How many moles are in $550.0 g$ of lead(II) sulfate?

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Solution

Answer
$1.814 m o l e s P b S O_{4}$
Explanation:
You can go from moles of a given substance to grams or vice versa by using the molar mass of said substance.
The molar mass tells you the mass of 1 mole of a given substance
The molar mass tells you the mass of 1 mole of a given substance
$M o l a r m a s s = \frac{m a s s i n g r a m s}{1 m o l e o f s u b s t a n c e}$
In your case, lead(II) sulfate, $P b S O_{4}$ has a molar mass of $303.2626 g m o l^{- 1}$ which means that every mole of lead(II) sulfate has a mass of 303.2626 g.
To convert from grams to moles, set up the molar mass as a conversion factor with the mass of 1 mole as the denominator
$\frac{1 m o l e P b S O_{4}}{303 .2626 g}$
You will have
$550.0 g . \frac{1 m o l e P b S O_{4}}{303.2626 g} = 1.814 m o l e s P b S O_{4}$
The answer is rounded to four sig figs.

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