How many grams of oxygen are in $6.15times10^23$ formula units of $(NH_4)_2SO_4$ ?

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Answer 1 · 2017-08-01T02:23:48

$65.4$ $"g O"$

We're asked to find the mass, in $"g"$ , of oxygen in $6.15xx10^23$ $"formula units (NH"_4")"_2"SO"_4$ .

To do this, we recognize that there are $4$ $"atoms O"$ per formula unit of ammonium sulfate:

$6.15xx10^23cancel("formula units (NH"_4")"_2"SO"_4)((4color(white)(l)"atoms O")/(1cancel("formula unit (NH"_4")"_2"SO"_4)))$

$= 2.46xx10^24$ $"atoms O"$

Now, we use Avogadro's number to find the number of moles of $"O"$ :

$2.46xx10^24cancel("atoms O")((1color(white)(l)"mol O")/(6.022xx10^23cancel("atoms O"))) = ul(4.085color(white)(l)"mol O"$

Finally, we use the molar mass of oxygen ( $15.999$ $"g/mol"$ ) to convert from moles to grams:

$4.085cancel("mol O")((15.999color(white)(l)"g O")/(1cancel("mol O"))) = color(red)(ul(65.4color(white)(l)"g O"$

How many grams of oxygen are in 6.15times10^236.15times10^23 formula units of (NH_4)_2SO_4(NH_4)_2SO_4?