# Question ccb14

Oct 31, 2015

$0 , 1833 m o l$.
$1 , 104 \times {10}^{23}$ particles.
71,11% oxygen by mass.

#### Explanation:

Moles = Mass divided by molar mass.
$\therefore n = \frac{16 , 5 g}{2 \times 1 + 12 \times 2 + 16 \times 4 g / m o l} = 0 , 1833 m o l$

One mole contains an Avagadro number of particles, ie $6 , 023 \times {10}^{23}$ particles.
Hence by ratio and proportion, 0,1833 moles will contain
$0 , 1833 \times 6 , 023 \times {10}^{23} = 1 , 104 \times {10}^{23}$ particles.

In 1 mole of ${H}_{2} {C}_{2} {O}_{4}$, which has a mass of 90g, exactly 64g of this 90g is oxygen. this corresponds to a percent of 64/90xx100=71,11%#.