Question

unit 2 - lec 3 hw - mole conversions
1 mol = 6.02 x 10²³ atoms
(\frac{1\text{ mole}}{6.02x10^{23}\text{ atoms}}) or (\frac{6.02x10^{23}\text{ atoms}}{1\text{ mole}})
mole - atom conversions

1. how many moles are 2.55 x 10²⁷ atoms of zinc (zn)?
2. how many atoms are in 30.0 moles of calcium (ca)?
3. how many atoms are in 0.45 moles of carbon (c)?
4. how many moles are 7.8 x 10²⁴ atoms of helium?

1 mol = grams (molar mass) (get this from the periodic table) (\frac{1\text{ mole}}{g}) or (\frac{g}{1\text{ mole}})

Explanation:

Step1: Recall mole - atom conversion factor

We know that 1 mole of any substance contains $6.02\times 10^{23}$ atoms. The conversion factors are $\frac{1\ mol}{6.02\times 10^{23}\ atoms}$ and $\frac{6.02\times 10^{23}\ atoms}{1\ mol}$.

Step2: Solve problem 1

To find the number of moles of Zn given $2.55\times 10^{27}$ atoms of Zn, we use the conversion factor $\frac{1\ mol}{6.02\times 10^{23}\ atoms}$. So, $n=\frac{2.55\times 10^{27}\ atoms}{6.02\times 10^{23}\ atoms/mol}=4236\ mol$.

Step3: Solve problem 2

To find the number of atoms in 30.0 moles of Ca, we use the conversion factor $\frac{6.02\times 10^{23}\ atoms}{1\ mol}$. So, $N = 30.0\ mol\times6.02\times 10^{23}\ atoms/mol = 1.806\times 10^{25}\ atoms$.

Step4: Solve problem 3

To find the number of atoms in 0.45 moles of C, we use the conversion factor $\frac{6.02\times 10^{23}\ atoms}{1\ mol}$. So, $N=0.45\ mol\times6.02\times 10^{23}\ atoms/mol = 2.709\times 10^{23}\ atoms$.

Step5: Solve problem 4

To find the number of moles of He given $7.8\times 10^{24}$ atoms of He, we use the conversion factor $\frac{1\ mol}{6.02\times 10^{23}\ atoms}$. So, $n=\frac{7.8\times 10^{24}\ atoms}{6.02\times 10^{23}\ atoms/mol}=12.96\ mol$.

Answer:

1. $4236\ mol$
2. $1.806\times 10^{25}\ atoms$
3. $2.709\times 10^{23}\ atoms$
4. $12.96\ mol$