Chemistry 9701/23 — May/June 2011

Background Concept

The key idea is the relationship between the cubic units of volume. A decimetre is 10 cm, so a cubic decimetre is $(10 \text{ cm})^3 = 1000 \text{ cm}^3$. Equivalently, $1 \text{ dm}^3 = 1 \text{ L} = 1000 \text{ mL} = 1000 \text{ cm}^3$. Converting a volume from dm³ to cm³ therefore always means multiplying by 1000.

Understanding the Question

The stem tells us that a 'typical' ant contains $7.5 \times 10^{-6} \text{ dm}^3$ of solution A (the 50% methanoic acid solution injected on stinging). Part (a)(i) asks for this volume in cm³. The instruction at the top — give all numerical answers to two significant figures — applies here.

Approach

This is a pure unit conversion. Take the given volume in dm³ and multiply by 1000 to express it in cm³.

Step-by-Step Reasoning

$7.5 \times 10^{-6} \text{ dm}^3 \times 1000 = 7.5 \times 10^{-3} \text{ cm}^3$. The number 7.5 already has two significant figures, so the answer is already correctly expressed. The mark scheme gives exactly this value.

Key Takeaways

$1 \text{ dm}^3 = 1000 \text{ cm}^3$; converting between cubic units requires cubing the linear conversion factor.

Common Mistakes

• Multiplying by 10 instead of 1000 (confusing dm and dm³).
• Forgetting to convert at all and leaving the answer in dm³.
• Expressing the answer to the wrong number of significant figures.

Things to Be Careful About

The question explicitly demands two significant figures. Also note that this value ($7.5 \times 10^{-3} \text{ cm}^3$) is carried forward into parts (a)(ii), (a)(iii) and (b); the mark scheme allows error carried forward (ecf) if a slip is made here.

Background Concept

Solution A is 50% by volume pure methanoic acid. A percentage by volume means that out of every 100 cm³ of solution, 50 cm³ is the pure acid. To find the volume of pure acid in a given volume of solution, multiply the solution volume by the percentage expressed as a fraction (50/100 = 0.5).

Understanding the Question

We now know the volume of solution A in one ant ($7.5 \times 10^{-3} \text{ cm}^3$ from (a)(i)). We must find the volume of pure methanoic acid within it, given that solution A is 50% by volume acid.

Approach

Multiply the volume of solution A by 50/100.

Step-by-Step Reasoning

Volume of pure acid $= 7.5 \times 10^{-3} \times \frac{50}{100} = 3.75 \times 10^{-3} \text{ cm}^3$. To two significant figures this is $3.8 \times 10^{-3} \text{ cm}^3$. The mark scheme notes that the unrounded value $3.75 \times 10^{-3}$ is also accepted for the calculation in (a)(iii).

Key Takeaways

Percentage by volume: volume of component = (percentage/100) × total volume.

Common Mistakes

• Dividing by 50 instead of multiplying (getting the fraction upside down).
• Rounding 3.75 to 3.7 or 4.0 instead of 3.8.
• Using the wrong base volume (e.g. the dm³ value instead of the cm³ value).

Things to Be Careful About

Two significant figures are required. The unrounded $3.75 \times 10^{-3}$ is used in (a)(iii) to give a slightly different (still accepted) answer — the mark scheme allows ecf.

Background Concept

To find how many ants are needed to produce 1 dm³ of pure methanoic acid, we divide the target volume by the volume of pure acid per ant. Both volumes must be in the same units: $1 \text{ dm}^3 = 1000 \text{ cm}^3$.

Understanding the Question

We want the number of ants whose combined pure acid equals 1000 cm³ (1 dm³). Each ant contributes $3.8 \times 10^{-3} \text{ cm}^3$ of pure methanoic acid (from (a)(ii)).

Approach

Divide 1000 cm³ by the volume of pure acid per ant.

Step-by-Step Reasoning

Number of ants $= \frac{1000}{3.8 \times 10^{-3}} = 263157.9 \approx 2.6 \times 10^5$. If the unrounded value $3.75 \times 10^{-3}$ is used, the result is $266666.7 \approx 2.7 \times 10^5$; the mark scheme accepts both. The answer is a pure number with no unit.

Key Takeaways

Number of items = total quantity ÷ quantity per item. When dividing by a number in scientific notation, handle the powers of ten carefully.

Common Mistakes

• Dividing the wrong way round (giving a tiny fraction instead of a large number).
• Forgetting to convert 1 dm³ to 1000 cm³.
• Rounding errors in the division.

Things to Be Careful About

The answer must be to two significant figures. The mark scheme explicitly allows the alternative $2.7 \times 10^5$ if the unrounded volume from (a)(ii) is used.

Background Concept

This part combines two percentage calculations. First, only 80% of the solution A present in the ant is injected in a sting. Second, of that injected solution, 50% by volume is pure methanoic acid. The two percentages are applied successively: final volume = initial volume × 0.80 × 0.50.

Understanding the Question

From (a)(i) we know one ant holds $7.5 \times 10^{-3} \text{ cm}^3$ of solution A. A sting injects 80% of this. We need the volume of pure methanoic acid actually injected, i.e. 50% of the injected solution.

Approach

Step 1: find the volume of solution A injected (80% of $7.5 \times 10^{-3}$). Step 2: find the volume of pure acid in that injected volume (50% of the result).

Step-by-Step Reasoning

Volume of solution A injected $= \frac{80}{100} \times 7.5 \times 10^{-3} = 6.0 \times 10^{-3} \text{ cm}^3$.

Volume of pure methanoic acid injected $= \frac{50}{100} \times 6.0 \times 10^{-3} = 3.0 \times 10^{-3} \text{ cm}^3$.

Both values already have two significant figures. The mark scheme awards one mark for each percentage step, with ecf allowed on (a)(i) and on the first step of (b)(i).

Key Takeaways

Successive percentages multiply: final = initial × 0.80 × 0.50. Always identify clearly which quantity each percentage is being applied to.

Common Mistakes

• Applying only one of the two percentages.
• Applying the 50% to the total volume in the ant rather than to the injected volume.
• Rounding prematurely between the two steps.

Things to Be Careful About

Two significant figures throughout. The mark scheme allows ecf on (a)(i) and on the first part of (b)(i), so a slip earlier need not lose all marks.

Background Concept

Density relates mass and volume: $\text{density} = \frac{\text{mass}}{\text{volume}}$, so $\text{mass} = \text{volume} \times \text{density}$. Given the density of pure methanoic acid ($1.2 \text{ g cm}^{-3}$) and the volume of pure acid injected ($3.0 \times 10^{-3} \text{ cm}^3$ from (b)(i)), the mass follows directly.

Understanding the Question

We must convert the volume of pure methanoic acid injected in one sting into a mass, using the density provided in the stem of part (b).

Approach

Multiply the injected volume of pure acid by the density. Check that the units cancel: $\text{cm}^3 \times \text{g cm}^{-3} = \text{g}$.

Step-by-Step Reasoning

Mass $= 3.0 \times 10^{-3} \text{ cm}^3 \times 1.2 \text{ g cm}^{-3} = 3.6 \times 10^{-3} \text{ g}$. The value 3.6 has two significant figures, matching the requirement.

Key Takeaways

mass = volume × density; always check unit cancellation to confirm the physical quantity being calculated.

Common Mistakes

• Dividing volume by density instead of multiplying.
• Using the volume of solution A rather than the volume of pure acid.
• Forgetting the unit (g).

Things to Be Careful About

Two significant figures. The mark scheme allows ecf on (b)(i), so a wrong volume there can still earn this mark if used correctly.

Background Concept

Methanoic acid, $\text{HCO}_2\text{H}$, is the simplest carboxylic acid. Carboxylic acids react with hydrogencarbonates (bicarbonates) such as sodium hydrogencarbonate to give a carboxylate salt, water and carbon dioxide:

$\text{RCOOH} + \text{NaHCO}_3 \rightarrow \text{RCOONa} + \text{H}_2\text{O} + \text{CO}_2$

This is the basis of the 'fizzing' (CO₂ release) when bicarbonates are added to acids, and why sodium hydrogencarbonate is used to treat stings — it neutralises the injected acid.

Understanding the Question

We must write a balanced equation for the reaction between methanoic acid and sodium hydrogencarbonate. No state symbols are demanded, but the equation must balance in all atoms.

Approach

Identify the products: the sodium salt of the acid (sodium methanoate), water and carbon dioxide. Then balance the equation atom by atom.

Step-by-Step Reasoning

The acid is $\text{HCO}_2\text{H}$ and the hydrogencarbonate is $\text{NaHCO}_3$. The salt formed is sodium methanoate, $\text{HCO}_2\text{Na}$ (the H on the -COOH group is replaced by Na). The other products are water and carbon dioxide:

$\text{HCO}_2\text{H} + \text{NaHCO}_3 \rightarrow \text{HCO}_2\text{Na} + \text{H}_2\text{O} + \text{CO}_2$

Check atoms: C: 2 left, 2 right; H: 3 left, 3 right; O: 5 left, 5 right; Na: 1 left, 1 right. The equation is balanced with coefficients all 1.

Key Takeaways

Acid + hydrogencarbonate → salt + water + carbon dioxide. The general pattern applies to all carboxylic acids.

Common Mistakes

• Writing $\text{Na}_2\text{CO}_3$ instead of $\text{NaHCO}_3$.
• Omitting CO₂ or H₂O from the products.
• Writing an unbalanced equation (e.g. forgetting the Na on the salt).

Things to Be Careful About

The salt is sodium methanoate, $\text{HCO}_2\text{Na}$ (also written HCOONa). The equation must balance in all atoms — the mark is awarded for the correctly balanced equation.

Background Concept

This is a reacting-masses (stoichiometry) calculation. The balanced equation from (c)(i) shows a 1:1 mole ratio: 1 mol of $\text{HCO}_2\text{H}$ reacts with 1 mol of $\text{NaHCO}_3$. Using molar masses, we convert the given mass of methanoic acid to the equivalent mass of sodium hydrogencarbonate. Because the ratio is 1:1, the masses scale in proportion to the molar masses: $\text{mass of NaHCO}_3 = \text{mass of HCO}_2\text{H} \times \frac{M_r(\text{NaHCO}_3)}{M_r(\text{HCO}_2\text{H})}$.

Understanding the Question

A bee sting injects $5.4 \times 10^{-3} \text{ g}$ of methanoic acid. We must find the mass of $\text{NaHCO}_3$ needed to neutralise it, i.e. to react completely with it. The stem of part (c) establishes that sodium hydrogencarbonate is the treatment for stings.

Approach

1. Calculate $M_r(\text{HCO}_2\text{H})$ and $M_r(\text{NaHCO}_3)$.
2. Use the 1:1 mole ratio from the balanced equation.
3. Scale the given mass by the ratio of molar masses.

Step-by-Step Reasoning

$M_r(\text{HCO}_2\text{H}) = 12 + 1 + 16 + 16 + 1 = 46$ (C + H + O + O + H).

$M_r(\text{NaHCO}_3) = 23 + 1 + 12 + (3 \times 16) = 84$ (Na + H + C + 3O).

The equation is 1:1, so $46 \text{ g HCO}_2\text{H} \equiv 84 \text{ g NaHCO}_3$. Therefore:

$5.4 \times 10^{-3} \text{ g HCO}_2\text{H} \equiv \frac{84 \times 5.4 \times 10^{-3}}{46} = 9.86 \times 10^{-3} \text{ g NaHCO}_3$

To two significant figures: $9.9 \times 10^{-3} \text{ g}$.

Key Takeaways

Reacting masses scale with molar masses in the mole ratio given by the balanced equation. When the ratio is 1:1, mass scales directly as $M_r$(product)/$M_r$(reactant).

Common Mistakes

• Using the wrong molar mass (e.g. forgetting one oxygen in either compound).
• Using a wrong mole ratio (the equation is 1:1).
• Rounding to the wrong number of significant figures (e.g. giving 9.86 rather than 9.9).

Things to Be Careful About

Two significant figures: $9.9 \times 10^{-3} \text{ g}$. The mark scheme credits the molar-mass equivalence (46 ≡ 84) and the final calculation separately, so show both steps clearly.