Tetraphosphorus decaoxide

Question

How many particles are in 2.70 L of tetraphosphorus decaoxide (P₄O₁₀)?

Answer 1

First, you must know that "particles" hints the use of avogadro's number (6.022*10²³ particles/mole).

Then you must calculate the molar mass of P₄O₁₀, which is 283.89 g/mole.

1. 4 x 30.97 g/mole of P) + (10 x 15.99 g/mol of O) = 283.89 g/mole of P₄O₁₀

With this known information you can divide (6.022*10²³ particles/mole) / (283.89 g/mole of P₄O₁₀) to cancel the moles. This will leave you with 2.12*10²¹ particles/g of P₄O₁₀

To only have particles remaining you must know that the density of P₄O₁₀ is 2.39 g/cm³. Knowing that 1cm³=1mL we can convert the 2.70L to 2700mL which then means 2700mL=2700cm³.

With this information we can calculate the grams of P₄O₁₀ by multiplying (2.39 g/cm³) x (2700cm³) = 6453g of P₄O₁₀.

Lastly, we multiply (2.12*10²¹ particles/g of P₄O₁₀) x (6453g of P₄O₁₀) = 1.37particles of P₄O₁₀ in 2.70 L.