# How many particles are in 6.84 grams of Li_2Co_3?

## It is 73.890 g/mol.

$6.84 \cdot g$ represents $9.26 \times {10}^{-} 2 \cdot m o l$ of $\text{lithium carbonate}$.
And thus this molar quantity represents $2 \times 9.26 \times {10}^{-} 2 \cdot m o l$ of $\text{lithium atoms}$, $9.26 \times {10}^{-} 2 \cdot m o l$ of $\text{carbon atoms}$, and $3 \times 9.26 \times {10}^{-} 2 \cdot m o l$ of $\text{oxygen atoms}$.
To get the molar quantity of $\text{lithium carbonate}$ I simply took the quotient, $\text{mass"/"molar mass}$, i.e. (6.84*g)/(73.89*g*mol^-1)=??*mol.