# How many total atoms are in 0.410 g of P_2O_5?

Oct 2, 2016

Approx. $126 \times {10}^{20} \cdot \text{atoms of phosphorus and oxygen.}$

#### Explanation:

${P}_{2} {O}_{5}$ has a molar mass of $141.94 \cdot g \cdot m o {l}^{-} 1$.

And thus there is a molar quantity of $\frac{0.410 \cdot g}{141.95 \cdot g \cdot m o {l}^{-} 1}$ $=$ $2.89 \times {10}^{-} 3 \cdot m o l$.

Because there are $6.022 \times {10}^{23}$ particles in a mole by definition, there are thus,

$2.89 \times {10}^{-} 3 \cdot m o l \times 6.022 \times {10}^{23} \cdot m o {l}^{-} 1 \times 7 \cdot \text{atoms}$

$=$ $\text{how many atoms}$.

Note the actual molecule is ${P}_{4} {O}_{10}$; had I worked out the problem with this formulation I would have got precisely the same result.