Question

molecules→moles

1. how many moles are in 2.78 x 10^33 molecules of krypton
2. how many molecules are in 3.2 moles of cal₂
3. how many moles are in 1.65 x 10^18 molecules of potassium oxide

moles→grams

1. how many grams are in 9.2 moles of gold
2. how many moles are in 289.3 grams of titanium (ii) bromide

Explanation:

Step1: Use Avogadro's number ($N_A = 6.022\times10^{23}\text{ molecules/mol}$)

For the first problem:
$$n=\frac{N}{N_A}=\frac{2.78\times 10^{38}\text{ molecules}}{6.022\times10^{23}\text{ molecules/mol}}$$

Step2: Calculate the value

$$n=\frac{2.78}{6.022}\times10^{38 - 23}\text{ mol}\approx4.62\times10^{14}\text{ mol}$$

For the second problem:

Step1: Use the formula $N = n\times N_A$

$$N=3.2\text{ mol}\times6.022\times 10^{23}\text{ molecules/mol}$$

Step2: Calculate the value

$$N=(3.2\times6.022)\times10^{23}\text{ molecules}\approx1.93\times10^{24}\text{ molecules}$$

For the third problem:

Step1: Use Avogadro's number formula $n=\frac{N}{N_A}$

$$n=\frac{1.65\times 10^{18}\text{ molecules}}{6.022\times10^{23}\text{ molecules/mol}}$$

Step2: Calculate the value

$$n=\frac{1.65}{6.022}\times10^{18 - 23}\text{ mol}\approx2.74\times10^{-6}\text{ mol}$$

For the Moles $\to$ Grams first problem:

Step1: Find the molar mass of Gold ($Au$, $M = 196.97\text{ g/mol}$)

$$m=n\times M$$
$$m = 9.2\text{ mol}\times196.97\text{ g/mol}$$

Step2: Calculate the value

$$m=(9.2\times196.97)\text{ g}\approx1812.12\text{ g}$$

For the Moles $\to$ Grams second problem:

Step1: Find the molar mass of Titanium (II) bromide ($TiBr_2$, $M=(47.87 + 2\times79.90)\text{ g/mol}=207.67\text{ g/mol}$)

$$n=\frac{m}{M}$$
$$n=\frac{289.3\text{ g}}{207.67\text{ g/mol}}$$

Step2: Calculate the value

$$n=\frac{289.3}{207.67}\text{ mol}\approx1.39\text{ mol}$$

Answer:

1. Approximately $4.62\times10^{14}$ mol
2. Approximately $1.93\times10^{24}$ molecules
3. Approximately $2.74\times10^{-6}$ mol

Moles $\to$ Grams 1: Approximately $1812.12$ g
Moles $\to$ Grams 2: Approximately $1.39$ mol