Question

A chemical reaction is carried out using a 42.65-g sample of lead(II) nitrate, `Pb(NO_3)_2` ?

a.) Calculate the molar mass of lead(II) nitrate.
b.) Calculate the number of moles of lead(II) nitrate in the sample
c.) Calculate the number of formula units of lead(II) nitrate in the sample

Answer 1

Use:
the periodic table,
`n=m/M` formula,
and `n*6.02E23` formula.

Explanation:

a) Molar mass (found on periodic table)
Pb: 207.2 g/mol
N: 14.01 g/mol ( times 2 because 2 nitrogen in formula)
O: 16.00 g/mol ( times 6 becuase 6 oxygens in formula)

Add all of these up and you get: 331.22 g/mol for molar mass

b) Use this formula:
`n=m/M`
`n=42.65/331.22` (grams divided by grams/ mol= mol)
`n=0.128766... mol`

c) Use this formula:
`units= n*6.02E23` `units= 0.12877*6.02E23`
`units=7.7517E22`
(E means x10)

Answer 2

a.) Molar mass `"Pb(NO"_3)_2:` `"331.208 g/mol"`

b.) Moles `"Pb(NO"_3)_2:"` `"0.1288 mol Pb(NO"_3")"_2`

c.) Formula units `"Pb(NO"_3)_2:"` `7.756xx"10"^22` formula units `"Pb(NO"_3")"_2)`

Explanation:

a.) Molar mass `"Pb(NO"_3)_2"`.

In order to calculate the molar mass of a compound, multiply the subscript of each element by its molar mass (atomic weight on the periodic table in g/mol). No subscript is understood to be `1`. If you have a polyatomic ion inside parentheses, multiply the subscript outside the parentheses by the subscript of each element inside the parentheses.

Molar mass `"Pb(NO"_3)_2:` `(1xx"207.2 g/mol Pb") + (2xx1xx"14.007 g/mol N") + (2xx3xx"15.999 g/mol O")="331.208 g/mol Pb(NO"_3)_2"`

b.) Moles `"Pb(NO"_3)_2"`

In order to determine moles, divide the given mass by the molar mass. Since molar mass is a fraction, g/mol, I prefer to divide by multiplying by the reciprocal, mol/g.

Moles `"Pb(NO"_3")"_2` `=` `42.65"g Pb(NO"_3")"_2xx(1"mol Pb(NO"_3")"_2)/(331.208"g Pb(NO"_3")"_2"` `=` `"0.1288 mol Pb(NO"_3")"_2`

c.) Formula units `"Pb(NO"_3)_2"`

One mole of anything is `6.022xx10^23` anythings, including formula units. In order to determine the moles formula units `"Pb(NO"_3")"_2`, multiply the moles by `6.022xx10^23` formula units/mol.

Formula units `"Pb(NO"_3")"_2` `=` `0.1288"mol Pb(NO"_3")"_2xx(6.022xx10^23"formula units Pb(NO"_3")"_2)/(1"mol Pb(NO"_3")"_2)=7.756xx"10"^22` formula units `"Pb(NO"_3")"_2)`