First, we will need the atomic weight of the compounds:
B=10.8110 u
H=1.00784 u
Now, we will calculate the molecular weight of the molecule B4H10:
[tex]\begin{gathered} B\colon\text{ 10.8110 x 4 = 43.244} \\ H\colon\text{ 1.00784 x 10 = 10.0784} \\ \text{Total weight = 43.244 + 10.0784 = 53.3224 g/mol} \end{gathered}[/tex]
Now that we have the molecular weight we proceed to calculate the number of molecules present in 130 g of sample:
[tex]\text{Mol of B4H10 = }\frac{130\text{ g}}{53.3224\text{ g/mol}}\text{ = 2.4380 mol}[/tex]
So, we have 2.4380 mol of B4H4 in 130 g of sample, each molecule has 4 atoms of boron, which means that we will have:
2.4380 x 4 = 9.7520 atoms of B
The answer will be: 9.7520 atoms of B, here we have the answer with 4 significant digits.
