Question

A 0.4230-5 sample of impure sodium nitrate was heated, converting all the sodium nitrate to 0.2864 g of sodium nitrite and oxygen gas. Determine the percent of sodium nitrate in the original sample.

Short answer

The percent of sodium nitrate in the originalsample is 83.40%

Step-by-step solution

Definition

The molar mass of a material is defined as the mass of the substance contained in one mole of the substance, and it isequal to the total mass of the component atoms. It may be stated mathematically as follows:
Molar mass =$\frac{Weighting}{numberofmoles}$

Determining the percentage of Sodium nitrate

We are given animpure amount of sodium nitrate and a mass of product in this issue, and we are requested tocompute the percent of sodium nitrate in the original combination.

To begin, we'll return the mass of sodium nitrite created in the reaction to gram of sodium nitrate.

To create 0.2864 g sodium nitrite (NaNO₃) the mass of sodium nitrate (NaNO₂) required is:

$\left(0.2864gNaN{O}_2\times \frac{1molNaN{O}_3}{1molNaN{O}_2}\times \frac{85.00gNaN{O}_2}{1molNaN{O}_3}\right)=0.3528gNaN{O}_3$The percentage of sodium nitrate

$$\begin{gathered} =\frac{0.3528g}{0.4230g}\times 100 \\ =83.40\% \end{gathered}$$

Therefore the percentage of (NaNO₃) is 83.40% .