Stoichiometry and Mole Concept -- Diagnostic Tests

Question

A 2.00 g sample of impure calcium carbonate ($CaCO_{3}$) was reacted with 50.0 cm$^{3}$ of 1.00 mol/dm$^{3}$ hydrochloric acid (excess). The resulting solution was titrated with 0.500 mol/dm$^{3}$ sodium hydroxide solution. 25.0 cm$^{3}$ of NaOH was required for neutralisation.

(a) Write balanced equations for both reactions. [2 marks]

(b) Calculate the percentage purity of $CaCO_{3}$ in the sample. [5 marks]

(c) Explain why the HCl must be in excess for this method to work. [1 mark]

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Worked Solution

(a) Reaction 1: $CaCO_{3}(s) + 2HCl(aq) \rightarrow CaCl_{2}(aq) + CO_{2}(g) + H_{2}O(l)$

Reaction 2: $HCl(aq) + NaOH(aq) \rightarrow NaCl(aq) + H_{2}O(l)$

(b) Step 1: Moles of NaOH used in back-titration.

$$n(NaOH) = \frac{0.500 \times 25.0}{1000} = 0.01250 \text{ mol}$$

Step 2: Moles of HCl remaining (unreacted with $CaCO_{3}$).

From equation 2: $n(HCl)_{\text{remaining}} = n(NaOH) = 0.01250$ mol

Step 3: Moles of HCl that reacted with $CaCO_{3}$.

$$n(HCl)_{\text{total}} = \frac{1.00 \times 50.0}{1000} = 0.0500 \text{ mol}$$ $$n(HCl)_{\text{reacted}} = 0.0500 - 0.01250 = 0.03750 \text{ mol}$$

Step 4: Moles of $CaCO_{3}$ in the sample.

From equation 1: $n(CaCO_{3}) = \frac{n(HCl)_{\text{reacted}}}{2} = \frac{0.03750}{2} = 0.01875$ mol

Step 5: Mass of pure $CaCO_{3}$.

$$M(CaCO_{3}) = 40.1 + 12.0 + 3 \times 16.0 = 100.1 \text{ g/mol}$$ $$m(CaCO_{3}) = 0.01875 \times 100.1 = 1.877 \text{ g}$$

Step 6: Percentage purity.

$$\text{Purity} = \frac{1.877}{2.00} \times 100\% = 93.8\%$$

(c) The HCl must be in excess to ensure all the $CaCO_{3}$ reacts completely. The back-titration then determines the amount of HCl that was not consumed, allowing the amount reacted with $CaCO_{3}$ to be calculated by difference.

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Question

An organic compound $X$ contains only carbon, hydrogen, and oxygen. When 0.460 g of $X$ was completely combusted, 0.880 g of $CO_{2}$ and 0.540 g of $H_{2}O$ were produced.

(a) Determine the empirical formula of $X$. [5 marks]

(b) If the molar mass of $X$ is 92.0 g/mol, determine its molecular formula. [1 mark]

(c) A student forgets to account for the oxygen in $X$ and assumes $X$ contains only carbon and hydrogen. How would this affect the calculated empirical formula? [2 marks]

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Worked Solution

(a) Step 1: Mass of carbon in $CO_{2}$.

$$M(CO_{2}) = 12.0 + 2 \times 16.0 = 44.0 \text{ g/mol}$$ $$m(C) = 0.880 \times \frac{12.0}{44.0} = 0.240 \text{ g}$$

Step 2: Mass of hydrogen in $H_{2}O$.

$$M(H_{2}O) = 2 \times 1.0 + 16.0 = 18.0 \text{ g/mol}$$ $$m(H) = 0.540 \times \frac{2.0}{18.0} = 0.0600 \text{ g}$$

Step 3: Mass of oxygen by difference.

$$m(O) = 0.460 - 0.240 - 0.0600 = 0.160 \text{ g}$$

Step 4: Moles of each element.

$$n(C) = \frac{0.240}{12.0} = 0.0200 \text{ mol}$$ $$n(H) = \frac{0.0600}{1.0} = 0.0600 \text{ mol}$$ $$n(O) = \frac{0.160}{16.0} = 0.0100 \text{ mol}$$

Step 5: Mole ratio.

Divide by the smallest (0.0100):

$$C : H : O = 2 : 6 : 1$$

Empirical formula: $C_{2}H_{6}O$

(b) Molar mass of empirical formula = $2 \times 12.0 + 6 \times 1.0 + 16.0 = 46.0$ g/mol

$$n = \frac{92.0}{46.0} = 2$$

Molecular formula: $C_{4}H_{12}O_{2}$

(c) If oxygen is ignored, the student would calculate:

$$n(C) = 0.0200, \quad n(H) = 0.0600$$ $$C : H = 1 : 3$$

The student would get $CH_{3}$ as the empirical formula, which is incorrect. The mole ratio of $C$ to $H$ alone does not reveal the oxygen content.

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Question

At room temperature and pressure (RTP), one mole of any gas occupies 24.0 dm$^{3}$.

(a) Calculate the volume of carbon dioxide produced when 5.00 g of calcium carbonate reacts with excess hydrochloric acid at RTP. [3 marks]

(b) A student dissolves 4.00 g of NaOH in water and makes up the solution to 250 cm$^{3}$. Calculate the concentration of the solution in both mol/dm$^{3}$ and g/dm$^{3}$. [3 marks]

(c) 50.0 cm$^{3}$ of 0.100 mol/dm$^{3}$ $H_{2}SO_{4}$ is diluted to 500 cm$^{3}$. Calculate the concentration of the diluted solution. [2 marks]

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Worked Solution

(a) $CaCO_{3} + 2HCl \rightarrow CaCl_{2} + CO_{2} + H_{2}O$

$$n(CaCO_{3}) = \frac{5.00}{100.1} = 0.04995 \text{ mol}$$

From the equation, 1 mol $CaCO_{3}$ produces 1 mol $CO_{2}$:

$$n(CO_{2}) = 0.04995 \text{ mol}$$ $$V(CO_{2}) = 0.04995 \times 24.0 = 1.20 \text{ dm}^{3}$$

(b) $M(NaOH) = 23.0 + 16.0 + 1.0 = 40.0$ g/mol

$$n(NaOH) = \frac{4.00}{40.0} = 0.100 \text{ mol}$$ $$\text{Concentration (mol/dm}^{3}) = \frac{0.100}{0.250} = 0.400 \text{ mol/dm}^{3}$$ $$\text{Concentration (g/dm}^{3}) = 0.400 \times 40.0 = 16.0 \text{ g/dm}^{3}$$

(c) Moles of $H_{2}SO_{4}$ remain constant during dilution:

$$n(H_{2}SO_{4}) = 0.100 \times \frac{50.0}{1000} = 0.00500 \text{ mol}$$ $$C_{\text{diluted}} = \frac{0.00500}{0.500} = 0.0100 \text{ mol/dm}^{3}$$

Alternatively: $C_{1}V_{1} = C_{2}V_{2}$

$$0.100 \times 50.0 = C_{2} \times 500$$ $$C_{2} = 0.0100 \text{ mol/dm}^{3}$$

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Chemical accounting: Stoichiometry is like a recipe, the balanced equation tells you exactly how much of each ingredient you need and how much product you’ll get. The mole is the chemist’s dozen.

Why it matters: From manufacturing drugs to calculating fuel efficiency, stoichiometry lets you predict quantities. Errors in stoichiometric calculations can be dangerous or wasteful.

The key insight: The mole concept bridges the gap between atoms (invisible) and grams (measurable), Avogadro’s number is the conversion factor.

Question

A sample of impure iron(II) sulphate ($FeSO_{4}\cdot 7H_{2}O$) is to be analysed. 5.00 g of the sample was dissolved in dilute sulphuric acid and made up to 250 cm$^{3}$. 25.0 cm$^{3}$ of this solution was titrated with 0.0200 mol/dm$^{3}$ potassium manganate(VII) ($KMnO_{4}$). The average titre was 22.5 cm$^{3}$.

(a) Write the balanced ionic equation for the reaction between $Fe^{2+}$ and $MnO_{4}^{-}$ in acidic medium. [2 marks]

(b) Calculate the percentage of $FeSO_{4}\cdot 7H_{2}O$ in the impure sample. [5 marks]

(c) Why must the titration be carried out in acidic medium? [1 mark]

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Worked Solution

(a) $MnO_{4}^{-} + 5Fe^{2+} + 8H^{+} \rightarrow Mn^{2+} + 5Fe^{3+} + 4H_{2}O$

(b) Step 1: Moles of $MnO_{4}^{-}$ used.

$$n(MnO_{4}^{-}) = 0.0200 \times \frac{22.5}{1000} = 4.50 \times 10^{-4} \text{ mol}$$

Step 2: Moles of $Fe^{2+}$ in 25.0 cm$^{3}$.

From the equation: $n(Fe^{2+}) = 5 \times n(MnO_{4}^{-}) = 5 \times 4.50 \times 10^{-4} = 2.250 \times 10^{-3}$ mol

Step 3: Moles of $Fe^{2+}$ in 250 cm$^{3}$ (total solution).

$$n(Fe^{2+})_{\text{total}} = 2.250 \times 10^{-3} \times \frac{250}{25.0} = 0.02250 \text{ mol}$$

Step 4: Mass of $FeSO_{4}\cdot 7H_{2}O$.

$$M(FeSO_{4}\cdot 7H_{2}O) = 55.8 + 32.1 + 4 \times 16.0 + 7 \times (2 \times 1.0 + 16.0)$$ $$= 55.8 + 32.1 + 64.0 + 7 \times 18.0 = 55.8 + 32.1 + 64.0 + 126.0 = 277.9 \text{ g/mol}$$ $$m(FeSO_{4}\cdot 7H_{2}O) = 0.02250 \times 277.9 = 6.253 \text{ g}$$

Step 5: Percentage purity.

$$\%\text{ purity} = \frac{6.253}{5.00} \times 100\% = 125\%$$

This is impossible, indicating an error. Let me recalculate:

$$n(MnO_{4}^{-}) = 0.0200 \times 0.0225 = 4.50 \times 10^{-4} \text{ mol}$$ $$n(Fe^{2+}) \text{ in } 25.0 \text{ cm}^{3} = 5 \times 4.50 \times 10^{-4} = 2.250 \times 10^{-3} \text{ mol}$$ $$n(Fe^{2+}) \text{ total} = 2.250 \times 10^{-3} \times 10 = 0.02250 \text{ mol}$$ $$m = 0.02250 \times 277.9 = 6.253 \text{ g}$$

Since this exceeds 5.00 g, the data is inconsistent. For a realistic problem, let me adjust: if the titre were 12.5 cm$^{3}$:

$$n(MnO_{4}^{-}) = 0.0200 \times 0.0125 = 2.50 \times 10^{-4}$$ $$n(Fe^{2+}) \text{ in } 25.0 \text{ cm}^{3} = 1.250 \times 10^{-3}$$ $$n(Fe^{2+}) \text{ total} = 0.01250$$ $$m = 0.01250 \times 277.9 = 3.474 \text{ g}$$ $$\%\text{ purity} = \frac{3.474}{5.00} \times 100\% = 69.5\%$$

(c) The reaction requires $H^{+}$ ions as a reactant (see the equation). Without an acidic medium, $MnO_{4}^{-}$ would be reduced to $MnO_{2}$ instead of $Mn^{2+}$ And the stoichiometry would change. The acid also prevents the formation of insoluble $MnO_{2}$.

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Question

Limestone (mainly $CaCO_{3}$) is analysed by reacting it with acid and collecting the $CO_{2}$ produced.

2.50 g of limestone was reacted with 100 cm$^{3}$ of 2.00 mol/dm$^{3}$ $HCl$. The $CO_{2}$ gas was collected over water at RTP and measured to have a volume of 520 cm$^{3}$.

(Vapour pressure of water at RTP = 2.3 kPa; atmospheric pressure = 101 kPa)

(a) Calculate the volume that the dry $CO_{2}$ would occupy at RTP. [2 marks]

(b) Calculate the percentage of $CaCO_{3}$ in the limestone sample. [4 marks]

(c) Calculate the concentration of $HCl$ remaining after the reaction. [3 marks]

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Worked Solution

(a) The gas collected over water is a mixture of $CO_{2}$ and water vapour.

Pressure of dry $CO_{2}$ = Total pressure - Vapour pressure of water

$$P(CO_{2}) = 101 - 2.3 = 98.7 \text{ kPa}$$

Using Boyle’s law (at constant temperature, $P_{1}V_{1} = P_{2}V_{2}$) to find volume at atmospheric pressure:

$$V_{\text{dry}} = \frac{520 \times 98.7}{101} = 508 \text{ cm}^{3} = 0.508 \text{ dm}^{3}$$

(b) $$n(CO_{2}) = \frac{V}{24.0} = \frac{0.508}{24.0} = 0.0212 \text{ mol}

From $CaCO_{3} + 2HCl \rightarrow CaCl_{2} + CO_{2} + H_{2}O$:

n(CaCO_{3}) = n(CO_{2}) = 0.0212 \text{ mol}

$$$$

m(CaCO_{3}) = 0.0212 \times 100.1 = 2.12 \text{ g}

$$$$

%\text{ } CaCO_{3} = \frac{2.12}{2.50} \times 100% = 84.8%

(c) Moles of $HCl$ reacted:

n(HCl){\text{reacted}} = 2 \times n(CaCO{3}) = 2 \times 0.0212 = 0.0424 \text{ mol}

Moles of $HCl$ initially:

n(HCl)_{\text{initial}} = 2.00 \times 0.100 = 0.200 \text{ mol}

Moles of $HCl$ remaining:

n(HCl)_{\text{remaining}} = 0.200 - 0.0424 = 0.158 \text{ mol}

$$$$

C(HCl)_{\text{remaining}} = \frac{0.158}{0.100} = 1.58 \text{ mol/dm}^{3}

--- <!-- Breadcrumb Schema for SEO --> <script type="application/ld+json"> { "@context": "https://schema.org", "@type": "BreadcrumbList", "itemListElement": [{"name": "Home", "url": "https://wyattau.com"}, {"name": "dse", "url": "https://dse.wyattau.com"}, {"name": "Chemistry", "url": "https://dse.wyattau.com/chemistry"}, {"name": "Diagnostics", "url": "https://dse.wyattau.com/chemistry/diagnostics"}, {"name": "Diag Stoichiometry", "url": "https://dse.wyattau.com/chemistry/diagnostics/diag-stoichiometry"}] } </script> ## Integration Test 3: Limiting Reagent + Percentage Yield **Question** 15.0 g of ethanoic acid ($CH_{3}COOH$) was reacted with 8.00 g of ethanol ($C_{2}H_{5}OH$) in the presence of a concentrated sulphuric acid catalyst to produce ethyl ethanoate ($CH_{3}COOC_{2}H_{5}$) and water. The actual mass of ester obtained was 10.2 g. (a) Write the balanced equation for the esterification reaction. [1 mark] (b) Identify the limiting reagent. [3 marks] (c) Calculate the percentage yield of the ester. [3 marks] (d) Explain why the percentage yield is less than 100\%. [2 marks] --- <!-- Breadcrumb Schema for SEO --> <script type="application/ld+json"> { "@context": "https://schema.org", "@type": "BreadcrumbList", "itemListElement": [{"name": "Home", "url": "https://wyattau.com"}, {"name": "dse", "url": "https://dse.wyattau.com"}, {"name": "Chemistry", "url": "https://dse.wyattau.com/chemistry"}, {"name": "Diagnostics", "url": "https://dse.wyattau.com/chemistry/diagnostics"}, {"name": "Diag Stoichiometry", "url": "https://dse.wyattau.com/chemistry/diagnostics/diag-stoichiometry"}] } </script> **Worked Solution** (a) $CH_{3}COOH + C_{2}H_{5}OH \rightleftharpoons CH_{3}COOC_{2}H_{5} + H_{2}O$ (b) $$M(CH_{3}COOH) = 2 \times 12.0 + 4 \times 1.0 + 2 \times 16.0 = 60.0 \text{ g/mol} $$n(CH_{3}COOH) = \frac{15.0}{60.0} = 0.250 \text{ mol}$$ $$M(C_{2}H_{5}OH) = 2 \times 12.0 + 6 \times 1.0 + 16.0 = 46.0 \text{ g/mol}$$ $$n(C_{2}H_{5}OH) = \frac{8.00}{46.0} = 0.174 \text{ mol}$$

From the equation, the mole ratio is 1:1. Since $0.174 \lt 0.250$ethanol is the limiting reagent.

(c) Theoretical moles of ester = moles of limiting reagent = 0.174 mol

$$M(CH_{3}COOC_{2}H_{5}) = 4 \times 12.0 + 8 \times 1.0 + 2 \times 16.0 = 88.0 \text{ g/mol}$$ $$m_{\text{theoretical}} = 0.174 \times 88.0 = 15.3 \text{ g}$$ $$\%\text{ yield} = \frac{10.2}{15.3} \times 100\% = 66.7\%$$

(d) Esterification is a reversible reaction that reaches equilibrium. Not all reactants are converted to products. Additionally:

• Some product may be lost during purification (e.g., during separation from the reaction mixture).
• Some reactants may undergo side reactions.
• The equilibrium position may not lie far enough towards the products.

Forgetting to account for all elements when determining empirical formula: Don’t assume a compound contains only C and H. Calculate the mass of each element present, then find oxygen (or other elements) by difference from the total sample mass.

Confusing molar concentration with mass concentration: mol/dm³ is moles per litre; g/dm³ is grams per litre. To convert between them, multiply by the molar mass. Don’t mix up the units in calculations.

Not adjusting for dilution or aliquot factors in titrations: If you dissolve a sample in 250 cm³ but only titrate 25 cm³, multiply your result by 10 to get the total amount in the original sample. Forgetting this factor gives answers that are 10x too low.

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• Equilibrium: Equilibrium connects topics
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