Percentage of NaNO3 in 5.37g Mixture: 1.61g Sodium

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To determine the percentage by mass of NaNO3 in a 5.37g mixture containing 1.61g of sodium, the calculation involves dividing the sodium mass by the total mass and multiplying by 100, resulting in approximately 29.98% sodium in the mixture. The molar masses of NaNO3 (85 g/mol) and Na2SO4 (142 g/mol) are provided for further calculations. An equation can be devised using the moles of each compound to solve for the amounts of NaNO3 and Na2SO4 present in the mixture. By setting up the appropriate equations based on the total mass and sodium content, the composition of the mixture can be accurately determined. This approach effectively combines stoichiometry with percentage calculations to analyze the mixture's components.
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a mixture of NaNO3 and Na2SO4 of mass 5.37g contains 1.61g of sodium, what is percentage by mass of NaNO3 in the mixture?
 
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Hmm. Sodium? Which formula is that?
 
Um...Na?
 
You have already solved the question, just do the dividing operation and multiply by 100 to learn the percentage.
 
Divide what by what? both compunds contain Na...
 
1.61/5.37*100=29.98% total sodium is present in the mixture.

Na=23, N=14, O=16, S=32 and NaNO3 is 48+14+23=85 g/mol, whereas Na2SO4 is 46+32+64=142 g/mol.

Devise an equation, in which x moles of nitrate and y moles of sulfate is present, and make it equal to 5.37. Use mole amounts to learn x and y by making them equal to 1.61. It shouldn't be hard...
 

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