Answer
$1.45 \times 10^{-1} mol$ of ammonium ions in $6.955$ grams of ammonium carbonate.
Work Step by Step
The compound ammonium carbonate has a formula of $(NH_{4})_{2}CO_{3}$. Here's the breakdown of the molecular mass of each atom in the compound:
Nitrogen: $2 atoms \times 14 g/mol = 28 g/mol$
Hydrogen $8 atoms \times 1 g/mol = 8 g/mol$
Carbon: $1 atom \times 12 g/mol = 12 g/mol$
Oxygen $3 atoms \times 16 g/mol = 48 g/mol$
So in total, the molar mass of the compound adds up to $96 g/mol$.
Therefore we can use the conversion factor: $(6.955 grams / 1) \times (1 mol / 96 grams)$ to divide out the moles: $6.955 grams \div 96 g/mol = 0.07244 grams$ or $ 7.24 \times 10^{-2} grams$
Then, multiply by the two ammonium ions to get: $(7.24 \times 10^{2}) \times 2 = 14.48 \times 10^{-2}$ or $1.45 \times 10^{-1} mol$ (ammonium ions)
