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Define the term "mole of a substance"

Chemistry

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1. (a) Define the term "mole of a substance"
The mole is the SI unit for the amount of substance. It is defined as the amount of substance that contains exactly 6.02214076×10236.02214076 \times 10^{23} elementary entities (such as atoms, molecules, ions, or electrons). This specific number is known as Avogadro's number (NAN_A).

1. (b) 1.50 g of ethyl ethanoate was hydrolysed with 50.0 cm³ of 0.5 M NaOH solution and then made up to 250 cm³ with distilled water in a volumetric flask.

  • A: The number of moles of ethyl ethanoate in 1.50 g. (RMM: CH₃CO₂C₂H₅ = 88)

  1. Identify the given mass and molar mass.

    Mass of ethyl ethanoate = $1.50 \text{ g}$  
Molar mass (RMM) of ethyl ethanoate = $88 \text{ g/mol}$

  2. Calculate the number of moles using the formula

    Moles=MassMolar Mass\text{Moles} = \frac{\text{Mass}}{\text{Molar Mass}}. Moles of ethyl ethanoate=1.50 g88 g/mol \text{Moles of ethyl ethanoate} = \frac{1.50 \text{ g}}{88 \text{ g/mol}} Moles of ethyl ethanoate≈0.017045 mol \text{Moles of ethyl ethanoate} \approx 0.017045 \text{ mol} Moles of ethyl ethanoate=0.0170 mol (3 s.f.) \text{Moles of ethyl ethanoate} = \boxed{0.0170 \text{ mol (3 s.f.)}}

    • B: The number of moles of NaOH in 50 cm³ of 0.5 M solution.

  3. Identify the given concentration and volume.

    Concentration of NaOH = $0.5 \text{ M}$ (or $0.5 \text{ mol/L}$)  
Volume of NaOH solution = $50 \text{ cm}^3$

  4. Convert the volume from cm3\text{cm}^3 to L\text{L}.

    $$
\text{Volume in L} = 50 \text{ cm}^3 \times \frac{1 \text{ L}}{1000 \text{ cm}^3} = 0.050 \text{ L}
$$

  5. Calculate the number of moles using the formula

    Moles=Concentration×Volume\text{Moles} = \text{Concentration} \times \text{Volume}. Moles of NaOH=0.5 mol/L×0.050 L \text{Moles of NaOH} = 0.5 \text{ mol/L} \times 0.050 \text{ L} Moles of NaOH=0.025 mol \text{Moles of NaOH} = \boxed{0.025 \text{ mol}}

    • C: The percentage yield when 1.50 g of ethyl ethanoate reacts completely with sodium hydroxide to produce 0.63 g of ethanol. (RAM: C = 12.0, H = 1.0, O = 16.0, Na = 23.0)

  6. Write the balanced chemical equation for the hydrolysis of ethyl ethanoate.

    $$
\text{CH}_3\text{COOC}_2\text{H}_5 \text{(aq)} + \text{NaOH (aq)} \rightarrow \text{CH}_3\text{COONa (aq)} + \text{C}_2\text{H}_5\text{OH (aq)}
$$
The molar ratio between ethyl ethanoate and ethanol is $1:1$.

  7. Calculate the molar mass of ethanol (C2H5OH\text{C}_2\text{H}_5\text{OH}).

    $$
\text{Molar mass of C}_2\text{H}_5\text{OH} = (2 \times 12.0) + (6 \times 1.0) + (1 \times 16.0)
$$
$$
\text{Molar mass of C}_2\text{H}_5\text{OH} = 24.0 + 6.0 + 16.0 = 46.0 \text{ g/mol}
$$

  8. Calculate the theoretical moles of ethanol produced from 1.50 g of ethyl ethanoate.

    From part (b) A, moles of ethyl ethanoate = $0.017045 \text{ mol}$.  
Since the ratio is $1:1$, theoretical moles of ethanol = $0.017045 \text{ mol}$.

  9. Calculate the theoretical mass of ethanol.

    $$
\text{Theoretical mass of ethanol} = \text{Moles} \times \text{Molar Mass}
$$
$$
\text{Theoretical mass of ethanol} = 0.017045 \text{ mol} \times 46.0 \text{ g/mol}
$$
$$
\text{Theoretical mass of ethanol} \approx 0.78407 \text{ g}
$$

  10. Calculate the percentage yield.

    Actual yield of ethanol = $0.63 \text{ g}$ (given)
$$
\text{Percentage yield} = \frac{\text{Actual Yield}}{\text{Theoretical Yield}} \times 100\%
$$
$$
\text{Percentage yield} = \frac{0.63 \text{ g}}{0.78407 \text{ g}} \times 100\%
$$
$$
\text{Percentage yield} \approx 80.35 \%
$$
$$
\text{Percentage yield} = \boxed{80.4\% \text{ (3 s.f.)}}
$$
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