of carbon. To compare numbers, compare moles: $n = m / M$ ⁠.
Feedback for a right answer
Water: $\dfrac{1}{18} \approx 0.056$ ⁠ mol, about $3.3 \times 10^{22}$ ⁠ molecules. The lighter the particle, the more of them fit in a gram.

Hint
$n = m / M$ ⁠ with $m$ ⁠ fixed.

Worked solution
1. $n = \dfrac{1}{M}$ ⁠ for one gram.
2. Water: $0.056$ ⁠ mol; glucose: $0.0056$ ⁠; gold: $0.0051$ ⁠.
3. The most moles, and so the most particles: water.

Lesson 4 of 4 · About 25 minutes

Formulas from masses

How much of a compound’s mass belongs to each element: percentage composition and an element’s mass in a sample, and how measured masses become an empirical formula and a molecular formula.

Open the lesson

The 6 steps and their answers

Step 1 of 6
Open this step
A share of the mass
Practice · Calculate
Question
What percentage of water’s mass is oxygen?

Correct answer
88.81 ± 0.3
Setup, choices and feedback
The Scene at this step
Substance
Water (H₂O)fixed
Mass
18 g
Sample / Particles
Particles
Other settings
Readings/nodes/bench/props/readings = "molar-mass"
Other settings
Element/nodes/bench/props/element = "O"
Accepted within
± 0.3
Computed from the Scene at this step’s setting
The chosen element's share of the mass, per cent; NaN if none/nodes/bench/outputs/elementPercent
Wrong answers it recognises
- 33.3 ± 0.5
  Misconception: Works out the percentage by mass from the share of the atoms, as if all atoms weighed the same.
  Feedback from the catalogue
  A percentage by mass weighs each atom by its $A_r$ ⁠. In water only one atom in three is oxygen, but oxygen is $16$ ⁠ times heavier than hydrogen, so it is $88.8\%$ ⁠ of the mass, not $33\%$ ⁠. The sum: the element’s $A_r$ ⁠ times its number of atoms, divided by the whole formula’s $M_r$ ⁠.
Feedback for a right answer
$88.8\%$ ⁠: $\dfrac{16.00}{18.02} \times 100$ ⁠. Only one atom in three is oxygen, but almost nine tenths of the mass.

Hint
$\dfrac{16.00}{18.02} \times 100$ ⁠.

Worked solution
1. In the molecule: one oxygen, $16.00$ ⁠; the whole molecule, $18.02$ ⁠.
2. $\%\,\mathrm{O} = \dfrac{16.00}{18.02} \times 100\%$ ⁠.
3. $\approx 88.8\%$ ⁠.
If you answered $33.3\%$ ⁠, you counted atoms: one in three. A percentage by mass weighs each atom by its $A_r$ ⁠.
Step 2 of 6
Open this step
An element in a sample
Practice · Calculate
Question
How many grams of carbon are in 50 g of chalk, CaCO₃?

Correct answer
6 g ± 1%
Setup, choices and feedback
The Scene at this step
Substance
Chalk (CaCO₃)fixed
Mass
50 gfixed
Sample / Particles
Sample
Other settings
Readings/nodes/bench/props/readings = "molar-mass"
Other settings
Element/nodes/bench/props/element = "C"
Accepted within
± 1%
Computed from the Scene at this step’s setting
Mass of the chosen element in the sample; NaN if none/nodes/bench/outputs/elementMass
Any unit of the same quantity is accepted; it is converted before checking.
Wrong answers it recognises
- 10 g ± 1%
  Misconception: Works out the percentage by mass from the share of the atoms, as if all atoms weighed the same.
  Feedback from the catalogue
  A percentage by mass weighs each atom by its $A_r$ ⁠. In water only one atom in three is oxygen, but oxygen is $16$ ⁠ times heavier than hydrogen, so it is $88.8\%$ ⁠ of the mass, not $33\%$ ⁠. The sum: the element’s $A_r$ ⁠ times its number of atoms, divided by the whole formula’s $M_r$ ⁠.
Feedback for a right answer
6.0 g: carbon is $12.0\%$ ⁠ of chalk’s mass, and $12\%$ ⁠ of 50 g is 6 g.

Hint
Carbon’s share of the mass is $\dfrac{12.01}{100.09}$ ⁠.

Worked solution
1. In each formula unit: carbon $12.01$ ⁠ out of $100.09$ ⁠.
2. Its share of the mass: $\dfrac{12.01}{100.09} \approx 0.120$ ⁠.
3. $m(\mathrm{C}) = 0.120 \times 50 = 6.0$ ⁠ grams.
If you got 10 g, you divided the mass by five because one of the five atoms is carbon. But the atoms differ in mass.
Step 3 of 6
Open this step
Atoms of one element
Practice · Calculate
Question
How many moles of oxygen atoms are in 25 g of chalk?

Correct answer
0.7494 mol ± 1%
Setup, choices and feedback
The Scene at this step
Substance
Chalk (CaCO₃)fixed
Mass
25 gfixed
Sample / Particles
Particles
Other settings
Readings/nodes/bench/props/readings = "molar-mass"
Other settings
Element/nodes/bench/props/element = "O"
Accepted within
± 1%
Computed from the Scene at this step’s setting
Moles of the chosen element's atoms in the sample; NaN if none/nodes/bench/outputs/elementMoles
Any unit of the same quantity is accepted; it is converted before checking.
Wrong answers it recognises
- 0.2498 mol ± 1%
  Misconception: Mixes up the number of molecules or formula units with the number of atoms in them.
  Feedback from the catalogue
  Mind what you are counting. A mole of water is a mole of molecules, but three moles of atoms, since each molecule has three. A mole of CaCO₃ holds three moles of oxygen atoms. Multiply the number of units by the number of atoms of the kind asked about in each one.
Feedback for a right answer
0.75 mol: $\dfrac{25}{100.09} \approx 0.25$ ⁠ mol of chalk, and three oxygens in each formula unit.

Hint
First how many moles of CaCO₃, then how many times that.

Worked solution
1. $n(\mathrm{CaCO_3}) = \dfrac{25}{100.09} \approx 0.250$ ⁠ mol.
2. Each unit has $3$ ⁠ oxygen atoms.
3. $n(\mathrm{O}) = 3 \times 0.250 = 0.75$ ⁠ mol.
If you answered $0.25$ ⁠, you gave the amount of chalk, not the amount of oxygen atoms in it.
Step 4 of 6
Open this step
The simplest ratio
Practice · Choose
Question
A sample of a compound holds 2.4 g of carbon and 0.6 g of hydrogen, and nothing else. What is its empirical formula?

Correct answer
- Correct CH₃
Setup, choices and feedback
The Scene at this step
Substance
Ethanol (C₂H₅OH)
Mass
46 g
Sample / Particles
Particles
Other settings
Element/nodes/bench/props/element = "C"
Choices
Shown in a random order.
- Correct CH₃
- Wrong C₄H
  Misconception: Takes the ratio of the elements’ masses as the ratio of their atoms.
  Feedback from the catalogue
  A formula counts atoms, not grams. To go from masses to numbers of atoms, divide each mass by the element’s $A_r$ ⁠ to get moles. Only the ratio of the moles is the ratio of the atoms. 2.4 g of carbon and 0.6 g of hydrogen are $0.2$ ⁠ and $0.6$ ⁠ mol: CH₃, not C₄H.
- Wrong C₂H₆
  Feedback
  That is a right ratio, but not the simplest: it can be halved. It may be the molecular formula, but to know you need the molecule’s mass.
- Wrong CH₄
  Feedback
  Methane is CH₄, but here the ratio of moles is $1 : 3$ ⁠.
Feedback for a right answer
CH₃: $\dfrac{2.4}{12.01} = 0.20$ ⁠ mol of carbon and $\dfrac{0.6}{1.008} = 0.60$ ⁠ mol of hydrogen, a ratio of $1 : 3$ ⁠. The substance is ethane, C₂H₆, but the masses alone give only the ratio.

Hint
Divide each mass by the element’s $A_r$ ⁠.

Worked solution
1. Carbon: $\dfrac{2.4}{12.01} \approx 0.20$ ⁠ mol.
2. Hydrogen: $\dfrac{0.6}{1.008} \approx 0.60$ ⁠ mol.
3. Divide by the smaller, $0.20$ ⁠: carbon $1$ ⁠, hydrogen $3$ ⁠.
4. The empirical formula: CH₃.
If you chose C₄H, you used the ratio of the masses, $2.4 : 0.6 = 4 : 1$ ⁠. But a carbon atom is $12$ ⁠ times heavier than a hydrogen atom.
Step 5 of 6
Open this step
Empirical and molecular
Practice · Choose
Question
The molecular formula of glucose is C₆H₁₂O₆. What is its empirical formula?

Correct answer
- Correct CH₂O
Setup, choices and feedback
The Scene at this step
Substance
Glucose (C₆H₁₂O₆)
Mass
180 g
Sample / Particles
Particles
Other settings
Element/nodes/bench/props/element = "C"
Choices
Shown in a random order.
- Correct CH₂O
- Wrong C₆H₁₂O₆
  Feedback
  That is the molecular formula. The empirical one is the simplest ratio: all the numbers can be divided by six.
- Wrong C₃H₆O₃
  Feedback
  A right ratio, but not the simplest: it can be divided by three again.
- Wrong CHO
  Misconception: Counts each element in a formula once, leaving out the subscripts.
  Feedback from the catalogue
  The small number below a symbol says how many atoms of that element are in the formula unit, and it belongs only to the symbol before it. H₂O has two hydrogens, and CaCO₃ three oxygens. When working out a formula mass, multiply each $A_r$ ⁠ by its number of atoms before adding.
Feedback for a right answer
CH₂O: divide $6 : 12 : 6$ ⁠ by six to get $1 : 2 : 1$ ⁠. Formaldehyde, acetic acid and glucose all share this empirical formula.

Hint
Find the greatest common factor of $6$ ⁠, $12$ ⁠ and $6$ ⁠.

Worked solution
1. The ratio in glucose: carbon $6$ ⁠, hydrogen $12$ ⁠, oxygen $6$ ⁠.
2. The greatest common factor: $6$ ⁠.
3. $6 : 12 : 6 = 1 : 2 : 1$ ⁠, that is CH₂O.
Step 6 of 6
Open this step
Back to the molecule
Practice · Choose
Question
A sugar has the empirical formula CH₂O, and its molecule’s mass was measured as $M_r = 180$ ⁠. What is its molecular formula?

Correct answer
- Correct C₆H₁₂O₆
Setup, choices and feedback
The Scene at this step
Substance
Glucose (C₆H₁₂O₆)
Mass
180 g
Sample / Particles
Sample
Choices
Shown in a random order.
- Correct C₆H₁₂O₆
- Wrong CH₂O
  Feedback
  CH₂O has a mass of only $30$ ⁠, not $180$ ⁠.
- Wrong C₃H₆O₃
  Feedback
  C₃H₆O₃ has a mass of $90$ ⁠: half of what is needed.
- Wrong C₁₈₀H₃₆₀O₁₈₀
  Misconception: Turns a ratio or a division upside down, such as M ÷ m for the moles.
  Feedback from the catalogue
  Check with the units: g ÷ g/mol = mol, so $n = m / M$ ⁠. If you divided the other way, $M / m$ ⁠, the unit is 1/mol, which is not an amount. And in “how many times” questions, divide the thing asked about by the thing it is compared with.
Feedback for a right answer
C₆H₁₂O₆: CH₂O has a mass of $30.03$ ⁠, and $\dfrac{180}{30} = 6$ ⁠. It is the formula of glucose, and the readings in the Scene show its formula mass, $180.16$ ⁠.

Hint
Work out the mass of CH₂O, and divide $180$ ⁠ by it.

Worked solution
1. $M_r(\mathrm{CH_2O}) = 12.01 + 2 \times 1.008 + 16.00 \approx 30.03$ ⁠.
2. $\dfrac{180}{30.03} \approx 6$ ⁠.
3. Multiply each number in the empirical formula by six: C₆H₁₂O₆.
C₁₈₀H₃₆₀O₁₈₀ multiplies by $180$ ⁠ itself rather than by the ratio of the masses.

Across the unit · 10 misconceptions

Misconception catalogue

Each common mistake this unit recognises, in a few words. Open one for what the learner reads when an answer matches it, and where it is asked.

Index

Relative masses

Too light to weigh

Question

Correct answer

The Scene at this step

Wrong answers it recognises

Hint

Worked solution

Formula mass

Question

Correct answer

The Scene at this step

Wrong answers it recognises

Hint

Worked solution

How many atoms of each kind

Question

Correct answer

The Scene at this step

Wrong answers it recognises

Hint

Worked solution

Counting atoms

Question

Correct answer

The Scene at this step

Wrong answers it recognises

Hint

Worked solution

Find it

Question

Goal

The Scene at this step

One setting that reaches it

Hint

Worked solution

Counting by weighing

Counting by weighing

Question

Correct answer

The Scene at this step

Wrong answers it recognises

Hint

Worked solution

The same number, a different mass

Question

Correct answer

The Scene at this step

Choices

Once the prediction is locked in, the Scene changes

Worked solution

Molar mass

Question

Correct answer

The Scene at this step

Wrong answers it recognises

Hint

Worked solution

Weighing out a mole

Question

Goal

The Scene at this step

One setting that reaches it

Hint

Worked solution

The same mass, a different number

Question

Correct answer

The Scene at this step

Choices

Hint

Worked solution

Grams, moles and particles

From grams to moles

Question

Correct answer

The Scene at this step

Wrong answers it recognises

Hint

Worked solution