How to calculate the percentage composition

Calculate the composition in percentage of the single atoms and $H_2O$ in the chemical compound $Na_2B_4O_7\cdot 10H_2O.$

SOLUTION:
First we consider the atomic weight of the species:
$\begin{cases} Na=23\\ B=10.8\\ O=16\\ H_2O=18\\ \end{cases}$

The molecular weight of $Na_2B_4O_7\cdot 10H_2O$ is:

$MW=2\cdot 23+4\cdot 10.8+7\cdot 16+10\cdot 18=381.2gr/mol$
So one mole of $Na_2B_4O_7\cdot 10H_2O$ has a mass of 381.2gr.

The percentage of the atoms in the compound is, in general:
$\%X=\frac{nX\cdot AtomicWeight}{AtomicMass}$

We can now calculate:
$\%Na=\frac{2\cdot 23gr/mol}{381.2gr/mol}\cdot 100=12.06\%$

$\%B=\frac{4\cdot 10.8gr/mol}{381.2gr/mol}\cdot 100=11.33\%$

$\%O=\frac{7\cdot 16gr/mol}{381.2gr/mol}\cdot 100=29.38\%$

$\%H_2O=\frac{10\cdot 18gr/mol}{381.2gr/mol}\cdot 100=47.22\%$

---

PROBLEM 1:
Compute the percentage of K, Fe, C, N and $H_2O$ in $K_4Fe(CN)_6 \cdot 5H_2O$ crystals.