Question

Calculate the number of Li+ ions in 9.18 g Li3PO4.

Calculate the grams of O in 9.18g Li3PO4.

Answer 1

Solution.

1.

Since lithium orthophosphate contains 3 lithium ions, the amount of lithium ion substance will be equal to:

$n(Li3PO4) = \frac{9.18}{115.79} = 0.06 \ mole$

$n(Li^+) = n(Li3PO4) \times 3 = 0.18 \ mole$

2.

In order to find the mass of oxygen in lithium orthophosphate, it is necessary to find the mass fraction of the element (oxygen) and multiply it by the mass of the salt:

$w(O) = \frac{4 \times 16.00}{115.79} \times 100 \% = 55.27 \%$

$m(O) = \frac{9.18 \times 55.27}{100} = 5.07 \ g$

Answer:

1.

n(Li+) = 0.18 mole

2.

m(O) = 5.07 g