Answer to Question #138087 in Chemistry for LM

Question

stomach relief medications contains the compound Sodium hydrogen carbonate; an over the counter bottle of the medication contains 350g of Sodium hydrogen carbonate.
Calculate the number of particles in this sample of Sodium hydrogen carbonate.
Calculate the number of oxygen ions present in this mass.

Answer 1

The number of the particles of the sodium hydrogen carbonate can be calculated from the number of the moles $n$ of NaHCO₃ using the Avogardro's number $N_A = 6.022\cdot10^{23}$ mol^-1:

$N = n\cdot N_A$ .

The number of the moles of NaHCO₃ equals its mass $m=350$ g, divided by its molar mass $M = 84.01$ g/mol:

$n = \frac{m}{M} = \frac{350 \text{ g}}{84.01\text{ g/mol}} = 4.17$ mol.

Thus:

$N = 4.17\text{ mol} \cdot 6.022\cdot10^{23}\text{ mol}^{-1} = 2.51\cdot10^{24}$ .

In every NaHCO₃ unit, there are 3 ions of oxygen. Therefore, the number of oxygen ions present in this mass is three times the number of NaHCO₃ molecules:

$N(O) = 3N = 7.53\cdot10^{24}$ .

Answer: In 350 g of sodium hydrogen carbonate, there are 2.51x10²⁴ molecules of NaHCO₃ and 7.53x10²⁴ ions of oxygen.