# Question d85b1

Jul 29, 2015

Number of moles: $0.05996$
Number of formula units: $3.661 \cdot {10}^{22}$

#### Explanation:

Every time you're asked to determine how many moles of a substance you get in a certain mass of that substance, you have to use molar mass.

Molar mass simply tells you what the mass of 1 mole of a substance is. In your case, lead (II) nitrate has a molar mass of 331.21 g/mol, which means that every mole of lead (II) nitrate has a masss of 331.21 g.

SInce you have much less than 331.21 g, you can expect the number of moles you get to be smaller than one.

19.86color(red)(cancel(color(black)("g"))) * "1 mole"/(331.21color(red)(cancel(color(black)("g")))) = color(green)("0.05996 moles")

Now for the number of molecules, which in this case is actually the number of formula units.

Since lead (II) nitrate is an ionic compound, it does not exist as individual molecules, but rather as a giant crystal structure in which $P {b}^{2 +}$ cations and $N {O}_{3} {\text{}}^{-}$ anions are intertwined in a $1 : 2$ ratio.

Now, to get the number of formula units, which represents the empirical formula of lead (II) nitrate, you must use Avogadro's number.

One mole of any substance is known to contain exaclty $6.022 \cdot {10}^{23}$ atoms, molecules, or formula units of that substance - this is known as Avogadro's number. In your case, since you're dealing with less than one mole, you can expect the number of formula units to be smaller that Avogadro's number.

0.05996color(red)(cancel(color(black)("moles"))) * (6.022 * 10^(23)"f units")/(1color(red)(cancel(color(black)("mol")))) = color(green)(3.611 * 10^(22)"f units"#

Both answers are rounded to four sig figs, the number of sig figs you gave for the mass of the compound.