Answer :
Let's tackle each question step-by-step:
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]
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]
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]
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]
- (i) Calcium chloride [tex](CaCl_2)[/tex]:
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].
(a) The empirical formula for H₂CO₂:
- Empirical formula is the simplest form, so it stays [tex]H_1C_1O_2[/tex].
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]
- (a) Mercury(II) oxide [tex](HgO)[/tex]:
These steps break down each calculation to help you understand how to determine compositions and formulas.