High School

1. Calculate the percentage of nitrogen in ammonium nitrate (NH₄NO₃).

2. A compound contains 86.88% of carbon and 13.12% of hydrogen, and the molecular weight is 345 g. What will be the empirical and molecular formula of the compound?

3. Find the percentage composition of carbon, calcium and oxygen in calcium carbonate?

4. Calculate the molecular masses of the following compounds:
(i) Calcium chloride
(ii) Sodium carbonate
(iii) Potassium chlorate
(iv) Ammonium sulphate
(Given, atomic masses: N = 14u; Ca = 40 u; Cl = 35.5 u; Na = 23 u; C = 12 u; O = 16 u; S = 32 u)

5. Caffeine has the following composition: 49.48% of carbon, 5.19% of hydrogen, 16.48% of oxygen and 28.85% of nitrogen. The molecular weight is 194.19 g/mol. Find out the molecular formula and its molecular mass.

6. (a) The molecular formula of an organic acid is H₂CO₂. What is its empirical formula?
(b) A compound has the following percentage composition: Hydrogen 2.2%, Carbon 26.6% and Oxygen 71.2%. Calculate the empirical formula of the compound. If its molecular weight is 90, find its molecular formula.
(Given, atomic masses: H = 1, C = 12, O = 16)

7. Calculate the percentage of oxygen in the following:
(a) HgO
(b) K₂Cr₂O₇
(c) Al₂(SO₄)₃
(Given, atomic masses: Hg = 200 u; O = 16 u; K = 39 u; Cr = 52 u; Al = 27 u; S = 32 u)

Answer :

Let's tackle each question step-by-step:


  1. Calculate the percentage of nitrogen in ammonium nitrate [tex](NH_4NO_3)[/tex]:


    • The molecular formula for ammonium nitrate is [tex]NH_4NO_3[/tex].

    • The atomic masses are: [tex]N = 14\, u[/tex], [tex]H = 1\, u[/tex], [tex]O = 16\, u[/tex].

    • Total mass of ammonium nitrate:
      [tex]= (2 \times 14) + (4 \times 1) + (3 \times 16) = 28 + 4 + 48 = 80 \, u[/tex]

    • Nitrogen mass in [tex]NH_4NO_3[/tex]:
      [tex]= 2 \times 14 = 28 \, u[/tex]

    • Percentage of nitrogen:
      [tex]= \left(\frac{28}{80}\right) \times 100 = 35\%[/tex]



  2. Empirical and molecular formula of the compound:


    • Given percentages: Carbon [tex]= 86.88\%[/tex] and Hydrogen [tex]= 13.12\%[/tex].

    • Assume [tex]100\, g[/tex] of the compound, so we have [tex]86.88\, g[/tex] C and [tex]13.12\, g[/tex] H.

    • Convert masses to moles:
      [tex]\text{Moles of C} = \frac{86.88}{12} = 7.24 \, \text{mol}[/tex]
      [tex]\text{Moles of H} = \frac{13.12}{1} = 13.12 \, \text{mol}[/tex]

    • Find the simplest ratio:
      [tex]\text{Ratio of C to H} = \frac{7.24}{7.24} : \frac{13.12}{7.24} \approx 1:1.812
      \approx 5:9[/tex]

    • Empirical formula: [tex]C_5H_9[/tex]

    • Determine molecular formula using molecular weight [tex]345 \, g/mol[/tex]:

      • Molar mass of empirical formula [tex]C_5H_9 = (5 \times 12) + (9 \times 1) = 69\, g/mol[/tex]

      • Molecular formula factor:(
        = \frac{345}{69} = 5
        )



    • Molecular formula: [tex](C_5H_9) \times 5 = C_{25}H_{45}[/tex]



  3. Percentage composition of carbon, calcium, and oxygen in calcium carbonate [tex](CaCO_3)[/tex]:


    • Atomic masses: [tex]Ca = 40\, u,\ C = 12\, u,\ O = 16\, u[/tex]

    • Molar mass of [tex]CaCO_3:[/tex]
      [tex]= 40 + 12 + (3 \times 16) = 100\, u[/tex]

    • Percent calculations:
      [tex]\text{%C} = \left(\frac{12}{100}\right) \times 100 = 12\%[/tex]
      [tex]\text{%Ca} = \left(\frac{40}{100}\right) \times 100 = 40\%[/tex]
      [tex]\text{%O} = 100 - 12 - 40 = 48\%[/tex]



  4. Calculate the molecular masses of the following compounds:


    • (i) Calcium chloride [tex](CaCl_2)[/tex]:
      [tex]= 40 + (2 \times 35.5) = 111\, u[/tex]

    • (ii) Sodium carbonate [tex](Na_2CO_3)[/tex]:
      [tex]= (2 \times 23) + 12 + (3 \times 16) = 106\, u[/tex]

    • (iii) Potassium chlorate [tex](KClO_3)[/tex]:
      [tex]= 39 + 35.5 + (3 \times 16) = 122.5\, u[/tex]

    • (iv) Ammonium sulphate [tex](NH_4)_2SO_4[/tex]:
      [tex]= (2 \times (14 + 4)) + 32 + (4 \times 16) = 132\, u[/tex]



  5. Molecular formula and mass of caffeine:


    • Percentage composition: [tex]C = 49.48\%[/tex], [tex]H = 5.19\%[/tex], [tex]O = 16.48\%[/tex], [tex]N = 28.85\%[/tex].

    • Assume [tex]100\, g[/tex] of sample for simplification:
      [tex]\text{Moles of C} = \frac{49.48}{12} = 4.123 \, \text{mol}[/tex]
      [tex]\text{Moles of H} = \frac{5.19}{1} = 5.19 \, \text{mol}[/tex]
      [tex]\text{Moles of O} = \frac{16.48}{16} = 1.03 \, \text{mol}[/tex]
      [tex]\text{Moles of N} = \frac{28.85}{14} = 2.061 \, \text{mol}[/tex]

    • Simplest ratio for empirical formula (e.g., compared to [tex]N[/tex]) translates to [tex]C_5H_6N_2O[/tex].

    • Molecular weight [tex]= 194.19 \, g/mol[/tex] yields actual formula factor:
      [tex]\frac{194.19}{97} = 2\text{ which suggests: } C_8H_{10}N_4O_2[/tex]

    • Therefore, the molecular formula is [tex]C_8H_{10}N_4O_2[/tex].



  6. (a) The empirical formula for H₂CO₂:


    • Empirical formula is the simplest form, so it stays [tex]H_1C_1O_2[/tex].



  7. Calculate the percentage of oxygen in the following:


    • (a) Mercury(II) oxide [tex](HgO)[/tex]:
      [tex]\text{Total mass} = 200 + 16 = 216 \, u[/tex]
      [tex]\text{%O} = \left(\frac{16}{216}\right) \times 100 \approx 7.4\%[/tex]

    • (b) Potassium dichromate [tex](K_2Cr_2O_7)[/tex]:
      [tex]\text{Total mass} = (2 \times 39) + (2 \times 52) + (7 \times 16) = 294 \, u[/tex]
      [tex]\text{%O} = \left(\frac{112}{294}\right) \times 100 \approx 38.1\%[/tex]

    • (c) Aluminium sulphate [tex](Al_2(SO_4)_3)[/tex]:
      [tex]\text{Total mass} = (2 \times 27) + (3 \times 32) + (12 \times 16) = 342 \, u[/tex]
      [tex]\text{%O} = \left(\frac{192}{342}\right) \times 100 \approx 56.1\%[/tex]




These steps break down each calculation to help you understand how to determine compositions and formulas.

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