The 5.00 g is a mixture of sodium bicarbonate and an unknown substance, making it impure.

Balanced decomposition equation:

$$\ce{2NaHCO3(s) -> NaCO3(s) + H2O(g) + CO2(g)}$$

There are 0.06 mol of $\ce{NaHCO3}$ in a pure 5.00 g sample of sodium bicarbonate.
Theoretically, 0.0298 moles of $\ce{H2O}$ and $\ce{CO2}$ would be lost durin the decomposition of 5.00 g of sodium bicarbonate.
0.0298 mol converted to grams is:
0.536 g of $\ce{H2O}$
1.31 g of $\ce{CO2}$


1 Answer 1


First there is a big gotcha. We'll let $x$ be the grams of sodium bicarbonate and $y$ be the grams of the impurity. Since we were given no information about the impurity we must assume that the impurity neither loses mass nor gains mass when heated with the sodium carbonate. So: $$x + y = 5.00$$

Well the balanced equation is really:

$\ce{2NaHCO3(s) -> Na2CO3(s) + H2O(g) + CO2(g)}$

so 2 moles of $\ce{NaHCO3}$ yields 1 mole of $\ce{Na2CO3}$. Thus the mass conversion factor is: $$\dfrac{0.5*\text{MW(}\ce{Na2CO3}\text{)}}{\text{MW(}\ce{NaHCO3}\text{)}} = \dfrac{0.5*106.0}{84.0}= 0.6310$$

So we now also have: $$0.6310x + y = 4.00$$

Subtracting the second equation from the first we get: $$0.3690x=1.000$$ so $$x=2.71g$$ and $$\text{Purity} = \dfrac{2.71}{5} = 54.2\%$$ $$\text{Impurity} = 100\% - 54.2\% = 45.8\%$$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.