What is the mass of nitrogen in a 50.0 g sample of sodium nitrite ($NaN{{O}_{2}}$)?
(A) 20.2 g
(B) 16.4 g
(C) 10.1 g
(D) 8.23 g
(E) 23.4 g
Answer
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Hint: Calculate the total mass of the compound, sodium nitrite. Now determine the number of nitrogen atoms in one molecule of sodium nitrite and then find the total mass of nitrogen as well. Now take the ratio of total mass of nitrogen atoms and total mass of compound. With this ratio you can determine the mass of nitrogen present in 50 g of sample and thus answer the question.
Complete step by step solution:
We will identify the atoms present in the molecule and then determine the total mass of the compound.
The atoms present in sodium nitrite are, sodium, nitrogen and oxygen.
Mass of sodium = 23 g
Mass of nitrogen = 14 g
Mass of oxygen = 16 g
Total mass of the compound, sodium nitrite = (1 x 23) + (1 x 14) + ( 2 x 16) = 69 g
Total mass of nitrogen in the compound = (1 x 14) = 14 g
Ratio = 0.202
This means that in every 100 g of sodium nitrite, the mass of nitrogen present is 20.2 g.
So, in 50 g of compound, the mass of nitrogen is 10.1 g.
Therefore, the correct answer is option (C).
Note: It is important to know that although we calculated the ratio for 1 mole of the compound, the same value can be used for weight calculation. This is because the ratio will remain the same and unitless.
Complete step by step solution:
We will identify the atoms present in the molecule and then determine the total mass of the compound.
The atoms present in sodium nitrite are, sodium, nitrogen and oxygen.
Mass of sodium = 23 g
Mass of nitrogen = 14 g
Mass of oxygen = 16 g
Total mass of the compound, sodium nitrite = (1 x 23) + (1 x 14) + ( 2 x 16) = 69 g
Total mass of nitrogen in the compound = (1 x 14) = 14 g
Ratio = 0.202
This means that in every 100 g of sodium nitrite, the mass of nitrogen present is 20.2 g.
So, in 50 g of compound, the mass of nitrogen is 10.1 g.
Therefore, the correct answer is option (C).
Note: It is important to know that although we calculated the ratio for 1 mole of the compound, the same value can be used for weight calculation. This is because the ratio will remain the same and unitless.
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