DP Chemistry SL S1.4 Counting Particles by Mass: The Mole

DP Chemistry SL S1.4 Counting Particles by Mass: The Mole - Full notes

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DP Chemistry SL | S1.4 Counting Particles by Mass: The Mole

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S1.4

Counting Particles by Mass: The Mole

Full notes for SL · IB-style

Contents

1	The mole and the Avogadro constant
2	Relative masses and molar mass
3	Converting between mass, moles and particles
4	Empirical and molecular formulas
5	Concentration of solutions
6	Avogadro's law and molar volume
7	The Avogadro constant in practice
8	Molar mass calculations
9	Mass, moles and molar mass
10	Percentage composition by mass
11	Empirical formulas from mass data
12	Molecular formulas
13	Combustion analysis
14	Concentration of solutions
15	Dilution
16	Concentrations of ions
17	Standard solutions and volumetric technique
18	Parts per million and percentage concentrations
19	Molar volume of gases (introduction)
20	Water of crystallization
21	Mixed review: worked problems
22	Additional exam-style practice
23	Common misconceptions and links
24	Exam-style question with markscheme
25	Check your understanding
26	Summary and topic map

These are full teaching notes. Read each section in order, try the Think about it question before you read on, and do every Quick check before moving to the next section.

Before you start

Rate yourself on each statement now, and again after you've worked through the notes. Shade R (not yet), A (getting there) or G (confident).

I can…	R	A	G
define the mole and use the Avogadro constant
calculate relative formula mass and molar mass
convert between mass, amount (moles) and number of particles
determine empirical formulas from percentage composition or combustion data and molecular formulas from molar mass
calculate concentrations in mol dm−3 and g dm−3, and carry out dilution calculations
apply Avogadro's law to gas volumes

There is no additional HL content in this subtopic. SL and HL students study the same material.

Guiding questions

How do chemists count particles too small to see? How does the mole connect mass, particles and volume?

How this topic is examined

S1.4 calculations appear throughout Papers 1 and 2. Expect mole conversions, empirical formula determination and concentration calculations, often as steps in longer stoichiometry problems (R2.1).

Show units at every step and give answers to an appropriate number of significant figures.

Command term	What the examiner wants
State / Identify	a brief answer, no explanation
Outline / Describe	a brief or detailed account, without reasons
Explain	a detailed account with causes, reasons or mechanisms
Calculate / Determine	a numerical answer with working, correct significant figures and units
Deduce / Predict	reach a conclusion from given information
Draw / Sketch	a clear labelled diagram, structure or graph
Evaluate / Discuss	weigh strengths and limitations; reach a judgement

How to use these notes

Each section teaches one part of the syllabus in full. You'll see these boxes throughout; each has one job:

Box	What it's for
Think about it	a question to get you thinking before a section; the answer comes in the section
Key point	the one idea you must not forget
Common mistake	an error examiners see every year, and how to avoid it
In the real world	where the chemistry shows up outside the classroom
Try this	something quick you can do yourself to test the idea
Worked example	full solutions with IB-style marks (M1 method, A1 answer, R1 reasoning)
Quick check	short questions to test yourself at the end of each section

1 The mole and the Avogadro constant

A chemist's dozen, but much bigger.

Think about it

A single drop of water contains about 1.7 × 1021 water molecules. How can chemists count particles in such huge numbers?

Atoms and molecules are far too small and numerous to count individually, so chemists use a counting unit called the mole (symbol mol). One mole of a substance contains exactly 6.02 × 1023 elementary entities (atoms, molecules, ions or formula units). This number is the Avogadro constant, NA = 6.02 × 1023 mol−1.

The amount of substance (n) is measured in moles. The number of particles N = n × NA. Just as a dozen eggs means 12 eggs, a mole of carbon atoms means 6.02 × 1023 carbon atoms.

WORKED EXAMPLE 1 Moles and particles

Calculate the number of molecules in 0.250 mol of water, and the number of hydrogen atoms.

Solution

Molecules = 0.250 × 6.02 × 1023 = 1.51 × 1023A1

Each H2O has 2 H atoms: H atoms = 2 × 1.51 × 1023 = 3.01 × 1023A1

Quick check

1. What is the Avogadro constant?

2. How many atoms are in 2.0 mol of helium?

Answers: 1. 6.02 × 1023 mol−1 2. 1.2 × 1024

2 Relative masses and molar mass

The mass of one mole equals the relative mass in grams.

The relative formula mass (Mr) of a compound is the sum of the relative atomic masses of all the atoms in its formula. For water, Mr = 2(1.01) + 16.00 = 18.02. Like Ar, it has no units.

The molar mass (M) is the mass of one mole of a substance, in g mol−1. It is numerically equal to Ar or Mr: the molar mass of water is 18.02 g mol−1. This works because the mole was defined so that one mole of carbon-12 has a mass of exactly 12 g.

Substance	Formula	Calculation	Molar mass / g mol−1
Carbon dioxide	CO2	12.01 + 2(16.00)	44.01
Sodium chloride	NaCl	22.99 + 35.45	58.44
Calcium hydroxide	Ca(OH)2	40.08 + 2(16.00 + 1.01)	74.10
Glucose	C6H12O6	6(12.01) + 12(1.01) + 6(16.00)	180.18

Quick check

1. Define molar mass.

2. Calculate the molar mass of H2SO4.

Answers: 1. mass of one mole of a substance, in g mol−1 2. 98.09 g mol−1

3 Converting between mass, moles and particles

One equation does most of the work.

n = m ÷ M (moles = mass in g ÷ molar mass in g mol−1)

WORKED EXAMPLE 2 Mass and moles

(a) Calculate the amount of substance in 11.0 g of CO2. (b) Calculate the mass of 0.150 mol of NaOH (M = 40.00 g mol−1).

Solution

(a) n = 11.0 ÷ 44.01 = 0.250 molA1

(b) m = n × M = 0.150 × 40.00 = 6.00 gA1

WORKED EXAMPLE 3 Mass of one molecule

Calculate the mass of a single molecule of water in grams.

Solution

Mass = M ÷ NA = 18.02 ÷ 6.02 × 1023M1

= 2.99 × 10−23 gA1

Common mistake

Using the mass number of an isotope instead of Ar from the data booklet, or forgetting to multiply by the number of atoms in the formula.

Quick check

1. How many moles are in 4.00 g of NaOH?

2. What is the mass of 2.00 mol of O2?

Answers: 1. 0.100 mol 2. 64.00 g

4 Empirical and molecular formulas

The simplest ratio and the actual formula.

The empirical formula is the simplest whole-number ratio of atoms of each element in a compound. The molecular formula gives the actual number of atoms of each element in a molecule. For glucose, the molecular formula is C6H12O6 and the empirical formula is CH2O. The molecular formula is a whole-number multiple of the empirical formula.

Finding the empirical formula from percentage composition (or masses): (1) assume 100 g, so percentages become grams; (2) divide each mass by Ar to get moles; (3) divide all by the smallest number of moles; (4) multiply to obtain whole numbers if necessary.

Finding the molecular formula: divide the molar mass by the empirical formula mass; multiply the empirical formula by this whole number.

WORKED EXAMPLE 4 Empirical and molecular formula

A compound contains 40.0% carbon, 6.7% hydrogen and 53.3% oxygen by mass. Its molar mass is 180 g mol−1. Determine its empirical and molecular formulas.

Solution

Moles: C 40.0 ÷ 12.01 = 3.33; H 6.7 ÷ 1.01 = 6.63; O 53.3 ÷ 16.00 = 3.33M1

Divide by 3.33: C 1, H 2, O 1: empirical formula CH2OA1

Empirical formula mass = 30.03; 180 ÷ 30.03 = 6M1

Molecular formula C6H12O6A1

Empirical formulas can also be found from combustion analysis: a known mass of an organic compound is burned completely, and the masses of CO2 and H2O produced are measured. All the carbon ends up in CO2 and all the hydrogen in H2O; any remaining mass is oxygen.

WORKED EXAMPLE 5 Combustion analysis

Burning 0.230 g of a compound containing C, H and O gave 0.440 g CO2 and 0.270 g H2O. Find the empirical formula.

Solution

n(C) = n(CO2) = 0.440 ÷ 44.01 = 0.0100 mol; mass C = 0.120 gM1

n(H) = 2 × 0.270 ÷ 18.02 = 0.0300 mol; mass H = 0.0303 gM1

mass O = 0.230 − 0.120 − 0.030 = 0.080 g; n(O) = 0.00500 molM1

Ratio C : H : O = 2 : 6 : 1: C2H6OA1

Quick check

1. Define empirical formula.

2. The empirical formula is CH2 and M = 56 g mol−1. What is the molecular formula?

Answers: 1. simplest whole-number ratio of atoms 2. C4H8

5 Concentration of solutions

How much solute in a given volume.

A solution consists of a solute dissolved in a solvent. Concentration is the amount of solute per unit volume of solution, usually in mol dm−3 (sometimes written M). It can also be given in g dm−3. Note: 1 dm3 = 1000 cm3 = 1 litre.

c = n ÷ V (concentration in mol dm−3 = moles ÷ volume in dm3)

A standard solution has an accurately known concentration. It is made by dissolving an accurately weighed mass of solute and making up to a precise volume in a volumetric flask.

Dilution: adding solvent does not change the amount of solute, so c1V1 = c2V2.

WORKED EXAMPLE 6 Concentration calculations

(a) 2.92 g of NaCl is dissolved to make 250.0 cm3 of solution. Calculate the concentration. (b) What volume of this solution is needed to make 100.0 cm3 of 0.0500 mol dm−3 NaCl?

Solution

(a) n = 2.92 ÷ 58.44 = 0.0500 mol; V = 0.2500 dm3M1

c = 0.0500 ÷ 0.2500 = 0.200 mol dm−3A1

(b) V1 = c2V2 ÷ c1 = 0.0500 × 100.0 ÷ 0.200 = 25.0 cm3A1

Common mistake

Using cm3 in c = n ÷ V. Convert cm3 to dm3 by dividing by 1000.

Quick check

1. Calculate the moles in 25.0 cm3 of 0.100 mol dm−3 HCl.

2. Convert 0.500 mol dm−3 NaOH to g dm−3.

Answers: 1. 2.50 × 10−3 mol 2. 20.0 g dm−3

6 Avogadro's law and molar volume

Equal volumes of gases contain equal numbers of particles.

Avogadro's law: equal volumes of all gases, at the same temperature and pressure, contain equal numbers of particles. So the volume of a gas is proportional to the amount in moles, regardless of which gas it is.

The molar volume of an ideal gas at standard temperature and pressure (STP: 273.15 K, 100 kPa) is 22.7 dm3 mol−1. So n = V ÷ 22.7 (V in dm3) at STP.

Avogadro's law allows volumes of reacting gases to be compared directly using the coefficients in an equation: in N2(g) + 3H2(g) → 2NH3(g), 1 volume of nitrogen reacts with 3 volumes of hydrogen to give 2 volumes of ammonia.

WORKED EXAMPLE 7 Gas volumes

Calculate the volume of 0.400 mol of CO2 at STP, and the volume of oxygen needed to burn 10.0 dm3 of methane (CH4 + 2O2 → CO2 + 2H2O).

Solution

V = 0.400 × 22.7 = 9.08 dm3A1

Ratio CH4 : O2 = 1 : 2, so O2 = 20.0 dm3A1

Quick check

1. State Avogadro's law.

2. What is the molar volume at STP?

Answers: 1. equal volumes of gases at same T and p contain equal numbers of particles 2. 22.7 dm3 mol−1

7 The Avogadro constant in practice

Counting particles.

One mole contains 6.02 × 1023 particles (the Avogadro constant, L or NA). Number of particles N = n × L. Always state what type of particle is being counted: atoms, molecules, ions or electrons.

WORKED EXAMPLE 8 Counting atoms (standard)

How many hydrogen atoms are in 0.25 mol of methane, CH4?

Solution

Molecules: 0.25 × 6.02 × 1023 = 1.505 × 1023M1

H atoms: × 4 = 6.02 × 1023A1

WORKED EXAMPLE 9 Ions (standard)

How many ions are in 0.10 mol of CaCl2?

Solution

3 ions per formula unit: 0.30 molM1

0.30 × 6.02 × 1023 = 1.81 × 1023A1

Quick check

1. How many molecules are in 2.5 mol of water?

2. How many moles of oxygen atoms are in 0.20 mol of Al2(SO4)3?

Answers: 1. 1.51 × 1024 2. 2.4 mol

8 Molar mass calculations

Adding relative atomic masses.

Molar mass M (g mol−1) is numerically equal to the relative formula mass. For hydrated salts, include the water of crystallization: M(CuSO4·5H2O) = 159.62 + 5(18.02) = 249.72 g mol−1.

WORKED EXAMPLE 10 Molar masses (basic)

Calculate M for Ca(NO3)2 and (NH4)2SO4.

Solution

40.08 + 2(14.01 + 48.00) = 164.10 g mol−1A1

2(18.05) + 32.07 + 64.00 = 132.17 g mol−1A1

Quick check

1. Calculate M for Na2CO3·10H2O.

2. Calculate M for C6H12O6.

Answers: 1. 286.19 g mol−1 2. 180.18 g mol−1

9 Mass, moles and molar mass

n = m/M.

n = m/M links mass and amount. Rearrange as m = nM or M = m/n. Give answers to an appropriate number of significant figures, matching the data.

WORKED EXAMPLE 11 Mass to moles (basic)

Calculate the amount of 12.6 g of nitric acid, HNO3.

Solution

M = 63.02 g mol−1A1

n = 12.6/63.02 = 0.200 molA1

WORKED EXAMPLE 12 Identifying a substance (standard)

0.150 mol of an element has mass 9.53 g. Identify it.

Solution

M = 9.53/0.150 = 63.5 g mol−1M1

CopperA1

Quick check

1. Find the mass of 0.0350 mol of K2Cr2O7.

2. 0.500 mol of a gas has mass 22.0 g. Suggest its identity.

Answers: 1. 10.3 g 2. M = 44.0: CO2 (or N2O, C3H8)

10 Percentage composition by mass

Element fractions.

% of element = (number of atoms × Ar ÷ M) × 100. This helps compare fertilizers or ores.

WORKED EXAMPLE 13 Fertilizers (exam level)

Compare the percentage of nitrogen in NH4NO3 and (NH2)2CO (urea).

Solution

NH4NO3: 28.02/80.06 = 35.0%A1

Urea: 28.02/60.07 = 46.6%A1

Urea supplies more nitrogen per kilogramR1

Quick check

1. Find the percentage of iron in Fe2O3.

2. Find the percentage of water in CuSO4·5H2O.

Answers: 1. 69.9% 2. 36.1%

11 Empirical formulas from mass data

Simplest whole-number ratio.

Convert masses (or percentages) of each element to moles, divide by the smallest, and convert to whole numbers (multiply if ratios like 1.5 or 1.33 appear).

WORKED EXAMPLE 14 From percentages (standard)

A compound contains 40.0% C, 6.7% H and 53.3% O. Find its empirical formula.

Solution

Moles: 3.33, 6.65, 3.33M1

Ratio 1 : 2 : 1A1

CH2OA1

WORKED EXAMPLE 15 Multiplying ratios (exam level)

A compound contains 0.84 g Fe and 0.36 g O. Find its empirical formula.

Solution

Fe 0.01504 mol, O 0.0225 molM1

Ratio 1 : 1.5 = 2 : 3A1

Fe2O3A1

Quick check

1. A hydrocarbon contains 85.6% carbon. Find its empirical formula.

2. A compound contains 2.40 g Mg and 1.60 g O. Find its empirical formula.

Answers: 1. CH2 2. MgO

12 Molecular formulas

Multiples of the empirical formula.

Molecular formula = (empirical formula)n, where n = M ÷ M(empirical formula). M is found from mass spectrometry, gas volumes or other data.

WORKED EXAMPLE 16 Molecular formula (standard)

A compound with empirical formula CH2O has M = 180 g mol−1. Find its molecular formula.

Solution

M(CH2O) = 30.03; n = 6M1

C6H12O6A1

Quick check

1. Empirical formula CH, M = 78. Find the molecular formula.

2. Empirical formula C2H5, M = 58. Find the molecular formula.

Answers: 1. C6H6 2. C4H10

13 Combustion analysis

From CO2 and H2O masses.

When an organic compound burns completely, all its carbon becomes CO2 and all its hydrogen becomes H2O. The masses of these products give the masses of C and H; any remaining mass is oxygen.

WORKED EXAMPLE 17 Combustion (exam level)

0.230 g of a compound of C, H and O burns to give 0.440 g CO2 and 0.270 g H2O. Find its empirical formula.

Solution

C: 0.440/44.01 = 0.0100 mol (0.120 g); H: 2 × 0.270/18.02 = 0.0300 mol (0.0302 g)M1 A1

O: 0.230 − 0.150 = 0.080 g = 0.0050 molA1

Ratio 2 : 6 : 1: C2H6OA1

Quick check

1. 0.180 g of a hydrocarbon gives 0.549 g CO2. Find the mass of hydrogen and the empirical formula.

Answers: 1. C 0.1498 g (0.01247 mol); H 0.0302 g (0.0300 mol): ratio 1 : 2.4 = 5 : 12: C5H12

14 Concentration of solutions

c = n/V.

Concentration c (mol dm−3) = n/V with V in dm3 (1 dm3 = 1000 cm3). Mass concentration (g dm−3) = c × M.

WORKED EXAMPLE 18 Preparing a solution (standard)

What mass of NaOH is needed to make 250.0 cm3 of 0.200 mol dm−3 solution?

Solution

n = 0.200 × 0.2500 = 0.0500 molM1

m = 0.0500 × 40.00 = 2.00 gA1

WORKED EXAMPLE 19 Concentration (basic)

Find the concentration of 5.85 g NaCl in 500 cm3.

Solution

n = 5.85/58.44 = 0.100 molA1

c = 0.100/0.500 = 0.200 mol dm−3A1

Quick check

1. Find the amount of H2SO4 in 25.0 cm3 of 0.150 mol dm−3 solution.

2. Convert 0.250 mol dm−3 NaCl into g dm−3.

Answers: 1. 3.75 × 10−3 mol 2. 14.6 g dm−3

15 Dilution

c1V1 = c2V2.

Diluting a solution adds solvent but not solute, so the amount stays constant: c1V1 = c2V2.

WORKED EXAMPLE 20 Dilution (standard)

What volume of 2.00 mol dm−3 HCl is needed to make 500 cm3 of 0.150 mol dm−3 HCl?

Solution

V1 = 0.150 × 500/2.00M1

= 37.5 cm3A1

Quick check

1. 25.0 cm3 of 1.20 mol dm−3 solution is diluted to 200 cm3. Find the new concentration.

2. By what factor is a solution diluted when 10 cm3 is made up to 250 cm3?

Answers: 1. 0.150 mol dm−3 2. 25

16 Concentrations of ions

Account for the formula.

When an ionic compound dissolves, the concentration of each ion depends on the formula: in 0.10 mol dm−3 Na2SO4, [Na+] = 0.20 mol dm−3 and [SO42−] = 0.10 mol dm−3.

WORKED EXAMPLE 21 Ion concentration (standard)

Find [Cl−] in 0.25 mol dm−3 AlCl3.

Solution

3 Cl− per formula unitM1

0.75 mol dm−3A1

Quick check

1. Find the total ion concentration in 0.050 mol dm−3 Ca(NO3)2.

Answers: 1. 0.15 mol dm−3

17 Standard solutions and volumetric technique

Accuracy in the lab.

A standard solution has an accurately known concentration. To prepare one: weigh the solid accurately by difference, dissolve in a little distilled water in a beaker, transfer quantitatively (with rinsings) to a volumetric flask, make up to the mark with the bottom of the meniscus on the line, and invert to mix.

WORKED EXAMPLE 22 Errors (standard)

A student overshoots the graduation mark when making up a standard solution. Effect on concentration?

Solution

Volume too largeR1

Concentration lower than intendedA1

Common mistake

Adding the solid directly to a volumetric flask and filling to the mark. The solid should be dissolved first in a beaker, and all rinsings transferred.

18 Parts per million and percentage concentrations

Very dilute solutions.

For trace amounts, concentrations are given in parts per million (ppm): 1 ppm = 1 mg of solute per kg of solution (≈ 1 mg dm−3 for dilute aqueous solutions). Percentage by mass = (mass of solute/mass of solution) × 100.

WORKED EXAMPLE 23 ppm (standard)

Drinking water contains 0.25 mg of lead in 500 cm3. Express this in ppm.

Solution

0.50 mg dm−3M1

≈ 0.50 ppmA1

In the real world

The WHO guideline limit for fluoride in drinking water is 1.5 ppm; many countries add fluoride at about 0.7 ppm to reduce tooth decay.

Quick check

1. Express 3.0 ppm of nitrate as mol dm−3 (M(NO3−) = 62.01).

Answers: 1. 4.8 × 10−5 mol dm−3

19 Molar volume of gases (introduction)

22.7 dm3 at STP.

At STP (273 K, 100 kPa), one mole of any ideal gas occupies 22.7 dm3. Equal volumes of gases at the same temperature and pressure contain equal numbers of molecules (Avogadro's law).

WORKED EXAMPLE 24 Gas volume (standard)

Find the volume of 4.40 g CO2 at STP.

Solution

n = 0.100 molA1

V = 2.27 dm3A1

Quick check

1. Find the mass of 1.00 dm3 of O2 at STP.

Answers: 1. 1.41 g

20 Water of crystallization

Finding x in a hydrate.

Heating a hydrated salt to constant mass removes the water. The mass lost is the water; comparing moles of anhydrous salt and water gives x in salt·xH2O.

WORKED EXAMPLE 25 Hydrate (exam level)

2.50 g of hydrated magnesium sulfate MgSO4·xH2O gives 1.22 g of anhydrous MgSO4. Find x.

Solution

Water 1.28 g = 0.0710 mol; MgSO4 1.22/120.38 = 0.01013 molM1 A1

x = 7.01: x = 7A1

Common mistake

Not heating to constant mass. If some water remains, the calculated value of x will be too low.

Quick check

1. 5.00 g of BaCl2·xH2O loses 0.738 g of water. Find x.

Answers: 1. BaCl2 4.262 g = 0.02047 mol; water 0.04096 mol: x = 2

21 Mixed review: worked problems

Bringing the ideas together.

WORKED EXAMPLE 26 Mixed problem (exam level)

A compound of sodium, sulfur and oxygen contains 29.1% Na, 40.5% S. (a) Find its empirical formula. (b) Calculate the number of oxygen atoms in 1.00 g.

Solution

(a) O 30.4%; moles 1.266, 1.263, 1.900: ratio 2 : 2 : 3: Na2S2O3M1 A1

(b) 1.00/158.1 = 6.33 × 10−3 mol × 3 × 6.02 × 1023 = 1.14 × 1022M1 A1

Quick check

1. How many moles of ions are in 2.00 g of MgCl2?

Answers: 1. 0.0210 × 3 = 0.0630 mol

22 Additional exam-style practice

Paper 1 and Paper 2.

WORKED EXAMPLE 27 Paper 1 question (standard)

Which sample contains the most atoms? A. 1.0 g H2 B. 1.0 g He C. 1.0 g CH4 D. 1.0 g H2O

Solution

Atoms: A 0.99 mol; B 0.25; C 0.312; D 0.166M1

AA1

WORKED EXAMPLE 28 Paper 2 question (exam level)

25.0 cm3 of a solution of sodium carbonate contains 0.318 g Na2CO3. (a) Find its concentration. (b) Find [Na+]. (c) This solution is diluted to 100 cm3. Find the new [Na2CO3].

Solution

(a) 0.318/105.99 = 0.00300 mol: 0.120 mol dm−3A1 A1

(b) 0.240 mol dm−3A1

Quick check

1. Calculate the number of electrons in one mole of water molecules.

Answers: 1. 10 × 6.02 × 1023 = 6.02 × 1024

23 Common misconceptions and links

Students often think…	Actually…
one mole always has the same mass	molar mass depends on the substance
volumes in cm3 can go straight into c = n ÷ V	convert to dm3
empirical formula is always different from molecular formula	they can be the same (e.g. H2O)

How S1.4 connects to the rest of the course

Topic	Connection
S1.2 Nuclear atom	relative atomic mass
S1.5 Ideal gases	pV = nRT
R2.1 How much?	reacting masses and titrations
R3.1 Proton transfer	concentration and pH

24 Exam-style question with markscheme

Try this under exam conditions (about 15 minutes) before reading the markscheme.

Text A. A hydrocarbon contains 85.6% carbon by mass. Its molar mass is 56.1 g mol−1.

(a) Determine the empirical formula. [3]

(b) Determine the molecular formula. [2]

(d) Calculate the volume occupied by 2.81 g of the gas at STP. [2]

Markscheme

Part	Answer	Marks
(a)	C 85.6 ÷ 12.01 = 7.13; H 14.4 ÷ 1.01 = 14.3	M1
	ratio 1 : 2	M1
	CH2	A1
(b)	56.1 ÷ 14.03 = 4	M1
	C4H8	A1
(c)	n = 2.81 ÷ 56.1 = 0.0501 mol	M1
	0.0501 × 6.02 × 1023 = 3.02 × 1022	A1
(d)	0.0501 × 22.7	M1
	= 1.14 dm3	A1

Examiner's comments. In (a), the hydrogen percentage must be found by difference (14.4%).

25 Check your understanding

Section A: multiple choice

1. The Avogadro constant is

A. 6.02 × 1023 mol−1 B. 6.02 × 10−23 C. 22.7 dm3 D. 8.31 J K−1 mol−1

2. Moles in 9.0 g of water are

A. 2.0 mol B. 0.25 mol C. 0.50 mol D. 9.0 mol

3. The empirical formula of C2H4 is

A. C2H4 B. CH2 C. CH D. C4H8

4. Concentration of 0.10 mol in 500 cm3

A. 0.05 mol dm−3 B. 50 mol dm−3 C. 0.0002 mol dm−3 D. 0.20 mol dm−3

5. Molar volume of a gas at STP

A. 24.0 dm3 B. 22.7 dm3 mol−1 C. 1.00 dm3 D. 100 dm3

6. Diluting a solution changes

A. its concentration but not the moles of solute B. the moles of solute C. nothing D. the molar mass

Section B: structured

7. Calculate the mass of 3.01 × 1023 molecules of CO2. [2]

8. Describe how to prepare 250.0 cm3 of 0.100 mol dm−3 CuSO4 solution. [4]

9. Explain the difference between empirical and molecular formulas. [2]

Answers

1. A 2. C 3. B 4. D 5. B 6. A

7. 0.500 mol × 44.01 = 22.0 g.

8. Weigh 0.0250 mol (3.99 g anhydrous); dissolve in water; transfer to volumetric flask with rinsings; make up to mark; mix.

9. Empirical: simplest ratio; molecular: actual numbers of atoms; multiple of empirical.

26 Summary and topic map

Idea	Remember
Mole	6.02 × 1023 particles
Molar mass	g mol−1 = Ar or Mr
n = m ÷ M	mass and moles
Formulas	empirical (simplest ratio), molecular (actual)
Concentration	c = n ÷ V; c1V1 = c2V2
Gases	Avogadro's law; 22.7 dm3 mol−1 at STP

Before the exam, make sure you…

•show units at each step
•convert cm3 to dm3
•set out empirical formula calculations in columns
•use correct significant figures

The Mole on one page

Cover this page with a sheet of paper and try to rebuild it from memory. Whatever you can't recall is what to revise next.

Figure 1. Topic map for The Mole.

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