Question

A silver ring is composed of 1.81×1023 atoms. Calculate the mass of the ring in grams.

Answer 1

Solving with Moles

Mole equation:
`n=(m)/(MM)`

• n = number of moles (mol)
• m = mass (g)
• MM = molar mass (g/mol)

To find mass given number of atoms:

1. Divide number of atoms by the number of atoms in the chemical formula ⇒ find number of molecules
2. Divide number of molecules by Avogadro's number (
   `6.02*10^(23)`) ⇒ find number of moles (n)
3. Solve for m using moles equation

Solving the Question

We're given:

• Ag (silver)
• Atoms =
  `1.81*10^(23)` atoms
• m = ?

In the chemical formula, which is Ag, there is only 1 atom.

Divide
`1.81*10^(23)` atoms by 1 atom to get the number of molecules in the silver ring:

`1.81*10^(23)/ 1\\=1.81*10^(23)`

Therefore, there are
`1.81*10^(23)` molecules in the silver ring.

Now, divide
`1.81*10^(23)` molecules by Avogadro's number to find n:

`(1.81*10^(23))/(6.02*10^(23))\\\\=(1.81)/(6.02)\\\\=(1.81)/(6.02)\\\\= 0.30066`

Therefore, the sample has 0.30066 mol.

Finally, solve for the mass using the moles equation:

`n=(m)/(MM)`

⇒ Rearrange the equation:

`m=n*MM`

⇒ MM of Ag = 107.87 g/mol

⇒ Plug in given information:

`m=0.30066* 107.87\\m=32.4`

Therefore, the mass of the ring is 32.4 g.

Answer

32.4 g