Question

A 0.4230 g sample of impure sodium nitrate (contains sodium nitrate plus inert ingredients) was heated, converting all the sodium nitrate to 0.1080 g of sodium nitrite and oxygen gas. Determine the percent of sodium nitrate in the original sample.

Answer 1

Answer : The percent of sodium nitrate in the original sample is, 31.35 %

Explanation : Given,

Mass of impure
`NaNO_3` = 0.4230 g

Mass of
`NaNO_2` = 0.1080 g

Molar mass of
`NaNO_3` = 84.99 g/mole

Molar mass of
`NaNO_2` = 68.99 g/mole

First we have to calculate the moles of
`NaNO_2`.

`\text{Moles of }NaNO_2=\frac{\text{Mass of }NaNO_2}{\text{Molar mass of }NaNO_2}=(0.1080g)/(68.99g/mole)=0.00156moles`

Now we have to calculate the moles of
`NaNO_3`.

The balanced chemical reaction is,

`NaNO_3\rightarrow NaNO_2+(1)/(2)O_2`

From the balanced reaction we conclude that

As, 1 mole of
`NaNO_2` react to give 1 mole of
`NaNO_3`

As, 0.00156 mole of
`NaNO_2` react to give 0.00156 mole of
`NaNO_3`

Now we have to calculate the mass of
`NaNO_3`.

`\text{Mass of }NaNO_3=\text{Moles of }NaNO_3* \text{Molar mass of }NaNO_3`

`\text{Mass of }NaNO_3=(0.00156mole)* (84.99g/mole)=0.1326g`

Now we have to calculate the percent of sodium nitrate in the original sample.

`\% \text{ of }NaNO_3=\frac{\text{Mass of }NaNO_3}{\text{Mass of impure }NaNO_3}* 100`

`\% \text{ of }NaNO_3=(0.1326g)/(0.4230g)* 100=31.35\%`

Therefore, the percent of sodium nitrate in the original sample is, 31.35 %