Question

determine the quantity of atoms of vanadium in 1.28 grams of vanadium.
starting amount
add factor
answer
reset
6.022 × 10²³
1.51 × 10²³
g v/mol
50.94
1.28
atoms v
1.51 × 10²²
7.71 × 10²³
g v
6.022 × 10²⁴
mol v
0.0251
kg v
1
3.93 × 10²⁵
1000

Explanation:

Step1: Identify starting amount

$1.28\ \text{g V}$

Step2: Convert grams to moles

Use molar mass of V ($50.94\ \text{g V/mol V}$):
$$1.28\ \text{g V} \times \frac{1\ \text{mol V}}{50.94\ \text{g V}}$$

Step3: Convert moles to atoms

Use Avogadro's number ($6.022 \times 10^{23}\ \text{atoms V/mol V}$):
$$1.28\ \text{g V} \times \frac{1\ \text{mol V}}{50.94\ \text{g V}} \times 6.022 \times 10^{23}\ \frac{\text{atoms V}}{\text{mol V}}$$

Step4: Calculate final value

$$\frac{1.28 \times 6.022 \times 10^{23}}{50.94} = 1.51 \times 10^{22}$$

Answer:

$1.51 \times 10^{22}$ atoms of vanadium

Filled setup for the tool:
Starting amount: $1.28\ \text{g V}$
Conversion factor: $\frac{1\ \text{mol V}}{50.94\ \text{g V}} \times \frac{6.022 \times 10^{23}\ \text{atoms V}}{1\ \text{mol V}}$