Given 54g of (NH4)2Fe(SO4)2.5H2O find: a) The number of atoms in the compound, b) The number molecules of H2O, c) The number of Ammonium ions in the compound, d) The number of sulphate ions in the compound.

$$\text{Molar Mass} = 2 \times (\text{N} + 4\text{H}) + \text{Fe} + 2 \times (\text{S} + 4\text{O}) + 5 \times (2\text{H} + \text{O})$$ $$\text{Molar Mass} = 2 \times (14.007 + 4 \times 1.008) + 55.845 + 2 \times (32.06 + 4 \times 15.999) + 5 \times (2 \times 1.008 + 15.999)$$ $$\text{Molar Mass} = 2 \times (14.007 + 4.032) + 55.845 + 2 \times (32.06 + 63.996) + 5 \times (2.016 + 15.999)$$ $$\text{Molar Mass} = 2 \times 18.039 + 55.845 + 2 \times 96.056 + 5 \times 18.015$$ $$\text{Molar Mass} = 36.078 + 55.845 + 192.112 + 90.075$$ $$\text{Molar Mass} = 374.11 \, \text{g/mol}$$

$$\text{Moles} = \frac{\text{Mass}}{\text{Molar Mass}}$$ $$\text{Moles} = \frac{54 \, \text{g}}{374.11 \, \text{g/mol}} \approx 0.14434 \, \text{mol}$$

$$\text{Molecules of compound} = \text{Moles} \times N_A$$ $$\text{Molecules of compound} = 0.14434 \, \text{mol} \times 6.022 \times 10^{23} \, \text{molecules/mol}$$ $$\text{Molecules of compound} \approx 8.691 \times 10^{22} \, \text{molecules}$$

a) The number of atoms in the compound
First, count the total number of atoms in one molecule of $(\text{NH}_4)_2\text{Fe}(\text{SO}_4)_2 \cdot 5\text{H}_2\text{O}$:
$\text{N}: 2 \times 1 = 2$
$\text{H}: (2 \times 4) + (5 \times 2) = 8 + 10 = 18$
$\text{Fe}: 1 \times 1 = 1$
$\text{S}: 2 \times 1 = 2$
$\text{O}: (2 \times 4) + (5 \times 1) = 8 + 5 = 13$
Total atoms per molecule = $2 + 18 + 1 + 2 + 13 = 36$ atoms.
Now, multiply the number of molecules of the compound by the total atoms per molecule:

$$\text{Total atoms} = 8.691 \times 10^{22} \, \text{molecules} \times 36 \, \text{atoms/molecule}$$ $$\text{Total atoms} \approx 312.876 \times 10^{22} \approx \boxed{\text{3.13} \times 10^{24} \, \text{atoms}}$$

b) The number of molecules of H₂O
From the chemical formula, there are 5 molecules of $\text{H}_2\text{O}$ for every molecule of the compound.

$$\text{Molecules of H}_2\text{O} = \text{Molecules of compound} \times 5$$ $$\text{Molecules of H}_2\text{O} = 8.691 \times 10^{22} \times 5$$ $$\text{Molecules of H}_2\text{O} \approx 43.455 \times 10^{22} \approx \boxed{\text{4.35} \times 10^{23} \, \text{molecules}}$$

c) The number of Ammonium ions in the compound
From the chemical formula, there are 2 $\text{NH}_4^+$ ions for every molecule of the compound.

$$\text{Number of NH}_4^+ \text{ ions} = \text{Molecules of compound} \times 2$$ $$\text{Number of NH}_4^+ \text{ ions} = 8.691 \times 10^{22} \times 2$$ $$\text{Number of NH}_4^+ \text{ ions} \approx 17.382 \times 10^{22} \approx \boxed{\text{1.74} \times 10^{23} \, \text{ions}}$$

d) The number of sulphate ions in the compound
From the chemical formula, there are 2 $\text{SO}_4^{2-}$ ions for every molecule of the compound.

$$\text{Number of SO}_4^{2-} \text{ ions} = \text{Molecules of compound} \times 2$$ $$\text{Number of SO}_4^{2-} \text{ ions} = 8.691 \times 10^{22} \times 2$$ $$\text{Number of SO}_4^{2-} \text{ ions} \approx 17.382 \times 10^{22} \approx \boxed{\text{1.74} \times 10^{23} \, \text{ions}}$$