Question

A 13 g sample of P4010 contains how many

molecules of phosphorus pentaoxide?
Answer in units of molec.

Answer 1

`\large \boxed{5.5 * 10^(22)\text{ molecules of P$_(2)$O}_(5)}`

Step-by-step explanation:

You must calculate the moles of P₄O₁₀, convert to moles of P₂O₅, then convert to molecules of P₂O₅.

1. Moles of P₄O₁₀

`\text{Moles of P$_(4)$O}_(10) = \text{13 g P$_(4)$O}_(10) * \frac{\text{1 mol P$_(4)$O}_(10)}{\text{283.89 g P$_(4)$O}_(10)} = \text{0.0458 mol P$_(4)$O}_(10)`

2. Moles of P₂O₅

P₄O₁₀ ⟶ 2P₂O₅

The molar ratio is 2 mol P₂O₅:1 mol P₄O₁₀

`\text{Moles of P$_(2)$O}_(5) = \text{0.0458 mol P$_(4)$O}_(10) * \frac{\text{2 mol P$_(2)$O}_(5)}{\text{1 mol P$_(4)$O}_(10)} = \text{0.0916 mol P$_(2)$O}_(5)`

3. Molecules of P₂O₅

`\text{No. of molecules} = \text{0.0916 mol P$_(2)$O}_(5) * \frac{6.022 * 10^(23)\text{ molecules P$_(2)$O}_(5)}{\text{1 mol P$_(2)$O}_(5)}\\\\= \mathbf{5.5 * 10^(22)}\textbf{ molecules P$_(2)$O}_(5)\\\text{There are $\large \boxed{\mathbf{5.5 * 10^(22)}\textbf{ molecules of P$_(2)$O}_(5)}$}`