   Chapter 3, Problem 156AE

Chapter
Section
Textbook Problem

# A 0.4230-g 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.

Interpretation Introduction

Interpretation: A sample of impure sodium nitrate was heated to produced sodium nitrite and oxygen gas The percent of sodium nitrate in the original sample is to be calculated.

Concept introduction: The number of moles of a substance is calculated by the formula,

Numberofmoles=GivenmassofthesubstanceMolarmassofthesubstance

To determine: The percent of sodium nitrate in the original sample.

Explanation

Given

The mass of the impure sodium nitrate (NaNO3) sample is 0.4230g .

The mass of sodium nitrite (NaNO2) formed is 0.2864g .

The molar mass of NaNO3 =Na+N+3O=(23+14+(3×16))g/mol=85g/mol

The molar mass of NaNO2 =Na+N+2O=(23+14+(2×16))g/mol=69g/mol

Formula

The number of moles of a substance is calculated by the formula,

Numberofmoles=GivenmassofthesubstanceMolarmassofthesubstance

Substitute the value of the given mass and the molar mass of NaNO2 in the above expression.

NumberofmolesofNaNO2=0.2864g69g/mol=4.15×103mol

The mass of NaNO3 is assumed to be “ x

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