Question

calculate the number of atoms of each type in 5.3g of Na2CO3.

Answer 1

Moles of Na2CO3=5.3/106=0.05
so moles of Na=0.05×2=0.1
moles of C=0.05×1=0.05
moles of O=0.05×3=0.15
we can now find the no. of atoms of each type by multiplying the no. of moles by Avagadro's constant.
Hope it helps:-)

Answer 2

Given:

Mass of the sodium carbonate = 5.3 g

Molar mass of   sodium carbonate $=105.99\ \frac{g}{mol} mol$

Moles of  sodium carbonate $= \frac{\text { Mass }}{\text { Molar mass }}=\frac{5.3 g}{105.98\ \mathrm{g} / \mathrm{mol}}=0.05\ moles$

1 mole sodium carbonate contains:

2 moles of Na    

1 mole of C

3 moles of O  

Hence 0.05 moles of sodium carbonate will have:  

$2 \times 0.05=0.1 \text { moles of } \mathrm{Na}$

$1 \times 0.05=0.05 \text { moles of } \mathrm{C}$

$3 \times 0.05=0.15 \text { moles of } \mathrm{O}$

$1 \text { mole }=6.022 \times 10^{23}\ atoms$

Hence

$\mathrm{Na}=0.1 \times 6.022 \times 10^{23} \text { atoms }=0.6022 \times 10^{23}\ atoms$

$\begin{array}{l}{C=0.05 \times 6.022 \times 10^{23} \text { atoms }=0.30 \times 10^{23}} \\ {\text { atoms }}\end{array}$

$\begin{array}{l}{\mathrm{O}=0.15 \times 6.022 \times 10^{23} \text { atoms }} \\ {=0.90 \times 10^{23} \text { atoms }}\end{array}$