Question

How many grams would 3.37 x 10 22 atoms of sulfur be?

Answer 1

Task:

How many grams would 3.37 x 10 22 atoms of sulfur be?

Solution:

According to Avogadro Law: 1 mol of substance contains $6.02 \cdot 10^{23}$ atoms (molecules, ions).

$$n(\text{mol}) = N / N_A$$

n – the amount of sulfur (mol)

N – the quantity of atoms of sulfur

$N_A$ – Avogadro constant ($6.02 \cdot 10^{23}$)

$$n(\text{mol}) = m(g) / \text{MW}(g/\text{mol})$$ $$m(g) / \text{MW}(g/\text{mol}) = N / N_A$$

MW(S) = 32 g / mol (from the periodic table of elements)

The mass of sulfur is

$$m(S) = N \cdot \text{MW} / N_A = 3.37 \cdot 10^{22} \cdot 32 / (6.02 \cdot 10^{23}) = 1.79 \text{ g}$$

Answer: $m(S) = 1.79 \text{ g}$