What is the mass of #1.2*10^18# formula units of calcium chloride?

Question

Mateo Aguilar · Accepted Answer

#2.2 * 10^(-4)# #"g"#

The thing to keep in mind about formula units is that you need #6.022 * 10^(23)# of them to have exactly #1# mole of an ionic compound #-># think Avogadro's constant here.

In this case, you know that #6.022 * 10^(23)# formula units of calcium chloride are needed in order to have exactly #1# mole of calcium chloride.

Moreover, you know that calcium chloride has a molar mass of #"110.98 g mol"^(-1)#, which means that #1# mole of calcium chloride has a mass of #"110.98 g"#.

So if #1# mole of calcium chloride contains #6.022 * 10^(23)# formula units and has a mass of #"110.98 g"#, you can say that #6.022 * 10^(23)# formula units of calcium chloride have a mass of #"110.98 g"#.

This means that your sample has a mass of

#1.2 * 10^8 color(red)(cancel(color(black)("f. units CaCl"_2))) * "110.98 g"/(6.022 * 10^(23)color(red)(cancel(color(black)("f. units CaCl"_2)))) = color(darkgreen)(ul(color(black)(2.2 * 10^(-4) quad "g")))#

The answer is rounded to two sig figs, the number of sig figs you have for the number of formula units.

Wyatt Atkins · Answer

undefined To calculate the mass of $1.2 \times 10^{18}$ formula units of calcium chloride ($CaCl_2$), you need to know the molar mass of calcium chloride and then multiply it by the number of formula units.

The molar mass of calcium chloride ($CaCl_2$) is:

$$ \text{Ca} : 40.08 \, \text{g/mol} $$
$$ \text{Cl} : 35.45 \, \text{g/mol} \times 2 = 70.90 \, \text{g/mol} $$

So, the molar mass of $CaCl_2$ is $40.08 + 70.90 = 110.98 \, \text{g/mol}$.

Now, to find the mass of $1.2 \times 10^{18}$ formula units of $CaCl_2$:

$$ \text{Mass} = \text{Number of formula units} \times \text{Molar mass} $$

$$ \text{Mass} = 1.2 \times 10^{18} \times 110.98 \, \text{g/mol} $$

$$ \text{Mass} = 1.33176 \times 10^{20} \, \text{grams} $$