Question

Calculate the number of Li+ ions in 6.24 g Li3PO4
.

Answer 1

M(Li₃PO₄) = 116 g/mol

m(Li₃PO₄) = 6.24 g

$n = \frac{m}{M}$

$n(Li3PO4) = \frac{6.24}{116} = 0.053\; moles$

1 mole of Li₃PO₄ contains 3 moles of Li+ ions.

0.053 moles of Li₃PO₄ contains $3 \times 0.053 = 0.159$ moles of Li⁺ ions.

Proportion:

1 mole of Li⁺ ions – $6.022 \times 10^{23} \;ions$

0.159 moles of Li⁺ ions – x

$x = 0.159 \times 6.022 \times 10^{23} ions = 0.957 \times 10^{23}\; ions$

Answer: 9.5×10²² ions