Question

A 0.4272 g sample of an element contains 2.241 x 10 ^21 atoms . what is the symbol of the element?

Answer 1

Answer:

Likely $\rm In$ (indium.)

Explanation:

Number of atoms: $N = 2.241 \times 10^{21}$.

Dividing, $N$, the number of atoms by the Avogadro constant, $N_A \approx 6.023 \times 10^{23} \; \rm mol^{-1}$, would give the number of moles of atoms in this sample:

$\begin{aligned} n &= \frac{N}{N_{A}} \\ &\approx \frac{2.241 \times 10^{21}}{6.023 \times 10^{23}\; \rm mol^{-1}} \approx 3.72 \times 10^{-3}\; \rm mol \end{aligned}$.

The mass of that many atom is $m = 0.4272\; \rm g$. Estimate the average mass of one mole of atoms in this sample:

$\begin{aligned}M &= \frac{m}{n} \\ &\approx \frac{0.4272\; \rm g}{3.72 \times 10^{-3}\; \rm mol} \approx 114.82\; \rm g \cdot mol^{-1}\end{aligned}$.

The average mass of one mole of atoms of an element ($114.82\; \rm g \cdot mol^{-1}$ in this example) is numerically equal to the average atomic mass of that element. Refer to a modern periodic table and look for the element with average atomic mass $114.82$. Indium, $\rm In$, is the closest match.