Question

grams to atoms
5.11 g b
8.02 g s
1.5 g k
0.2550 g pt
1.37×10⁻¹⁰ g au

Explanation:

Step1: Recall molar - mass and Avogadro's number

To convert grams to atoms, we use the molar - mass of the element and Avogadro's number ($N_A = 6.022\times10^{23}\text{ atoms/mol}$). The molar - mass of an element can be found on the periodic table.

Step2: For boron (B)

The molar - mass of B is approximately $M_{B}=10.81\text{ g/mol}$. The number of moles of B, $n_{B}=\frac{m_{B}}{M_{B}}=\frac{5.1\text{ g}}{10.81\text{ g/mol}}\approx0.472\text{ mol}$. The number of atoms of B, $N_{B}=n_{B}\times N_A = 0.472\text{ mol}\times6.022\times10^{23}\text{ atoms/mol}\approx2.84\times 10^{23}\text{ atoms}$.

Step3: For sulfur (S)

The molar - mass of S is approximately $M_{S}=32.06\text{ g/mol}$. The number of moles of S, $n_{S}=\frac{m_{S}}{M_{S}}=\frac{8.02\text{ g}}{32.06\text{ g/mol}}\approx0.25\text{ mol}$. The number of atoms of S, $N_{S}=n_{S}\times N_A=0.25\text{ mol}\times6.022\times10^{23}\text{ atoms/mol}=1.51\times 10^{23}\text{ atoms}$.

Step4: For potassium (K)

The molar - mass of K is approximately $M_{K}=39.10\text{ g/mol}$. The number of moles of K, $n_{K}=\frac{m_{K}}{M_{K}}=\frac{1.5\text{ g}}{39.10\text{ g/mol}}\approx0.0384\text{ mol}$. The number of atoms of K, $N_{K}=n_{K}\times N_A = 0.0384\text{ mol}\times6.022\times10^{23}\text{ atoms/mol}\approx2.31\times 10^{22}\text{ atoms}$.

Step5: For platinum (Pt)

The molar - mass of Pt is approximately $M_{Pt}=195.08\text{ g/mol}$. The number of moles of Pt, $n_{Pt}=\frac{m_{Pt}}{M_{Pt}}=\frac{0.2550\text{ g}}{195.08\text{ g/mol}}\approx0.00131\text{ mol}$. The number of atoms of Pt, $N_{Pt}=n_{Pt}\times N_A=0.00131\text{ mol}\times6.022\times10^{23}\text{ atoms/mol}\approx7.89\times 10^{20}\text{ atoms}$.

Step6: For gold (Au)

The molar - mass of Au is approximately $M_{Au}=196.97\text{ g/mol}$. The number of moles of Au, $n_{Au}=\frac{m_{Au}}{M_{Au}}=\frac{1.37\times 10^{-10}\text{ g}}{196.97\text{ g/mol}}\approx6.96\times 10^{-13}\text{ mol}$. The number of atoms of Au, $N_{Au}=n_{Au}\times N_A=6.96\times 10^{-13}\text{ mol}\times6.022\times10^{23}\text{ atoms/mol}\approx4.19\times 10^{11}\text{ atoms}$.

Answer:

Boron: Approximately $2.84\times 10^{23}$ atoms
Sulfur: $1.51\times 10^{23}$ atoms
Potassium: Approximately $2.31\times 10^{22}$ atoms
Platinum: Approximately $7.89\times 10^{20}$ atoms
Gold: Approximately $4.19\times 10^{11}$ atoms