How much Sodium nitrate is in the original sample

Question

A 0.4230 g sample of impure sodium nitrate (contains sodium nitrate plus inert ingredients) was heated, converting all the sodium nitrate to 0.1958 g of sodium nitrite and oxygen gas. Determine the percent of sodium nitrate in the original sample.

Answer 1

First, lets write the balanced equation for the reaction:

2NaNO₃ ==> 2NaNO₂ + O₂

Next, using the molar mass of NaNO2 and NaNO3, we can find the moles of NaNO3 originally present.

molar mass NaNO2 = 69.0 g / mol

molar mass NaNO3 = 85.0 g / mol

0.1958 g NaNO2 x 1 mol / 69 g = 0.002838 mols

0.002838 mols NaNO2 x 2 mols NaNO3 / 2 mol NaNO2 = 0.002838 mols NaNO3

Now convert this to grams:

0.002838 mols NaNO3 x 85 g / mol = 0.241 g NaNO3

Finally, calculate %

0.241 g / 0.4230 g (x100%) = 57.0%