Quantitative Chemistry -- Diagnostic Tests

Chemistry is the science of change — how atoms combine, react, and transform into new substances.

Question:

A sample of impure calcium carbonate ($\text{CaCO}_3$) weighing $2.50\,\text{g}$ is reacted with exactly $50.0\,\text{cm}^3$ of $1.00\,\text{mol dm}^{-3}$ hydrochloric acid (an excess). The reaction is:

$\text{CaCO}_3(s) + 2\text{HCl}(aq) \to \text{CaCl}_2(aq) + \text{CO}_2(g) + \text{H}_2\text{O}(l)$

After the reaction is complete, the excess HCl is titrated with $0.500\,\text{mol dm}^{-3}$ sodium hydroxide solution. $24.0\,\text{cm}^3$ of NaOH is required for complete neutralisation.

(a) Calculate the percentage purity of the calcium carbonate sample.

(b) Calculate the volume of $\text{CO}_2$ produced at room temperature and pressure ($24.0\,\text{dm}^3\,\text{mol}^{-1}$).

Solution:

(a) Step 1: Moles of NaOH used in the titration

$n(\text{NaOH}) = \frac{0.500 \times 24.0}{1000} = 0.0120\,\text{mol}$

Step 2: Moles of excess HCl

$\text{NaOH} + \text{HCl} \to \text{NaCl} + \text{H}_2\text{O}$ (1:1 ratio)

$n(\text{HCl}_{\text{excess}}) = 0.0120\,\text{mol}$

Step 3: Moles of HCl that reacted with $\text{CaCO}_3$

$n(\text{HCl}_{\text{total}}) = \frac{1.00 \times 50.0}{1000} = 0.0500\,\text{mol}$

$n(\text{HCl}_{\text{reacted}}) = 0.0500 - 0.0120 = 0.0380\,\text{mol}$

Step 4: Moles of $\text{CaCO}_3$ in the sample

From the equation, 2 mol HCl react with 1 mol $\text{CaCO}_3$:

$n(\text{CaCO}_3) = \frac{0.0380}{2} = 0.0190\,\text{mol}$

Step 5: Mass of pure $\text{CaCO}_3$

$m(\text{CaCO}_3) = 0.0190 \times (40.1 + 12.0 + 3 \times 16.0) = 0.0190 \times 100.1 = 1.902\,\text{g}$

Step 6: Percentage purity

$\text{Percentage purity} = \frac{1.902}{2.50} \times 100 = 76.1\%$

(b) From the equation, 1 mol $\text{CaCO}_3$ produces 1 mol $\text{CO}_2$:

$n(\text{CO}_2) = 0.0190\,\text{mol}$

$V(\text{CO}_2) = 0.0190 \times 24.0 = 0.456\,\text{dm}^3 = 456\,\text{cm}^3$

Question:

A student collects $0.154\,\text{g}$ of an unknown gas in a gas syringe at a temperature of $77\,^\circ\text{C}$ and a pressure of $98.5\,\text{kPa}$. The volume of gas collected is $72.0\,\text{cm}^3$.

(a) Calculate the molar mass of the gas. ($R = 8.31\,\text{J K}^{-1}\text{ mol}^{-1}$)

(b) The gas is known to be one of: \text{N}_2$$\text{O}_2$$\text{CO}$$\text{NO}_2Or $\text{C}_3\text{H}_8$. Identify the gas.

Solution:

(a) Using $pV = nRT$Ensuring consistent SI units:

• $p = 98.5\,\text{kPa} = 98500\,\text{Pa}$
• $V = 72.0\,\text{cm}^3 = 72.0 \times 10^{-6}\,\text{m}^3 = 7.20 \times 10^{-5}\,\text{m}^3$
• $T = 77 + 273 = 350\,\text{K}$

$n = \frac{pV}{RT} = \frac{98500 \times 7.20 \times 10^{-5}}{8.31 \times 350} = \frac{7.092}{2908.5} = 2.438 \times 10^{-3}\,\text{mol}$

$M = \frac{m}{n} = \frac{0.154}{2.438 \times 10^{-3}} = 63.2\,\text{g mol}^{-1}$

(b) Calculated molar masses:

• $\text{N}_2 = 28.0\,\text{g mol}^{-1}$
• $\text{O}_2 = 32.0\,\text{g mol}^{-1}$
• $\text{CO} = 28.0\,\text{g mol}^{-1}$
• $\text{NO}_2 = 46.0\,\text{g mol}^{-1}$
• $\text{C}_3\text{H}_8 = 44.0\,\text{g mol}^{-1}$

None of these exactly match $63.2\,\text{g mol}^{-1}$. However, note that $\text{NO}_2$ dimerises: $2\text{NO}_2 \rightleftharpoons \text{N}_2\text{O}_4$. The molar mass of $\text{N}_2\text{O}_4$ is $92.0\,\text{g mol}^{-1}$. The experimental value of $63.2$ is between $46.0$ ($\text{NO}_2$) and $92.0$ ($\text{N}_2\text{O}_4$), consistent with an equilibrium mixture of $\text{NO}_2$ and $\text{N}_2\text{O}_4$ at this temperature. The gas is $\text{NO}_2$ (existing as an equilibrium mixture with its dimer).

Question:

Ethanol can be produced from ethene via the following two-step process:

Step 1: $\text{C}_2\text{H}_4 + \text{H}_2\text{SO}_4 \to \text{C}_2\text{H}_5\text{HSO}_4$ (yield = 95%)

Step 2: $\text{C}_2\text{H}_5\text{HSO}_4 + \text{H}_2\text{O} \to \text{C}_2\text{H}_5\text{OH} + \text{H}_2\text{SO}_4$ (yield = 90%)

(a) Calculate the overall percentage yield of ethanol from ethene.

(b) Calculate the atom economy of Step 2.

(c) Starting with $28.0\,\text{kg}$ of ethene, calculate the mass of ethanol actually produced.

Solution:

(a) The overall yield for consecutive steps is the product of the individual yields:

$\text{Overall yield} = 0.95 \times 0.90 = 0.855 = 85.5\%$

(b) Atom economy = \frac{\text{M_r of desired product}}{\text{Sum of M_r of all products}} \times 100

Desired product: $\text{C}_2\text{H}_5\text{OH}$ ($M_r = 2 \times 12.0 + 6 \times 1.0 + 16.0 = 46.0$)

Other product: $\text{H}_2\text{SO}_4$ ($M_r = 2 \times 1.0 + 32.1 + 4 \times 16.0 = 98.1$)

$\text{Atom economy} = \frac{46.0}{46.0 + 98.1} \times 100 = \frac{46.0}{144.1} \times 100 = 31.9\%$

Note: The low atom economy is because $\text{H}_2\text{SO}_4$ is regenerated (it appears on both sides), so in practice the $\text{H}_2\text{SO}_4$ acts as a catalyst and is recycled, making the effective atom economy much higher in industrial processes.

(c) M_r(\text{C}_2\text{H}_4) = 28.0$$M_r(\text{C}_2\text{H}_5\text{OH}) = 46.0

Theoretical mass of ethanol from $28.0\,\text{kg}$ ethene (1:1 stoichiometry):

$m_{\text{theoretical}} = 28.0 \times \frac{46.0}{28.0} = 46.0\,\text{kg}$

Actual mass:

$m_{\text{actual}} = 46.0 \times 0.855 = 39.3\,\text{kg}$

Question:

$0.480\,\text{g}$ of a hydrocarbon (containing only carbon and hydrogen) is completely burned in oxygen. The products are passed through concentrated $\text{H}_2\text{SO}_4$Which increases in mass by $0.720\,\text{g}$And then through limewater ($\text{Ca(OH)}_2$ solution), which produces $2.20\,\text{g}$ of white precipitate ($\text{CaCO}_3$).

(a) Calculate the empirical formula of the hydrocarbon.

(b) $0.120\,\text{g}$ of the hydrocarbon vapour occupies $49.5\,\text{cm}^3$ at $100\,^\circ\text{C}$ and $1.01 \times 10^5\,\text{Pa}$. Determine the molecular formula of the hydrocarbon.

(c) Draw two possible structural isomers consistent with this molecular formula and identify which is the more stable isomer, explaining your reasoning.

Solution:

(a) All hydrogen in the hydrocarbon becomes $\text{H}_2\text{O}$Absorbed by $\text{H}_2\text{SO}_4$:

$m(\text{H}_2\text{O}) = 0.720\,\text{g}$ $n(\text{H}_2\text{O}) = \frac{0.720}{18.0} = 0.0400\,\text{mol}$ $n(\text{H}) = 2 \times 0.0400 = 0.0800\,\text{mol}$

All carbon in the hydrocarbon becomes $\text{CO}_2$Which reacts with limewater:

$\text{Ca(OH)}_2 + \text{CO}_2 \to \text{CaCO}_3 + \text{H}_2\text{O}$

$n(\text{CaCO}_3) = \frac{2.20}{100.1} = 0.0220\,\text{mol}$ $n(\text{C}) = n(\text{CaCO}_3) = 0.0220\,\text{mol}$

Ratio C : H = $0.0220 : 0.0800 = 1 : 3.64$.

Multiplying to find the simplest integer ratio: $0.0220 : 0.0800 = 5 : 18.2$. The ratio is approximately $5 : 18$Giving an empirical formula of $\text{C}_5\text{H}_{18}$. However, this is not a standard hydrocarbon formula (alkanes follow $\text{C}_n\text{H}_{2n+2}$So $\text{C}_5\text{H}_{12}$ would be the alkane with 5 carbons). The discrepancy is within expected experimental rounding error.

(b) Using $pV = nRT$:

$n = \frac{1.01 \times 10^5 \times 49.5 \times 10^{-6}}{8.31 \times 373} = \frac{5.000}{3099.6} = 1.613 \times 10^{-3}\,\text{mol}$

$M = \frac{0.120}{1.613 \times 10^{-3}} = 74.4\,\text{g mol}^{-1}$

The measured molar mass of $74.4\,\text{g mol}^{-1}$ is closest to $\text{C}_5\text{H}_{12}$ (pentane, $M_r = 72.1\,\text{g mol}^{-1}$). Comparing with nearby hydrocarbons: \text{C}_4\text{H}_{10} = 58$$\text{C}_5\text{H}_{12} = 72$$\text{C}_6\text{H}_{14} = 86. The molecular formula is $\text{C}_5\text{H}_{12}$ (pentane). The small discrepancy ($74.4$ vs $72.1$) arises from rounding in the experimental data.

(c) Two structural isomers of $\text{C}_5\text{H}_{12}$:

• Pentane ($\text{CH}_3\text{CH}_2\text{CH}_2\text{CH}_2\text{CH}_3$): straight chain
• 2-methylbutane ($\text{CH}_3\text{CH}(\text{CH}_3)\text{CH}_2\text{CH}_3$): branched

Pentane is the straight-chain isomer and has a higher boiling point due to greater surface area for van der Waals interactions. 2-methylbutane is more branched and has a slightly lower boiling point. Both are stable alkanes; pentane is the least sterically hindered.

Question:

$5.00\,\text{g}$ of hydrated barium chloride ($\text{BaCl}_2 \cdot x\text{H}_2\text{O}$) is heated strongly until all the water of crystallisation is removed, leaving $4.26\,\text{g}$ of anhydrous $\text{BaCl}_2$.

(a) Determine the value of $x$ in the formula $\text{BaCl}_2 \cdot x\text{H}_2\text{O}$.

(b) The enthalpy of solution of anhydrous $\text{BaCl}_2$ is $-13.2\,\text{kJ mol}^{-1}$While the enthalpy of solution of $\text{BaCl}_2 \cdot 2\text{H}_2\text{O}$ is $+8.8\,\text{kJ mol}^{-1}$. Use a Hess”s law cycle to calculate the enthalpy change for the reaction:

$\text{BaCl}_2(s) + 2\text{H}_2\text{O}(l) \to \text{BaCl}_2 \cdot 2\text{H}_2\text{O}(s)$

Solution:

(a) Mass of water lost: $5.00 - 4.26 = 0.74\,\text{g}$

$n(\text{H}_2\text{O}) = \frac{0.74}{18.0} = 0.0411\,\text{mol}$

$n(\text{BaCl}_2) = \frac{4.26}{137.3 + 2 \times 35.5} = \frac{4.26}{208.3} = 0.02045\,\text{mol}$

$x = \frac{n(\text{H}_2\text{O})}{n(\text{BaCl}_2)} = \frac{0.0411}{0.02045} = 2.01 \approx 2$

The formula is $\text{BaCl}_2 \cdot 2\text{H}_2\text{O}$.

(b) Hess’s law cycle:

$\text{BaCl}_2(s) + 2\text{H}_2\text{O}(l) \xrightarrow{\Delta H} \text{BaCl}_2 \cdot 2\text{H}_2\text{O}(s)$

Route 1 (direct): $\Delta H = ?$

Route 2 (via aqueous solution):

• $\text{BaCl}_2(s) \to \text{BaCl}_2(aq)$: $\Delta H_1 = -13.2\,\text{kJ mol}^{-1}$
• $\text{BaCl}_2 \cdot 2\text{H}_2\text{O}(s) \to \text{BaCl}_2(aq) + 2\text{H}_2\text{O}(l)$: $\Delta H_2 = +8.8\,\text{kJ mol}^{-1}$

Route 2 gives: $\text{BaCl}_2(s) + 2\text{H}_2\text{O}(l) \to \text{BaCl}_2(aq)$ (reverse of the second step, so $-\Delta H_2$)

$\Delta H = \Delta H_1 + (-\Delta H_2) = -13.2 - 8.8 = -22.0\,\text{kJ mol}^{-1}$

The hydration of $\text{BaCl}_2$ to form the dihydrate releases $22.0\,\text{kJ mol}^{-1}$.

Question:

$25.0\,\text{cm}^3$ of a solution containing both $\text{HCl}$ and $\text{CH}_3\text{COOH}$ is titrated with $0.100\,\text{mol dm}^{-3}$ $\text{NaOH}$ using phenolphthalein indicator. $20.0\,\text{cm}^3$ of $\text{NaOH}$ is required to reach the end point. In a separate experiment, $25.0\,\text{cm}^3$ of the same solution is titrated with $0.100\,\text{mol dm}^{-3}$ $\text{NaOH}$ but using a pH meter. The pH curve shows the first equivalence point at $10.0\,\text{cm}^3$ and the second at $20.0\,\text{cm}^3$.

(a) Calculate the concentration of $\text{HCl}$ in the original solution.

(b) Calculate the concentration of $\text{CH}_3\text{COOH}$ in the original solution.

(c) Explain why the pH at the second equivalence point is greater than 7, and explain the significance of the first equivalence point.

Solution:

(a) The first equivalence point at $10.0\,\text{cm}^3$ corresponds to neutralisation of the strong acid $\text{HCl}$ (since $\text{HCl}$ is neutralised first as it fully dissociates, and the $\text{pH}$ rises steeply first around this point due to the strong acid being consumed before the weak acid):

$n(\text{HCl}) = n(\text{NaOH})_{\text{first eq. pt.}} = 0.100 \times \frac{10.0}{1000} = 1.00 \times 10^{-3}\,\text{mol}$

$[\text{HCl}] = \frac{1.00 \times 10^{-3}}{25.0/1000} = 0.0400\,\text{mol dm}^{-3}$

(b) The second equivalence point at $20.0\,\text{cm}^3$ corresponds to total acid neutralised. The additional $\text{NaOH}$ between the two equivalence points neutralises the $\text{CH}_3\text{COOH}$:

$n(\text{CH}_3\text{COOH}) = 0.100 \times \frac{20.0 - 10.0}{1000} = 1.00 \times 10^{-3}\,\text{mol}$

$[\text{CH}_3\text{COOH}] = \frac{1.00 \times 10^{-3}}{25.0/1000} = 0.0400\,\text{mol dm}^{-3}$

(c) At the second equivalence point, all $\text{HCl}$ and $\text{CH}_3\text{COOH}$ have been neutralised. The solution contains $\text{NaCl}$ (neutral from the strong acid-strong base reaction) and $\text{CH}_3\text{COONa}$ (a salt of a weak acid and strong base). The acetate ion ($\text{CH}_3\text{COO}^-$) is a weak base that hydrolyses in water:

$\text{CH}_3\text{COO}^-(aq) + \text{H}_2\text{O}(l) \rightleftharpoons \text{CH}_3\text{COOH}(aq) + \text{OH}^-(aq)$

This produces $\text{OH}^-$ ions, making the solution alkaline ($\text{pH} \gt 7$).

The first equivalence point (at $10.0\,\text{cm}^3$) represents the point where all $\text{HCl}$ has been neutralised but the $\text{CH}_3\text{COOH}$ has not yet been titrated. The solution at this point contains $\text{CH}_3\text{COOH}$ and $\text{NaCl}$So the pH is determined by the weak acid alone ($\text{pH}$ approximately equal to $\frac{1}{2}\text{p}K_a - \frac{1}{2}\log[\text{HA}]$).

Question: In a three-step synthesis, the percentage yields are 85%, 72%, and 90% respectively. If $10.0\,\mathrm{g}$ of starting material is used, calculate the mass of final product obtained.

Solution:

Overall yield $= 0.85 \times 0.72 \times 0.90 = 0.551 = 55.1\%$ (1 mark).

Mass of final product $= 10.0 \times 0.551 = 5.51\,\mathrm{g}$ (assuming 1:1 molar ratio in each step; if the molar ratios differ, the calculation must account for the molar mass changes at each step) (1 mark).

This illustrates the importance of high yields in each step of a multi-step synthesis. Even with relatively good individual yields (72—90%), the overall yield drops to 55%, meaning nearly half the starting material is lost.

Question: A student calculates the volume of gas produced using $V = nRT/p$ with n = 0.050\,\mathrm{mol}$$R = 8.314$$T = 298And $p = 100$. They obtain $V = 123.7$. Identify the error and give the correct answer in $\mathrm{cm}^3$.

Solution:

The student used $p = 100$ without units. If they intended $100\,\mathrm{kPa}$They needed to convert to pascals: $p = 100000\,\mathrm{Pa}$ (1 mark).

$V = \frac{nRT}{p} = \frac{0.050 \times 8.314 \times 298}{100000} = \frac{123.9}{100000} = 1.24 \times 10^{-3}\,\mathrm{m}^3 = 1.24\,\mathrm{dm}^3 = 1240\,\mathrm{cm}^3$

The student’s answer of $123.7$ is actually correct numerically but lacks units. If they meant $\mathrm{dm}^3$Their answer is close. The key error was likely not tracking units through the calculation (1 mark).

Using the wrong molar mass in empirical formula calculations: The empirical formula is the simplest whole-number ratio of moles of each element. Always use the correct atomic masses from the periodic table and divide all mole values by the smallest to get the ratio. Students often forget to multiply to get whole numbers when the ratio is fractional (e.g., 1 : 1.5 becomes 2 : 3).

Forgetting to convert between units in gas calculations: The ideal gas equation $pV = nRT$ requires consistent SI units: pressure in Pa (not kPa or atm), volume in m$^3$ (not dm$^3$ or cm$^3$). A common error is using $p = 100$ when it should be $p = 100\,000$ Pa, or forgetting that $1\,\text{m}^3 = 1000\,\text{dm}^3$.

Confusing atom economy with percentage yield: Atom economy measures how many atoms from the reactants end up in the desired product (a measure of waste). Percentage yield measures how much product you actually get compared to the theoretical maximum. A reaction can have 100% atom economy but a low yield, or vice versa.

• Organic Chemistry: Organic chemistry covers carbon-based compounds and their reactions
• Physical Chemistry: Physical chemistry underpins reaction rates, energetics, and equilibrium
• Atomic Structure: Atomic structure determines chemical bonding and reactivity