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Both questions ask the same thing in opposite ways.

One gives you $"16.26 mg"$ containing $1.66xx10^20$ $"atoms"$. That gives you a mass and a quantity, which is enough to convert to the atomic units of $"g/mol"$.

$("16.26" cancel"mg")/(1.66xx10^20 cancel"atoms") xx "1 g"/("1000" cancel"mg") xx (6.0221413xx10^23 cancel"atoms")/"mol"$

$=$ $color(blue)("58.988 g/mol")$,

which is close enough to cobalt, whose accepted value is $"58.933 g/mol"$, and our value is off by only $0.093%$.

The other gives you $"1 million"$ silver atoms weighing $1.79xx10^(-16) "g"$. That gives you a quantity and a mass, which is enough to convert to the atomic units of $"g/mol"$, again.

$(1.79xx10^(-16) "g")/(10^6 cancel"Ag atoms")xx(6.0221413xx10^23 cancel"atoms")/"mol"$

$=$ $color(green)("107.796 g/mol")$

which is close enough to the accepted value of $"107.868 g/mol"$ ($0.066%$ difference).

In either case you still get $\mathbf("g/mol")$.

WHAT IS AMU?

You are not required to use the conversion

$"1 g"/(6.0221413xx10^(23) "atoms") = "1 amu"$

$= 1.660599xx10^(-24) "g",$

unless you want to. It just depends on whether you are talking about $"1 mol"$ of atoms or $"1 atom"$, and you need to choose which one is more convenient for you.

CONVERTING BETWEEN G/MOL and AMU

You can still interconvert between $"g/mol"$ and $"amu"$...

$(58.988 cancel"g")/cancel"mol"xxcancel"1 mol"/(6.0221413xx10^23 cancel"atoms") xx "1 amu"/(1.660599xx10^(-24) cancel"g")xxcancel("1 atom")$

$~~ color(blue)(58.988)$ $color(blue)("amu for one atom")$

i.e. The experimental of $"1 atom"$ of cobalt according to the given information is $"58.988 amu"$, while the molar mass calculated at the top of the answer is $"58.988 g/mol"$.

So, $"amu"$ is just a way of giving the mass of one atom in atomic units ($"1 amu"$ $ne$ $"1 g"$), while $"g/mol"$ gives you the mass of $\mathbf("1 mol")$ of atoms.

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