Sodium nitrate in the original sample in %

Question

A 0.4230 g sample of impure sodium nitrate (contains sodium nitrate plus inert ingredients) was heated, converting all the sodium nitrate to 0.2500 g of sodium nitrite and oxygen gas. Determine the percent of sodium nitrate in the original sample.

sodium nitrate in the original sample = ____%

Answer 1

2NaNO₃ ===> 2NaNO₂ + O₂ ... balanced equation

Each mole of NaNO₃ produces 1 mole of NaNO₂ (1:1 mole ratio)

moles of NaNO₂ produced = 0.2500 g x 1 mol NaNO₂/69 g = 0.003623 moles

moles NaNO₃ present = 0.003623 moles NaNO₃

mass NaNO₃ = 0.003623 mol x 84.99 g/mol = 0.3079 g NaNO₃

% NaNO₃ = 0.3079 g/0.4230 g (x100%) = 72.80%