# How many atoms are present in 21.59 g of C_6H_12O_6?

Feb 10, 2016

We need to (i) work out the molar quantity of glucose; and (ii) convert this molar quantity into an actual number of the constituent elements.

#### Explanation:

$\text{Moles of glucose}$ $=$ $\frac{21.59 \cdot \cancel{g}}{180.15 \cdot \cancel{g} \cdot m o {l}^{-} 1}$ $=$ $0.120$ $m o l$.

Now in one MOLECULE of glucose there are $6$ $\times$ $C$, and $6$ $\times$ $O$, and $12$ $\times$ $H$ $=$ $24$ $\text{atoms}$ in total.

We have a $0.120$ $m o l$ quantity, i.e. $0.120$ $m o l$ $\times$ ${N}_{A}$, (where ${N}_{A}$ $=$ $\text{Avogadro's Number}$ $=$ $6.022$ $\times$ ${10}^{23}$ $m o {l}^{-} 1$), is the number of MOLECULES.

So $\text{number of atoms}$ $=$ $24 \times 0.120 \cdot m o l \times {10}^{23}$ $m o {l}^{-} 1$ $=$ ??