Question

solve the problems. show your work and use significant figures. round atomic mass numbers to two decimal places.

1. how many atoms are in 3.0 g of gold?
2. how many molecules are in 60.0 g of magnesium citrate, $\ce{c_{6}h_{6}mgo_{7}}$?

Explanation:

For problem 5:

Step1: Find molar mass of Au

Atomic mass of Au = $196.97$ g/mol

Step2: Calculate moles of Au

$\text{Moles of Au} = \frac{\text{Mass}}{\text{Molar Mass}} = \frac{3.0\ \text{g}}{196.97\ \text{g/mol}} \approx 0.01523\ \text{mol}$

Step3: Calculate number of atoms

Use Avogadro's number $N_A = 6.022 \times 10^{23}\ \text{atoms/mol}$
$\text{Atoms} = \text{Moles} \times N_A = 0.01523\ \text{mol} \times 6.022 \times 10^{23}\ \text{atoms/mol} \approx 9.17 \times 10^{21}$

For problem 6:

Step1: Calculate molar mass of $\text{C}_6\text{H}_8\text{MgO}_7$

Atomic masses: C=$12.01$ g/mol, H=$1.01$ g/mol, Mg=$24.31$ g/mol, O=$16.00$ g/mol
$\text{Molar Mass} = (6 \times 12.01) + (8 \times 1.01) + 24.31 + (7 \times 16.00)$
$= 72.06 + 8.08 + 24.31 + 112.00 = 216.45\ \text{g/mol}$

Step2: Calculate moles of $\text{C}_6\text{H}_8\text{MgO}_7$

$\text{Moles} = \frac{60.0\ \text{g}}{216.45\ \text{g/mol}} \approx 0.2772\ \text{mol}$

Step3: Calculate number of molecules

$\text{Molecules} = \text{Moles} \times N_A = 0.2772\ \text{mol} \times 6.022 \times 10^{23}\ \text{molecules/mol} \approx 1.67 \times 10^{23}$

Answer:

1. $\boldsymbol{9.17 \times 10^{21}}$ atoms
2. $\boldsymbol{1.67 \times 10^{23}}$ molecules