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Quantitative Aspects of Chemical Change Worksheet

Chemistry

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QUESTION 1

1.1 The mole is defined as the amount of substance that contains exactly 6.022×10236.022 \times 10^{23} elementary entities (atoms, molecules, ions, or other particles), which is Avogadro's number.

1.2

  1. Calculate the molar mass of H2SO4\text{H}_{2}\text{SO}_{4}.

    Molar mass of H=1.01 g/mol\text{H} = 1.01 \text{ g/mol}
    Molar mass of S=32.07 g/mol\text{S} = 32.07 \text{ g/mol}
    Molar mass of O=16.00 g/mol\text{O} = 16.00 \text{ g/mol}

    M(H2SO4)=(2×1.01)+32.07+(4×16.00)M(\text{H}_{2}\text{SO}_{4}) = (2 \times 1.01) + 32.07 + (4 \times 16.00) M(H2SO4)=2.02+32.07+64.00M(\text{H}_{2}\text{SO}_{4}) = 2.02 + 32.07 + 64.00 M(H2SO4)=98.09 g/molM(\text{H}_{2}\text{SO}_{4}) = 98.09 \text{ g/mol}
  2. Calculate the number of moles of H2SO4\text{H}_{2}\text{SO}_{4}.

    n=mMn = \frac{m}{M} n(H2SO4)=9.8 g98.09 g/moln(\text{H}_{2}\text{SO}_{4}) = \frac{9.8 \text{ g}}{98.09 \text{ g/mol}} n(H2SO4)=0.0999 moln(\text{H}_{2}\text{SO}_{4}) = 0.0999 \text{ mol}
  3. Calculate the number of H2SO4\text{H}_{2}\text{SO}_{4} molecules.

    Number of molecules = n×NAn \times N_{A}
    Number of H2SO4\text{H}_{2}\text{SO}_{4} molecules = 0.0999 mol×6.022×1023 molecules/mol0.0999 \text{ mol} \times 6.022 \times 10^{23} \text{ molecules/mol}
    Number of H2SO4\text{H}_{2}\text{SO}_{4} molecules = 6.0159×1022 molecules6.0159 \times 10^{22} \text{ molecules}

  4. Calculate the number of oxygen atoms.

    Each H2SO4\text{H}_{2}\text{SO}_{4} molecule contains 4 oxygen atoms.
    Number of oxygen atoms = Number of H2SO4\text{H}_{2}\text{SO}_{4} molecules ×4\times 4
    Number of oxygen atoms = 6.0159×1022×46.0159 \times 10^{22} \times 4
    Number of oxygen atoms = 2.41×1023 atoms\boxed{\mathbf{2.41 \times 10^{23} \text{ atoms}}}

    1.3.1 The percentage hydrogen will REMAIN THE SAME.

    1.3.2 The law applied is the Law of Definite Proportions (or Law of Constant Composition).

    1.4.1 Concentration is defined as the amount of solute (in moles) dissolved per unit volume of solution (in dm3\text{dm}^{3} or liters).

    1.4.2

  5. Calculate the molar mass of NaOH\text{NaOH}.

    Molar mass of Na=22.99 g/mol\text{Na} = 22.99 \text{ g/mol}
    Molar mass of O=16.00 g/mol\text{O} = 16.00 \text{ g/mol}
    Molar mass of H=1.01 g/mol\text{H} = 1.01 \text{ g/mol}

    M(NaOH)=22.99+16.00+1.01M(\text{NaOH}) = 22.99 + 16.00 + 1.01 M(NaOH)=40.00 g/molM(\text{NaOH}) = 40.00 \text{ g/mol}
  6. Calculate the number of moles of NaOH\text{NaOH}.

    n=mMn = \frac{m}{M} n(NaOH)=4.4 g40.00 g/moln(\text{NaOH}) = \frac{4.4 \text{ g}}{40.00 \text{ g/mol}} n(NaOH)=0.11 moln(\text{NaOH}) = 0.11 \text{ mol}
  7. Determine the volume of the solution.

    C=nVC = \frac{n}{V} V=nCV = \frac{n}{C} V=0.11 mol0.22 mol/dm3V = \frac{0.11 \text{ mol}}{0.22 \text{ mol/dm}^{3}} V=0.50 dm3V = \boxed{\mathbf{0.50 \text{ dm}^{3}}}
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