# How many atoms are present in a 13.9*g mass of "carbon disulfide"?

First we calculate the quotient $\text{Mass of stuff"/"Molar mass of stuff}$...........
And this quotient gives......$\frac{13.9 \cdot g}{76.14 \cdot g \cdot m o {l}^{-} 1} = 0.188 \cdot m o l$ with respect to $C {S}_{2}$.
Since each mole/molecule of $C {S}_{2}$ contains 1 mole of carbon, and 2 moles of sulfur, there are $3 \times 0.188 \cdot m o l \times 6.022 \times {10}^{23} \cdot m o {l}^{-} 1 = 3.39 \times {10}^{23}$ individual atoms.......