DP Chemistry HL R2.1 How Much? The Amount of Chemical Change

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DP Chemistry HL | R2.1 How Much? The Amount of Chemical Change

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R2.1

How Much? The Amount of Chemical Change

Full notes for HL · IB-style

Contents

1	Chemical equations and conservation of mass
2	Mole ratios and reacting masses
3	Limiting and excess reactants
4	Gas volumes and solutions in reactions
5	Titrations
6	Percentage yield and atom economy
7	Balancing equations: a method
8	Reacting masses: three steps
9	Limiting and excess reactants
10	Percentage yield
11	Atom economy
12	Solution stoichiometry
13	Acid–base titrations
14	Titration technique and errors
15	Back titration
16	Redox titration calculations (introduction)
17	Gas volumes in stoichiometry
18	Multi-step syntheses
19	Mixed review: worked problems
20	Additional exam-style practice
21	Purity of reactants
22	Stoichiometry in industry and medicine
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
write and balance chemical equations with state symbols
use mole ratios to calculate reacting masses
identify the limiting reactant and calculate the amount of excess
calculate volumes of gases and concentrations of solutions in reactions
carry out titration calculations, including back titration
calculate percentage yield and atom economy and explain their importance in green chemistry

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

Guiding questions

How do chemists predict the amounts of substances that react and form? How efficient are chemical reactions?

How this topic is examined

R2.1 is the core of quantitative chemistry and is tested in almost every paper: reacting masses, limiting reactants, titrations and yield calculations, often in multi-step Paper 2 questions.

Follow the same route every time: convert to moles, use the mole ratio, convert to the quantity required.

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 Chemical equations and conservation of mass

Atoms are rearranged, never created or destroyed.

Think about it

When magnesium burns in air, the white ash weighs more than the original ribbon. Does this break the law of conservation of mass?

In a chemical reaction, atoms are rearranged to form new substances; no atoms are created or destroyed. So the total mass of reactants equals the total mass of products (the law of conservation of mass). Magnesium ash weighs more than magnesium because oxygen from the air has combined with it.

A balanced equation has the same number of atoms of each element on both sides. Balance by changing the coefficients (numbers in front of formulas), never the subscripts in formulas. Include state symbols: (s), (l), (g), (aq).

The coefficients give the mole ratio in which substances react and form: 2H2(g) + O2(g) → 2H2O(l) means 2 mol of hydrogen react with 1 mol of oxygen to form 2 mol of water.

WORKED EXAMPLE 1 Balancing an equation

Balance: Al + O2 → Al2O3.

Solution

Need 3 O on each side and even numbers: 4Al + 3O2 → 2Al2O3A1

Check: Al 4 = 4; O 6 = 6M1

Common mistake

Changing subscripts to balance an equation (writing H2O2 instead of 2H2O). That changes the substance.

2 Mole ratios and reacting masses

Mass to moles, ratio, moles to mass.

To calculate the mass of a product or reactant: (1) convert the known mass to moles (n = m/M); (2) use the mole ratio from the balanced equation; (3) convert moles of the unknown to mass (m = nM).

WORKED EXAMPLE 2 Reacting masses

Calculate the mass of calcium oxide produced by heating 25.0 g of calcium carbonate: CaCO3(s) → CaO(s) + CO2(g).

Solution

n(CaCO3) = 25.0 ÷ 100.09 = 0.250 molM1

Ratio 1 : 1, so n(CaO) = 0.250 molM1

m(CaO) = 0.250 × 56.08 = 14.0 gA1

Quick check

1. What mass of water forms from 4.0 g of H2 burning?

2. Why must equations be balanced before calculations?

Answers: 1. 36 g (2 mol H2 → 2 mol H2O) 2. the coefficients give the mole ratio

3 Limiting and excess reactants

One reactant runs out first.

Reactants are often not present in the exact ratio of the equation. The limiting reactant is completely used up and determines the maximum amount of product. The other reactant is in excess: some is left over.

To identify the limiting reactant: calculate moles of each reactant; divide each by its coefficient in the equation; the smallest value indicates the limiting reactant. Then use the limiting reactant for all product calculations.

WORKED EXAMPLE 3 Limiting reactant

5.00 g of aluminium reacts with 10.0 g of chlorine: 2Al + 3Cl2 → 2AlCl3. Identify the limiting reactant and calculate the mass of AlCl3 formed.

Solution

n(Al) = 5.00 ÷ 26.98 = 0.185 mol; ÷ 2 = 0.0927M1

n(Cl2) = 10.0 ÷ 70.90 = 0.141 mol; ÷ 3 = 0.0470: Cl2 is limitingM1

n(AlCl3) = 0.141 × 2/3 = 0.0940 molM1

m = 0.0940 × 133.33 = 12.5 gA1

Quick check

1. Define limiting reactant.

2. Why are reactants often added in excess?

Answers: 1. the reactant completely used up, limiting product amount 2. to ensure the more valuable reactant reacts completely

4 Gas volumes and solutions in reactions

Using molar volume and concentration in stoichiometry.

For gases, use molar volume (22.7 dm3 mol−1 at STP) or pV = nRT (S1.4, S1.5). For gases at the same temperature and pressure, volume ratios equal mole ratios (Avogadro's law). For solutions, use n = cV (V in dm3).

WORKED EXAMPLE 4 Gas volume from a reaction

Calculate the volume of hydrogen at STP produced when 0.243 g of magnesium reacts with excess HCl: Mg + 2HCl → MgCl2 + H2.

Solution

n(Mg) = 0.243 ÷ 24.31 = 0.0100 mol = n(H2)M1

V = 0.0100 × 22.7 = 0.227 dm3 (227 cm3)A1

WORKED EXAMPLE 5 Solution stoichiometry

What volume of 0.200 mol dm−3 NaOH neutralizes 25.0 cm3 of 0.150 mol dm−3 H2SO4? H2SO4 + 2NaOH → Na2SO4 + 2H2O.

Solution

n(H2SO4) = 0.150 × 0.0250 = 3.75 × 10−3 molM1

n(NaOH) = 2 × 3.75 × 10−3 = 7.50 × 10−3 molM1

V = 7.50 × 10−3 ÷ 0.200 = 0.0375 dm3 = 37.5 cm3A1

5 Titrations

Finding unknown concentrations accurately.

A titration determines the concentration of a solution by reacting it with a standard solution of known concentration. A known volume of one solution is measured with a pipette into a conical flask with a few drops of indicator; the other is added from a burette until the indicator changes colour at the end point (equivalence point).

Good practice: rinse the burette with the solution it will contain and the pipette with its solution; read the burette at eye level from the bottom of the meniscus; swirl the flask; add dropwise near the end point; repeat until concordant results (within 0.10 cm3) are obtained, and average them.

A back titration is used when the substance being analysed is insoluble or reacts slowly (such as calcium carbonate in an antacid tablet or eggshell). A known excess of acid is added; then the unreacted acid is titrated with a standard alkali. The amount that reacted = amount added − amount left over.

Figure 1. Titration apparatus: burette, conical flask and white tile.

WORKED EXAMPLE 6 Back titration

A 0.500 g eggshell sample was added to 50.0 cm3 of 0.200 mol dm−3 HCl (excess). The excess HCl required 21.0 cm3 of 0.100 mol dm−3 NaOH. Calculate the percentage of CaCO3 in the eggshell.

Solution

HCl added = 0.200 × 0.0500 = 0.0100 mol; HCl left = NaOH = 0.100 × 0.0210 = 0.00210 molM1

HCl reacted = 0.00790 mol; CaCO3 + 2HCl, so n(CaCO3) = 0.00395 molM1

m = 0.00395 × 100.09 = 0.395 g; % = 0.395 ÷ 0.500 × 100 = 79.1%A1

Quick check

1. What are concordant results?

2. When is a back titration used?

Answers: 1. titres within 0.10 cm3 2. insoluble or slowly reacting substance

6 Percentage yield and atom economy

How efficient is a reaction?

The theoretical yield is the maximum mass of product calculated from the limiting reactant. The actual yield is the mass obtained. Percentage yield = actual yield ÷ theoretical yield × 100. Yields are less than 100% because of incomplete reactions, side reactions, losses during transfer and purification, and reversible reactions.

Atom economy measures how much of the reactants ends up in the desired product: atom economy = (molar mass of desired product ÷ total molar mass of all reactants) × 100. Reactions with high atom economy (such as addition reactions, 100%) produce less waste. Atom economy is a key principle of green chemistry, along with using renewable feedstocks, avoiding hazardous substances and reducing energy use.

WORKED EXAMPLE 7 Yield and atom economy

Ethanol is made by (1) hydration of ethene: C2H4 + H2O → C2H5OH; and (2) fermentation: C6H12O6 → 2C2H5OH + 2CO2. Compare the atom economies. In route (1), 28.0 g ethene gave 41.4 g ethanol. Calculate the percentage yield.

Solution

Route (1): all atoms in product: 100%A1

Route (2): 2(46.08) ÷ 180.18 × 100 = 51.1%A1

Theoretical yield (1): 1.00 mol × 46.08 = 46.1 g; % yield = 41.4 ÷ 46.1 × 100 = 89.8%A1

Quick check

1. Why are percentage yields less than 100%?

2. Why do addition reactions have 100% atom economy?

Answers: 1. incomplete reaction, side reactions, losses 2. all reactant atoms end up in the single product

7 Balancing equations: a method

Conservation of atoms.

Balance by adjusting coefficients, never subscripts. Start with the most complex substance, balance elements that appear only once on each side, leave O and H (or free elements) until last, and check every element and the total charge for ionic equations.

WORKED EXAMPLE 8 Balancing (standard)

Balance Fe2O3 + CO → Fe + CO2.

Solution

Fe2O3 + 3CO → 2Fe + 3CO2A2

WORKED EXAMPLE 9 Ionic equation (standard)

Balance MnO4− + Fe2+ + H+ → Mn2+ + Fe3+ + H2O.

Solution

MnO4− + 5Fe2+ + 8H+ → Mn2+ + 5Fe3+ + 4H2OA2

Quick check

1. Balance Al + O2 → Al2O3.

2. Balance C6H14 + O2 → CO2 + H2O.

Answers: 1. 4Al + 3O2 → 2Al2O3 2. 2C6H14 + 19O2 → 12CO2 + 14H2O

8 Reacting masses: three steps

Mass → moles → moles → mass.

1. Convert the known mass to moles (n = m/M). 2. Use the mole ratio from the balanced equation. 3. Convert the moles of the required substance back to mass.

WORKED EXAMPLE 10 Iron extraction (standard)

What mass of iron is produced from 1.00 kg of Fe2O3?

Solution

n(Fe2O3) = 1000/159.70 = 6.262 molA1

n(Fe) = 12.52 molA1

m = 12.52 × 55.85 = 699 gA1

Quick check

1. What mass of CO2 forms when 10.0 g CaCO3 decomposes?

2. What mass of oxygen is needed to burn 4.60 g of ethanol?

Answers: 1. 4.40 g 2. n = 0.0998 mol; 3 × 0.0998 × 32.00 = 9.58 g

9 Limiting and excess reactants

Which runs out first?

Divide the moles of each reactant by its coefficient: the smallest value identifies the limiting reactant, which determines the amount of product. The other reactant is in excess.

WORKED EXAMPLE 11 Limiting reactant (exam level)

5.00 g of aluminium reacts with 20.0 g of chlorine: 2Al + 3Cl2 → 2AlCl3. Identify the limiting reactant and the mass of AlCl3.

Solution

Al: 0.1853/2 = 0.0927; Cl2: 0.2821/3 = 0.0940M1

Al is limitingA1

n(AlCl3) = 0.1853 mol: 24.7 gA1

Quick check

1. 4.00 g H2 reacts with 16.0 g O2. Which is limiting, and what mass of water forms?

2. Find the mass of excess reactant left in the example above.

Answers: 1. O2 limiting: 18.0 g water 2. Cl2 used 0.2780 mol; 0.0041 mol (0.29 g) Cl2 left

10 Percentage yield

Actual versus theoretical.

Yields are below 100% because of incomplete reactions (equilibrium), side reactions, loss during transfer, filtration and purification, and impure reactants.

WORKED EXAMPLE 12 Yield (standard)

Theoretical yield of aspirin is 3.20 g; 2.45 g is obtained. Calculate the percentage yield.

Solution

2.45/3.20 × 100M1

= 76.6%A1

Quick check

1. A reaction should give 15.0 g of product but gives 12.3 g. Find the yield.

2. Give two reasons why yields are below 100%.

Answers: 1. 82.0% 2. incomplete reaction; losses during purification

11 Atom economy

Green chemistry measure.

WORKED EXAMPLE 13 Comparing routes (exam level)

Ethanol can be made by hydration of ethene (C2H4 + H2O → C2H5OH) or fermentation (C6H12O6 → 2C2H5OH + 2CO2). Calculate both atom economies.

Solution

Hydration: 100%A1

Fermentation: 92.14/180.18 × 100 = 51.1%M1 A1

Quick check

1. Find the atom economy for making iron: Fe2O3 + 3CO → 2Fe + 3CO2.

Answers: 1. 111.7/(159.7 + 84.0) = 45.8%

12 Solution stoichiometry

Volumes and concentrations.

For reactions in solution: n = cV (V in dm3); use the mole ratio; convert back to the required quantity.

WORKED EXAMPLE 14 Precipitation (standard)

What mass of BaSO4 forms when 25.0 cm3 of 0.200 mol dm−3 BaCl2 reacts with excess Na2SO4?

Solution

n = 0.00500 molA1

m = 0.00500 × 233.4 = 1.17 gA1

Quick check

1. What volume of 0.100 mol dm−3 NaOH reacts with 0.490 g of H2SO4?

Answers: 1. n(H2SO4) = 0.00500: 0.0100 mol NaOH = 100 cm3

13 Acid–base titrations

Finding unknown concentrations.

In a titration, a burette delivers a solution (titrant) into a known volume of another (from a pipette) until the end point, shown by an indicator. Use concordant titres (within 0.10 cm3) to calculate the mean.

WORKED EXAMPLE 15 Titration (exam level)

25.0 cm3 of NaOH is neutralized by a mean titre of 21.40 cm3 of 0.100 mol dm−3 HCl. Find [NaOH].

Solution

n(HCl) = 2.140 × 10−3 molA1

n(NaOH) = 2.140 × 10−3A1

c = 2.140 × 10−3/0.0250 = 0.0856 mol dm−3A1

Quick check

1. 25.0 cm3 of 0.0500 mol dm−3 H2SO4 needs 22.5 cm3 of KOH. Find [KOH].

Answers: 1. n(KOH) = 2 × 1.25 × 10−3: 0.111 mol dm−3

14 Titration technique and errors

Reliable results.

Rinse the burette with the titrant and the pipette with the solution it will deliver; rinse the conical flask with distilled water only. Read the bottom of the meniscus at eye level. Add titrant dropwise near the end point and swirl. Repeat until concordant.

Error	Effect on calculated concentration of the solution in the flask
burette rinsed with water	titre too large: too high
air bubble in burette tip	titre reading too large: too high
overshooting end point	too high
flask wet with water	no effect

WORKED EXAMPLE 16 Concordant titres (standard)

Titres: 22.45, 22.10, 22.15, 22.30 cm3. Find the mean of concordant titres.

Solution

Concordant: 22.10, 22.15A1

Mean 22.13 cm3A1

15 Back titration

When direct titration is impossible.

For insoluble bases (such as CaCO3 in limestone), add a known excess of acid, then titrate the leftover acid with alkali. Acid reacted = acid added − acid left.

WORKED EXAMPLE 17 Limestone (exam level)

1.00 g of limestone is dissolved in 50.0 cm3 of 1.00 mol dm−3 HCl. The excess acid needs 32.0 cm3 of 1.00 mol dm−3 NaOH. Find the percentage of CaCO3.

Solution

HCl added 0.0500; left 0.0320; reacted 0.0180 molM1 A1

n(CaCO3) = 0.00900 mol = 0.901 gA1

90.1%A1

16 Redox titration calculations (introduction)

Using mole ratios.

Potassium manganate(VII) titrations are self-indicating (pink at the end point). MnO4− reacts with Fe2+ in the ratio 1 : 5.

WORKED EXAMPLE 18 Iron tablets (exam level)

Iron(II) from a tablet needs 18.2 cm3 of 0.0100 mol dm−3 KMnO4. Find the mass of iron.

Solution

n(MnO4−) = 1.82 × 10−4A1

n(Fe2+) = 9.10 × 10−4A1

m = 0.0508 g (50.8 mg)A1

17 Gas volumes in stoichiometry

Combining with molar volume.

At STP, 1 mol of gas = 22.7 dm3; at other conditions use pV = nRT. Equal volumes of gases react in the ratio of their coefficients.

WORKED EXAMPLE 19 Volume of gas (standard)

Find the volume of H2 at STP when 0.600 g of Mg reacts with excess HCl.

Solution

n(Mg) = 0.02468 mol = n(H2)A1

V = 0.560 dm3A1

Quick check

1. Find the volume of O2 at STP needed to burn 1.00 mol of propane.

Answers: 1. 5 mol: 114 dm3

18 Multi-step syntheses

Overall yield.

For a multi-step synthesis, the overall yield is the product of the yields of each step (as decimals). A five-step synthesis with 80% yield per step has an overall yield of only 33%.

WORKED EXAMPLE 20 Overall yield (standard)

Three steps have yields 90%, 75% and 60%. Find the overall yield.

Solution

0.90 × 0.75 × 0.60M1

= 40.5%A1

Quick check

1. How many steps at 90% each give an overall yield below 50%?

Answers: 1. 7 steps (0.97 = 0.478)

19 Mixed review: worked problems

Bringing the ideas together.

WORKED EXAMPLE 21 Mixed problem (exam level)

2.00 g of Mg reacts with 50.0 cm3 of 2.00 mol dm−3 HCl. (a) Identify the limiting reactant. (b) Find the volume of H2 at STP. (c) Find the mass of excess reactant left.

Solution

(a) Mg 0.0823 mol; HCl 0.100/2 = 0.0500: HCl limitingM1 A1

(b) n(H2) = 0.0500: 1.14 dm3A1

Quick check

1. Find the atom economy for making CaO from CaCO3.

Answers: 1. 56.08/100.09 = 56.0%

20 Additional exam-style practice

Paper 1 and Paper 2.

WORKED EXAMPLE 22 Paper 1 question (standard)

What amount of O2 is needed to burn 2 mol of C2H6? A. 3.5 mol B. 5 mol C. 7 mol D. 14 mol

Solution

2C2H6 + 7O2: CA1

WORKED EXAMPLE 23 Paper 2 question (exam level)

Aspirin is made from salicylic acid (M = 138.12): C7H6O3 + C4H6O3 → C9H8O4 + CH3COOH. From 2.00 g salicylic acid, 1.95 g of aspirin (M = 180.17) is obtained. (a) Find the theoretical yield. (b) Find the percentage yield. (c) Find the atom economy.

Solution

(a) 0.01448 × 180.17 = 2.61 gM1 A1

(b) 74.7%A1

21 Purity of reactants

Accounting for impurities.

If a reactant is impure, multiply its mass by the percentage purity before converting to moles. Conversely, the purity of a sample can be calculated from the amount of product formed.

WORKED EXAMPLE 24 Impure sample (standard)

A 5.00 g sample of impure CaCO3 gives 1.98 g of CO2 on heating. Find the purity.

Solution

n(CO2) = 0.0450 mol = n(CaCO3)A1

m = 4.50 g: 90.0%A1

22 Stoichiometry in industry and medicine

Getting quantities right.

Industrial plants use stoichiometry to optimize feedstock use and minimize waste; pharmacists calculate doses and concentrations; environmental chemists calculate pollutant masses. Errors can be costly or dangerous.

In the real world

In the Haber process, nitrogen and hydrogen are fed in a 1 : 3 ratio by volume to match the stoichiometry and avoid wasting the more expensive hydrogen.

Quick check

1. A patient needs 5.0 mg kg−1 of a drug; they weigh 64 kg. The drug solution is 20 mg cm−3. What volume is needed?

Answers: 1. 320 mg: 16 cm3

23 Common misconceptions and links

Students often think…	Actually…
the mass of the excess reactant determines the product	the limiting reactant does
changing subscripts balances equations	only coefficients are changed
high yield means high atom economy	they measure different things

How R2.1 connects to the rest of the course

Topic	Connection
S1.4 The mole	n = m/M, n = cV, molar volume
R3.1 Proton transfer	acid-base titrations
R3.2 Electron transfer	redox titrations
R2.3 How far?	reversible reactions limit yield

24 Exam-style question with markscheme

Try this under exam conditions (about 18 minutes) before reading the markscheme.

Text A. 2.50 g of iron reacts with 3.20 g of sulfur: Fe(s) + S(s) → FeS(s). The mass of FeS obtained was 3.50 g.

(a) Identify the limiting reactant. [2]

(b) Calculate the theoretical yield of FeS. [2]

(d) Calculate the mass of excess reactant remaining. [2]

(e) Suggest two reasons why the yield is less than 100%. [2]

Markscheme

Part	Answer	Marks
(a)	n(Fe) = 2.50 ÷ 55.85 = 0.0448; n(S) = 3.20 ÷ 32.07 = 0.0998	M1
	Fe limiting	A1
(b)	0.0448 × 87.92	M1
	= 3.94 g	A1
(c)	3.50 ÷ 3.94 × 100 = 88.8%	A1
(d)	S used 0.0448 mol; left 0.0550 mol	M1
	× 32.07 = 1.76 g	A1
(e)	incomplete reaction; side reaction with O2; losses	A2

Examiner's comments. In (a), comparing masses instead of moles is a common error.

25 Check your understanding

Section A: multiple choice

1. The limiting reactant

A. is completely used up B. is in excess C. determines the rate D. is always the solid

2. Atom economy of an addition reaction is

A. 50% B. 0% C. 100% D. depends on yield

3. Concordant titres agree within

A. 1.0 cm3 B. 0.10 cm3 C. 5 cm3 D. 0.001 cm3

4. Moles of NaOH in 20.0 cm3 of 0.100 mol dm−3

A. 2.00 B. 0.200 C. 2.00 × 10−2 D. 2.00 × 10−3

5. Volume of 0.500 mol of gas at STP

A. 22.7 dm3 B. 11.4 dm3 C. 0.500 dm3 D. 45.4 dm3

6. Percentage yield equals

A. actual ÷ theoretical × 100 B. theoretical ÷ actual × 100 C. product ÷ reactants × 100 D. moles × 100

Section B: structured

7. Explain how to identify a limiting reactant. [2]

8. Describe how to carry out an accurate titration. [4]

9. Compare percentage yield and atom economy. [3]

Answers

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

7. Convert to moles; divide by coefficients; smallest is limiting.

8. Rinse burette and pipette; indicator; add to end point; eye-level reading; repeat to concordance.

9. Yield: actual vs theoretical; atom economy: proportion of reactant mass in product; green chemistry.

26 Summary and topic map

Idea	Remember
Conservation of mass	balanced equations; coefficients only
Reacting masses	mass → moles → ratio → mass
Limiting reactant	smallest moles ÷ coefficient
Gases and solutions	22.7 dm3; n = cV
Titrations	concordant results; back titration
Efficiency	percentage yield; atom economy; green chemistry

Before the exam, make sure you…

•always convert to moles first
•identify the limiting reactant before calculating products
•use dm3 for volumes
•show every step

How Much? 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 2. Topic map for How Much?.

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