Question

there are $1.5 \times 10^{24}$ atoms of carbon in a sample of aluminum acetate, $\ce{al(c_{2}h_{3}o_{2})_{3}}$. how many moles of al are in the sample? 0.42 moles al 6.6 moles al 9.3 moles al 2.8 moles al

Explanation:

Step1: Find moles of C atoms

Use Avogadro's number ($N_A = 6.022\times10^{23}\ \text{atoms/mol}$). Moles of C, $n_C=\frac{\text{number of C atoms}}{N_A}=\frac{1.5\times10^{24}}{6.022\times10^{23}}\approx2.49\ \text{mol}$

Step2: Determine C:Al mole ratio

In $\text{Al}(\text{C}_2\text{H}_3\text{O}_2)_3$, each formula unit has 6 C atoms (3 acetate groups, 2 C each) and 1 Al atom. So mole ratio $n_{\text{Al}}:n_{\text{C}} = 1:6$

Step3: Calculate moles of Al

$n_{\text{Al}}=\frac{n_{\text{C}}}{6}=\frac{2.49}{6}\approx0.42\ \text{mol}$

Answer:

0.42 moles Al